QQAD 2008 Compilation

------------------------------------------------------Quantitative Question # 001 ------------------------------------------------------ Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ?

(1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing

Solution:

Given a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312 => (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 312 - 3S + 4 = 256. Let a-1 = A, b-1 = B, c-1 = C, d-1 = D => we have non-negative integers A, B, C, D such that A^2 + B^2 + C^2 + D^2 = (A+B+C+D)^2 (the LHS <= RHS always) and can only be true if all but one number is zero => three among a, b, c, d are 1 and fourth number is 17.

Thus (3) holds true - (2) can guarantee the values for c and d. For a=b we have exactly 2 determinable values - thus (1) is not true.

=> Choice (5) is the right answer

------------------------------------------------------Quantitative Question # 002------------------------------------------------------ Consider a string of n 7s, 7777....77, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 777 + 7 = 791 could be obtained from five 7s in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?

(1) 105 (2) 106 (3) 108 (4) 109 (5) 111

Solution:

Let the arithmetic expression on the left side that has + signs in the LHS has p 7s, q 77s and q 777s => we have to find n such that p+2q+3r = n and 7p + 77q + 777r = 7000 => p + 11q + 111r =

1000. Solving we get 9(q+12r) = 1000-n; now n can vary from 1000 to 1 out of which n = 1, 10, 19, 37 are excluded as p <0 for these values => total such n are [999/9 + 1] - 4 = 108.


------------------------------------------------------Quantitative Question # 003------------------------------------------------------ Let f(x) be a function such that f(x).f(y) - f(xy) = 3(x+y+2). Then f(4) (equals )

(1) can not be determined (2) 7 (3) -8 (4) either 7 or -8 (5) none of these

Solution:

Given f(x).f(y) - f(xy) = 3(x+y+2). Put x=y=0 => and let f(0) = p => p^2 - p - 6 = 0 => p = 3, -2 = f(0).

In f(x).f(y) - f(xy) = 3(x+y+2), put y = 0 => for f(0) = 3 we get f(x) = x+3, for f(0) = -2 we get f(x) = -(3x+4)/2.

Check for both f(x) = x+3 and -(3x+4)/2 on putting in the original functional equation f(x).f(y) - f(xy) = 3(x+y+2). We see f(x) = -(3x+4)/2 fails and f(x) = x+3 holds true.


------------------------------------------------------Quantitative Question # 004------------------------------------------------------

N students are seated at desks in an m x n array, where m, n >= 3. Each student shakes hands with the students who are adjacent horizontally, vertically or diagonally. If there are 81 handshakes, what is N?

(1) 28 (2) 27 (3) 30 (4) 25 (5) 24

Solution:

Students in the corner shake hands 3 times, those on the sides 5 times and those in the middle 8 times. So the total number of handshakes is (4•3 + (2m-4+2n-4)5 + (m-2)(n-2)8)/2 = (12 + 10m + 10n - 40 + 8mn - 16m - 16n + 32)/2 = (16mn - 12m - 12n + 8)/4 = (4m - 3)(4n - 3)/4 - 1/4 = 81, so (4m-3)(4n-3) = 325 = 13.25. hence N = mn = 4•7 = 28.


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A particle moves around a circle (once) such that its displacement from the initial point in given time t is t(6-t) meters where t is the time in seconds after the start. The time in which it completes one-sixth of the distance is

(1) 0.60 s (2) 0.88 s (3) 1 s (4) 1.12 s (5) none of these

Solution:

The displacement from the starting position can be max when the particle (P) is at diametrically opposite position of S => when t(6-t) is max then the value of t(6-t) = 2R where R is the radius of the circle => t = 3 and R = 9/2.

Now, when the particles covers 1/6th of the distance => the angle subtended by SP at the center of the circle is 360/6 = 60 degrees => length of SP is R = 9/2 = t(6-t) => t = 0.88


Quantitative Question # 006------------------------------------------------------

In a soccer tournament n teams play against one another exactly once. The win fetches 3 points, draw 1 each and loss 0. After all the matches were played, it was noticed that the top team had unique number of maximum points and unique least number of wins. What can be the minimum possible value of n?

(1) 5 (2) 6 (3) 7 (4) 8 (5) none of these--------------------------------------------------------

Solution:

Total number of matches played =n(n-1)/2Total matches per team =n-1For getting the least number of teams,the difference in number of wins will be minimum ..i.e 1Assume All other teams (except winning team) have same number of wins.If some team other Y than the top team X had 2 more wins than X => Y had 6 points more than X from the wins => X would need 7 more draws than X => n will be > 9 here.Assuming K wins for winning team and n-1-k draws and k+1 wins for other teamswe have ,Total matches(based on wins and draws =k + (n-1-k) + (n-1) (k+1)=(n-1) (k+2)This should be equal to n(n-1)/2 (total matches played btw n teams )==> k= (n-4)/2------------------------------------(A)So from this we know n is even,now the points that the winning team gets =3k +n-1-k =2k+ n-1points other team gets (maximum) = 3(k+1) + 1=3k+4Given that 2k+n -1 > 3k+4 ==> n > k+5substituting k=(n-4)/2we get 2n > n+6 or n > 6 -------------------------(B)==> Since n is even, 8 is the answer

Example for n = 8 as possibility.

Name the 8 teams X, A, B, C ,D, E, F, and G

X defeats A and B, and draws against C ,D, E, F, and G for 2 wins and 11 points. A defeats B, D, and E for 3 wins and 9 points. B defeats C, D, and F for 3 wins and 9 points. C draws against X and defeats A, E, and G for 3 wins and 10 points. D draws against X and defeats C, E, and G for 3 wins and 10 points. E draws against X and defeats B, C, and F for 3 wins and 10 points. F draws against X and defeats A, D, and G for 3 wins and 10 points. G draws against X and defeats A, B, and E for 3 wins and 10 points.

Alternate solution:

Let X be the desired team. Now let n denote the number of soccer teams which participated in this tournament, so there are n(n-1)/2 matches played among these teams, now note that each game gives at least 2 points in total, so sum of the points of all teams is greater than n(n-1) so the average is at least (n-1). Note that X has played (n-1) games, and also we know that X has the most total points, so the total points of X is more than (n-1) so X must have won at least one game.X has won the fewest number of games, so every other team must have won at least 2 games => every other team has at least 6 points => X must draw at least 4 games in order to have the most total points, but note that if team Y had a draw with team X then the total points of Y would be at least 7, hence X must have drawn at least 5 games => n >= 7.We show that n > 7: if n = 7 then let S denote the set of all teams, X has won a single match and has drew matches, so the total points of X is equal to 8, now note that every other team has won exactly two games, and drew at most one game => every other team has lost at least games, so there are at least 18 looses in => there must be at least 18 wins in ,but we know that the number of wins in S is equal to 1+ 6*2 = 13,which is a contradiction.

Example for n = 8 as possibility.

Name the 8 teams X, A, B, C ,D, E, F, and G

X defeats A and B, and draws against C ,D, E, F, and G for 2 wins and 11 points. A defeats B, D, and E for 3 wins and 9 points. B defeats C, D, and F for 3 wins and 9 points. C draws against X and defeats A, E, and G for 3 wins and 10 points. D draws against X and defeats C, E, and G for 3 wins and 10 points. E draws against X and defeats B, C, and F for 3 wins and 10 points. F draws against X and defeats A, D, and G for 3 wins and 10 points. G draws against X and defeats A, B, and E for 3 wins and 10 points.



In a quadrilateral ABCD sides AB and CD are equal with <A = 150˚, <B = 44˚, and <C = 72˚. Perpendicular bisector of the segment AD meets meets the sides BC at point X. Then m(<AXD) is

(1) 42˚ (2) 58˚ (3) 64˚ (4) 78˚ (5) none of these--------------------------------------------------------

Solution:

Let P be a point on BC such that AP = AB => <BAP = 180˚ – 2*44˚ = 92˚Let Q be a point on BC such that DQ = CD => <CDQ = 180˚ – 2*72˚ = 36˚

Let E be the intersection point of AP, DQ.

We have <EAD = 150˚ -92˚ = 58˚ and <EDA = 94˚ – 36˚ = 58˚

So the triangle EAD is isosceles => EA = ED. Also AB = CD => AP = DQ

We'll prove that E lies on BC.

Assume that E doesn’t lie on BC

The line BC bisects the plane into two half-planes If E is on the same half-plane with A, D then EP = AP - EA and EQ = DQ - EDIf E is on the other half-plane with A, D then EP = EA - AP and EQ = ED - DQ

At any case, from we get EP = EQ which yields that EPQ is isosceles. But this is not possible because the lines AP and DQ make different angles with line BC.

We got in a contradiction because we assumed that E does not lie on BC. So the only possible case is that E lies on BC => <AXD = 64˚

Alternate solution:

Let XQ be the perpendicular bisector of AD such that Q lies on AD. Let <AXD = 2x => <AXQ = <DXQ = x.

=> <BAX = 60˚+x and <XDC = 4˚+x => <AXB = 76˚ -x and < DXC = 104˚-x. By sine rule DC/DX = AB/AX = sin(104-x)˚/sin72˚ = sin(76-x)˚/sin44˚ => x = 32˚



Let S be the set of first 14 natural numbers. A special subset of S is a subset S' which satisfies the following three properties

a) S' has exactly 8 elementsb) If x belonging to S is even, then x is in S' if and only if x/2 is in S'c) If y belonging to S is odd, then y is in S' if and only if (y+15)/2 is in S'

Let X denotes elements of S that cannot be the part of special subset. Then n(X) (i.e. number of elements in X) equals

(1)2 (2) 3 (3) 5 (4) 6 (5) none of these--------------------------------------------------------

Solution:

This is a problem that was raw and required playing with numbers.

The sets satisfying conditions (b) and (c) are {1, 2, 4, 8}, {3, 6, 9, 12}, {7, 11, 13, 14} and {5, 10}.

By condition (a) we can have 3 combinations of union of 2 sets each from above 3 sets having 4 elements each => set {5, 10} is excluded.



A shokeeper sells 2 different brands of an article, one for Rs 10 and other for Rs 12 each. One day he left the shop in the hands of his management graduate son who confused the two brands and sold them at each other's price. Due to this, the shopkeeper lost Rs 40 which amounted to 10% fall in his expected profit. If shopkeeper's expected profit was one among 10%, 15%, 20%, or 25% then

(1) the number of cheaper brand sold at higher cost was 60(2) the expected profit was 20%(3) if the profit per article of expensive brand is 40 paise more than that of cheaper brand => the least number of articles of expensive brand he had to sell to recover his loss would be 34 (4) at least two of the foregoing(5) none of the foregoing

Solution:

For first article SP = 10, sold "a" items, profit per item = (Pa), sold at 12 Rs per item

For Second article, SP = 12, sold "b" items, profit per item = (Pb), sold at 10 Rs per item

expected profit = a(Pa) + b(Pb)actual profit = a(Pa +2) + b(Pb -2)

expected - actual = 40 => b = a+20also, 40 is 10% of expected profit => expected profit = 400 rs.now, expected profit = total cost * profit percentage/100total cost = 10a + 12b - 400 = 22a + 240 -400 = 22a -160if profit% = 10, 22a = 4160 [ a is not an integer]if profit% = 15, 22a = 400 * 100/15 + 160 [ a is not an integer]if profit% = 20, 22a = 2160 [ a is not an integer]if profit% = 25, 22a = 1760(a has to be a positive integer since it is the number of items sold, so only profit = 25% would be possible).expected profit = 25%

also,a = 1760/22 = 80=> b = 100

For option (3)

80(Pa) + 100 (Pb) = 40 (expected profit)put Pb = Pa + 0.40 to get Pa = 2, and Pb = 2.40

=> to make up the lost profit of Rs 40, he needs to sell 40/2.4 = 16.66 => 17 items of the higher price, which is less than 34.



The total ordered pair of positive integers (p, q) such that the roots of equations x^2 -px + p + q - 3 = 0 and x^2 -qx + p + q - 3 = 0 are also positive integers are

(1) 2 (2) 3 (3) 4 (4) 6 (5) none of these--------------------------------------------------------

Solution:

Let a, b be the roots of 1st equation and c, d be the roots of 2nd equation

=> ab = cd = p+q-3 => (ab+cd)/2 = p+q-3 = (a+b) + (c+d) - 3=> ab + cd = 2(a + b + c + d) - 6 => (a-2)(b-2) + (c-2)(d-2) = 2

Now, given that a, b, c, d are all integers such that ab = cd => 1.1 + 1.1 = 2, -1.-1 + -1.-1 = 2, 1.2 + 0.x = 2, 0.y + 1.2 = 2 => we get (a, b, c, d) as (1, 1, 1, 1), (3, 3, 3, 3) and (2, 6, 3, 4)

Thus, (p, q) = (2, 2), (6, 6), (7, 8) -> these are unordered pairs, an extra permutation (8, 7) of (7, 8) willgive 4 ordered pairs.


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Let ABC be a triangle and D and E be internal points on AB and AC respectively. The segment s joining BE and CD divide the triangle into 4 regions. If all the 4 regions have equal area then the triangle ABC is

(1) not possible (2) obtuse-angled (3) equilateral (4) isosceles with < A = 36˚ (5) none of these

Solution:

Let F be the point of intersection of AE and CD. Given area of each of BFC = CFE = EFDA = DFB = A => Area of ABE = Area of EBC => BE is median. Similarly, CD is a median => F is the centeroid. But Area CFE = Area BFC shows CF is median in triangle ECB => F divides BE in 1:1 which is a contradiction as centeroid divideds the median in the ratio 1:2.


--------------------------------------------------------Solution to Quantitative Question # 012--------------------------------------------------------

How many eight letter words exist that are composed of Xs and Ys, and which contain neither three consecutive Xs nor three consecutive Ys?

(1) 74 (2) 66 (3) 76 (4) 68 (5) none of these

Solution:

Let F(n) denote the number of n-letter words consisting of Xs and Ys and not containing three consecutive identical letters. Thus F(1) = 2 since the one-letter words X and Y meet the no-triple-repeat condition. Since all four two-letter possibilities XX, XY, YX and YY meet the condition, we have F(2) = 4.

Assuming that n >= 3, how can we construct an n-letter word with no triple repeats? One way would be to start with an (n － 2)-letter word w and to append either XX if w ends in Y, or YY if w ends in X. Since

there are F(n － 2) possibilities for w, this allows us to construct F(n － 2) words of length n, and these are all of the n-letter words with no triple repeats and having a repeated letter at the end. Similarly, we can start with an (n － 1)-letter word w and append either X if w ends in X, or Y if w ends in X.

This yields F(n － 1) acceptable words, and these are all the acceptable n-letter words whose last two letters are different. We thus have a total of F(n － 2) + F(n － 1) n-letter words with no triple repeats, and thus F(n) = F(n － 2) + F(n － 1). Now F(1) = 2 and F(2) = 4, so F(3) = 2 + 4 = 6, F(4) = 4 + 6 = 10, F(5) = 6 + 10 = 16,F(6) = 10 + 16 = 26 , F(7) = 16 + 26 = 42, F(8 ) = 26 + 42 = 68.


Let x and y be positive real numbers such that x^3 + y^3 = 4x^2. A is the maximum value of x + y.

Let a, b, c be real such that a+b+c = 5 and ab + bc + ca = 3. B is the largest possible value of c.

Then A + B lies in the range

(1) [7, 8) (2) [8, 9) (3) [9, 10) (4) [10, 11) (5) none of these

Solution:

Let u take the two problems of finding A and B one by one.

Put y = xk in x^3 + y^3 = 4x^2 => x^2(x + k^3x) = 4x^2 => x(1+k^3) = 4. We need to maximize x + xk = x(1+k) = 4(1+k)/(1+k^3) = 4/[(k-1/2)^2 + 3/4)] = 16/3 = A

For the second part, please see that (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16 => c can be maximized when a = b = 1/3 and c = 13/3.

Alternatively: a+b+c = 5 and ab + bc + ca = 3, replace a by (5-b-c) in second equation and form a quadratic in b=> b^2 + (c-5)b + (c^2-5c+3) = 0, again for real b Discriminant >0=> -3c^2+10c+13 >= 0 => c <= 13/3


Let x be the number of base systems in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number. Then x in base 7 is a

(1) odd but not prime (2) prime (3) even and perfect square (4) even but not perfect square (5) none of these

Solution:

Let ABC be the largest 3 digit perfect square in base 6 (A, B, C <= 5) => in base 10 ABC will be 36A + 6B + C and this number is a perfect square. Since 5*(36 + 6 + 1) = 215 => 36A + 6B + C = 14^2 = 196 => ABC = 524.

Now, 524 is written as a 2 digit number CD in base b => b*C + D = 524 where C, D < b and C is a positive integer => b ranges from 23 to 524 => x = 502 = 1315 in base 7.

The properties (odd/even/prime/perfect square) of a number N in base b is same as that of decimal conversion of N => x is even in base 7 also.

Choice (4) is the right answer

How many (a, b, c) satisfy log(2ab) = loga*logb, log(bc) = logb*logc, log(2ac) = loga*logc ?

(1) none (2) 1 (3) 2 (4) 4 (5) none of these

Solution:

log20 = (loga - 1)*(logb - 1), 1 = (logb - 1)*(logc - 1), log20 = (logc - 1)*(loga - 1)

Multiplying 3 equations and taking square root we get+/-log20 = (loga - 1)*(logb - 1)*(logc - 1)=> logc = 1 +/-1, loga = 1 +/-log20, logb = 1 +/-1Thus, (a, b, c) = (200, 100, 100) or (1/2, 1, 1)


A PaGal is stationed at C, 60 meters directly west of a CAT located at M. The CAT is trying to escape running at 7m/s in a direction 30 degrees east of north. The PaGal an expert in geometry runs at 13m/s in a suitable straight line path that will intercept the CAT as quickly as possible.Then which among is following is true?

(1) It takes 15/2 seconds for the PaGaL to catch CAT(2) If CAT chooses a different direction to escape, irrespective of the direction, all interception points lie on a circle.(3) If the CAT is intercepted after running a distance of x meters in a particular direction andif the CAT had been intercepted after it had run a distance of y meters in the oppositedirection then min (x+y) is 14√30(4) All of the above(5) Exactly 2 of the above

Solution:

Let's try to draw the figure for this. Assume M as the origin (0, 0) where CAT is initially and let the PaGal be at P (-60, 0). Let both of run such that PaGal intercepts the CAT after t seconds T => MT = 7t and PT = 13t, but given that < PMT = 120 degrees. Using cosine rule in the triangle, we have cos(120˚) = (PM^2 + MT^2 - PT^2)/(2.PM.PT) = -1/2 = (3600 + 49t^2 - 169t^2)/(2.60.7t) => t = 15/2.

Let (x,y) be locus of point where the CAT and PaGal meets. Point M is considered to be the origin.(A) = angle from x-axis by which CAT runs off(B) = angle from x-axis by which PaGal runs off.(T) = time at which they CAT is busted.

Equations for CAT : x=7tcos(A) , y=7tsin(A)Equations for PaGal : x=13tcos(B) - 60 , y=13tsin(B)Since A,B,T are variables for locus. Eliminating them we get the equation(x-24.5)^2 + y^2 = 2070.25, which is equation of a circle with centre (24.5,0) and radius = sqrt(2070.25) = 45.5.

Part (c) is easy to see. In the above equation, substitute x= 0, calculate y, and double it to get 14√30 m.

http://2.PM.PT/


Let R(x) be the remainder when x^16 + x^8 + x^6 + x^4 + x^2 + 1 is divided by x^3 - 1. Let D(x) be the divisor (less than degree 4) of x^6 + 4x^3 + 8. Then which among the following is true?

(1) The sum of the coefficients of R(x) and D(x) is equal(2) The sum of the absolute value of coefficients of R(x) and D(x) is equal(3) R(x) > D(x) for all non-positive x(4) at least 2 of the above (5) none of the above

Solution:

X^16 + x^8 + x^6 + x^4 + x^2 + 1 = Q(x).(x^3 - 1) + R(x), where R(x) has degree lesser than that of the quotient. Thus if we put x^3 = 1 in both LHS and RHS we get, x + x^2 + 1 + x + x^2 + 1 = 2(x^2 + x + 1) = R(x)

For D(x), as we know that a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) x^6 + 4x^3 + 8 has a quadratic factor D(x) as x^2 - 2x + 2 as we can put a = x^2, b = -2x and c = 2. Want to try factorizing x^6 + 5x^3 + 8?


Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of

(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined

Let the CP of two vodkas be Rs 100 and Rs 100x and individual profit in Rs on them being A and B. => (A+2B)/3 = 10/100*(100+200x)/3 and (2A+B)/3 = 20/100*(200 + 100x)/3. solving we get A = (70+20x)/3 and B = (20x-20)/3=> profit percentages on each is (70+20x)/3 and (20x-20)/3x. When they are increased to 4/3 and 5/3 times respectively and mixed in the ratio 1:1 we get total profit % as (4/3*100*(70+20x)/3 +

5/3*100x*(20x-20)/3x)/(100+100x) = 100*(20x+20)/(100+100x) = 20 => choice (b) is the right answer.

Two spherical balls lie on the ground touching. If one of the balls has a radius of 8 cm, and the point of contact is 10 cm above the ground, what is the radius of the other ball?

(1) 18 cm (2) 40/3 cm (3) 25/2 cm (4) 13 cm (5) none of the these

Solution:

This is a simple problem on similar triangles. Please refer pages 99-100 in the discussions thread for figure and explanation.


N girls and 2N boys played a chess tournament. Every player played every other player exactly once. The boys won 7/5 times as many matches as the girls (and there were no draws). Then which among the following is definitely false? (Assume 1 point for a win and 0 for a loss)

(a) Boys pocketed prime number of points against girls(b) Girls always won twice or more matches than boys won against them(c) The sum of the scores of top 3 individual players was not between 25 and 33(d) The sum of the scores of top 3 individual players was 69(e) none of the foregoing

Total number of matches among boys were 2nC2, among girls were nC2 and between boys and girls were n*2n. Please note that 2nC2 + nC2 + 2n^2 = 3nC2. Assume 1 point for a win and 0 for a loss => Girls pocketed nC2 points amongst themselves and boys pocketed 2nC2 points among themselves. Let boys take k points from their matches against girls => girls take 2n^2 - k from their matches gainst boys. => 2nC2 + k = 7/5*(nC2 + 2n^2 - k), solving we get 8k = n(5n+1). for n = 3, k = 6. For n = 8, k = 41, For n = 11, k = 77. (a) can be true as for n = 8, k = 41. (b) can be true as can be seen for for n = 3, 8, 11, ... (c) is true as for n = 3, top 3 can score 8+7+6 = 21 points and for n = 11, when 33 matches are

played top 3 will always score more than 16+15+14 = 45. (d) is true, For n = 11, we can have the top 3 score as 23+23+23 = 69 => choice (e) is the right answer.

The question is followed by two statements X and Y. Answer each question using the following instruction:

Choose 1 if the question can be answered by X onlyChoose 2 if the question can be answered by Y onlyChoose 3 if the question can be answered by either X or Y Choose 4 if the question can be answered by both X and YChoose 5 if the question can be answered by neither X nor Y

The positive integers are such that p < q ≤ r < s < 100, ps = qr and √s - √p ≤ 1. What is the value of p?

(X) The last digit of s is either 1, 2 or 3(Y) 50 < p and r < 90

Solution:

Given that s-p > r -q as p < q <= r < s=> (s-p)^2 > (r-q)^2 => (s-p)^2 + 4sp > (r-q)^2 + 4rq [becuase sp = rq, we add 4sp in LHS and 4rq in RHS]=> (s+p)^2 > (r+q)^2=> s+p >= r+q+1 [as all numbers are integers] -> (1)

Suppose √s - √p = 1 (the other possibility is √s - √p < 1 that we will see later)=> s + p - 2√sp = 1 => s+p = 1 + 2√qr [becuase sp = rq] but (1) tells that √qr >= q + r => r = q [By AM-GM rule on positive numbers] and p+s = 2q + 1.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, x) = 1 and gcd(b, x) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1].

If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

=> In all s and p are perfect squares. Now take X -> only possible s is 81 => p = 64Now take Y, p > 50 => p can be 64 or 81 but if p = 81 then s = 100 (not possible as s < 100) => p can only be 64. The information on r is required to cross-check if our data in hand is correct and it indeed is as √64.81 = 72.


Amrutesh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Amrutesh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Amrutesh walks?

(a) 48/25 (b) 12/5 (c) 12 (d) 15 (e) none of the foregoing

Let M be the endpoint of the altitude on the hypotenuse. Since we are dealing with right triangles,triangle MAC ~ triangle ABC, so AM = 12/5. Let N be the endpoint he reaches on side AC.Triangle MAC ~ trangle NAM,So , MN/AM =4/5 . This means that each altitude that he walks gets shorter by a factor of 4/5 . The total distance is thus (12/5) /(1- 4/5) =12 => choice (c) is the right answer.

For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y), f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

Solution:

f(x+2, y+1) = f(f(x+1, y), f(x, y)) [Given] -> (1)f(x+1, 0) = f(x, 1) Given -> (2)put x=x-1 and y=0 in (1) we getf(x+1,1) = f(f(x,0),f(x-1,0)) = f(f(x-1,1),f(x-2,1)) = f(x,2) -> (3)

also f(x+2,2) = f(f(x+1,1),f(x,1) = f(f(x,2),f(x-1,2)) = f(x+1,3) -> (4) [using 3]going further we can see that f(x+n, n) = f(x+n-1, n+1) for positive

integer n -> (5)[can be proved using Mathematical Induction]

=> f(f(2,3), f(2,2)) = f(f(3,2),f(2,2) = f(4,3) [using 4]But f(4, 3) = f(3, 4) [using 5]

=> Choice (3) and (4) are the right answer

The minimum possible value of the largest of ab, 1-a-b+ab, and a+b-2ab if 0 <= a <= b <=1 is

(a) 4/9 (b)1/9 (c)5/9 (d)1/3 (e) none of the foregoing

Let s = a + b, p = ab, so a and b are (s+/- root(s^2-4p))/2 . Since a and b are real , s^2 - 4p>=0 . If one of the three quantities is less than or equal to 1/9, then at least one of the others is at least 4/9 by the pigeonhole principle since they add up to 1. Assume that s-2p < 4=9, then s^2 - 4p < (4/9 + 2p)^2 - 4p , and since the left side is non-negative we get 0<= p^2 -(5/9) p +4/81 =(p-1/9)(p-4/9).This implies that either p<=1/9 or p>=4/9 , and either way we're done. This minimum is achieved if a and b are both 1/3, so the answer is 4/9 => choice (b) is the right answer.

A point at the intersection of two or more grid lines is called a lattice point. S is a 5 x 5 array of lattice points. How many squares have their vertices in S?

(1) 30 (2) 36 (3) 44 (4) 50 (5) none of these

Solution:

For 3X3 array we have in all 5 squares (when joined vertically or horizontally) such that we have 2^2 squares of side 1X1, and 1 square of size 2X2. But when mid-points of the 2X2 square is joined, we have another square whose all vertices are on the perimeter of the array of 3X3. Please note that both the 2X2 square have all the vertices on the perimeter of 3X3 array. Thus, in all we have 4 + 2 = 6 such squares for 3X3.

For 4X4 array we have 9 squares of size 1X1, 3 squares (whose vertices lie on the perimeter of the array) of size 3X3 and 4*2 squares of size 2x2 [4 is conventional we all know, we double it as the midpoints of the edges joining will also form an equal number of

squares] => In all 9 + 8 + 3 squares.

Now for our problem in hand 5X5:We have 16 ((n-1)^2, when array is nXn) squares of size 1X1, 4 (n-1, when array is nXn) squares (whose vertices lie on the perimeter of the array) of size 3X3, 18 squares (2*(n-2)^2 of size 2X2) and 12 squares (3*(n-3)^2 of size 3X3) [for 3X3 we have trepled as done in 4X4 logic above]=> In all 4 + 16 + 18 + 12 = 50 squares.


Let f be a factor of 120, then the number of positive integral solutions of xyz = f is

(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the foregoing

Let k be such that k = 120/f. Then, the number of positive integral solutions of xyz = f is same as that of number of positive integral solutions of xyzk=120=2^3.3.5We can assign 3 and 5 to unknown quantities in 4*4 ways. We can assign all 2 to one unknown in 4C1 ways, to two unknowns in (4C2)(2) and to three unknown in 4C3 ways.Hence, the number of required solutions=4*4*[4C1 + (4C2)(2) +4C3] =4*4*20 =320 => choice (c) is the right answer.

Let a sequence S(n) be defined for positive integers n such that S(0) = 1 and S(1) = 1. If S(n+2) = 2S(n+1) + S(n), and S(n+1)/S(n) approaches a finite number R as n -> Infinity, then R equals

(1) √2 + 1 (2) 5/2 (3) √5 (4) 3√3/2 (5) none of these

Solution:

Given S(n+2) = 2S(n+1) + S(n); divide both the sides by S(n+1) => S(n+2)/S(n+1) = 2 + S(n)/S(n+1).

Also given that S(n+1)/S(n) -> R as n -> Infinity => S(n+2)/S(n+1) -> R as n -> Infinity

Thus, R = 2 + 1/R, since R > o => R = √2 + 1


Apple, Bombardier, Chat.sun and Doomsayer are to compile this year's Quant Question A Day. They can finish this work together in a certain number of integer days. However, they work two in a day and it is found that the compilation is completed when (Apple, Bombardier), (Bombardier, Chat.sun) and (Chat.sun, Doomsayer) worked for respectively 5, 9 and 4 days or 7, 6 and 5 days. They could not have all together done the work in

(a) 8 days (b) 9 days (c) 10 days (d) 11 days (e) none of these

Let Apple, Bombardier, Chat.sun and Doomsayer do 1/a ,1/b,1/c and 1/d parts of work per day respectively . Then , 5(1/a +1/b) +9(1/c +1/b) +4(1/c +1/d) =1 ; 7(1/a +1/b) +6(1/c +1/b) +5(1/c +1/d) =1and 1/a +1/b + 1/c +1/d = 1/n (say)

=>12( 1/c +1/b) = 1-3/n which implies n>3 and 4( 1/a +1/b) = 1- 7/n which implies n>7 .

Consequently, 1/c +1/d = 1/n - ( 1/a +1/b) = 1/n -1/4(1-7/n) =(11-n)/4nwhich implies n<11; So, from the given options n=11 is not possible => choice (d) is the right answer.

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ABCD is a square, point E is inside triangle ACD and point F is inside triangle ACB. < EAF = <ECF = 45˚. If DE = 3, and BF = 4, then EF equals

(1) 5 (2) 2√3 (3) 7/2 (4) 24/7 (5) none of these

Solution:

Reflect the point B in the lines AF and CF; let the reflections be R and S, respectively. It follows from the reflection that AR=AB=AD, and by adding up the angles at vertex A we have < EAR = < DAE. Hence the triangles AER and AED are congruent. Similarly, the triangles CES

and CED are also congruent.

It follows from the congruences that the following angles are equal< CSF = <CBF,< CSE = < CDE, < ERA = < EDA, < CSE = < CDE

=> FR = FS = FB = 4 and ER = ES = ED = 3 . The triangles FER and FES are congruent because they have pairwise equal sides. Thus their angles at R and S are equal => the sum of these two angles equals the sum of the angles of the square ABCD at B and D, => < FRE = < FSE = 90˚. Hence with the Pythagorean theorem, EF = 5.

Alternative Solution:

Let's use co-ordinate for our problem. Let A (0, 0), B (a, 0), C (a, a), D (0, a) and E = (p, q) and F = (r, s). Thus, the equations of line AE is qx - py, line AF is sx - ry, line CE is (a-q)x - y(a-p) = a(p-q), line CF is (a-s)x - y(a-r) = a(r-s).Let the notation tanI denotes Inverse of tan, as we know tanI (<EAF ) = 1 => tanI(q/p) - tanI(s/r) = 1 i.e. the angle between line AE and AF -> (1)Similarly, 1 + tanI[(a-q)/(a-p)] = tanI[(a-s)/(a-r)] -> (2)

From 1 we get qr = sp+pr+qs -> (3) from 2 we get that 2a^2 - 2ar -2qa + pr + qr + qs - ps = 0 -> (4)

We know that p^2 + (q-a)^2 = 9 and (r-a)^2 + s^2 = 16 and we have to find square root of (p-r)^2 + (q-s)^2, which is indeed √25 [from (3) and (4)]


Through T, the mid-point of the side QR of a triangle PQR, a straight line is to meet PQ produced to S and PR at U, so that PU = PS. If the length of UR = 2 cm, then the length of QS is

(a) 3/2 cm ( b) 2 cm (c) 5/2 cm (d) 3 cm (e) none of the foregoing

If <PSU = <PUS =Q , in triangle UTR the sine of < opposite to UR and TR is the same as Sine of < opposite to QS and QT in triangle QST correspondingly.Since, TR=QT =>QS=UR=2 => choice (b) is the right answer.

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The number of those subsets of {1, 2, 3, 4, 5, 6} such that the equation x+y = 7 has no solution in it is

(1) 18 (2) 21 (3) 27 (4) 36 (5) none of these

Solution:

Consider 3 pairs (n, 7-n) where n is 1, 2, 3. Thus, for each pair we have 3 possibilities, both aren't in the set, only n is in the set, only 7-n is in the set => in all 3^3 = 27 possibilities.


A new Ducati is designed for the Indian market such that its mileage at a particular speed follows a certain relationship with that speed. Also , the speed decreases linearly with the mass of the rider while the petrol consumption per km increases linearly with the mass of rider .Ideally , when the mass of the rider is negligible, the speed is 100km/hr and mileage is 100km/l .When the speed of the Ducati is 50 km/hr , the mileage is 50 km/l.When the speed of the Ducati is 75 km/hr , the mileage will be

(a) 60 km/l (b) 67 km/l (c) 72 km/l (d) 75 km/l (e) Cannot be determined

Speed = 100-k1*massPetrol Consumption per KM = c2+k2*massor Mileage = 100/(1+k2*mass)

When speed = 50KMPH, K1*mass = 50when mileage = 50KMPL, k2*mass = 1k1 = 50*k2

When speed = 75kmph, k1*mass = 25 => k2*mass = 0.5mileage = 100/1.5 = 67km/l => choice (b) is the right answer.

In a test taken by 100 students, 60 cleared cut-off in DI, 44 cleared cut-off in mathematics, 38 cleared cut-off in English and 27 students cleared cut-off in GK. 20 students cleared cut-off in all 4 sections. How many maximum students could have failed to clear the cut-off in all four sections?

(1) 38 (2) 41 (3) 47 (4) 50 (5) none of these

Solution:

In all we have 40 fails in DI, 56 fails in quant, 62 fails in VA/RC and 73 fails in GK, and 20 fails in none.

=> 1 Fails + 2 Fails + 3 Fails + 4 Fails = 80 (students).

At most we can have 40 students who fail all 4 sections [considering this as true]1 Fails (x) + 2 Fails (y) + 3 Fails (z) = 40and we still have 0 fails in DI, 16 fails in quant, 22 fails in quant and 33 fails in GK.=> x + 2y + 3z = 16 + 22 + 33 = 71.

Solving for this, we get 1 solution set (x, y, z) as (1, 24, 15) which is feasible=> 40 can be the maximum people who have 4 fails.


A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ?

(a) 364 (b) 231 (c) 455 (d) 472 (e) None of the foregoing

V(66,14 ) = V(14,66 )=(33/26)* V(14,52) =(33/26) * (26/19)* V(14, 38) = (33/19) *(19/12)* V(14, 24) = (33/12) *(12/5)* V(14,10) = (33/5)* (7/2)*V(10, 4) = (231/10)* (5/3) *V(4, 6) = (77/2)* 3 *V(4, 2) = (231/2)*2* V(2, 2) = 231*2 = 462 => choice (e) is the right answer.

Let ABCD be a rectangle with AB = a, and BC = b. Suppose x is the length of the radius of the circle passing through A and B and touching CD, and y be the length of the circle passing through B and C and touching AD. If x + y ≥ k.(a+b) for all a and b, then k =

(1) √3/2 (2) 5/8 (3) 1/√2 (4) 1/2 (5) none of these

Solution:

Let O be the centre of the circle which touches CD and passes through the points A and B, E be the point of tangency and F be the point at which OE meets AB. Since OE is perpendicular to DC and AB || DC => OE is perpendicular to AB. Also, OF = b-x, and FB = a/2. In triangle OFB, OB^2 = OF^2 + FB^2 => x^2 = (b-x)^2 + (a/2)^2 => x = b/2 + a^2/8b. Silmilarly, y = a/2 + b^2/8a

=> x+y = (a+b)/2 + (a^3 + b^3)/8ab >= (a+b)/2 + (a+b)/8


A cargo ship circles a lighthouse at a distance 20 km with speed 1500 km/h. A torpedo launcher fires a missile towards the ship from the lighthouse at the same speed and which moves so that it is always on the line between the lighthouse and the ship. How long does it take to hit?

(a) 37.7 secs (b) 56.57 secs (c) 75.43 secs (d) 94.29 secs (e) 113.14 secs

Let M be the position of the ship at the moment the missile is fired. Let V be the point a quarter of the way around the circle from M (in the direction the ship is moving).Take the point at which Lighthouse is situated to be U. Then the missile moves along the semi-circle on diameter UV and hits the ship at V.

To see this take a point T on the quarter circle and let the line UT meet the semi-circle at R. Let Z be the center of the semicircle. The angle VZR is twice the angle VUR, so the arc VT is the same length as the arc VR. Hence also the arc UR is the same length as the arc MT.

so time needed= 10 *pi*3600 /1500 =75.43 => choice (c) is the right answer.

Vineet has Rs 600 with him. Each day he buys either beer for Rs 100 or vodka for Rs 200 or whisky for Rs 200. In how many ways can Vineet spend all his money?

(1) 20 (2) 24 (3) 30 (4) 32 (5) none of these

Solution:

Let us use the notation to say Vineet can spend Rs 100n in S(100n) ways.

If he buys beer on day 1 then he can spend next Rs 100n-100 in S(100n-100) ways.If he buys vodka on day 1 then he can spend next Rs 100n-200 in S(100n-200) ways.If he buys whisky on day 1 then he can spend next Rs 100n-200 in S(100n-200) ways.

=> S(100n) = S(100n-100) + 2.S(100n-200).=> S(600) = S(500) + 2.S(400), but S(100) = 1 and S(200) = 3.=> S(300) = S(200) + 2.S(100) = 5, => S(400) = S(300) + 2.S(200) = 11=> S(500) = S(400) + 2.S(300) = 21 => S(600) = S(500) + 2.S(400) = 43


The sum of all the divisors of 19^88 - 1 which are of the form (2â).(3^b) with a, b > 0 is

(1) 168 (2) 224 (3) 360 (4) 744 (5) 1080

Solution:

Here we need to find the max power of 2 and 3 each that divides 19^88 - 1.

Now 19^88 - 1 can be written as (19^11 -1 ).(19^11 + 1).(19^22 + 1).(19^44 + 1).

Since 19 is of the form 3k+1 => each of 19^11 + 1, 19^22 + 1, 19^44 + 1 is of the form 3k+2 and hence not div by 3. But 19^11 - 1 can be written as (18+1)^11 - 1 = 18^11 + 11C1(18)^10 + .... 11C10(18) => it' div by 9 but not by 27 => max power of 3 is 2.

Similarly, each of 19^11 + 1, 19^22 + 1, 19^44 + 1 is of the form 2k+2 and hence div by 2 but not by 4. However, 19^11 - 1 = (20-1)^11 - 1 = 20^11 - 11C1(20)^10 + .... + 11C10(20), which means it' div by 4 but not by 8 => max power of 2 that divides our number is 5.

=> sum of the divisors is (2 + 2^2 + 2^3 + 2^4 + 2^5).(3 + 3^3) = 62.12 = 744.


Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

(a) 24 (b) 23 (c ) 22 (d) 20 (e)21

Each angle in degrees is 180(p-2)/p.

180-{360}/{p}=k, so 360/p has to be an integer.Factors of 360 = 2^3.3^2.5^1So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3So, 24 -2 =22 => choice (c) is the right answer.

Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X onlyChoose 2 if the question can be answered by Y onlyChoose 3 if the question can be answered by either X or Y Choose 4 if the question can be answered by both X and YChoose 5 if the question can be answered by neither X and Y

Let x and y be positive real numbers. Is x^2 + y^2 < 1?

(X) y^3 + y <= x - x^3(Y) |x| + |y| < √2 and |x| < 1 and |y| < 1

Solution:

y^3 + y <= x - x^3 <= x(1-x^2), buy y + y^3 is positive => 0 < x < 1Also, y^3 + x^3 + y < x => 0 < y < x < 1.

Now, x^2 + y^2 <= (x-y)/(x+y) + xy, we will do some reverse engineering hereIf (x-y)/(x+y) + xy < 1 => x^2 + y^2 < 1. We will prove that (x-y)/(x+y) + xy < 1 or 1 - 2y/(x+y) + xy < 1 or 2y/(x+y) > xy or 2/(x+y) > x or 2 > xy + x^2 which is true as both x, y < 1. Hence, (X) is sufficient.

For (Y), drawing graph tells us some values in 1st quadrant remain that doesn't satisfy x^2 + y^2 < 1 e.g. x = 0.8 and y = 0.6, but x = 0.1 and y = 0.1 satisfy => (Y) is not sufficient.


Consider a pair (x,y) of natural numbers satisfying x + y^2 + g^3 = xyg, where g is the greatest common divisor of x and y .Then , how many such pairs are possible?

(a) 2 (b) 3 (c) 4 (d) 5 (e) 6

Put x = Mg, y = Ng, so that M and N are coprime. Then: M + N^2g + g^2 = MNg^2. So g must divide M. Put M = M'g, then M' + N^2 + g = M'Ng^2.

So M' = (N^2 + g)/(Ng^2 - 1). Hence M'g^2 = N + (N+g^3)/(Ng^2 - 1). So (N + g^3)/(Ng^2 - 1) is an integer. If g = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. That gives the solutions (x, y) = (5, 2) and (5, 3). So assume g > 1. Since (N + g^3)/(Ng^2 - 1) is an integer and positive, we must have N + g^3 >= Ng^2 - 1, so N <= (g^3 + 1)/(g^2 - 1). If g = 2, then N <= 3. Then N = 1 gives the solution (x, y) = (4, 2), N = 2 gives (N + g^3)/(Ng^2 - 1) non-integral and hence no solution, N = 3 gives the solution (x, y) = (4, 6).

So assume g > 2. Then (g^3 + 1)/(g^2 - 1) < g + 1. Hence N ≤ g. Hence M' = (N^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(g^2 - 1), since N ≥ 1 and hence M' < 2 (since g > 2). So M' = 1. So N is a root of the quadratic N^2 - g^2N + g+1 = 0. But g^4 - 4(g+1) > g^4 - 4g^2 + 1 = (g^2 - 1)^2 and < g^4, so g^4 - 4(g+1) cannot be a square and hence N cannot be integral. So there are no solutions with g > 2 => choice (c) is the right answer

A cylinder of radius √6 cm and height 3√3 cm is inscribed inside a cube such that the axis of cylinder is along a diagonal of the cube. The length of side of the cube is

(1) 6 cm (2) 7 cm (3) 8 cm (4) 9 cm (5) none of these

Solution:

The cylinder will touch the face of the cube on its diagonal. Let x be the angle between its diagonal and the diagonal of the cube along the axis of the cylinder. Then sinx = 1/√3, and let l be the length of the side of the cube => √3 l = 3√3 + 2√6 cotx => l = 7 cm.


LMN is a triangle. MO is the angle bisector. The point P on LM is such that <LNP = (2/5) < LNM. MO and NP meet at Z . PO = ON = NZ. Find angle LPO.


Let's frst draw the triangle and assuming <LNM=x , Find out other angles as the conditions given in question are enuf to find out all the angles formed in the Triangle.

<LNP=(2/5)x ; <PNM=(3/5)x ; <NZO=<NOZ={90-(1/5)x} ....... as ON=NZ

<NZO=<PZN={90-(1/5)x} ...... vertical opposite angles<OPN=<ONP=(2/5)x ..... as PO=ON, <NZO+<OZP=180 ==>> <OZP={90+(1/5)x}<OZP=<MZN={90+(1/5)x} .... vertical opposite angles , <POZ={90-(3/5)x} .... Sum of angles in a triangle=180

<LOP=(4/5)x ..... Sum of angles in straight line=180<OMN={90-(4/5)x} ..... Sum of angles in a Triangle=180<OMN=<LMO={90-(4/5)x} ..... As MO is the angle bisector.<MPN=x .... Sum of angles in a triangle = 180<LPO={180-(7/5)x} ..... Sum of angles in straight line=180<MLN=(3/5)x .... Sum of angles in Triangle=180

Now take triangles LMN and LPO

(LM / MN ) = ( LO / ON ) ..... angle bisector theoremBut ON=OP ... Given==>> ( LM / MN ) = ( LO / OP ) And , <L is common .

==>> Both Triangles are similar.

==>> <LNM=<LPO ==>> x=180 - (7/5)x==>> x=<LNM=75 .... (1)

While required angle = <LPO = 180-(7/5)x = 180-(7/5)*75==>> <LPO = 180 - ( 7*15 )<LPO = 180 - 105 = 75==>> <LPO = 75 deg => choice (e) is the right answer.

Consider a rectangular purse of dimension 8 cm X 9 cm. What could be the maximum radius of two identical coins which can be completely put inside the purse without overlapping?

(1) 2 cm (2) 2.25 cm (3) 2.5 cm (4) 2.75 cm (5) none of these

Solution:

Let the line joining the centres of two coins be inclined at an angle x to AB => 2R + 2Rcosx = 9, 2R + 2Rsinx = 8 => (9-2r)^2 + (8-2R)^2 = (2R)^2 => R = 5/2.


N! is defined for non negative integers as N!=N*(N-1)*(N-2)* ...3*2*1 . The number of positive integers which divide (2^5 )! are

(a) 2^13.3^3.5^2 (b) 2^8.3^2.5^2 (c) 2^11.3^2.5 (d) 2^8.3^3.5^3 (e) None of the foregoing

32! is divisible by [32/2] + [32/4] + [32/8] + [32/16] +[32/32] = 31 powers of 2, and

[32/3] + [32/9] + [32/27] = 14 powers of 3. It is divisible by 5^7, 7^4, 11^2, 13^2, 17 , 19, 23, 29, 31.

In other words, 32! = 2^31 .3^14 . 5^7 .7^4 .11^2 •13 ^2 •17•19.23 .29.31 So it has (31+1) (14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1) = 2^13 . 3^3 . 5^2 => choice (a) is the right answer.

If r is the root of f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0, then which among the following is also a root of f(x) = 0?

(1) -1/r (2) (1+r)/(1-r) (3) (r-1)/(r+1) (4) All of these (5) Exactly two of these

Solution:

If r is the root of f(x) = 0, then f(-1/x) = f(x) tells us that -1/r is also the root off(x) = 0 -> (1)

When we are checking that for any r, f(r) = 0 we want to check if f((1+r)/(1-r)) = 0 or notthen (1+x)/(1-x) = y and x = (y-1)/(y+1), on substitution in f(x) we getf(y) = y^4 + ay^3 - 6y^2 -ay + 1 which is similar to our starting equation -> (2)

The 3rd choice follows from (1) and (2).

Alternative way

put x=-1/xf(-1/x)=f(x)=x^4 + ax^3 - 6x^2 - ax + 1 = 0

so if r is root, -1/r is also if z is then -1/z is.clearly also roots taken two at a time is -6so rz-1-1-r/z-z/r+1/rz=-6r(z-1/z)-1/r(z-1/z)=-4(r-1/r)(z-1/z)=-4now there is no need to go furtherz=(1+r)/(1-r) satisfies


The sum of the infinite series 1/3 + 2/21 + 3/91 + 4/273 + ... is given by which of the following?

(a) 1/4 (b) 1/2 (c) 3/4 (d) 1 (e) 3/2

General term T(n) = n/(n^2-n+1)(n^2+n+1) = 1/2 * (1/(n^2 - n + 1) - 1/(n^2 + n + 1)) Hence sum = 1/2 *(1 - 1/3 + 1/3 - 1/7 + 1/7 - ........) = 1/2 *(1) = 1/2 => choice (b) is the right answer.

In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

(1) 736 (2) 768 (3) 792 (4) 840 (5) 876

Solution:

Since the order of vowels will always remain the same despite these occupying different positions -> if we assume each vowel as X then our question is same as asking "arrange JPTRXXX" => in all 7!/3! ways


Shravya invests some amount of money in a firm M .This amount grows upto 5000 in 2 years and upto 5500 in 3 years on R% compound inetrest . Then she goes to another firm N and borrows Rs 7000 at a compound interest of R%. At the end of each year she pays back Rs 3000 to firm N. Then, the amount she should pay to firm N at the end of 3 years to clear all the dues is?


CI of 3rd year =5500-5000 =500Rate, R = 500*100/5000 =10%At the end of 3 years she pays = 7000[1+11/100]^3 - 3000[(1+10/100)^2 + (1+10/100)] = 2387 => choice (a) is the right answer.

The number of ordered (x, y) such that 1/√x + 1/√y = 1/√20 is

(1) 1 (2) 3 (3) 5 (4) 7 (5) none of these

Solution:

The given equation transforms to 1/y = (x+20-4√5.√x)/20x => √5.√x is rational.

=> 5x = (5a)^2, similarly 5y = (5b)^2

=> 1/a + 1/b = 1/2 => (a-2)(b-2) = 4 => (a, b) = (3, 6), (4, 4), (6, 3)


If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are (a)36 ( b)33 (c)66 (d)60 (e) None of the foregoing

Take b=a+k (where k is any integer)We boil down to |(c^2-k^2)*(2a+k-c)|=15Cases can only be out of the following : (+/-1,+/-15) ,(+/-15,+/-1) , (+/-3,+/-5) , (+/-5,+/-3)

(c,k) can be - > (+/-1,0) ,(0,+/-1),(+/-4,+/-1),(+/-1,+/-4),(+/-8+/-7),(+/-7,+/-8 ),(+/-2,+/-1),(+/-1,+/-2) ,(+/-3,+/-2),(+/-2,+/-3)

And each case for (c,k) will yield 2 values of a and for a given a and c , b is uniquely determined by k

Total c,k from above is (2+2) +(4+4)+(4+4)+(4+4) +(4+4) =36Since for each (c,k) we get two values of a .

Hence total (a,b,c) =2*36 =72 => choice (e) is the right answer.

In a triangle ABC, AB = AC, <A = 80˚ and S is the circumcentre. Bisectors of angles ACS and ABS meet BS and CS respectively at X and Y. Then < AXY equals

(1) 20˚ (2) 40˚ (3) 50˚ (4) 60˚ (5) 100 ˚

Solution:

Let BY and CX meet at P, at point on AS => SA = SB => <ABS = 40˚ => <ABP = 20˚ => <APB = 120˚, similarly <APC = 120˚ => <BPC = 120˚ => triangle ABP is congruent to triangle XBP => PA = PX, similarly PA = PY => <APX = <XPY = <YPA => PA = PX = PY => AXY is equilateral.

Hence, <AXY = 60˚


Consider a scalene triangle PQR.Points S, T and U are selected on sides QR, PR, and PQ respectively. The lines PS, QT, and RU meet at point Z. If area(PUZ) = 126, area(UQZ) = 63, and area(RTZ) = 24, the area of triangle PQR is ?

(a) 324 (b) 351 (c) 360 (d) 364 (e) 378

Consider triangle PQT and PQZ and PZTThe area of PQZ = 1/2 QZ *h = 189the area of PZT= 1/2 ZT *h =x so area of PZQ /area of PZT = 189/x = QZ/ZT ----------(1)Then take triangle QTR, similarly area of QZR / area of RTZ = y /24 = QZ/QT ------(2)from 1 and 2---- 189/x = QZ/ZT = y /24

Then take triangle PUR and URQ and apply the same procedure so we get the rel 24 +x /126 = y /63

Solve and get the value of x and y..So get total as 351 => choice (b) is the right answer

Pavan had 6 friends in a B-school. At a certain restaurant, he met each of them 12 times, every 2 of them 6 times, every 3 of them 4 times, every 4 of them 3 times, every 5 twice and all 6 only once. Pavan had dined out alone 8 times without meeting any of them. How many times had he dined out altogether?

(1) 36 (2) 22 (3) 32 (4) 26 (5) none of these

Solution:

Let set A, B, C, D, E, F are the dining instances when Pavan has dined with his six friends respectively.

n(A u B u C u D u F) = 6c1 * (no. of times he met each) - 6c2 * (no of times he met two) + 6c3 * (no. of times he met three) - .... - 6c6 * (no. of times he met all six)= 6 * 12 - 15 * 6 + 20 * 4 - 15 *3 +6 *2 - 1 * 1 = 28

Therefore total number of times is 28 + 8 = 36


There are three runners viz, Nishant , Deepak and Mohit who jog on the same path. Nishant goes jogging every two days. Deepak goes jogging every four days. Mohit goes jogging every seven days. If it’s the first day that they started this routine, what is the total number of days that each person will jog by himself in the next seven weeks?

(a) 12 (b) 13 (c) 14 (d) 15 (e)16

Nishant jobs a total of 25 days, Deepak a total of 13 days, and Mohit a total of 7 days. We know that Nishant jogs on days 1,3,5,7,.....49, and Deepak jobs on days 1,5,9,.....49, and Mohit jogs on days 1, 8, 15, 22,...43. Obviously when Deepak jogs, Nishant will always be jogging also, so Deepak jogs 0 days alone. Nishant jogs 25-13=12 days without Deepak also jogging, but Mohit jogs on 4 odd numbered days, 2 of which Deepak also jogs, giving us 12-4+2=10 days that Nishant jogs alone. Nishant jogs on every odd numbered day, so Mohit jogs alone only on even numbered days. Because there are 4 odd numbered days, as stated above, there are 3 even numbered days, so Mohit jogs alone on 3 days.

Answer: Nishant: 10Deepak: 0Mohit: 3=> choice (b) is the right answer.

Let |2x-1| - 3|x+1| = a has two real solutions p and q satisfying 2 <= |p-q| <= 10, then Max(a) - Min(a) equals (a) 19/3 (b) 15/2 (c) 35/6 (d) 11/2 (e) none of these

Solution:

Let f(x) = |2x-1| - 3|x+1|We have f(x) = x+4 if x <= -1,-5x-2 if -1 < x < 1/2,-x-4 if x >= 1/2.

Draw the graph of f(x) now and see that the nodes are (-1, 3) and (1/2, -9/2). The

shape of this graph is almost inverted V with V-vertex at (-1, 3) and sligh right shift from (1/2, -9/2).

The equation f(x) = a has no solution if a > 3 because for the line parallel to x-axis doesn't intersect the graph for y > 3.

The equation f(x) = a has 2 solutions when -9/2 <= a <= 3 and the solutions are determined by the intersection of line y = a and the lines y = x+4 and y = -5x-2. The solutions are (a-4, a) and (-(a+2)/5, a).

We are given that solutions p and q are such that 2 <= |p-q| <= 10 => 2 <= |a-4 + (a+2)/5| <= 10 subject to constraint on a => -9/2 <= a <= 4/3Similarly, the equation f(x) = a has 2 solutions when a < -9/2, and the solutions are determined by the intersection of line y = a and the lines y = x+4 and y = -x-4. The intersection points would be (a-4, a) and (-a-4, a).We are given that solutions p and q are such that 2 <= |p-q| <= 10 => -5 <= a <= -9/2.

Combining the above we get -5 <= a <= 4/3.


Consider a polynomial function P(y) =y^3+2y^2+5 and one another polynomial function, Q(y)=y^4 -3y^2+2y+1. Let there be two more functions S(y) and T(y), that satisfy gcd(P(y), Q(y) ) =S(y)*P(y) + T(y)*Q(y). Now consider a function M(y) = T(y) - S(y). Then, the product of all the roots of M(y) is given by


First stage division gives ...Q(y) = P(y)(y-2) + y^2 -3y +11y^2 -3y +11 = Q(y) - (y-2)P(y)And then from the second stage,P(y) = (y^2-3y+11)(y+5) + 4y-50= {Q(y) - (y-2)P(y)}(y+5) + 4y-50From the third stage of division, 4(y^2 -3y +11)= (4y-50)(y+19/2) +519=>519 = 4[Q(y) - (y-2)P(y) ] - [(y^2+3y-9)]P(y) - (y+5) Q(y)] [y+19/2]=>519 = 4 Q(y) - (4y-8 )P(y) - (y^2+3y-9)[y+19/2] P(y) + (y+5) Q(y) [y+19/2]=-[4y-8+y^3+19y^2/2 +3y^2 +57/2y -9y-171/2]P(y) + [y^2+19/2y +5y+95/2 +4]Q(y)=-P(y)/2 [2y^3 + 25y^2 +47y -187] + Q(y)/2[ 2y^2+29y+103]=>519 = [-(2y^3 + 25y^2 +47y -187)/2]P(y) + [(2y^2+29y+103)/2]Q(y)So, T(y) - S(y)= (2y^3 +27y^2 + 76y -84)/2 = y^3 +27/2y^2 + 38y -42So, product of roots =42 => choice (a) is the right answer.

Let f(k) be defined on integer k as f(k) = [k](3) + [2k](5) + [3k](7) - 6k, where [k](2n+1) denotes the multiple of (2n+1) closest to k. How many values can f(k) assume?

(1) 4 (2) 7 (3) 10 (4) 13 (5) none of these

Solution:

[k](n) = k + (n-1)/2 - (k+ (n-1)/2) mod n => f(k) = 6 - (k+1) mod 3 - (2k+2) mod 5 - (3k+3) mod 7

=> k +1 = a mod 3, where 0 <= a < 2: 2k +2 = b mod 5, where 0 <= b < 5: 3k +3 = c mod 7, 0 <= c < 7 => -6 <= f(k) <= 6

=> k = a-1 mod 3, k = -2b -1 mod 5, k = -2c -1 mod 7: By chinese remainder thoerem this has a solution for any choice of a, b, c => f(k) takes all values in [-6, 6].


A circle passes through the vertex C of rectangle ABCD and touches its sides AB and AD at P and Q respectively. If the distance from C to the line segment PQ is equal to 4 units, then the area of the rectangle ABCD in sq. units (is)

(a) 20 (b) can not be determined (c) 16 (d) greater than 20 (e) none of the foregoing

Let M be the feet of perpendicular from C to PQ. Now, by alternate segment theorem we have=> < CQM = < CPB and < CPM = < CQD. Thus, right triangles CQM and CPB are similar and also CPM and CQD. Thus, CQ/CP = CM/CB and CP/CQ = CM/CD. Thus, CB.CD = CM^2 = 16. => choice (c) is the right answer.

Two cars A and B started from P and Q respectively towards each other at the same time. Car A was travelling at a speed of 54km/h but due to some problem reduced its speed by 1/3rd after travelling for 60 minutes. Car B was travelling at a speed of 36km/h. Had the technical problem in car A had arisen 30 minutes later, they would have met at a distance which is (1/30*PQ) more than towards Q than where they met earlier(PQ > 120km). Anothet car C starts from P, 90 minutes after car B started at Q, and car C travels towards Q with a speed of 36km/h, at what distance from P will cars B and C meet?

(a) 63 km (b) 48 km (c) 40.5 km (d) none of the foregoing

Solution:

Two cars meet at mid-point of PQ if they were at same speed. If the starting point of A shifts by l1 towards Q and that of B by l2 towards Q, where l2 < l1), the meeting point shifts by (l1 - l2)/2 towards Q.=> meeting point shifts by (27-18)/2 km towards Q => PQ = 30*(4.5) km = 135 km

B covers 3/2*36 = 54 km from Q when C starts.Cars B and C are (135 - 54)km apart when C starts => PR = 81km. If C and B meet

http://CB.CD/

=> they meet in the mid-way as they havesame speed.


Maximums needed to calculate the volume of a rectangular room. He multiplied the length and the breadth correctly but the breadth had been incorrectly jotted down, it was one-third larger than what it should have been. To compensate for this, he reduced the height by one-third, then multiplied it on. He figured this was okay since the breadth was equal to the height. He then found his volume was off by 20 m^3. What was the actual volume?

(a) 160 (b) 120 (c) 210 (d) 159 (e) None of the forgoing

Actual Volume = L*B*H. Given, L*B*H - L*4/3B*2/3H = 20 => 1/9*LBH = 20. => choice (e) is the right answer.

N people vote for one of 27 candidates. Each candidate's vote % is atleast one less than his/her number of votes. What is the smallest possible value of N?

(a) 108 (b) 127 (c) 134 (d) 162 (e) none of these

Solution:

If a candidate has just 2 votes, then 2/n <= 1/100, so n >= 200. If a candidate has3 votes, then n >= 150.

So in a minimal solution each candidate must have at least 4 votes. If all have atleast 5, then n >= 135.

If a candidate has 4, then 4/n <= 3/100, so n >= 134. This can be achieved: 1candidate has 4 votes, the other 26 have 5 each. Then 5/134 = 3.7%, 4/134 = 2.99%.


Which among the following have the same graph?

I) y = x-2 II) y = (x^2-4)/(x+2) III) y(x+2) = x^2-4 IV) y = (√x-√2)*(√x+√2)

(a) I and III only (b) II and IV only (c) I, III, IV (d) II and III only (e) none of the forgoing

y = x-2, will be a straight-line with slope 45 deg and making intercept of 2 and -2 on X and Y axis respectively. II) Same as I except for a breakpoint at x = -2. III) at x = -

2, we will also have a line parallel to the Y axis. IV) Graph is restricted to first and fourth quadrants only.

=> choice (e) is the right answer.

Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X onlyChoose 2 if the question can be answered by Y onlyChoose 3 if the question can be answered by either X or YChoose 4 if the question can be answered by both X and YChoose 5 if the question can be answered by neither X and Y

For positive reals x, y, z, Is 1/x + 1/y + 1/z <= 1?

(X) For every quadrilateral with sides a, b, c, d, x.a^2 + y.b^2 + z.c^2 > d^2(Y) √x + √y +√ z >= √x.√y.√z

Solution:

(X)Take the 3 sides of the quadrilateral as 1/x, 1/y, 1/z -> since the sum of 3 sides > fourth side => 4th side can be taken as 1/x + 1/y + 1/z - 1/n where n > 0 and largeWe are given that, x/x^2 + y/y^2 + z/z^2 > (1/x + 1/y + 1/z - 1/n)^2 hold true for every nTake 1/x + 1/y + 1/z = p, and limiting case when n -> Infinity=> p > (p-1/n)^2 => p < 1.

(Y)Put 1/x = a, 1/y = b, 1/z = c => √(ab) + √(bc) + √(ca) >= 1=> 2√(ab) + 2√(bc) + 2√(ca) >= 2=> By AM-GM on positive numbers, (a+b+c) >= 1 => 1/x + 1/y + 1/z >= 1. Hence, (Y) is not enough but would have been enough if question asked was, Is 1/x + 1/y + 1/z < 1 or it was given that at least two of x, y, z are distinct?


All the digits of a 50 digit positive number are 4 except for the nth digit. If the number is divisible by 13 for some choice of that nth digit, then how many possible values can n have?

(a) 17 (b) 21 (c) 25 (d) 33 (e) none of them

The seed of 13 is 4 which means to check the divisibility by 13 of a n digit number, multiply the last digit by 4 and add the result to the initial (n-1) digits and see if it is div by 13 e.g. 182 -> 18+2*4 = 26. Thus, 182 is div by 13. 1001 -> 100 +1*4 = 104 -> 10 + 4*4 = 26, hence 1001 is div vy 13. It's also the same thing for 182 as saying that 2*(4^2) + 8*4 + 1 is div by 13 or for 1001 1*(4^3) + 1 is div by 13 - this comes from recurrence. Note that we can find the seed of every prime number. The seed of 7 is 5, the seed of 17 is 12, the seed of 19 is 2.

If 444...x...444 is div by 13 => 4*(4^49 + 4^48 + ... + 1) + (x-4)*4^(n-1) is div by 13, where nth digit is x and not 4.

4*(4^49 + 4^48 + ... + 1) = 4/3*(4^50-1).By Fermat's theorem 4^12 = 1 mod (13) => 4^50 = 3 mod (13). Thus, 4*(4^50 - 1) leaves remainder of 8 when div by 13. If 4/3*(4^50-1) leaves x by 13 => 3x%13 = 8 => x = 7. Thus 4/3*(4^50-1) leaves remainder 7 when div by 13.Now, 7 + (x-4)*4^(n-1) is div by 13.when n = 1, 7+(x-4) has to be div by 13, we have no such xwhen n = 2, 7+4*(x-4) has to be div by 13, we have no such xwhen n = 3, 7+3*(x-4) has to be div by 13, we have x = 6when n = 4, 7+12*(x-4) has to be div by 13, we have no such xwhen n = 5, 7+9*(x-4) has to be div by 13, x = 9when n = 6, 7+10*(x-4) has to be div by 13, x = 2when n = 7, 7+(x-4) has to be div by 13, we have no such x and the whole cycle of 6 repeats from here. In each cycle of 6 we have 3 desired n.

Till n = 50 we have 48*3/6 + 0 = 24 such numbers. => (e) is the right answer.

Consider two cones of heights 1 and 8 units having the same base radii. It is found that their height is increased by x keeping their vertex angle unchanged, their volume becomes equal. Then x equals

(1) 2/3 (2) 4/3 (3) 8/3 (4) 16/3 (5) none of these

Solution:

Suppose the radii of two cones to be r initially and r1 and r2 after increasing height by xsince the vertex angle are unchangedr/1 = r1/(1+x) => r1 = r(1+x)r/8 = r2/(8+x) => r2 = r(8+x)/8

now the volumes are equal=> 1/3*pi*r1*r1*h1 = 1/3*pi*r2*r2*h2=> [r(1+x)]^2 * (1+x) = [r(8+x)/8]^2*(8+x)=> x = 4/3


Let a, b, c be positive reals. (I) and (II) are independent statements.(I) Minimum value of a^3/4b + b/8c^2 + (1+c)/2a is p(II) a + b + 2c = 8 and a^2 + b^2 + 2c^2 = 25. Maximum possible value of c is q.

Then which among the following is |p-q|?

(a) 5/2 (b) 9/4 (c) 1/2 (d) 3/8 (e) none of the foregoing

(I) a^3/4b + b/8c^2 + (1+c)/2a = a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a. For positive reals AM >= GM and the equality occurs when all the numbers are equal. Thus, a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a >= 5(a^3/4b*b/8c^2*1/4a*1/4a*c/2a)^1/5 = 5/4. Thus, p = 5/4 when a^3/4b = b/8c^2 = 1/4a = c/2a.

(II) a+b=8-2c; a^2+b^2 >= 1/2*(a+b)^2 = 1/2(8-2c)^2 => 25-2c^2 = 1/2(8-2c)^2 => 1/2 <= c <= 7/2. Thus, q is 7/2 when a=b.=> (b) is the right answer.

The question is followed by two statements X and Y. Answer using the following instructions:

Choose 1 if the question can be answered by X onlyChoose 2 if the question can be answered by Y onlyChoose 3 if the question can be answered by either X or YChoose 4 if the question can be answered by both X and YChoose 5 if the question can be answered by neither X and Y

Two vessels A and B having different capacities are partly filled with spirit of different concentrations. If the content of A is poured into vessel B till it is full, then the % concentration of B increases by 5%. Is the difference in concentrations of spirits in containers more than 5%?

(X) initial level of spirit B is greater than that in A

(Y) If the content of B is poured into vessel A till it is full, then the % concentration of A decreases by 10%

Solution:

Pouring the content of A increases the concentration of B by 5% => A will at least have 5% more concentration than B. But since B is already partly filled => concentration of A - concentration of B > 5%. We pick the strongest option as our answer which is already implied in the question and we don't need to bother checking for (X) and (Y).


ABCD is a convex quadrilateral in which <(BAC) = <(CBD) = 30 deg, <(CAD) = 60 deg, <(CDB) = 15 deg. If E is the point of intersection of AC and BD, then measure of <(BEA) in degrees (is)

(a) can not be determined (b) 75 (c) 105 (d) 120 (e) none of the foregoing

Draw a circle passing O through each point B, C, and D. In any circle the angle subtended by an arc at the centre is twice the angle subtended at the perimeter. Since, <(BAC) = 2*(<(CDB)) and <(CAD) = 2*(<(CBD)) => A is the centre of the circle O.

Thus, AB = AD is the radius of circle O => <(ABD) = 1/2*(180-90) = 45 deg => <(BEA) = 105 deg => choice (a) is the right answer.

Let F be be mini 4X4 chessboard => it has 16 fields in all. In how many ways is it possible to select two fields of F such that the midpoint of the segment joining the centres of the two fields should also be the centre of a field?

(1) 15 (2) 18 (3) 24 (4) 32 (5) none of these

Solution:

The fields of the centre of the fields of the chessboard can be assigned coordinates.Let the centre of the 1st field in the left most bottom corner be (0, 0) =>the centres of the fields are of the form (x, y) where 0 <= x, y <= 3.The coordinate of the midpoint of the segment connecting the points (a,b) and (c,d) are (a+c)/2 and (b+d)/2, and these coincide with the coordinates of the centre of a field if and only if a+c and b+d are both even.

The point (a,b) may be the centre of any field out of the 16 fields of the mini chessboard. (a,b) having been selected, the number c can be any of the 2 numbers that have the same parity as a, and independently of that, d can also have 2 different values. In order to make the chosen points different, the number of the possible pairs (c,d) is 1 less than 2.2=4. Buthe order of the two selected points does not matter => the number of the appropriate number pairs is 1/2*(4*4)*3 = 24.

If our question was on NXN square, then it could have been solved in a jiffy with the above logic. For 6X6 the answer is 144. For 8X8, the answer is 480.


All India Pagalguy Meet of year 2008' is in June (which has 30 days), but Allwin forgot which day, so he asked around. Rohit said that the date was an odd number; Apurv claimed it was greater than 13. Divya declared it was not a perfect square, while Sonam swore it was a perfect cube. Finally, Grand-ma tells Allwin the date was less than one-fourth her (Grand-ma's)age, which Allwin knew to be 68. Yesterday Allwin learned that only one of them had told the truth! If the date of the All India PG Meet is D (numerical value), then

(a) D is uniquely determinable (b) D can have exactly 2 values (c) D doesn't exist (d) D has atleast 4 values (e) none of the foregoing

The question says that only one of them spoke the truth! Let's see if Rohit spoke the truth or not. If he has then the date D is an odd number <= 13 (since Apurv must have lied), and D > 17 (Grandma must have lied too in this case). Contradiction, and hence we have no such value of D. If Apurv spoke the truth, then D > 13 and D >= 17 (Grandma must have lied).

Since, Rohit and Divya also lie => D is a even perfect square. Not possible as D <= 30. Please check yourself that if Divya or Sonam speak truth then no value of D exists.Let's check if Grandma has spoken the truth or not. If she has then D < 17. Rohit lies => D is even, Divya lies => D is a perfect square, Sonam lies => D is not a perfect cube => D can be 4 or 16 but as Apurv also lies => D = 4 is the only possibility => choice (a) is the right answer.

Let ABCD be a rectangle and E be a point beyond C on AC extended. If < DEB = < CBE, and AB/BC = 3, then BE/CE equals

(1) √3: √2 (2) √2: 1 (3) √2 + 1: √ 3 (4) 4: √5 + 1 (5) √3+1: 2

Solution:

Treat A (0, 0), B (3, 0), C (3, 1), D (0, 1) and E is (3p, p) as it lies on AC. Let tanI denotes the Inverse of tan.

Slope of EB = tanI[p/(3p-3)] = 90 - µ (equal angle)Slope of ED = tanI[(p-1)/(3p)] = 90 -2*µ=> µ = tanI[(3p-3)/p], and 2*µ = tanI[(3p)/(p-1)].

Eliminiate µ using tan (2*µ) = 2tan(µ)/(1-tan^2(µ))and we get p as [11 +/- root(11)]/10.

We need to find square root of [(3p-3)^2 + p^2]/[(3p-3)^2 + (p-1)^2] = 2.

You all will be stunned to note that the ratio AB/BC was irrelevant and infact our asked ratio is always √2:1. Check this in case of a square.

Alternative Solution

Let F be the intersection of the lines BC and ED. The triangle BEF is isosceles, according to the given information. Let R be the point that is on the same side of line EB as F, and for which the triangle EBR is isosceles and right-angled, that is, BR=ER and < ERB=90 degrees.

Let, furthermore, T be the intersection of the circumscribed circle of rectangle ABCD with the line ED. BD is a diameter in the circle, therefore the segment BT is perpendicular to the line ED.As ACTD is a cyclic quadrilateral, < ETC= < EAD = < ECF. The angles of the triangles ECF and ETC are pairwise equal, thus the two triangles are similar, and EC:EF=ET:EC, that is, EC^2=EF.ET.

As < ERB = < ETB = < 90 degrees, the quadrilateral ERTB is also cyclic, and < RTE =< RBE =< ERF. The triangles ERF and ETR are also similar, thus with the above reasoning we have ER^2=EF.ET.We have obtained that EC^2=ER^2=EF.ET, that is, EC=ER, and hence our answer.


Let m be the largest positive term of an harmonic progression whose first two terms are 2/5 and 4/9.

http://EF.ET/

http://EF.ET/

http://EF.ET/

A real number r satisfying m/2-1/n < r <= m+1/n, for every positive integer n, is best described by:

(a) 1 < r < 5 (b) 2 < r <= 4 (c) 1 < r <= 5 (d) 2 <= r <= 4 (e) none of the foregoing

If 2/5 and 4/9 are the first two terms of the harmonic progression then 5/2 and 9/4 are the terms of the corresponding AP => common difference of the AP is -1/4 => we want 5/2 + (n-1)*(-1/4) as positive as well as minimum. Thus, n = 10 and the AP term is 1/4. m = 4. Please note that our HP consists of 10 terms only.Now, 2-1/n < r <= 4+1/n for all positive intger n. Putting n = 1 we get 1 < r <= 5. Putting n = 2 we get 3/2 < r < 9/2 etc. If r is more than 4 then r <= 4 + 1/n fails for some n (e.g. if r = 4.01 then n = 101). If r is less than 2 then 2-1/n < r fails for some n (e.g. if r = 1.99 then n = 100)Note that 1/n > 0 always for any choice of positive integer n. Thus, 2-1/n < 2, and 4 <= 4+1/n is true for all postive integers n.=> the best description of r contains 2 and 4 as well besides the range (2, 4).=> choice (d) is the right answer.

The number of real roots of the equation |1 - |x|| - (1.01)^(1.01x) = 0 is/are


The number of real solutions to |1 - |x|| - (1.01)^(1.01x) = 0 is


At x = 0 we have a solution. Let f(x) = |1 - |x|| - (1.01)^(1.01x) f(1) < 0, f(3) > 0 => we have odd number of solutions between (1, 3)but f(x) is increasing in (1, 3) => we have just 1 solution in (1, 3)f(-1) < 0 and f(-2) > 0 and f(x) is decreasing in (-2, -1) => 1 solution in this interval also.Also, f(1000) < 0 and f(3) > 0 and f(x) in (3, 1000) is x - 1 - (1.01)^(1.01x) which is a decreasing function in this interval => we have one more root in (3, 1000) => 4 roots in all

=> choice (e) is the right answer.

The set S has 5 elements. In how many ways can one select two (possibly identical) subsets of S whose union is S?

(a) 32 (b) 63 (c) 64 (d) 93 (e) 122

Let the subsets of S be A and B. For each element in S we have three choices (it can belong to either of A, B or both). That gives each pair of subsets twice except for the case A = B = S. Hence, we can select 2 subsets in (3^5 + 1)/2 ways.=> choice (e) is the right answer.

Katrina walks down an up-escalator and counts 150 steps. Priyanka walks up the same escalator and counts 75 steps. Katrina takes three times as many steps in a given time as Priyanka. How many steps are visible on the escalator?


Solution:

Let T be time Katrina takes to make 25 steps. Then Katrina takes 3T to make 75, and Priyanka takes 2T to make 150. Suppose the escalator has N steps visible and moves n steps in time T. Then Priyanka covers N + 2n = 150, N - 3n = 75. Hence N = 120, n = 15


Twinkle tells Raveena that she has got 3 kids and 2 of these kids are twins, and also that their ages are all integers. She tells Raveena the sum of the ages of her kids and also the product of their ages. Raveena says that she has insuficient information to determine the ages, but one possibility is that the twins are a prime number of years old. If Twinkle’s twins are teenagers and their age is not prime, then the sum of the ages of her kids is

(a) a prime number (b) is greater than 43 (c) a composite number(d) exactly 2 of the foregoing (e) still undeterminable

Let x,and y be the ages of Twinkle’s children. Since the twins are teenagers, x is an integerwith 13 <=x <=19. Furthermore, Let p, p and q be the solution that Raveena discovered with the twins having prime age p. Since Raveena knows the sum of the ages, we must have 2x + y = 2p + q. Similarly, she is given the product of the ages, so x^2y = p^2q.Multiply the first equation by p^2 and replace p^2q by x^2y to obtainp^2(2x + y) = p^2(2p + q) = 2p^3 + p^2q = 2p^3 + x^2y, and thus y(x−p)(x+p) = y(x^2−p^2) = 2p^2(x−p). Since x is different from p, we can divide by x−p andget y(x + p) = 2p^2. In particular, the integer x + p divides 2p^2. Since p is a prime, the positive divisors of 2p^2 are 1, p, p^2, 2, 2p and 2p^2. Furthermore, x + p > 1, 2, p and x + p not equal to 2p since x not equal to p. This leaves x + p = p^2 or x + p = 2p^2 and hence x = p^2 − p or x = 2p^2 − p.Since p is a prime and 13 <=x <= 19, the only possibility here is p = 3 and x = 2p^2 − p = 15. Using (x + p)y = 2p^2, it then follows that y = 1. Thus Twinkle’s children have ages 15, 15 and 1. Note that 2x + y = 31, so 2p + q = 31 and q = 25. As a check, we observe that (15)^2•1 = 3^2 * 25.

=> choice (a) is the right answer.Two budding mathematicians, Srikar and arbit_rageur, play a game. The computer selects some secret positive integer N < 60 (both Srikar and arbit_rageur know that , but that they don't know what the value of N is). The computer tells Srikar the unit digit of N, and it tells arbit_rageur the number of divisors of N. Then, Srikar and arbit_rageur have the following dialogue:

Srikar: I don't know what N is, and I'm sure that you don't know either. However, I know that N is divisible by at least two different primes.

arbit_rageur: Oh, then I know what the value of N is.

Srikar: Now I also know what N is.

Assuming that both Srikar and arbit_rageur speak truthfully and to the best of their knowledge, how many possible values of N are there?

(a)0 (b) 1 (c) 2 (d) 3 (e) none of these

Solution:

Srikar's first statement implies that has a last digit of 0, and must be divisible by 10. Since 10, 20, 30, 40, and 50 have 4, 6, 8, 8, and 6 factors, respectively, arbit_rageur can only be sure of the value of if he knows has 4 factors. Therefore, the only possible value of is 10.


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The question was that was 1st Last Year ---------------------------------------------

The equation |x-1| - |x-2| + |x-4| = m has exactly n real solutions for some real m. Then which among the following relations between m and n can not be true?

(a) m/n = 3/5 (b) m = n (c) m/n = 3/2 (d) m/n = 5/3 (e) m = n-1

Since 1, 2, 4 are the critical point, we divide the domain into 4 regions; 1) when x > 4 |x-1| - |x-2| + |x-4| = (x-1)-(x-2)+(x-4) = m => x-3 = m => m>12) when 2 < x <=4 |x-1|-|x-2|+|x-4| = (x-1)-(x-2)-(x-4) = m => 5-x = m => 1 <= m < 33) when 1 < x <=2 |x-1|-|x-2|+|x-4| = (x-1)+(x-2)-(x-4) = m => x+1 = m => 1 < m <= 34) when -Infinity < x <=1 |x-1|-|x-2|+|x-4| = -(x-1)+(x-2)-(x-4) = m => 3-x = m => 2 <= m <= InfinityClearly for n=4 we have 2 < m < 3; for n=3 we have m = 2,3; for n=2 we have 1 < m < 2; for n=1 we have m=1; for n=0 we have 1 > m. Looking at the choices (a) m = 2.4 and n = 4 satisfy (b) m = n = 3 satify (d) m = 10/3 and n = 2 satisfy (e) m = -1 and n = 0 satisfy

=> choice (c) is the right answer. Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X onlyChoose 2 if the question can be answered by Y onlyChoose 3 if the question can be answered by either X or YChoose 4 if the question can be answered by both X and YChoose 5 if the question can not be answered by combining X and Y also

If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even

Solution:

Put 2-a = m => 0 < m < 1, we will show that [(2-m)k/m] = [(2-m)[k/m] + (2-m)/2]

[(2-m)k/m] = [2k/m] - k, since k is an integer.

Let us take this case by case.

Case 1: when k/m is an integer = i[(2-m)k/m] = 2i - k, and [(2-m)[k/m] + (2-m)/2] = [2i -mi + 1-m/2] = [2i -k + 1-m/2] = 2i - k.

Case 2: when i-1/2 < k/m < i, where i is an integer[(2-m)k/m] = 2i - 1 - k, [(2-m)[k/m] + (2-m)/2] = [2i - m/2 - k] = 2i - 1 - k

Case 3: when i < k/m < i + 1/2, where i is an integerProve yourself here that [(2-m)k/m] = [(2-m)[k/m] + (2-m)/2]

=> X is sufficient for all k

Now Y adds 1/2 in X and depending on the value of a[k/(2-a)], we can have multiple (two) possible value of [ak/(2 - a)] for q.


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The question was that was 2nd Last Year ---------------------------------------------

When Katrina get's Swiss chocolcates, she swings in delight, equal to her total chocolates at that time. For instance, if Katrina gets 3 chocolates, then 7 chocolates and then 3 chocolates again, she at first makes 3 swings, then she makes 3+7 = 10 swings and then she makes 3+7+3 = 13 swings, making a total of 3+10+13 = 26 swings. If all of Katrina's Swiss chocolates are in a group of either 3 or 7 and Katrina makes 99 swings during the process, in how many different ways can she get the chocolates in that process ?


If a1, a2, a3, ... are the numbers in which Katrina gets chocolates in order then Katrina's total swings will be n*a1 + (n-1)*a2 + .... + an where a1, a2 etc. take the values 3 or 7. Given the fact n*a1 + (n-1)*a2 + .... + an = 99; if all a1, a2 etc are 3 then 3n(n+1)/2 = 99 => n(n+1) = 66, so we know n can atmost be 7. Taking a1, a2

as 7 we can see that n is greater than 4.OK, if it's 7a1 + 6a2 + ... + a7 = 99 then because 3 and 7 are each of the form 4k-1 we see 7a1 + 6a2 + ... + a7 = 99 reduces to 4*(7k1 + 6k2 + ...+ k7) = 99 + (1 + 2 +... +7) = 99 + 28, which is not possible. Similarly it's not for n=5. For 6 it's possible as 99 + (1 + 2 +...+ 6) = 120 is div by 4.

Now you are left with 6k1 + 5k2 + ...+ k6 = 30 where k1,k2 ... takes either 1 or 2

(k1, k2, k3, k4, k5, k6) = (1, 2, 1, 2, 1, 2), (2, 1, 1, 1, 2, 2), (1, 1, 2, 2, 2, 1), (1, 2, 2, 1, 1, 1), (2, 1, 1, 2, 1, 1

=> choice (c) is the right answer. Given p and q be positive such that 2 >= p-q, the min value of 2/(p+q) + q/2 is

(1) √2 - 1/2 (2) (√2 + 1)/2 (3) 1 (4) 1/√2 (5) none of these

Solution:

p+k = 2+q where k >= 0Thus, 2/(p+q) + q/2 = 2/(2+2q-k) + (2q+2-k)/4 + (k-2)/4.By AM-GM on positive integers, q can be generated based on kand our expression is >= sqrt(2) + (k-2)/4 which assumes min at k = 0


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The question was that was 52nd Last Year ---------------------------------------------

Let the equation 8^(x/(x+2)) = 6/3^x has n roots m(1), m(2), ..., m(n) such that m(1) >= m(2) >= ... >= m(n). Then the numerical value of m(1) - m(2) - ... - m(n) is

(a) 2+3log2/log3 (b) 2-3log2/log3 (c) 3+2log2/log3 (d) 3-2log2/log3 (e) none of the foregoing

The eqn becomes 2^(3x/(x+2)) . 3^x = 2*3 =6,or, 2^(3x/(x+2) - 1) . 3^(x-1) = 1.So, (2x - 2)/(x+2). log2 + (x-1).log3 = 0=> either x-1 = o => x = 1 or,x+2 = -2log2/log3,so, x = -2 -2log2/log3So, m1 - m2 = 1-(-2 - 2log2/log3)= 3+2log2/log3

=> choice (c) is the right answer. Nbangalorekar bought a 10 kg of water-melon in Bangalore that had 99% water. After the water-melon was left outdoors for a day, it was 95% water. What was the weight of the dehydrated water-melon?

(1) 9.6 kg (2) 9.5 kg (3) 2 kg (4) can not be determined (5) none of these

Solution:

Initially, 0.1 kg of dry part + 9.9 kg of water is there. After dehydration, 0.1 kg of dry part id 5% of total water-melon's weight => weight of dehydrated water-melon is 2 kg.


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The question was that was 53rd Last Year ---------------------------------------------Concentric circles radii 1, 2, 3, ... , 100 are drawn. The interior of the smallest circle is colored red and the annular regions are colored alternately green and red, so that no two adjacent regions are the same color. The total area of the green regions divided by the area of the largest circle is


Total Area of Green Regions = pi*(100^2-99^2+98^2-97^2+....+2^2-1^2)Area of the largest circle = pi*100^2=>Required ratio = (100^2-99^2+98^2-97^2+....+2^2-1^2)/100^2= {(100+99)(100-99)+(98+97)(98-97)+.....+(2+1)(2-1)}/100^2= (199+195+191+...+7+3)/100^2= 5050/100^2 = 101/20

=> choice (c) is the right answer. Given a trapezium ABCD with AB || CD , CD = 2AB and DB perpendicular to BC. Let E be the intersection of lines DA and CB, and F be the midpoint of DC. Which among the following is not true?

(1) ABFD is a rhombus(2) Triangle CDE is isosceles(3) If AF and BD meet at G, and GE and AB meet at H then the line DH bisects segment EB(4) At least two of the above(5) none of the above

Solution:

For (1) & (2)AB/CD = EB/EC = EA/AD => EB= BC and EA = ADNow, looking at CDE, we see that the BD is a median (EB = BC), and is also perpedicular to EC (DB at right angles to EC) => the median is also an altitude.

=> the triangle is isosceles with DE = DC (this proves (2))

Now, since A is the midpoint of DE and F of DC, AF||EC=> DB perpendicular to AF.Also, we know that AB = DF and AB||DF=>ABDF is a parallelogram with diagonals intersecting at right angle=> it is a rhombus.

For (3)We know that G is the midpoint of AF and BD.Consider the triangle EDB.EG is the median to BD, while BA is the median to ED. They intersect at the point H.=> the third median must also pass through this point.Joining the third vertex D to H and extending, DH is the median and hence bisects EB.=> (3) is true.


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The question was that was 54th Last Year ---------------------------------------------A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date?

(a) 420 (b) 560 (c) 630 (d) 720 (e) 840

Let no. of Fishes on earlier day = x60 fishes are tagged, out of which 25% have died/movedremaining fishes that are tagged = 45

No. of fishes on day of second observation = 5/4xNo. of fishes caught = 70No. of Fishes tagged = 3 = 4.2857% (3/70)Total No. fishes on Second day of Observation = 45/.042857 (45*70/3) = 1050

No of Fishes on earlier day = x = 4/5*1050 = 840

=> choice (e) is the right answer. 5 professors decide to hold daily meetings such that (i) at least one professor attend each day (ii) a different set of professors must attend on different days. (iii) on day N for each 1 <= d < N, at least one professor must attend who was present on day d. How many maximum days can meetings be held?

(1) 14 (2) 16 (3) 20 (4) 24 (5) none of these

Solution:

We need to find the largest possible number of subsets of {1, 2, 3, 4, 5} such that no 2 subsets are disjoint. Fix one element from the set to be presented in each subset and we can have 2^4 such possibilities.


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The question was that was 55th Last Year ---------------------------------------------The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the cost price of 6 kg oranges and 5 kg apples (is)

(a) can not be determined (b) Rs 318 (c) Rs 375 (d) Rs 384 (e) none of the foregoing

20 Oranges and 1KG of apple = 12 Apples and 1.2 KG of Oranges (50*1.2)

If the profit %ages are different, then it must be clear that the ratio of apples : Oranges in the 60Rs. sale and the 72Rs. sale must be the same.

Hence, we can safely say that 1.2 KG of oranges = 20 Oranges and 1KG of apples = 12 apples.

SP of 20 Oranges and 12 Apples = 72 => SP of 5 Oranges and 3 apples = 18CP of 1.2KG of Oranges and 1KG of Apples = 60 => CP of 6KG Oranges and 6KG apples = 300

=> choice (b) is the right answer. Square ABCD has side length 6. Circle Q is tangent to sides AB and BC, and is externally tangent to circle P. Circle P is tangent to sides CD and DA, and is externally tangent to circles O1 and O2. Circle O1 is tangent to side CD, circle O2 is tangent to side DA, and circles O1 and O2 are externally tangent to each other and to circle P. If the radius of circle P is twice the radius of circle Q, and if circles O1 and O2 both have radius r, then r is (upto 2 places of decimal)

(1) 0.29 (2) 0.36 (3) 0.47 (4) 0.54 (5) none of these

Solution:

Please refer the discussions thread for the solution with figure (post # 227).


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The question was that was 56th Last Year ---------------------------------------------

A word is a combination of 8 letters, each either A or B. Let x and y be 2 words differing in exactly 3 places. How many words differ from each of x and y in at least 5 places?

(a) 38 (b) 48 (c) 58 (d) 68 (e) 78

Exp--

X--_ _ _ * * * * *

Y--_ _ _ * * * * *

Case 1.

Lets assume that X and Y are same in the * positions. So our number now can have anything in _ positions if we assign a different alphabets in *.

S0, 2*2*2 =8

Case 2:

Let’s say only 4 positions are satisfied. In that case 1 will have to come from the first 3 positions.

There are 6 ways to do that and since there are 5 positions where we can change the alphabet we get 5*6=30

So total sum=38

=> choice (a) is the right answer. Given a dart board divided in two regions, one red, one green. If you hit the red region you get 5 points, if you hit the green region you get y > 2 points. If gcd(5, y) = 1 and let R be the the maximum number of points you can not get for a given choice of y, but can get R+1 points for same choice of y, then R can not be a

(1) prime (2) composite (3) perfect square (4) two of the foregoing (5) none of the foregoing

Solution:

By Chicken McNugget Theorem, if gcd(m, n) = 1, then the max R such that am + bn has no solution in non-negative integers is (m-1)(n-1) - 1. => R = 4y -5 for our choice of values here.

But 4y - 5 = 4k+3 for some k, and perfect squares are of the form 4k or 4k+1.


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Squares WXYZ, XPQR, PWMN are drawn externally on the sides of a triangle WXP. The line segments YZ, QR, MN, when extended, form a triangle W'X'P'. Find the area of W'X'P' if WXP is an equilateral triangle of side length 2.

(a) 6 +9(3)^1/2 (b)12 + 13(3)^1/2 (c) 9 + 13(3)^1/2 (d) > 12 + 9(3)^1/2 (e)None of the foregoing

Area of triangle W'X'P' is area of three squares + area of equi triangle +area of 3 Quads..calc area of squares and triangle WXP is no prob....area of quad is area of the two triangles formed in that quad...we get area of one such triangle as 2(3)^1/2..(using trigonometry...)there are six such triangles (contained in the three quads) and hence the total area is

Area of three squares = 12 (4*3) +area of equilateral triangle WXP = 3^1/2 +Area of all six triangles is 12(3)^1/2

=> total area is 12+13(3)^1/2

=> choice (b) is the right answer. If the base 8 representation of a perfect square is ab3c, where a is non-zero, then c equals

(1) 0 (2) 1 (3) 3 (4) 4 (5) can not be uniquely determinable

Solution:

The perfect square in base 10 is 512a + 64b + 24 + c. The perfect squares are of the form 4k, 8k+1 => c can be 0, 1 or 4. If c = 0 then 512a + 64b + 24 + c = 4(4t+2) for some t, which can't be a perfect square. When c = 4, 512a + 64b + 24 + c = 4(4r+3) for some r, which can't be a perfect square.


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The question was that was 58th Last Year ---------------------------------------------The following sequence1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17 ...has one odd number followed by two evens, then three odds, four evens, and so on. What number is the 2003rd term?

(a) 3942 (b) 3943 (c) 3944 (d)3945 (e) None of the foregoing

1,--2,4,--5, 7, 9,-- 10, 12, 14, 16,-- 17,19,21,23,25 ...

1. n(n+1)/2 63(64)/2 = 2016. i.e from 1st to 63rd set there will be in total 2016 terms.

2. Each set will cover (2n-1) consecutive integers.

Therefore 63 set will cover 125 consecutive numbers, 62nd set will cover 123 numbers.. so on.therefore 1+3+5+....+125 = 63*63 = 3969 consecutive integers are covered by the end of 63rd set.

3. 63rd set will contain 63 odd numbers, 3969 as the 2016th term.3969, 3967, 3965, 3963, 3961, 3959, 3957, 3955, 3953, 3951, 3949, 3947, 3945,

=> choice (b) is the right answer. Let 0 ≤ m ≤ n ≤ k ≤ 9 be three integers such that mn + nk + km = 60. The least possible value of m is

(1) 1 (2) 2 (3) 3 (4) 4 (5) none of these

Solution:

mn + nk + km + m^2 = 60 + m^2 = (m+k)(m+n).

m = 1 gives, 61 = (1+k)(1+n) -> not possible; when m = 2, 64 = (2+k)(2+n) => k = n = 6


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The question was that was 59th Last Year ---------------------------------------------Each day, Chetna is either happy or sad. If she is happy one day, then four times out of five she is happy the next day. If she is sad one day, then she is sad the next day one time out of three.In the long run, what are the chances that Chetna is happy on any given day?

(a) 8/15 (b) 15/13 (c) 10/13 (d) 4/15 (e) None of the foregoing

Use Bayes Theorem. It states that

P(A/B) = [P(B/A) * P(A)] / P(B)

where : P (B/A) : Probability of B ( sad ) given A(happy) = 2/3 ( Since she is sad 1 times out of 3 if she is sad one day - implies she is happy 2 times out of 3)P (A) = Probability of being happyP(B) = Probability of being sadP(A/B) : Probability of A (happy) given B = 1/5 ( similar logic as above )

this gives 2/3 = (1/5*P(A)) / P(B)or, P(A) / P(B) = 10/3

Now,We have the ratio of P(A) and P(B) as 10:3

So, P(A) = 10x and P(B) = 3x

Here, P(A) + P(B) = 1 (Since Chetna is either happy or sad every day)So,10x+3x =1 or 13x = 1Implies that,

P(A) = 10x/13x = 10/13and P(B) = 3x/13x = 3/13 or,P(Happy) = P (A) = 10/13

=> choice (c) is the right answer. If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =

(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3

Solution:

Multiply the 3 equations, and we get pqr + p+1/q + q+1/r + r+1/p + 1/pqr = 28/3 => pqr + 1/pqr = 2 => pqr = 1.


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The question was that was 60th Last Year ---------------------------------------------Let the net profit when pencil, pen and eraser are sold respectively at 20%,30% and 40% profits be 10% more than that when they are sold respectively at 40%,20% and 30% profits. Let K be the net profit when they are sold respectively at 30%, 40% and 20% profits. Then the least value of K will be?

(a) 10% (b) 15% (c) 20% (d) 25% (e) None of the foregoing

Let x, y, z be the prices of pencil, pen and eraser.2x+.3y+.4z = 1.1(.4x+.2y+.3z)7z +8y = 24x => z+y = 3x + z/8 ...(1)K = (0.3x+0.4y+0.2z)/(x+y+z) = 0.2+ (0.1x+0.2y)/(x+y+z) = 0.2+ (0.7x - 1.4z/8 )/(4x+z/8 )

equating 0.7x to 1.4z/8, we get z = 4xbut from equation (1), 24x > 7zand also, after simplifying the above fraction, since the Nr. of the fraction is less than the Dr., the value of the fraction increases with x.Least K = 0.2 + (4.9Z/24 - 1.4z/8 )/(7z/6+z/8 ) = 0.2 + 0.7/31 ~ 22.26%

=> choice (e) is the right answer. The numbers +1 and -1 are positioned at the vertices of a regular 12-gon so that all but one of the vertices are occupied by +1. It is permitted to change the sign of the numbers in any k successive vertices of the 12-gon. It is possible to shift the only -1 to the adjacent vertex if k =

(1) 3 (2) 4 (3) 6 (4) at least two of the foregoing (5) none

Solution:

The key word was "any successive k" -> Let us denote our vertices asA1, A2, A3, ..., A12 and let -1 be at A1. Let the 4 successive vertices be A1, A4, A7, A10.The product of the numbers on these vertices is -1 to start with, and if we are changing the sign of each of A1, A4, A7, A10 then also the product should remain -1. But when we say that we want to shift -1 to an adjacent vertex, then product on nodes A1, A4, A7, A10 will be 1, which is not possible.

Similarly, we can argue for k = 6 -> Take successive vertices as A1, A3, A5, A7, A9, A11.

For k = 3, we need even number of touches in 10 nodes and odd number of touches in 2 nodes to make our case happen. But, this means in all we require even number of touches, or in other words we should perform our operation of any 3 successive changes even number of times which in effect means we are back at k = 6, which we proved is not possible.


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The question was that was 61st Last Year ---------------------------------------------If in the expansion of (1+x)â (1–x)^b, the coefficient of x and x^2 are 3 and –6 respectively, then 'a' is


(1 + x)â(1 - x)^b = (1 + aC1.x + aC2.x^2 + ..)(1 - bC1.x + bC2.x^2 - ..)Co-effn of x: (a-b) = 3...(1), Co-effn of x^2: (b(b-1)/2 + a(a-1)/2 -ab) = -6...(2)=>(a-b)^2 - (a+b) = -12(a+b) = 21, Hence a = 12

=> choice (d) is the right answer. Bus A leaves the terminus every 20 minutes, it travels a distance 1 km to a circular road of length 10 km and goes clockwise around the road, and then back along the same road to to the terminus (a total distance of 12 km). The journey takes 20 minutes and the bus travels at constant speed. Having reached the terminus it immediately repeats the journey. Bus B does the same except that it leaves the terminus 10 minutes after Bus A and travels the opposite way round the circular road. The time taken to pick up or set down passengers is negligible. A man wants to catch a bus a distance 0 < x < 12 km from the terminus (along the route of Bus A). Let f(x) the maximum time his journey can take. The value of x for which f(x) is a maximum is

(1) 3 (2) 5 (3) 8 (4) 10 (5) none

Solution:

The bus takes 20 mins for 12 km or 5/3 min/km. For x <= 1, the best strategy is to wait up to 10 minutes for a returning bus, so f(x) = 10 + 5x/3. Similarly for 11 <= x <= 12, f(x) = f(12-x).For 1 < x <= 6, the worst case is that he just misses the right bus, so that the wrong bus comes 10 minutes later and the right bus 10 minutes after that. So he can wait 10 minutes and then travel 12-x km for a total time of 10 + 5(12 - x)/3 = 30 - 5x/3 or he can wait 20 minutes and then travel x km for a total time of 20 + 5x/3. The first is better for x >= 3. So, summarising:

f(x) = 10 + 5x/3 for 0 ≤ x <= 120 + 5x/3 for 1 < x <= 330 - 5x/3 for 3 < x <= 610 + 5x/3 for 6 < x <= 940 - 5x/3 for 9 < x < 1130 - 5x/3 for x ≥ 11

The maximum value of 25 minutes occurs at x = 3 and x = 9.


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The question was that was 62nd Last Year ---------------------------------------------On a regular Chess Board, the numbers of rectangles such that in each one of them the number of Black Houses is equal to the number of White Houses are

(a) 896 (b) 1036 (c) 1156 (d) 796 (e) 856

Only those rectangles with even number of houses satisfy this condition.Total ways in which length = 1 can be selected ->8Total ways in which length = 2 can be selected ->7......Total ways in which length = 8 can be selected ->1

So, Total ways in which all different length can be selected = 8+7+6+5+4+3+2+1

same holds for Breadth.So, Total no. of rectangles (by Multiplication Theorem) = (8+7+6+5+4+3+2+1)*(8+7+6+5+4+3+2+1)= 36^2=1296Total no. of rectangles with odd number of houses = (8+6+4+2)*(8+6+4+2)=400

=> the no. of rectangles satisfying the above condition is 1296-400 =896

=> choice (a) is the right answer. For which positive integer values of n the set {1, 2, 3, ..., n} can be split into n disjoint elements subsets {a, b, c, d} such that a = (b+c+d)/3?

(1) 6 (2) 12 (3) 16 (4) 36 (5) at least two of the foregoing

Solution:

Since we have 4 distinct elements in each subset, 'n' shud be divisible by 4.directly option (1) is eliminated.now let a1, a2, a3.... be the different values 'a' can take.so we get 4*(a1+a2+a3......)=n(n+1)/2 {since all subsets r disjoint}so n(n+1) shud be divisible by 8answer can be only option (3)


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The question was that was 62nd Last Year ---------------------------------------------Let V(t) =V(1-t)+V((1-t^2)^0.5), for all 0<t<1.Then V(1 - ((3)^0.5)/2) is equal to

(a) V((1 - ((3)^0.5)/2)^0.5) (b) V((1 - ((3)^0.5)/4)^0.5)(c) V(((3)^0.5 - (3/2))^0.5) (d) V(((3)^0.5 - (3/4))^0.5) (e) None of the foregoing

Putting t=1/2 we have:V(1/2) = V(1/2) + V((3)^0.5/2)canceling V(1/2) from both sides, V((3)^0.5/2) = 0 ..................... equation (1)

Now, V(1- (3)^0.5/2) = V((3)^0.5/2) + V(((3)^0.5 - 3/4)^0.5)= 0 + V(((3)^0.5 - 3/4)^0.5) [ using equation (1) ]= V(((3)^0.5 - 3/4)^0.5)

=> choice (d) is the right answer. Let ABC be a triangle and D and E be internal points on BC and AC respectively. BD/DC = EA/CE = 1/2. O is the intersection of AD and BE. If the area of triangle ABC is 2 sq. unit, then area of quadrilateral ODCE (in sq. unit) is

(1) 4/5 (2) 14/15 (3) 16/15 (4) 6/5 (5) none of these

Solution:

Area(ODCE) = ar(DCE) + ar(ODE) = 8/9 + ar(ODE)

ar(ODE) = 0.5*DE*OM (OM is the altitude of ODE)

OM = 2*(altitude of ABDE)/5= 2*(1/3 * altitude of ABC) /5 = 2(altitude of ABC)/15ar(ODE) = 8/45

ar(ODCE) = 16/15


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The question was that was 64th Last Year ---------------------------------------------Let PQRS be a convex quadrilateral with PQ=QR=RS with PR not equal to QS and T be the intersection point of its diagonals. If PT=ST then sum of the angles QPR and QSR in degrees?

(a) 30 (b) 60 (c) 90 (d) 120 (e) can not be determined

Let U be the reflection of R through PS. Thus, <(UPS) = <(TPS) = <(TSP) => PU || SQ.=> PQSU is an isosceles trapezium.So < PQS+< PRS=< PQS+< PUS=180< TPS+< TSP=< QTP=< SQR+< PRQso < QPR+ < RSP=2*(< SQR+< PRQ) ...... (1)< QPS+< RSP=360-< PQR-< QRS=180-< SQR-< PRQ ....... (2)from (1) and (2) < SQR+ < PRQ=60

=> choice (b) is the right answer. Katrina and Deepika have some marbles with each of them, such that the number of marbles with Deepika is thrice that with Katrina. If Katrina distributes her marbles equally among certain number of bags, then she is left with 31 extra marbles. If Katrina and Deepika were to pool the marbles and then distribute the total marbles equally among the same number of bags as Katrina did, they will be left with 16 marbles. The number of marbles with Deepika is the largest possible three digit number. How many bags are needed to equally divide all the marbles with Deepika, if the number of those bags is the smallest possible two digit number?

(1) 11 (2) 13 (3) 29 (4) 31 (5) none of these

Solution:

Let 3x and x be the number of marbles with deepika and katrina.Now using the conditions given in the problem the following two equations can be formed..x=nk+31 (Implies n>31)4x=np+16 (n is the number of bags)

Now n(p-4k)=108 and n >31..so=> n = 36, 54 or 108 as n has to be div by 108 and n > 31.

=> Number of marbles with Deepika is 3*(36*8 + 31) = 957 for n = 36; for n = 54, 108 we have lesser value.

957 = 3*11*29


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The question was that was 65th Last Year ---------------------------------------------In a class there are 100 students. A division of the students in n sections is good if: 1) the sections have different numbers of students2) for any partition of one of the sections in 2 smaller sections, among the (n+1) sections you get 2 with the same number of students (any section has at least 1 student). The positive difference between the maximal and minimal possible value of n such that the division is good is


Well, the language of the question wasn't perfect and people interpreted this question in different manners. Some assumed "one of the sections" as - some one, while others assumed it as each one. Then, "you get 2 with the same number", some assumed at least 2, while some assumed exactly 2. We give the answer of all possible 4 cases.Case 1: when the question meant each one + at least 2 Max = [1, 2, 3, ..., 12, 22]. Min = [1, 3, 5, ..., 19]. Thus, answer is 13-10 = option (a).Case 2: when the question meant each one + exactly 2 Max = [1, 3, 5, ..., 19]. Min = [1, 3, 5, ..., 19]. Thus, answer is 10-10 = option (e).Case 3: when the question meant some one + exactly 2 Max = [1, 3, 4, ..., 12, 24]. Min = [1, 3, 96]. Thus, answer is 12-3 = option (e).Case 4: when the question meant some one + at least 2 Max = [1, 2, 3, 4, ..., 12, 22]. Min = [1, 3, 96]. Thus, answer is 13-3 = option (e).

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If all palindromes (positive integers which is unchanged if you reverse the order of its digits) are written in increasing order, what is the possible number of prime values can the difference between successive palindromes take?

(1) 0 (2) 1 (3) 2 (4) 3 (5) none of these

Solution:

Let x be a palindrome and x' the next highest palindrome. If x < 101, then it is easy to see by inspection that x' - x = 1, 2 or 11, so the only prime differences are 2 and 11.

So assume x > 100. If x and x' have the same final digit, then their difference is divisible by 10 and hence not prime. So they must have different digits. Thus either x = d9...9d and x' = d'0...0d', where d < 9 and d' = d+1, or x' has one more digit than x and d = 9, d' = 1. In the first case x' - x = 11. In the second case x' - x = 2. So again the only prime differences are 2 and 11.


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The question was that was 66th Last Year ---------------------------------------------Let a, b, c be distinct non-zero integers such that -5 <= a, b, c <= 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have?


1/a + 1/b + 1/c = 1/(a+b+c) reduces to (a+b)(b+c)(c+a) = 0; if a+b = 0 then we can chose b in 10 ways (integers from -5 to 5 excluding 0) and c in 10-2 = 8 ways (since a, b and c are different). Also, b+c = 0, and c+a = 0 gives 2 more 80 each such solutions => option (e).

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Which among the following is true? (Mark the strongest option)

(1) Given 8 natural numbers, none greater than 15 => at least 3 pairs of them will have same positive difference(2) Among the numbers x, 2x, . . . , 5x, there is one that differs from an integer by at most 1/6(3) A warehouse contains 200 boots of size 41, 200 boots of size 42, and 200 boots of size 43. Of these 600 boots, 300 each are left and right boots => Implex can find among these boots at least 100 usable pairs(4) All of the above(5) Exactly two of the above

Solution:

We have 8C2 = 28 pairs of differences -> the values it can take is from 1 to 14. But 14 can't be more than once => at least 27 pairs of differences can be from 1 to 13 => 1 difference takes at least 3 values => (1) is true

(2) is left for student's to prove.

There are two categories into which we can fit the three sizes; those sizeswhich are more right boots than left boots, left = right, left > right.=> the two sizes lie in the same catgory. Let us say that sizes 41 and 42 have more right boots than left boots (an analogous argument will hold if two sizes have more left boots than right boots).

We have 300 boots in all, and at most 200 left boots in any one size. => sum of the left boots in any two sizes is at least 100. We have shown that there are at least 100 left boots in sizes 41 and 42, and each of these sizes contains more right boots than left boots. Hence, (3) is true.


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One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?


Let d=distance truck is in front of tunnel entrance, L=length of tunnel, x=Vikram's speed. Case 1: Vikram turns around and heads for entrance, a distance of L/4. Vikram and truck get to entrance at same time T1=d/80=(L/4)/x. Case 2: Vikram streaks for exit, a distance of 3L/4. Vikram and truck get to exit at same time T2=(L+d)/80=(3L/4)/x. Solving both equations for x and setting them equal, x=(L/4)*80/d=(3L/4)*80/(L+d) After simplifying, d=L/2, hence x=40.=> option (c).

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Let -2 < x < 3, 0 < y < 4, 2 < z < 5. If (3-x)(4-y)(5-z)(3x+4y+5z) achieves the maximum possible value then which among the following is not true?

(a) 3x+4y = 0 (b) |x| < |y| (c) z = 5/2 (d) two of the foregoing (e) none

Solution:

(3-x)(4-y)(5-z)(3x+4y+5z) = 1/60*(9-3x)(16-4y)(25-5z)(3x+4y+5z). Take 9-3x = A, 16-4y = B, 25-5z = C, 3x+4y+5z = D. A+B+C+D = 50 => ABCD is max when A=B=C=D. Solving we get x = -7/6, y = 7/8, z = 5/2.


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The question was that was 68th Last Year ---------------------------------------------A square sheet of paper ABCD is so folded that B falls on the mid point of M of CD. The crease will divide BC in the ratio

(a) 3 : 2 (b) 5 : 3 (c) 2 : 1 (d) 9 : 4 (e) none of the foregoing

Point B is coincided with the midpoint of CD(i.e., M). Let the crease cut the side BC at N. Assuming the length of the sides as 2a, CM=MD=a. Let NC=x, making BN=2a-x.Now when B is coincident with M, a rt triangle is formed: triangle BNC(or MNC, as B now lies on M)Applying Pythagoras' theorem,(2a-x)^2= x^2+a^2solving for x we get: x=3a/4

therefore 2a-x=5a/4BN:NC=5a/4:3a/4 or 5:3.=> option (b).

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In a chess tournament each player plays every other player once. A player gets 1 point for a win, ½ point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. The total number of players in the tournament can be

(a) 18 (b) 25 (c) 32 (d) 42 (e) 45

Solution:

Let x be the number of men and y be the number of women. Total number of matches being played in the tournament are (x+y)C2. Men play xC2 among themselves, women play yC2 amongst themselves, and men play xy against women.=> xC2 + xy + yC2 = (x+y)C2.But since, each contestant scores same number of points against men as against women => xC2 + yC2 = xy. Thus, 2xy = (x+y)C2 => 4xy = (x+y)*(x+y-1) => (x-y)^2 = x+y => Total number of contestants is a perfect square.

=> Choice (b) is the right answer

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The question was that was 70th Last Year ---------------------------------------------Akshay and John started to dig a canal 10 km long. It is calculated that if Akshay takes off for 3 days, then John has to dig 1 km more and if John takes off for 4 days, Akshay has to dig n km more. Which among the following is true?

(a) If Akshay and John do the work without absence and Akshay digs 6 km, then the work gets completed in a little over 7 days(b) n = 4/3(c) If the work gets completed in 7.5 days, when Akshay and John dug without any absence, then Akshay dug 5 km(d) all of the foregoing(e) none of the foregoing

Let W be the amount of work done in digging. Let Akshay, John do W/a and W/b parts of work per day respectively.Let N, p, q be the number of days in which the work can be done if Akshay, John do the work together without absence, with Akshay absent for 3 days, with John absent for 4 days respectively.

=> NW/a + NW/b = W, (p-3)W/a + pW/b = W and qW/a + (q-4)W/b = W.Also, pW/b - NW/b = W/10 and qW/a - NW/a = nW/10. => p = b/10 + N, q = an/10 + N=> (b/10 + N-3)/a + (b/10 + N)/b = 1; (na/10 + N)/a + (na/10 + N-4)/b = 1

=> (b/10 - 3)/a + 1/10 + N(1/a + 1/b) = 1 and n/10 + (na/10 - 4)/b + N(1/a+1/b) = 1, but N(1/a + 1/b) = 1=> b/10 - 3 + a/10 = 0; and n/10(1/a + 1/b) = 4 => n = 4/3 and a+b = 30

This answer is part (b)

Part (a)If Akshay digs 6 km then John digs 4 km => Amount of work done by them is in the ratio 3:2 => b:a = 2:3but a+b = 30 => a = 12, b= 18 => work is completed in 1/(1/12 + 1/8 ) = 7.2 days

Part (c)1/a + 1/b = 2/15 and a+b = 30 => a = b = 15 => Akshay dug 5 km.

Hence, choice (d) is the right answer

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A circle C1 of radius x touches other two circles C2 and C3 of radii y and z (both < x), the centres of 3 circles being on the line (C1 being in the middle). If the common tangents of C1, C2 and C1, C3 are perpendicular, then (1+ √(y/x))(1+√(z/x)) =

(a) √2 (b) √3 (c) 2 (d) can not be determined (e) none

Solution:

Let us draw the figure for this. Let common tangent to C1 and C2 be segement PQR where, P lies on C2, Q on C1. Let common tangent to C1 and C3 be STR where S lieson C3, T on C1 and R is the intersection point of perpendicular tangents.Let the centres of circles C1, C2, C3 be M, N, P respectively.The parallel to PQ from N meets MQ at T.

Look at this -> cos A = (x-y)/(x+y) [in triangle TMN], where A = < QMN, and sin A = (x-z)/(x+z) [as <TMP = 90 - A].

=> (x-y)^2/(x+y)^2 + (x-z)^2/(x+z)^2 = 1.=> (x-y)/(x+y) = 2√(yz)/(y+z)=> √(y/x) = (√x - √z)/(√x + √z)=> (1+ √(y/x))(1+√(z/x)) = 2

=> Choice (c) is the right answer

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The question was that was 71st Last Year ---------------------------------------------Consider a set P= {1,2,3...,11,12} of natural numbers. We define another set Q such that it contains no more than one out of any three consecutive natural numbers. How many subsets Q of P including the empty set are possible?

(a) 114 (b) 117 (c) 129 (d) 136 (e) 130

Let f(n) be number of subsets of {1,2, .....n} which contain no more than one out of any three consecutive natural numbers.

then f(n) = f(n-1) + f(n-3)

Logic is to include f(n-1) subsets and add n to all the subsets for f(n-3) to get all possible subsets for n

with f(1) =2, f(2) = 3 , f(3) =4.

Solving we get f(12)=129.=> option (c).

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Solution to Quantitative Question # 071-------------------------------------------------------- For a balanced diet Khiladi Akshay Kumar needs to take pulse between 1/2 kg and 1 kg for every 3 kg of rice. Let there be an unequal percent rise in the prices of rice and pulse. Assuming that Akshay doesn't take unbalanced diet and keeps his total amount of consumption of rice and pulse same as earlier, the maximum percent rise in the consumption of rice Akshay can make will be about

(a) 8% (b) 14% (c) 22% (d) can not be determined (e) none

Solution:

Since maximum rise is therewe need to assume that initially Akshay took 1 kg of pulse for every 3 kg of riceand after price rise he takes 1/2 kg of pulse for every 3 kg of rice

now initially he eats x kg rice and x/3 kg pulseLet akshay consumes y kg of rice after price rise so total consumption = y+y/6x+x/3=y+y/6y=8x/7this will lead to 14% rise if the prices rise by equal percentage.

=> Choice (b) is the right answer

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The question was that was 72nd Last Year ---------------------------------------------PQRST is a cyclic pentagon with QR = RS = ST.The diagonals PR and QT intersect at X.Y is the foot of the altitude from X to PQ. We have XP = 25,XS = 113, and XY = 15. The area of triangle PQT is approximately

(a)1544 (b)1600 (c) 1648 (d)1560 (e) None of the foregoing

Pythagoras gives PY = 20. We draw QS and PS, and construct the altitudeXZ to PS, with Z on PS, and altitude XX' to PT, with X' on PT. Because QR =RS = ST, angles QPR, RPS, and SPT are congruent. Because Z is on PS, trianglesXYP and XZP are congruent by AAS, so XZ = 15 and ZP = 20, from which Pythagoras gives ZS = 112, implying PS = 132.

Let k = <QPR, so <XPT = 2k, and <YPT = 3k. Because we have sink = 3/5 and cosk=4/5, so sin2k=24/25 and sin3k = 117/125we find that XX' = 24 using sin2k=24/25. By sine law ST : TQ : QS = 25 : 39 : 40.Now, area of PQT= (15*PQ+24*PT)/2Now by Ptolemy’s theorem on quadrilateral PQST:

PQ*ST+PT*SQ=PS*QTor, (25x)*PQ+(40x)*PT=132*(39x)5PQ+8PT=(132*39)/5

hence area =1544.4=1544(approx.) => option (a).

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Let 2 numbers be given in each OPTION below

(a) A = 31^11, B = 17^14(b) C = 2^100 + 3^100, D = 4^80(c) E = (1-1/4)(1-1/9)(1-1/16)….(1-1/100^2) , F = 1/2(d) G = 1/2 - 1/3 + 1/4 - 1/5 + .... + 1/100, H = 1/5

Which among the set constitutes the greater number of the two in the pair in (a) to (d)?

(1) (B, D, F, H) (2) (B, C, E, G) (3) (A, D, F, H) (4) (B, D, E, G) (5) none of these

Solution:

B > A. How? 31^11 < 32^11 = 2^55 < 2^56 = 16^14 < 17^14

D > C. How? 4^80 has [80log4 +1] = 49 digits. 3^100 has [100log3 + 1] = 48 digits. 2^100 has 31 digits.

Alternate method:

We will show that 4^80 > 2.(3^100). Remember, (1+x)^n >= 1+nx for x >= -1 and n >= 1. Thus, we need to show if (256/243)^20 > 2 -> look at this, 256/243 = 1+1/20 => (256/243)^20 > (1+1/20)^20 >= 2. We are dome here!

E > F. How? E = 1.3.2.4.3.5.4.6.5.7.6.8.7.9.8.10.9.11/(2.3.4.5.6.7.8.9..10)^2E = 11/20 = 0.55 and F = 0.50

G > H is simple. How? Club the two terms each i.e. (1/2-1/3) + (1/4-1/5) + ... 1/100 = 1/6 + 1/20 ... > 13/60 > 1/5.


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The question was that was 73rd Last Year ---------------------------------------------Maalchand had 100 pieces each of articles A and B. A was cheaper in price than B. Fearing police raid, Maalchand decides to sell all the pieces in two days. On the first day, he sold A and B at 10% and 30% profit respectively. Although, he made a net profit of 25%, the articles did not sell much and was not more than 25 pieces of each type. On second day, A and B were sold at 30% and 10% profit respectively. In the process, all the articles of Maalchand were sold, but his net profit for the second day was reduced to 20%. Which among the following is not true?

(a) the cost price of each piece of article A was less than 18% cheaper than each piece of article B(b) the total sale on the first day was 19.5% of the total sale(c) the net profit on the total sale was 20.9%(d) all of the foregoing(e) none of the foregoing

Let 'a' be the price per item and 'x' be the no: of article sold of article A on the first day.Similarly let 'b' be the price per item and 'y' be the no: of article sold of article B on the first day.

so,1.1ax+1.3by=1.25(ax+by)=> 3ax = by------------------(1)also for the second day,(100-x)*1.3a+(100-y)*1.1b=1.2[(100-x)a+(100-y)b]=>[(100-x)a = (100-y)b--------------------(2)from (1) and (2)y = 300x/(2x+100)since x,y<=25;only one integer value satisfies this i.e. x=10 and y=25,from (1) we get,30a = 25ba = .833b--------------------------(3)so,((b-a)/b)*100 = 16.7%so option (a) is true.Total sale on the first day= 11a+32.5b = 50aTotal sale on the second day= 117a+82.5b = 117a+99a = 216aratio of sale on first day to second day = 50/256 = 19.5%Hence (b) is also true.Also total sale = (216a+50a) = 266a,total cost = 100a+120a = 220a,hence net profit percent = (46/220)*100 = 20.9%

hence (c) is also true.=> option (e) is the correct answer.

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In a certain class of 300 students , the number of students who either do not study at home or do not attend classes is a third more than of those who either study at home or attend classes. the number of students who do not study at home but attend classes is two fifths more than those who study at home but do not attend classes, while the number of students who study at home as well as attend classes is half of those who neither study at home nor attend classes. If the number of students who only study at home or only attend classes is a third less than those who do either, then how many students who either do neither or do both?

(1) 120 (2) 150 (3) 180 (4) 210 (5) none of these

Solution:

Let a be the number of students who do not study at home and do not attend classes.Let b be the number of students who do not study at home but attend classes.Let c be the number of students who study at home and attend classes also.Let d be the number of students who study at home but do not attend classes.Given d+b = 2/3(b+c+d) and d+c+b = 3/4(a+b+d)if d+b = 2x => b+c+d = 3x, and a+b+d = 4xAlso, (a+b+d) + (c+b+d) - (b+d) = 300=> x = 60

=> a+c = 180


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The question was that was 74th Last Year ---------------------------------------------What is the smallest positive integer k for which there are at least 11 even and 11 odd positive integers m so that (m^3 +k)/(m+2) is an integer?


Notice that m^3 + 8 is divisible by m+2. Therefore, k−8 must be divisible by m+2 for the expressionto be an integer. If f is a factor of k − 8, m = f − 2 is a corresponding suitable m; we then needf >=3 to make m > 0. Thus k − 8 must have twelve each odd and even factors including 1 and 2. To make the number of odd and even factors equal in order to minimize k, the power of 2 in the prime factorization of k− 8 must be 1. Suppose the prime factorization of k− 8 is then 2^1•3â •5^b •7^c •11^d(larger prime factors will clearly not minimize k. Then (p+1)(q+1)(r+1)(s+1)>= 12. To minimize k, p >=q>= r>= s. We then examine values of (k-8)/2 to determine the

best (p, q, r, s). 3 • 5 • 7 • 11 = 1155, 3^2 • 5 • 7 = 315. Moving any more factors into smaller primes involves multiplying by 3^2/7 or 3^2/5 (or subsequent larger powers of 3), which increases the value. Therefore k− 8 = 2 •3^2 • 5 • 7, so k= 638.

=> option (c) is the correct answer.

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Yuvraj, Rohit and Mahender each had age (always considered an integer) less than 100, such that sum of the ages of any two of them is same as reverse of third's age. Which among the following can not be true?

(1) The sum of the ages of the three is always 99.(2) Yuvraj's age, strictly middle in age among the three, can assume 8 values(3) If Rohit was older than either of the others, the youngest he could be is 45(4) At least two of the above(5) none of the above

Solution:

Let the ages of Yuvraj, Rohit and Mahender be : A, B and C respectively.Given : A, B, C <= 99And some of any two is reverse of the 3rd => sum of two is 9's complement of the third => A+B = 99 - C=> A+B+C = 99.Hence, (1) is true.

Hence, the first and second digits of the sum of two have to be 9's complement of each other.Thus, we have only 10 possibilities :(1st Digit of A+B, 2nd Digit of A+B) = (SUM) ==> (C) - (Yuvraj,B)(0,9-0) = (09) ==> (90) --> Not a solution, since we can't split 9 further (only poss. is 9 and 0, but that is trivial, hence not valid)(1,9-1) = (18 ) ==> (81) - (9,9) Hence, Yuvraj cannot be 9(2,9-2) = (27) ==> (72) - (18,9)(3,9-3) = (36) ==> (63) - (27,9) {(18,18 ) isn't valid for Yuvraj}(4,9-4) = (45) ==> (54) - (27,18 ),(36,9)(5,9-5) = (54) ==> (45) - (36,18 ){(27,27) again isn't valid)(6,9-6) = (63) ==> (36) - In this case, Yuvraj is 36(7,9-2) = (72) ==> (27) - Again Yuvraj is 27(8,9-1) = (81) ==> (18 )- (27,54), And Yuvraj can be 18(9,9-9) = (90) ==> (09) --> Again not a valid solution...

But from the above deductions, for him to be strictly aged in between, he cannot take 8 different ages,=> (2) is false,

(3) is true is easy to check.


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The question was that was 75th Last Year ---------------------------------------------Every week the Pagalguy.com magazine publishes a list of the Top 20 contributors on Pagalguy.com. If the order is never the same in any two consecutive weeks, and no contributing member ever regains any lost popularity (i.e. no contributor rises in ranking once he/she starts to drop in ranking) how many consecutive weeks could the same twenty contributors remain on the Top 20 list?


If you look at the initial and the final weeks of the rankings, during this time, the contributor initially in the top position can drop at most 19 spots to position 20; the second ranked contributor can drop at most 18 spots to position 19, and so on. This can take at most 19 + 18 + ... + 2 + 1 = 190 weeks. Therefore there are at most 191 weeks in which the same contributors can appear on the charts. You can achieve this maximum, by switching only two contributors each week.

=> option (b) is the correct answer.

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N < 100 Miss Universe contestants of 2010 are standing in a circle and numbered from 1 to N. Starting counting from 1 initially, in succession, every second one is removed from the contest and eliminated, and the last one is declared the winner. If contestant number 25 wins eventually, then how many values can N take?

(1) 3 (2) 1 (3) 2 (4) 0 (5) 4

Solution:

Let the number of contestants be 2^n where n is non-negative => for n = 0, 1 the winner is 1. For n =2, 3 winner is 1. It is easy to prove that for any n here (2^n) contestants, the contestant number 1 is the winner always.

Let the number of contestants be 2^n + M where 0 < M < 2^n. Thus, after first M eliminations i.e. passing through 2M contestants and being at 2M+1 (which will be spared for elimination), we have 2^n contestants and (2M+1)th is at 1st position, and will be the winner. Thus, the winner for k number of contestants is 2(k-2^m) + 1 where m is largest integer such that k > 2^m.

Thus, for N = 12 + 16 = 28, 12 + 32 = 44, 12 + 64 = 76.


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Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back. The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat and yet someone else’s hat is r. Then p+q+r is


The no. of ways in which the 1st person doesn't pick up his own top-coat is 3. The 2nd person (whose top-coat the 1st one picked) can pick someone else's top-coat in 3 more ways. For the 3rd and 4rth person we have just 1 choice. Thus, in all when none among the 4 pick his/her own top-coat is 9. Thus p = 9*9 = 81.A person among the 4 who picks his own top-coat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own top-coat or not own hat] = not own top-coat + not-own hat - (not won top-coat and not own hat) = 2*6 + 2*6 - 2*2 = 20. Thus, q = 80.As explained in the 1st case, each person can pick someone else's top-coat in 9 ways. To each of these 9 ways we have 2 ways where he picks yet someone else's hat. Thus, r = 9*2 = 18

=> option (e) is the correct answer.

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A rectangular block L x 100 x H, with L ≤ 100 ≤ H, where L and H are integers, is cut into two non-empty parts by a plane parallel to one of the faces, so that one of the parts is similar to the original. How many possibilities are there for (L, H)?

(1) 10 (2) 12 (3) 20 (4) 24 (5) none of these

Solution:

We must cut the longest edges, so the similar piece has dimensions L x 100 x k for some 1 ≤ k < H. The shortest edge of this piece cannot be L, so it must be k. Thus L x 100 x H and k x L x 100 are similar. Hence H = 100^2/L, k = L^2/100. Now 100 = 2^4•5^4, so 100^2 has 25 factors, of which (25-1)/2 = 12 are < 100.


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The question was that was 77th Last Year ---------------------------------------------(i) The sum of real numbers x and y is 1. The maximum value of xy(x^3 + y^3) is p.(ii) Let the equation [n/2] + [n/4] = n has q possible solutions, where [n] denotes the greatest integer less than or equal to n e.g. [3.21] = 3.

What is the value of p*q?


(i) If n is positive then [n/2] + [n/4] can at the max be 3n/4. Thus, no solutions for positive n. For non-negative n, n = 0 is a trivial solution. For n < -5 LHS > RHS. For -5 <= n < 0, we have 3 solutions, n = -2, -3, -5. Thus p = 4.(ii) xy(x^3 + y^3) = xy(1-3xy) = 1/3*(3xy*(1-3xy)). Thus, 3xy + (1-3xy) is constant => max of 3xy*(1-3xy) occurs when 3xy = 1-3xy => xy = 1/6 => q = 1/12


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ABCDE is a pentagon such that < A = < B = < D = 120˚, and < C = < E. Let BC = 1 and CD = √3. If a circle can be inscribed in the pentagon, then its radius will be

(1) 4 - 2√3 (2) 3 - √3 (3) 3√3 - 4 (4) 4√3 - 6 (5) none of these

Solution:

The construction of the pentagon is very much valid here.Let's see if we can inscribe a circle in this pentagon.

tan (< DBC) = √3/1 => < DBC = 60 degrees.Let O be the centre of the circle and P and Q be tangency points of CB and AB respectively => By RHS, triangles OQB and OPB are congruent and hence < OBP = 60 degrees, but since < DBP = 60 degrees, we are effectively saying that DB passes through the centre of the circle.

What you get is R > 1 [3/2(√3-1)], when clearly it should be < 1. Thus, no such construction of the circle is possible.


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The question was that was 78th Last Year ---------------------------------------------A car gives the mileage of 60km/L, 50km/L and 40km/L when driven at the speeds of 40km/hr, 50km/hr and 60km/hr respectively. Assume that each car is driven only at the three speeds mentioned above. The car is driven for 3 hours using 2 Litres of petrol. The distance covered by the car

(a) > 120 km (b) < 120 km (c) = 120 km (d) can not be determined (e) none of the foregoing

Let the distance traveled with 40km/hr, 50km/hr, 60km/hr is A, B and C respectively. => A/40 + B/50+ C/60 = 3 and A/60 + B/50 + C/40 = 2. Eliminating A, we get 2B/5 + 5C/6 = 0 => B=C=0 => A = 120 km


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Let [x] denotes the greatest integer less than or equal to x. The number of positive reals x such that 9x/10 = [x]/( x - [x] ) is a

(1) perfect square (2) prime (3) perfect cube (4) perfect number (5) none of these

Solution:

Let x = I + f, where I is non-negative integer and 0 < f < 1 [for f = 0, 1 we have no solution]

Thus, what you get is 9(I+f)/10 = I/f. This is a quadratic in f whose only one root will be in (0, 1), as the product of the roots of 9f^2 + 9If - 10I = 0 is negative, the other root is negative.

The positive root is [-9I + root(81I^2 + 400I)]/18 and is < 1. Solving for this, we get eight legitimate I from 1 to 8.


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The question was that was 79th Last Year ---------------------------------------------Let f(x) be an algebric expression of odd degree (> 1). If the degree of f(x) + f(1-x) is at least two less than the degree of f(x) and the coefficients of x and x^2 are equal in magnitude but opposite in sign in the given expression, then the highest possible degree of f(x) is


Taking f(x) = ax^n + bx^n-1 + cx^n-2 + .... , where the third last coefficient and the second last coefficient are equal in magnitude but opposite in sign and n is odd. f(x) + f(1-x) has at least first 2 terms as 0 => 2b + na = 0. Thus, we can have such f(x) for any n, thus n-> Infinity.


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7 IIMs participate in a B-schools sports meet, where the use of expletives is officially prohibited. Each team from an IIM plays against the other exactly once. What is the possible minimum number of matches that could have been played so far such that among every 3 teams, atleast two have played against each other?

(1) 9 (2) 10 (3) 14 (4) 15 (5) none of these

Solution:

Treat each team as a point (vertex), and denote matches between two teams as the segment joining these two points. If we isolate these teams into a set of 4 and 3 each, and draw the complete graphs (a quadrilateral with diagonals also joined, and another disjoint triangle), then we have 4C2 + 3C2 = 9 edges (matches) in all. Our condition is satisfied. Can we do better than 9? The answer is NO, and left as an exercise for students to prove.


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The question was that was 80th Last Year ---------------------------------------------100 ex-students of the same batch meet for alumni meet in the campus of a B-school. The benches of Fr Prabhu Hall in the B-school are arranged in a rectangle of 10 rows of 10 seats each. All the 100 have different salaries. Each of them asks all his neighbours (sitting next to, in front of, or behind him, i.e. 4 members at most) how much they earn. They feel a lot of envy towards each other: a person is content with his salary only if he has at most one neighbour who earns more than himself. What is the maximum possible number of ex-students who are satisfied with their salaries?


Let us represent the ex-students with a square grid of 10x10 points, and label each point with the salary of the ex-student. Have a look at the figure in the document by following the undermentioned link.Let us draw arrows between neighbouring points such that the arrow is directed from the smaller to the larger number: www.pagalguy.com/forum/814259-post1030.html(Satisfied alumni are marked in green, and dissatisfied ones in red.) Let a be the number of satisfied EX-STUDENT's sitting in the corners, b the number of those sitting at the sides of the square, and c the number of those sitting inside. The number of arrows is 180. There is at most one arrow originating at any satisfied EX-STUDENT, and there will be at least one point where no arrow originates, the EX-STUDENT with the largest salary (obviously satisfied). Hence the number of arrows originating at satisfied EX-STUDENT's is at most a+b+c-1. There are at most (4-a).2 arrows from the 4-a dissatisfied EX-STUDENT's in the corners, at most (32-b).3 from the 32-b dissatisfied EX-STUDENT's along the sides, and at most (32-b).3 from those (64-c).4 sitting inside. The total number of arrows is thus 180 <=(a+b+c-1)+(4-a).2+(32-b).3+(64-c).4, that is, a+2b+3c <=179. The one with the lowest salary out of the 36 EX-STUDENT's around the circumference is necessarily dissatisfied, thus a+b<=35. It is also obvious that a <=4. By adding the inequalities, we have 3(a+b+c)=(a+2b+3c)+(a+b)+a <=179+35+4=218, that is, a+b+c <=72. Hence, the number of satisfied EX-STUDENT's cannot be greater than 72.

http://www.pagalguy.com/forum/814259-post1030.html

The diagram shows the case when there are exactly 72 EX-STUDENT's who are content with their salaries.

=> option (a) is the correct answer.

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In a triangle ABC, AC = 3AB. From C, CD is drawn perpendicular to the bisector of < A. If AD intersects BC at X, then AX/XD is

(1) 1:1 (2) √2: 1 (3) √2 + 1: 2 (4) √3 - 1: 1 (5) √3: √2

Solution:

Draw the exernal bisector of < A, meeting CB extended at E => < EAX = 90 degrees.

By internal and external angle bisector rules, we have EB = 2y, BX = y, XC = 3y as BX/XC = EB/EC = AB/AC = 3 => EX = XC.

Thus, we have congruent triangles AEX and DCX. Thus, AX = XD.

Alternate solution:

Let AB=x=>AC=3x, Also, <ADC=90

We also know AX= 2.AC.BC.cos(A/2)/(AC+BC)

AX=2.3x^2 cos(A/2)/4x, AX=3x cos(A/2)/2

From tria ADC, cos(A/2)=AD/AC=AD/3x

Simplifying, 2AX=AD


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The question was that was 81st Last Year ---------------------------------------------Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is

(a) 200 (b) 125 (c) 375 (d) 400 (e) 325

Let O be the point of intersection of diagonals PR and QS,and T be the point of intersection of PR and the circumcircle of triangle PQS.Extend SQ to meet the circumcircle of triangle PRS at U.

PO.OT=QO.OS and SO.OU=PO.OR

Let PR=2a and QS=2b.Because PT is a diameter of the circumcircle of triangle PQS, and SU is a diameter of the circumcircle of triangle PRS, the above equalities can be rewritten as

a(25-a)=b^2 and b(50-b)=a^2Hence, a=2b.It follows that 50b=5b^2=>b=10,a=20.Thus area of PQRS=1/2 .PR.QS=2ab=400

=> option (d) is the correct answer.

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Gripened comes back to India after earning good bucks from Korea and heads for a shop in Bangalore having expensive shirts. The price of each shirt is a multiple of ten in Rs and is marked in K (in Rs) i.e. multiple of thousand e.g. if the price is Rs 2670, then it's marked as 2.67 K. Gripened buys 4 shirts and goes to the counter for the bill. The counter-boy multiplies the marked price of 4 shirts and hands gripened the bill of 7.11 K. Gripened being smart realizes the mistake and asks for the new bill in which the marked price of each shirt is added. To Gripened's astonishment, the bill this time also comes as 7.11 K. What was the marked price in K of the least expensive shirt that Gripened bought?

(1) 1 (2) 1.2 (3) 1.5 (4) can not be determined (5) none of these

Solution:

Let the MP in Tens (Rs) of the four shirts be a, b, c, d. Then a, b, c, d are whole numbers with

a + b + c + d = 711 = 32 × 79 and (a/100).(b/100).(c/100).(d/100) = 711/100.

=> abcd = 711 × 10^6 = 2^6 × 3^2 × 5^6 × 79. Exactly one marked price is a multiple of 79, and at most three prices (in Tens Rs) are even or are a multiple of 5.

It is not possible for three prices to be a multiple of 25. Otherwise, the remaining price would be the multiple of 79, and the sum of the three remaining prices would also be a multiple of 79 as well as of 25.But 79 × 25 > 711, and this is not possible. Hence, at least one of the prices is a multiple of 5^3 = 125; this price is clearly not a multiple of 79.

Case 1: One of the prices is 5×79 = 395. Suppose that a = 5×79 = 395. Suppose that b is a multiple of 5^3 = 125. Since, not all four MPs can be a multiple of 5, one price, c, say, must be a multiple of 5^2 = 25. If (a, b) = (395, 125), then, modulo 25, a+b+c = 20. Since d can have only 2, 3, 5 as prime divisor, d = 16. But this leads to c = 175 = 7×5^2, which is not possible. If (a, b) = (395, 250), again d = 16 so that c = 50 = 2×2×5^2. But then abcd is not divisible by 3. Since a+b < 711, this exhausts the possibilities and Case 1 cannot occur.

Case 2. One of the prices, say a is one of the multiples 79, 158, 231, 316, 474 of 79 and another, say b is one of the multiples 125, 250, 375, 500, 625 of 125. Examining the cases and conducting an analysis similar to that of Case 1, we arrive at the

unique solution (a, b, c, d) = (316, 125, 150, 120) = (2^2 × 79, 5^3, 2 × 3 × 5^2, 2^3 × 3 × 5)

Therefore, the MP of four shirts is 1.2 K, 1.25 K, 1.50 K and 3.16 K.


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The question was that was 82nd Last Year ---------------------------------------------The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...up to infinity is

(a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

Each term in the summation is n/(n^4 + n^2 + n). Had each term be n/(n^4 + n^2 + 1), then the summation would become 1/2(1 - 1/3 + 1/3 - 1/7 + .... up to infinity) = 1/2. Note that n^4 + n^2 + 1 can be factored as (n^2+1)^2 - n^2 = (n^2+n+1)*(n^2-n+1).

n/(n^4 + n^2 + n) < n/(n^4 + n^2 + 1) for all n > 1. Thus, our answer is < 1/2. All options among the first 4 are >=1/2.


-----Each question is followed by 2 statements X and Y. Answer each question using the following instructions

Choose A if the question can be answered using X aloneChoose B if the question can be answered using Y aloneChoose C if the question can be answered using either X or (exclusive) YChoose D if the question can be answered using X and Y togetherChoose E if the question can not be answered using X and Y also

A PaGalian alphabet consists of consonants and vowels. The rule is that a finite sequence of letters in a word alternates between consonants and vowels. How many letters are there in the alphabet?

(X) There are 4800 five letter words

(Y) The difference between the number of five and four letter words is between 1500 and 1800

Solution:

Let the alphabet has c consonants and v vowels. Our 5 letter word can have c*v*c*v*c + v*c*v*c*v words. From X we have 4800 = c^2.v^3 + v^2.c^3 = c^2.v^2.(c+v) = 3.2^6.5^2 => c or v is a multiple of 5. Let c = 5, => v can be 2^2 = 4, doesn't satisfy; if c = 10, and v = 2, this fits in good, and we can have no other case here => c+v = 12. (X) is good enough to answer us!

What we have with (Y) is 1500 < F (c, v) = c^2.v^2(c+v-2) < 1800.Now, F(c, v) is increasing function w.r.t. c and v. Thus, if we take c = v, we have c^4.(2c-2) < 1800 => c can at most be 4.

Now, take c = 1, 2, 3, 4 each and we are done.For c = 1, 1500 < v^2.(v-1) < 1800 => v = 12 => alphabets can be 13For c = 2, 1500 < 4v^3 < 1800, but 4.7^3 < 1500 and 4.8^3 > 1800. No solution.For c = 3, 1500 < 9v^2.(v+1) < 1800, F(3, 5) < 1500 and F(3, 6) > 1800For c = 4, v = 4 satisfies => 8 alphabets => (Y) is not sufficient.

=> Choice (A) is the right answer

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The question was that was 82nd Last Year ---------------------------------------------The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...up to infinity is

(a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

Each term in the summation is n/(n^4 + n^2 + n). Had each term be n/(n^4 + n^2 + 1), then the summation would become 1/2(1 - 1/3 + 1/3 - 1/7 + .... up to infinity) = 1/2. Note that n^4 + n^2 + 1 can be factored as (n^2+1)^2 - n^2 = (n^2+n+1)*(n^2-n+1).

n/(n^4 + n^2 + n) < n/(n^4 + n^2 + 1) for all n > 1. Thus, our answer is < 1/2. All options among the first 4 are >=1/2.


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Problems 81, 82 and 83 were posed in QQAD 2008. 25 students solved at least one of the three. Amongst those who did not solve 81, twice as many solved 82 as 83. The number solving only 81 was one more than the number solving 81 and at least one other. The number solving just 81 equalled the number solving just 82 plus the number solving just 83. How many solved just 83?

(1) 2 (2) 5 (3) 6 (4) can not be determined (5) none of these

Solution:

Let a solve just 81, b solve just 82, c solve just 83, and d solve 82 and 83 but not 81. Then 25 - a - b - c - d solve 81 and at least one of 82 or 83. The conditions give: b + d = 2(c + d); a = 1 + 25 - a - b - c - d; a = b + c. Eliminating a and d, we get: 4b + c = 8. But d = b - 2c ≥ 0, so b = 6, c = 2.


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The question was that was 84th Last Year ---------------------------------------------The number of permutations of the set {1, 2, 3, 4} in which no two adjacent positions are filled by consecutive integers (increasing order)

(a) is a prime(b) is a composite number divisible by 3(c) does not exceed 17(d) is at most 19(e) None of the foregoing

The permutations where there are consecutive numbers are1234, 1243, 1342, 1423, 2134, 2341, 2314, 3124, 3412, 3421, 4123, 4312, 4231


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Let x, y, z be three distinct integers such that x < y < z and 3x^2 – (y+z)x + 2(y-z)^2 = 0. The minimum positive value of z will be

(1) 2 (2) 3 (3) 5 (4) 8 (5) none of these

Solution:

This is a completely raw problem where it's better to use your weirdo skills, substitution methods to get at the desired answer. Please read the solution by students on the thread. The thing to see quickly here is that x can not be negative, and if x = 0 then y = z. Thus, our 1st starting point is that z can at least be 3. Put z = 3 in our equation and we have valid x and y.


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The question was that was 85th Last Year ---------------------------------------------When Divya is one year younger than Twinkle will be when Divya is half as old as Twinkle will be when Divya is twice as old as Twinkle is now, Twinkle will be three times as old as Divya was when Twinkle was as old as Divya is now. One of them is in her teens and the ages are natural numbers. What is the sum of the ages?

(a) 42 (b) 44 (c) 48 (d) 52 (e) 54

DIVYA TWINKLE w-1 3z v w 2y 2v X(present ages) Y (present ages)

Z (some time in past) x (some time in past)

Here 2y-x = 2v-y, v = (3y-x)/2Similarly, w = (5y-3x)/2 ; z = 2x-y or 3z = 6x-3yand w-1 = (5y-3x-2)/2

Now,{(5y-3x-2)/2} – x = (6x-3y) – y =>13y = 17x+2

Here only at x = 19 and y =25 satisfies for one is a teenager.

=> option (b) is the correct answer.

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Let a, b, c be real such that a+b+c ≠ 0. If (a^5+b^5+c^5)/((a^2+b^2+c^2)(a^3+b^3+c^3)) = 5/6, then ab(a+b) + bc(c+a) + ca(c+a) - 3abc equals

(1) 1/2*(a^3 + b^3 + c^3)(2) (a^2 + b^2 + c^2)^2/(a+b+c)(3) a^3 + b^3 + c^3(4) 3*(a^3 + b^3 + c^3)(5) (a^4 + b^4 + c^4)/(a + b +c)

Solution:

Put a = 1 and b = 0 the equation (a^5+b^5+c^5)/((a^2+b^2+c^2)(a^3+b^3+c^3)) = 5/6 reduces to (c+1)^3*(c^2 - 3c + 1) = 0 => c^2 - 3c + 1 = 0 as a+b+c ≠ 0.We are being asked to find ab(a+b) + bc(b+c) + ca(c+a) - 3abc = c(1+c) [this is what we need to find]Looking at the choices (a) becomes 1/2*(1+c^3), by putting a = 1, b = 0.Multiplying both sides of c^2 - 3c + 1 = 0 by c + 1 we get c(1+c) = 1/2*(1+c^3)

The idea is that it might not strike you immediately as to how to resolve expressions, and here is when it helps putting some manageable values, that satisfies conditions of the problem. What we have got here in terms of c is something that need not be solved. It just needs to be equated with answer options quickly.

Also, try the same approach by putting a = -b, we will reach to (1) only.


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The question was that was 86th Last Year ---------------------------------------------Let there be Z number of rational triplets (g, h, i) for which g, h, i are the roots of x^3 + gx^2 + hx + i = 0. Then which of the following is the value of Z ?

(a)0 (b)1 (c)2 (d)3 (e) Cannot be determined

We require (1) g + h + i = -g, (2) gh + hi + ig = h, and (3) ghi = -i.From (3), either I = 0, or gh = -1. If i = 0, then (1) becomes h = -2g, and (2) becomes h(g - 1) = 0. Hence either g = h = 0, or g = 1, h = -2.

So assume i ≠ 0, and gh = -1. (1) becomes i = - h – 2g. Substituting in (2), we get: -1 - (2g+ h)(g + h) = h, so -g^2 –2g^4 + 3g^2 - 1 = -g, or 2g^4 – 2g^2 - g + 1 = 0. So g= 1, or 2g^3 + 2g^2 - 1 = 0 ….

The first possibility gives g = 1, h = -1, i = -1. Suppose g = m/n is a root of 2g^3 + 2g^2 - 1 = 0 with m, n relatively prime integers. Then 2m^3 + 2m^2n - n^3 = 0. So any prime factor of n must divide 2 and any prime factor of m must divide 1. Hence the only possibilities are g = 1, -1, 1/2, -1/2, and we easily check that these are not solutions. So 2g^3 + 2g^2 - 1 = 0 has no rational roots.


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A manufactured cloth piece comes in a fixed length, say L. The piece can be soldin at most two parts, however it becomes difficult to sell if the cloth is cut.A shopkeeper sells it with selling price which is directly proportional to the length x of cloth piece bought upto L/2 and after that (x >= L/2) he also charges the cost price of the remaining unsold part. If the selling price of cloth piece bought of length x >= L/2 is directly proportional to (x+L), then for x <= L/2, the profit % on the sale of the cloth is

(1) 50% (2) 75% (3) 100% (4) can not be determined (5) none

Solution:

SP = Ax (x <= L/2)= Bx + P(L-x) (x >= L/2)

At x = L/2, AL/2 = BL/2 + PL/2 => A = B + P --(1)

Using the info given: K(x+L) = Bx + P (L-x) = (B-P)x + P(L)A case that satisfies would be B = 2P => A = 3P

Hence, for x <= L/2, SP = Ax = 3Px, CP = Px=> Profit = 200%


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The question was that was 87th Last Year ---------------------------------------------Let x, y, z be prime numbers in the arithmetic progression such that x > y > z. Which among the following is always true?

(a) x – z is divisible by 12

(b) x + y is divisible by 8(c) xz - 1 is divisible by 7(d) atleast two of the foregoing(e) none of the foregoing

Take x = 7, y = 5 and z = 3 rules out first 4 options. However, if z were > 3 then x - z is always divisible by 12. Check yourself for that.


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Five students Implex, Slam, Sanyo, dewan and nbangalorekar are wearing caps of Blue or Green color without knowing the color of his own cap. It is known that the students wearing the Blue cap always speaks the truth while the ones wearing Green always tell lies. If the students make the following statements

Implex: I see 3 blue caps and one GreenSlam: I see 4 Green capsSanyo: I see 1 Blue cap and 3 Greendewan: I see 4 Blue caps

Then, which among the following (Student, Cap Color) combination is correct?

(1) (Implex, Blue) (2) (Slam, Green) (3) (dewan, Blue) (4) at least two of the foregoing (5) none of these

Solution:

Statement - dewan: I see 4 Blue capsIf dewan is wearing a Blue Cap that means, everyone is wearing a Blue Cap. Not Possible. So dewan is wearing a green cap.

Now come to statement: Implex: I see 3 blue caps and one GreenIf implex wears a Blue Cap then he sees i guy with dewan with green and rest with blue. so Slam wears Blue but he sees 4 green cap. Not possible again.So Implex also wears a Green Cap.

left cases -Implex___Slam___Sanyo___dewan___nbangalorekar ____CaseG_______G______B________G________B___________Satis ifes

G_______B______G________G________G_____Rejected(as sanyo is wearing green but still speaking truth.)

G_______B_______B________G______G___Rejected.

so only statement holds true. (2) (Slam, Green)


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The question was that was 88th Last Year ---------------------------------------------What is the minimum value of the expression 6x^2+3y^2-4xy-8x+6y+2?

(a) -13/7 (b) -2 (c) -8/5 (d) -7/3 (e) none of the foregoing

The expression 6x^2+3y^2-4xy-8x+6y+2 can be written as (14x^2 -12x - 3)/3 + 3(y - (4x-6)/6)^2 To minimize (y - (4x-6)/6) = 0 and 14x^2 -12x - 3 should be minimum.

14x^2 -12x - 3 is minimum for x = 3/7 => y = -5/7.

Alternatively: Differentiate F(x, y) = 6x^2+3y^2-4xy-8x+6y+2 = 0 w.r.t to x 1st and then y. Solve the 2 equations you get and the values of x and y give you the answer.


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If f(x) = x^2 - 2x then for how many distinct real α is f(f(f(f(α)))) = 3?

(1) 3 (2) 6 (3) 5 (4) 9 (5) none of these

Solution:

f(f(f(f(α))))=3

For f(α)=3, x=3 or -1Since 3 is repeated now, we just need to check for -1

For f(α)=-1, α=1

For f(α)=1, α=(1+√2) or (1-√2)

For f(α)=(1+√2), α=(1+√(2+√2)) or (1-√(2+√2))

For f(α)=(1-√2), α=(1+√(2-√2)) or (1-√(2-√2))

For each of the above highlighted values of α, the expression will give a result =3

The important aspect to notice here is that f(3)=3 and hence the moment we get answer as 3, the cycle will not affect it. Eg. α=1, after 2 cycles will give 3 and that will always give the answer as 3.


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The question was that was 89th Last Year ---------------------------------------------Allwin likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example, 89.34 is 89 and 99.54 is 100, and 115.5 is 116. Allwin always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true?

(a) The minimum possible number of courses is less than 15(b) The maximum possible number of courses is not more than 50(c) The minimum possible current rounded average is 95(d) Either of 126 or 127 can be the current rounded average(e) none of the foregoing

An + 75 = (A - d1)*(n+1); An + 75 + 83 = (A - d1 - d2)*(n+2) -> A is the current average. Let's simplify the expression above:75 + d1 = A - nd1 and 79 + (d1 + d2) = A - n/2(d1 + d2) => n*(d1 - d2) = 8 + 2d2 => (n*d1 – 8)/*(n+2) = d2

The fact to be noted here is d1 > d2 as adding 75 will bring down the average more than adding 83. Let's look at the range of both d1 and d2 now.d2 will lie in [1+x, 3-y] where both x and y -> 0 => x + y > 2=> d1 will lie in [1+a, 2-b] where both a and b -> 0 => a < x, 1 > y - b, a+b > For n = 10, we have (5d1 - 4)/6 = d2, for extreme values of d1 < 2 we have d2 < 1 => n > 10.For n = 11, we have (11d1 - 8)/13 = d2; but d2 > 1 => d1 > 21/11. After this we require to do some quick calculations.See yourself that An = 1083 satisfies our conditions where d1 is approx. 1.95Hence, choice (a) as well as choice (c) is correct.

The expression in pink tells us we can have n > 50 and d1 > d2. Hence, choice (b) is false.

No need to check further for choice (d) as we have our answer already. For completion sake we can verify that current round average 127 is obtained when to start with there are 3653 points from 28 course current round average 126 is obtained when to start with there are 4521 points from 35 courses

Hence, option (b) is the right answer

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In triangle ABC, M is the mid-point of BC. If < AMB = 45˚, and < ACM = 30˚, then < BAM is(1) 30˚ (2) 45˚ (3) < 30˚ (4) > 45˚ (5) none of these

Solution:

Take the point D on AC such that BD is perpendicular to AC. Why? Because in 30-60 right angle, we know a side is half the hypotenuse and the question gives a big HINT by telling that M is the mid-point of BC. Thus, there has to be something in us to explore this kind of construction.Join DM.

In triangle DBC, we already have angles BDC and ACB => < DBM = 60 degrees. By sine rule on DBC we have BD = BC/2 = BM => triangle BDM is equilateral.Thus, < DMB = 15 degrees, and it triangle ADM, < AMD = 60 -45 = 15 degrees.=> AD = BD = DM, D being the common point, we see a hope here! Convert the hope into action by treating D as centre of thecircle that passes through A, B and M.Use the well known rule that the angle subtended by the arc of a circle at the centre is twice of angle subtended at the circumfrence. Thus, 1/2* (< BDM) = 1/2*60 degrees = < BAM.


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The question was that was 90th Last Year ---------------------------------------------Let P(x) = nx + a where n and a are integers with n > 0. If the solution to 2^ P(x) = 5 is x = log10/log8, then 4n-3a is


Taking log on both the sides of 2^P(x) = 5 we get P(x) = log5/log2. x = log10/log8 = 1/3(1 + log5/log2) => n = -3 and a = 1.

Hence, option (e) is the right answer

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Let S be a subset of {1, 2, 3, ... , 15} such that no two subsets of S have the same sum. What is the largest possible sum for S?

(1) 55 (2) 58 (3) 61 (4) 63 (5) none of these

Solution:

To find S with sum of the elements having max sum, we can start with 15 and 14 safely. 13 can also be taken without any harm.Now, 15 = 15 (what a good identity!), 14 = 15-1, 13 = 15-2.The next number will be 15-4 as if next were 15-3, then 15-3 + 15 = 15-1 + 15-2 [condition violated].Now, we have 4 numbers with us. Each of the next number below 15-4 i.e. 15-5 can't be in our set as15-5 + 15 = 15-1 + 15-4, 15-6 can't also be there as 15-6 + 15 = 15-2 + 15-4.But we have no such problems with 15-7. The set till now is {15, 14, 13, 11, 8}. Any new element 15-x, where x > 7 can be combined with some of the elements in

{15, 14, 13, 11, 8} whose sum will be equal to some of the rest elements. We stop here!

Thus -> 0, 1, 2, 4, 7 became seed (for the namesake) elements which needed to be subtracted from 15.The next seed element will be 13 as any number between [8, 12] will violate condition.e.g. 8+0 = 7+1 or 11+0+1 = 7+4+1 -> note that we must have same number of summands on either side to do this kind of generation.The 1st few seed elements are 0, 1, 2, 4, 7, 13, 24, 46, 86.


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The question was that was 91st Last Year ---------------------------------------------For each positive integer p, let S(p) denotes the sum of the digits of p. For how many values of p is p+ S(p) + S(S(p)) = 2007?


Four numbers satisfy the given condition: 1977, 1980, 1983, 2001. numbers must be of the form of 19ab or 2001 (obvious)case 1: a+b<10above equation is1900+10a+b+(10+a+b)+(a+b+1)=200712a+3b=96solution is a=8;b=0 1980case 2: a+b>10equation is 1900+10a+b+(10+a+b)+(a+b+2-10)=2007that is 12a+3b=105solutions are a=8;b=3 1983a=7;b=7. 1977

Hence, option (d) is the right answer

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Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible?

(1) 36 (2) 60 (3) 72 (4) 84 (5) none of these

Solution:

Keeping one face of DIE1 constant, we can bring it in contact with each of the 6 faces of DIE2 in 4 possible ways. While bringing a face of DIE1 in contact with the same face of DIE2 (same number), each of the generated arrangements will be unique. But when bringing a face of DIE2 with a different face (different number), each arrangement would be repeated once.

This leads to the Number of possible arrangements = (6x4) + (6x5x4/2)= 24 + 60


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The question was that was 91st Last Year ---------------------------------------------For each positive integer p, let S(p) denotes the sum of the digits of p. For how many values of p is p+ S(p) + S(S(p)) = 2007?


Four numbers satisfy the given condition: 1977, 1980, 1983, 2001. numbers must be of the form of 19ab or 2001 (obvious)case 1: a+b<10above equation is1900+10a+b+(10+a+b)+(a+b+1)=200712a+3b=96solution is a=8;b=0 1980case 2: a+b>10equation is 1900+10a+b+(10+a+b)+(a+b+2-10)=2007that is 12a+3b=105solutions are a=8;b=3 1983a=7;b=7. 1977


Let the quadratic ax^2 + bx + c be such that a, b, c are distinct and each of a, b, c belong to {1, 2, 3, ..., n} such that x+1 divides ax^2 + bx + c.If the number of such quadratic polynomials are < 99, then max (n) is

(1) 14 (2) 15 (3) 16 (4) 18 (5) none of these

Solution:

For the set 1 to N, the sets (a,b,c) can be formed as:1 with (2 to N-1) will give 'c' less than or equal to N => N-1-2+1 = 2(N-2) sets2 with (3 to N-2) => 2(N-4) sets3 with (4 to N-3) => 2(N-6) sets and so on.(factor of two added to account for a and c getting interchanged)

We divide into two cases from here:case N = 2x (even)=> total number of sets = 2 [2 + 4 + .. + (2x-2)] = 2x(x-1)=> 2 (x^2 -x) < 99=> x = 7 (max)=> N = 14

case N = 2x+1=> total number of sets = 2 [1+3+..+(2x-1)] = 2 x^2=> 2.x^2 < 99=> x = 7 (max)=> N = 15


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The question was that was 93rd Last Year ---------------------------------------------Let 3 statements be made (P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10(Q) The remainder on dividing 16! + 89 by 323 is q(R) 46C23 leaves remainder r on division by 23

Then p+q+r equals


The notation a % b = c means that when a is divided by b, c is the remainder that is obtained.

(P) Check for p = 9; 10^12 % 13 = 1 by Fermat's theorem => 10^18 % 13 = 10^6, since 10^3 % 13 = -1 + 13 => 10^18 % 13 = 1 => 10^18 - 10^9 + 1 leaves remainder (1-(-1)+1=3) by p = 9; Check yourself the remainder when p = 8.When p = 7, as 10^12 % 13 = 1 => 10^14 % 13 = 9; also 10^6 % 13 = 1 => 10^7 % 13 = 10. Thus p = 7 satisfies our condition.

(Q) 323 = 17*19. (p-1)! + 1 is divisible by p when p is prime => 16! % 17 = -1 + 17. Also 18! % 19 = -1 + 19If 16! % 19 = x => 18! % 19 = 17*18x => 306x % 19 = 18 => 2x % 19 = 18.Thus, 16! leaves the remainder 16 by 17 and 9 by 19 => It leaves 237 by 323 => q = 3

(R) 2nCn = (nC0)^2 + (nC1)^2 + ... + (nCn)^2; when n is prime, each of the term in RHS except 1st and last is divisible by n => r = 2

Hence, option (a) is the right answer

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Yana and Urvashi normally work for 8 hours but do extra four hours on overtime. A job was done by 2 ladies together in 12 days in which Yana and Urvashi did overtime for few days (not necessarily same number of days). It is calculated that had Yana not done her overtime, then the work would have taken 15 days to complete and corresponding value for Urvashi is 20 days. How many days it would have taken for the work to complete had both ladies not done overtime?

(1) 20 (2) 23 (3) 30 (4) can not be determined (5) none of these

Solution:

Let the number of days required by yana to finish the work alone be x days of 8 hrs.Let the number of days required by urvashi to finish the work alone be y days of 8 hrs.Let the number of days of overtime by yana and urvashi be p and k respectively.So,in 12 days yana finishes 12/x+k/2x of work.In 12 days,urvashi finishes, 12/y+p/2y of work.So by condition 1, 12/x+12/y+k/2x+p/2y=1....(1)By condition 2, 15/x+15/y+p/2y=1.....(2)And,by condition 3, 20/x+20/y+k/2x=1.....(3)Adding 2 and 3 and subtracting 1 from the result,1/x+1/y=1/23.

So,if both work normally,they take 23 days.


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In a sports tournament, the number of European teams is 9 more than the number of Asian teams. Each team plays the other exactly once, the winner gets 1 point and the loser gets nothing. There were no draws during the tournament. If the total points earned by European teams in 9 times the total points earned by the Asian teams. What is the maximum possible score of the best Asian team?

(a) 9 (b) 11 (c) 21 (d) not be uniquely determined (e) none of the foregoing

Let the no. of Asian teams be x. Therefore no. of European teams will be x+9. Matches can be divided into three groups: 1) Those played b/w two Asians 2) Those played b/w two Europeans 3) Those played b/w an Asian and a EuropeanIn the 1st group no. of points won (by Asians) = x(x-1)/2In the 2nd group no. of points won (by Europeans) = (x+9)( x+8 )/2Let k be the points Asians gain in the 3rd group. Therefore those won by Europeans = x(x+9) - kUsing the given conditions we have9*[x(x-1)/2 + k] = (x+9)( x+8 )/2 + x(x+9) - kThis gives 3x^2 - 22x + 10k - 36 = 0=> Factorize it to get the following: (3x+5)(x-9) = -10k - 9Since RHS is negative, LHS too should be negative. Hence, x lies between 0 and 9. On checking, we find that the only integer solutions of the above equation occur when either x=8 and k=2 or when x=6 and k=6. The maximum no. of matches an Asian team can win if x = 6 and k = 6 is5 (won from other 5 Asian teams in 1st Group) + 6 (If the same Asian team is the one which wins all the Asian-European matches in 3rd Group) = 11

and if x = 8 and k = 2 it is 7 + 2 = 9So the maximum no. of points an Asian team can have is 11.

Hence, option (b) is the right answer

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An Elective on strategy is attended by 15 students sitting on adjacent chairs in a circle. The Prof. enters the class with 6 copies of material on BSC. The prof. wishes to distribute the copies in such a manner that each student gets at most 1 copy and anyone who hasn't should be able to read from his immediate (left or right) neighbour's copy. In how many ways can the copies be distributed?

(1) 60 (2) 75 (3) 90 (4) 125 (5) 150

Solution:

Suppose that you are one of the student (X); then there’s a 6/15 chance that you’ll get one of the copies. Given that you do get a copy, how many ways are there to distribute the rest? We need only multiply the answer to that question by 15/6 = 5/2 to answer the original question.

Going clockwise around the circle from student (X), write down the sizes of the gaps between students with copies. There are six such gaps, each of size 0–2, and the sum of their sizes must be 15 − 6 = 9. So the gap sizes are either 1, 1, 1, 2, 2, 2 in some order, or 0, 1, 2, 2, 2, 2 in some order. Remember, the copies are indistinguishable.

In the former case, 6!/3!3! = 20 orders are possible; In the latter, 6!/1!1!4! = 30. => there are 20 + 30 = 50 possibilities.

Multiplying this by 5/2, gives 125.


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Given p <= 4 is a positive real . Let A be the area of the bounded region enclosed by the curves y = 1 - |1-x| and y = |2x-p|. Then which among the following best describes A?

(a) 0 < A <= 1/2 (b) 1/6 <= A <= 1/4 (c) 0 < A <= 1/3 (d) 0 < A < 2/3 (e) 1/6 <= A <= 1/3

Case 1: 0 < p <= 1The area formed is of a triangle with vertices (p/3, p/3), (p/2, 0) and (p, p). Thus, the area is p^2/6 sq. units.Case 2: 1 <= p <= 3

The figures formed is a quadrilateral (p/3 ,p/3),(p/2,0), ((p+2)/3 , (4-p)/3) and (1,1).Thus, the area formed is 1/3 - 1/6*(p-2)^2Case 3: 3 <= p <= 4The area formed is the image of case 1. Area = (4-p)^2/6Thus, min(A) = 0 at p = 4 and max(A) = 1/3 at p = 2

Hence, option (c) is the right answer

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Let x= p/11 satisfies log(2x-3/4)/log x > 2, where p is an integer. Then the number of possible p is

(1) 2 (2) 6 (3) 3 (4) 5 (5) none of these

Solution:

(log(2x-3/4) – 2logx).log x > 0 solving Case 1: When x > 1 => (2x-3/4) > x^2.Case 2: When x < 1 => (2x-3/4) < x^2 && x > 3/8=> x lies in (3/8, ½) U (1, 3/2)


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If x, y, z are natural numbers such that y ≠ 1 and (x-y√3)/(y-z√3) is a rational number. Then which among the following is always true?

(a) x^2 + 4z^2 is composite (b) x^2 + y^2 + z^2 is non-prime (c) x^2 + 4z^2 is prime (d) both (a) and (b) (e) none of the foregoing

(x-y√3)/(y-z√3) = (x-y√3)*(y+z√3)/(y^2 - 3z^2) = (xy +√3(xz - y^2) -3yz)/(y^2 - 3z^2) is rational => y^2 = xz.

Let y = xr, z = xr^2 => x^2 + 4z^2 = x^2*(1+4r^4) = x^2*(2r^2 + 2r+1)(2r^2-2r+1). If x = 1 then r ≠ 1; Thus, x^2 + 4z^2 can be written as product of 2 natural numbers > 1 which are x(2r^2 + 2r+1) and x(2r^2 - 2r + 1).

Similarly, x^2 + y^2 + z^2 = x(r^2 + r + 1)*x(r^2 - r + 1) and can be written as product of 2 natural numbers > 1.


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How many natural numbers n are there such that out of all the positive divisors of number n (other than both 1 and n) the largest one is 15 times than the smallest one?

(1) 1 (2) 2 (3) 3 (4) There are no such numbers (5) Infinitely many

Solution:

(If S is the smallest asked factor and L the largest then S.L = N => 15S^2 = N. But S can take values less than or equal to the smallest prime factor of 15 => S = 2, 3.


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(Part A)Let three 3-digit integers abc, bca, cab when divided respectively by 10, 8 and 3 leave the same remainder 1. If 2 < b < 8, then a =

(Part B)A four digit number is formed by using 5, 6, 8 and 9 (each number being used once). How many of them will be divisible by 7?

(a) 3 (b) 4 (c) 5 (d) can not be determined (e) none of the foregoing

(Part A)abc leaves remainder of 1 when divided by 10 => c = 1. bca leaves remainder of 1 on dividing by 8 => 4b+2c+a = 8m+1 and similarly a+b+c = 3n+1. On solving we get 3b = 8m -1 - 3n => m = 2, 5. Also 4b+a = 39 means b > 7 => m = 5 rejected => b = 5-n => a+b = 3n => a = 4n-3 but b > 2 => n < 3. But 0 < a < 10 => n = 2 and hence a = 3.Hence, choice (a) is the right option(Part B)The concept of seeds can again be applied to this problem. The seed of 7 is 5 => if abcd is div by 7 then a+5b+25c+125d is div by 7 => (a-d) + 2(2c-b) is div by 7.

(a-d) + 2(2c-b) can take 0, 7, 14, 21.If (a-d) + 2(2c-b) were 21, we have just 1 solution i.e. c = 9, b = 8, a = 6, d = 5.If (a-d) + 2(2c-b) were 0, we have just 1 solution i.e. c = 5, b = 9, a = 6, d = 8.if (a-d) + 2(2c-b) were 7 => 2c-b=3 && a-d = 1 OR 2c-b=2 && a-d=3. It is easy to see one solution in each case.Similarly, we will have 2 solutions when (a-d) + 2(2c-b) = 14. Check yourself!Thus, in all 6 possible such numbers.

Hence, choice (e) is the right option------------------------------------------------------------

F is a fixed point in the plane. P, Q, R are points such that FP = 3, FQ = 5, FR = 7 and the area PQR is as large as possible. Then F must be (of PQR)

(1) incentre (2) orthocentre (3) circumcentre (4) centeroid (5) none

Solution:

Consider all points A' such that FA' = 3. They lie on a circle center F. The area of A'BC is BC/2 times the distance of A' from BC. That distance is maximal for A'F perpendicular to BC (because the distance is the distance of P from BC + FA'sin x, where x is the angle between A'F and BC). Hence AF must be perpendicular to BC. Similarly BF must be perpendicular to AC, so F must be the orthocenter.


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(Part A)Three articles are sold at the profit of 20%, 25% and 40% and the net profit comes out to be 30%. Had the article sold at 25% been sold at 15% profit and the article sold at 40% been sold at its cost price then the profit on the sale of three articles would have been

(Part B)Let the cost price of an article A is as much more than another article B as much that of B is more than that of another article C whose cost price is more than that of fourth article D by the same amount. If A, B, C, D are sold at 10%, 15%, 30% and 5% profit respectively, then the net profit percent on the sale of four articles together is

(a) 5% (b) 10% (c) 15% (d) can not be determined (e) none of the foregoing

(Part A)From 1st condition, C3 = C1 + 0.5C2 => new profit = (0.2C1+0.15C2)*100/(2C1+1.5C2) = 10%Hence, choice (b) is the right option(Part B)If Cd is the Cp of D and x be the common difference in CPs, then net profit = (60Cd+90x)/(4Cd+6x) = 15%

Hence, choice (c) is the right option

Consider two distinct positive integers x and y having integer arithmetic, geometric and harmonic means. The minimum value of |x - y| is

(1) 15 (2) 20 (3) 30 (4) 40 (5) 50

Solution:

Clearly both m and n must have the same parity (i.e. either both are odd or both are even). Let m = a+b, and n=a-b for integers a, b where a > b > 0.we have |m-n| = 2a, mn = a^2 - b^2 = p^2 is a perfect square. Thus, (p, b, a) form a pythagorean triplet.(1/2m + 1/2n)^-1 = (a^2 - b^2)/a => a divides both b^2 and p^2. From this we get the triplet (20, 15, 25).=> m = 40, n = 10.


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The question was that was 99th Last Year ---------------------------------------------(Part A)Two triangles of sides (2,2,a) and (3,3,b) are inscribed in a circle. If a^4+36b^2 = b^4 +16a^2, then what is the ratio of a/b?

(Part B)RST is a triangle and U, V, W are the feet of perpendiculars dropped from R, S, T on sides ST, TR and RS respectively. If SU=2, SW=3 and TV=4, then UV is

(a) 9/4 (b) 7/3 (c) 7/2 (d) 3/2 (e) None of the foregoing

(Part A solution )a^4+36b^2 = b^4 +16a^2 => a^2*(a^2-16) = b^2*(b^2-36) => a^2/b^2 = (36-b^2)/(16-a^2)=>a/b = root((36-b^2)/(16-a^2))-------------------(A)

Also , area of the triangle with side (2,2,a) = (a*2*2)/(4*R) = (a/4) * root(4*2^2 - a^2) {Where R is the circumradius}

=> R = 4/root(16 - a^2)------------(1)

Also from the second triangle with sides(3,3,b) we get

R = 9/root(36-b^2)------------------(2)

Equating (1) and (2) we get,(36-b^2)/(16-a^2)= 81/16

Using (A) we have a/b =9/4Hence, choice (a) is the right option(Part B)Triangles TVU & TSR are similar, 4/(TU+2)=UV/(3+RW)Triangles SUW & SRT are similar, 2/(3+RW) = 3/(2+TU)Hence UV=8/3Hence, choice (e) is the right option

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For how many pair of primes (p, q) does there exist natural number n such that (p^2+1).(q^2+1) = n^2+1?

(1) 2 (2) 3 (3) 5 (4) 6 (5) none of these

Solution:

Case 1: when each of p, q > 3 => p and q are of the form 6k+/-1.We need to find (p, q) such that p^2 + q^2 + p^2.q^2 is a perfect square. Each of p^2, q^2, p^2.q^2 is of the form 4t+1 => n^2 is of the form 4r+3, which is a contradiction as perfect squares are of the form 4r+1 or 4r.

Case 2: when p = 3, or p = 2, and q is any prime -> Solve this yourself.

Alternate Solution:

If 'p' and 'q' are both odd,(p^2+1)(q^2+1) = 4x = n^2+1which implies that n^2 = 4x-1.Since the squares are of form 4x or 4x+1, 'p' and 'q' are not both odd. Hence, one of the prime numbers in solution, if any, is 2.

Say, p = 2This gives us, 5(q^2+1) = n^2+1 which can be simplified to5q^2 = (n+2)(n-2)5q^2 can be factorized in two factors as (1,5q^2), (q,5q) or (5,q^2).For (1,5q^2) n=3 which means q=1, hence not possible. This also removes the set (q,5q).For (5,q^2) n=7 (as n=3 has already been ruled out). This gives q=3.

The only feasible pair (p, q) is (2, 3) and (3, 2)


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The question was that was 100th Last Year ---------------------------------------------(Part A)Consider the series S=1+ 1/2^2 + 1/3^2 + 1/4^2 +... +1/n^2.The value of S for any n is less than which of the following?

(Part B)

If x,y,z are positive real numbers representing the sides of a triangle,then (x^2+y^2+z^2)/(xy+yz+xz) is necessarily less than which of the following?

(a) 1.80 (b) 2.0 (c) 1.85 (d) 1.95 (e) None of the foregoing

(Part A)The roots of sinx/x are n*pi,viz +/- pi.+/-2pi and so on.Hence, sinx/x = (1-x/pi) (1+x/pi) (1-x/2pi) (1+x/2pi) (1-x/3pi) (1+x/3pi)...

Coeff of x^2 here= -(1/pi^2 + 1/4pi^2 + 1/9pi^2 + ...)

But we also know that, sinx = x-x^3/3! +...Here coeff of x^3 is -1/6

Hence, -(1/pi^2 + 1/4pi^2 + 1/9pi^2 + ...) =-1/6=>1+1/4+1/9+... = (pi^2)/6 Hence, all the choices from choice (a) to choice (d) are the right option(Part B)As x,y and z are sides of the triangle hence,

x-y < z ----------------(1)z-x < y ----------------(2)y-z < x ----------------(3)squaring and adding all this three equation we get:

(x-y)^2 + (z-x)^2 + (y-z)^2 < x^2 + y^2 + z^2=> 2*(x^2 + y^2 + z^2) - 2*(xy + yz + zx) < x^2 + y^2 + z^2=> x^2 + y^2 + z^2 < 2*(xy + yz + zx)=>(x^2+y^2+z^2)/(xy+yz+xz) < 2

Hence, choice (b) is the right option

The average value of |a - b| + |c - d| + |e - f| for all possible permutations a, b, c , d, e, f of 1, 3, 5, 7, 9, 11 is

(1) 21 (2) 18 (3) 12 (4) 14 (5) none of these

Solution:

By symmetry the average of |a - b| is independent of choices in which they are chosen (provided they are unequal). Suppose a = k. Then the average of |k - b| is (k-1 + k-3 + ... + 2 + 2 + 4 + ... + 11-k)/5 = 2*(k^2 -7k + 21)/5. So average of |a - b| is (1/6)*2*sigma (k^2 -7k + 21)/5. Hence required average is 2*sigma(k^2 -7k + 21)/10 = 2*(91 - 147 + 126)/10 = 14.


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The question was that was 101st Last Year ---------------------------------------------(Part A)An insect starts at vertex A of a certain cube and is trying to reach at vertex B, which is opposite A, in 5 or fewer steps where a step consists travelling along an edge from one vertex to another. The insect will stop as soon as it reaches B. The number of ways in which the insect can achieve its objective is

(Part B)A committee of 5 is to be chosen from a group of 9 people. How many ways it can be done if Vikram and kaizen must serve together or none at all, and Sunit and Pratyush refuse to serve with each other?

(a) 40 (b) 36 (c) 41 (d) 30 (e) 48

(Part A)Refer post # 3 http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.html Hence, choice (e) is the right option

(Part B)Refer post # 4 http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.html Hence, choice (c) is the right option

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A gathering of a certain number of families consists of people belonging to two generations only. It is known that the number of families is less than the number of girls, the number of girls is less than the number of boys and that the number of boys is less than the number of parents. If the minimum number of single parent families is two, then what is the minimum number of families, given that no family has more than 3 children?

(1) 3 (2) 4 (3) 5 (4) 7 (5) none of these

Solution:

Let no of families , girls , boys and parents be : F,G,B,P respectively.

Now , F<G<B<P

1) F cant be 2 as minimum number of single parent families is 2 and hence 2 parents.

F<G<B<P2) 3<4<5<6 : 6 parents not possible with 2 single parent families(2*1) and one family with two parents (1*2)

3) 4<5<6<7 : 7 parents not possible (2*1 + 2*2)

4) 5<6<7<8 : Possible (P = 2*1 + 3*2)

Hence the minimum no. of families = 5


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The question was that was 102nd Last Year ---------------------------------------------(Part A)Let f(x+1, y+1) = f(x, y) for all integers x and y. If f(x, 0) = x then f(0, 4) is

http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.html




(Part B)Let f(x+1, y) = 1 + f(y, x) for all real x and y. Then, f(x+2, y+2) - f(x, y) =

(a) -4 (b) 0 (c) 4 (d) 2 (e) none of the foregoing

(Part A)f(0, 4) = f(-1, 3) = f(-2, 2) = f(-3, 1) = f(-4, 0) = -4

Hence, choice (a) is the right option

(Part B)f(x+2,y+2) - f(x,y) = 1+f(y+2,x+1) - f(x,y) = 2+f(x+1,y+1) - f(x,y)= 3+f(y+1,x) - f(x,y) = 4+f(x,y) - f(x,y) = 4


Let f(x + f(x)) = x for all real x, and if f(ax + bf(x)) = cx + df(x), then whichamong the following is necessarily true?

(1) b = c (2) b = d+1 (3) a = d (4) at least 2 of the foregoing (5) none

Solution:

Substitute x = x + f(x) in f(x + f(x)) = x and we will have our answer. Please refer the discussions thread for more on this.


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The question was that was 103rd Last Year ---------------------------------------------(Part A)

The painting of a hotel can be done by 10 painters in 5 days. But, in reality there were few dabblers in the group of 10 painters and therefore the work took a little over 6 days for completion. A painter worked twice as fast as the dabbler and used 25% less paint than a dabbler used. If 5 litres extra paint was used than expected, then the total paint used in painting the hotel was (in litres)

(Part B)

A, B and C can do a work together in a certain number of days. If A leaves after half the work is done, then the work takes 4 more days for completion, but if B leaves after half the work is done, the work takes 5 more days for completion. If A takes 10 more days than B to do the work alone, then in how many days can C alone do the work?

(a) 45 (b) 52.5 (c) 60 (d) 65 (e) none of the foregoing

(Part A)Let the number of painters be x, then dabblers is 10-xNow x/50 + (10-x)/ 100 = slightly less than 1/6 .(2x+10-x)/100 = slightly less than 1/6x+10 = slightly less than 100/6x= slightly less than 6.67Since x has to be integer nearest integer is 6.That means the number of dabblers = 4.Let p be the amount of paint used by each dabbler, so by each painter = .75pExtra paint used = .25p * 4 = 5 litresSo p = 5 litres.

Dabblers will be using 5*4 = 20 litres of paint. Now as the painters are working at twice the efficiency so they will use twice the amount of 75% of paint used by dabbler in the same time.i.e. paint used by each painter = 2*(.75)*5 = 7.5 litres.Hence paint used by 6 painters = 7.5*6 = 45 litresSo total paints used = 45 + 20 = 65litresHence, choice (d) is the right option

(Part B)Let A alone can do the work in a days, B alone in b days and C alone in c days.Let A, B and C together complete the work in x days.So, 1/a+1/b+1/c = 1/x -----------(1)Then half of the work they will complete together in x/2 days. When A leaves, Rest half of the work is done by B and C together,(2/b+2/c) = 1/{(x/2)+4}Using (1) , 2*(1/x-1/a) = 1/{(x/2)+4}Solving we get, a = [x*{(x/2)+4}]/4----------------(2)On similar lines b = [x*{(x/2)+5}]/5 ------------------(3)Given that a = b+10.[x*{(x/2)+4}]/4 = [x*{(x/2)+5}]/5 + 10Solving we get x = 20.From (2) and (3) we geta = 70 and b = 60 Put these values in (1) we get c = 52.5 days.

Hence, choice (b) is the right option

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How many equilateral triangles of side 2/√3 are formed by the lines y = k, y = x(√3) + 2k, y = -x(√3) + 2k for |k| <= 10 where k is an integer?

(1) 600 (2) 660 (3) 720 (4) 780 (5) none of these

Solution:

Between the lines y = 10 and y = -10 we have lines parallel to them, but outside them we do not. Similarly for the lines y = x√3 ± 20, and for the lines y = -x√3 ± 20. Thus the area where triangles are formed is the hexagon bounded by these six lines. It has long diagonal length 40/√3 from -20/√3 (the intersection of y = 0, y = x√3 + 20 and y = -x√3 - 20) to 20/√3 (the intersection of y = 0, y = x√3 - 20 and y = -x√3 +

20). So we can regard it as made up of 6 equilateral triangles side 20/√3. Each of these is divided into equilateral triangles side 2/√3. Each has side 1/10 of the large triangle, so area 1/100, so there are 100 of them, or 600 in all. But there is a trap. There is a line of triangles outside each edge of the hexagon (with bases on the hexagon). Each edge has 10 triangles, so 60 in all.


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The question was that was 104TH Last Year ---------------------------------------------(Part A)A regular hexagon ABCDEF is inscribed in a circle with radius r. Another regular hexagon is formed (lying fully inside the circle with radius r) with 2 of its vertices internally on AB and another 2 on the circle. What is the ratio of the sides of the smaller hexagon to the larger hexagon?

(Part B)In circle O, chords AB and AC are drawn so that AB = AC. Chord PQ is drawn intersecting AB and AC at points D and E respectively. D lies internally between E and P. Given AD = EC = 1/3 and DB = AE = 1/4, what is the maximum value of |PD - QE|?


(Part A)Please refer sunit's post #122 here http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.html


(Part B)By chord theorem;PD*QD = AD * BD = (1/3)*(1/4) = 1/12 ----------------(1)QE*PE = AE * EC = (1/4)*(1/3) = 1/12 ----------------(2)from (1) and (2) we get,PD*QD = QE*PE => PD*(QE+ED) = QE*(PD+ED)=> PD*QE + PD*ED = QE*PD + QE*ED=> (PD - QE) * ED = 0 when ED not equal to zero |PD - QE| = 0when ED = 0 ; |PD - QE| will take its maximum value.

If ED = 0 then D and E coincide which is not possible as it is given that AD = 1/3 and AE = 1/4 so D and E cannot coincide. so ED cannot be equal to zero hence only possible value of |PD-QE| is 0 . hence maxm. value of |PD-QE| = 0.

Hence, choice (d) is the right option

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S is a set of positive integers containing 1 and 99. No elements are larger than 99. For every n in S, the arithmetic mean of the other elements of S is an integer. What is the largest possible number of elements of S? (1) 8 (2) 10 (3) 12 (4) 15 (5) none of these

Solution:

Let us tackle this problem with small example.Let we have 4 numbers in S, and let these be a, b, c, dThus, (a+b+c)/3, (b+c+d)/3, (a+c+d)/3 and (a+b+d)/3 are each integers.=> subtract each term from any other => each of a, b, c, d are equal modulo (3) i.e. each on division by 3 leave same remainder.Extending this when we have n numbers with us then each will be equal modulo (n-1).This was the funda part.

We start the calculation now -> Let we have n numbers [the modulo (n-1) is 1 as the 1st term is 1].1, 1+(n-1).f, 1+(n-1).2f, .... 1+(n-1).(n-1)fThe last term is 1+(n-1).(n-1)f = 99 => (n-1)^2.f = 98 which yields integer value of n for f = 2. => we can at max have 8 numbers in our set.


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The question was that was 105TH Last Year ---------------------------------------------(Part A)How many integral values of p are there for which the inequality 3 - |x-p| > x^2 is satisfied by at least one negative x?

(Part B)How many integral values of p are there for which the inequality x^2 + px + p^2 + 6p < 0 is satisfied for all x in (1, 2)?


(Part A)If x > p then 3-x+p > x^2 => p > -13/4. If x < p then 3-p+x > x^2 => p < 3. Thus, p lies in (-13/4, 3)


(Part B)If f(x) = x^2 + px + p^2 + 6p then f(1) < 0 && f(2) < 0 => p lies in ((-7-√45)/2, -4+2√3))


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For a positive integer i define p(i) as the product of base 4 digits of i e.g. 27 in base 4 is 123 => p(27) = 1*2*3 = 6. What is the numerical value of p(1) + p(2) + p(3) + ... + p(255)?

(1) 1496 (2) 1554 (3) 1592 (4) 1636 (5) 1684

Solution:

The sum of all p(i) where i has 1 digit in base 4 is (0+1+2+3) = 6The sum of all p(i) where i has 2 digits in base 4 is (0+1+2+3)(0+1+2+3) = 6^2The sum of all p(i) where i has 3 digits in base 4 is (0+1+2+3)(0+1+2+3)(0+1+2+3) = 6^3The sum of all p(i) where i has 4 digits in base 4 is (0+1+2+3)(0+1+2+3)(0+1+2+3)(0+1+2+3) = 6^4

Note that 255 = 3.(1+4+4^2+4^3) is the largest decimal number that can be expressed as 4 digit number in base 4


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The question was that was 106TH Last Year ---------------------------------------------(Part A)A tower stands on a horizontal plane. From a mound 14 m above the plane and at a horizontal distance of 48 m from the tower, an observer notices a loophole and finds that the portions of the tower above and below the loophole subtend equal angles. If the height of the loophole is 30 m, then the height of the tower is

(Part B)Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of 45˚ and 30˚ at the points A and B where the roads are nearest to it. If OA = 81 m and OB = 17 m, then what is the height of the tower?

(a) 56 (b) 64 (c) 78 (d) 92 (e) data insufficient

(Part A)Please refer post #148 herehttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.htmlHence, choice (c) is the right option

(Part B)Please refer post #145 herehttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007-quantitative-questions-day.html






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The inscribed circle of an isosceles triangle ABC is tangent to side AB at point D and bisects the segment CD. If CD = 6√2. Which among the following can not be true about ABC?

(a) The perimeter is 24(b) It's obtuse angled(c) The bisector segment of the smallest angle is 6√2(d) The perimeter is 28(e) none of the foregoing

Solution:

Let in triangle ABC the incircle meet AB at D, BC at F and AC at E.

Case 1: AB = AC

Now CE^2 = 6√2*3√2 = 36 => CE = CF = 6,Also in an isosceles triangle AF will contain the incentre, circumcentre , centroid and orthocentre.=> AF will bisect BC hence BF = 6 = BD.

In triangle CDB BD = 6 CD = 6√2 and BC = 12,=> cos(< B) = 3/4 -> FB / AB = 3/4 -> AB =8.=> the perimeter is 28 and angle A is obtuse (square of the largest side is greater than sum of square of other 2 sides) so it is an obtuse angled triangle. So (2) and (4) are true here.

Case 2: When AC = BC

we already have CE = BF = 6 and CD is also perpendicular to AB. Let AD = AE = a, applying pythagoreantheorem for triangle ADC, we get a = 3 => AC = BC = 9, and AB = 6=> (1) and (3) are also true.


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The question was that was 107TH Last Year ---------------------------------------------(Part A)Divya works for Google that claims only 4 working days. In the 1st week of joining, she went for movies after 4 working days. On Friday, she noticed that the fraction p of the line is in front of her, while 1/q of the line was behind her. On Saturday, the same fraction p of the line was in front of her, while 1/(q+1) of the line was behind her. On Sunday, the same fraction p of the line was in front of her, while 1/(q+2) of the line was behind her. For how many values of p is this possible?

(a) 3 (b) 1 (c) 2 (d) no such value (e) infinitely many

(Part B)What is the sum of the reciprocals of all the numbers in the form (2â)*(3^b), where a and b are non-negative integers?

(a) 3 (b) 3.5 (c) 2âˆš3 (d) âˆš3 + 1 (e) none of the foregoing

(Part A)Let p = (x-1)/x for some natural number x. Let q = 1/(x+y), Divya herself represents the line => 1 - ((x-1)/x + 1/(x+y)) = y/x(x+p) => x(x+p)/y is the length of the line. To make it an integer we can make x to be a multiple of 6. For q = x+1, we want to get an integer for y=1, 2, 3. More generally, for x(x+1), x(x+2)/2, and x(x+3)/3 to be integers we require x to be a multiple of 6.On Friday, there are 36k^2+6k people in the queue, on saturday 18k^2+6k and on Sunday 12k^2+6k.Hence, choice (e) is the right option

(Part B)Let the asked sum be x => 1+(1/2+1/4+1/8+...)+(1/3+1/9+1/27+...)+1/6(x) = x => 1+1+1/2+1/6(x) = x => x = 3


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The number of triplets (A, B, C) where A, B, C are subsets of {1, 2, 3} andA/B/C is NULL, A/B is not NULL, and B/C is not NULL is (A/B denotes intersection set of A and B)(1) 12 (2) 18 (3) 27 (4) 36 (5) 48

Solution:

Assume the Veinn-Diagram with 3 subsets A, B and C. As usual Universal set gets divided into 8 regions. The region A/B/C is NULL. Start from here.Any element can go in the 7 regions in 7 ways, 3 elements will go in 7^3 ways. Subtract some cases as given in the question and we will have our answer.

A - Total Ways 7 regions can be filled = 7*7*7 = 343B - Total ways A/B is null 6*6*6 = 216C - Total ways B/C is null 6*6*6 = 216D - Total ways A/B is null and B/C is Null = 5*5*5 = 125E= Total way A/B or B/C is null = B+C-D = 307=> our desired answer, A- E = 36


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The question was that was 107TH Last Year ---------------------------------------------(Part A)Divya works for Google that claims only 4 working days. In the 1st week of joining, she went for movies after 4 working days. On Friday, she noticed

that the fraction p of the line is in front of her, while 1/q of the line was behind her. On Saturday, the same fraction p of the line was in front of her, while 1/(q+1) of the line was behind her. On Sunday, the same fraction p of the line was in front of her, while 1/(q+2) of the line was behind her. For how many values of p is this possible?

(a) 3 (b) 1 (c) 2 (d) no such value (e) infinitely many

(Part B)What is the sum of the reciprocals of all the numbers in the form (2â)*(3^b), where a and b are non-negative integers?

(a) 3 (b) 3.5 (c) 2âˆš3 (d) âˆš3 + 1 (e) none of the foregoing

(Part A)Let p = (x-1)/x for some natural number x. Let q = 1/(x+y), Divya herself represents the line => 1 - ((x-1)/x + 1/(x+y)) = y/x(x+p) => x(x+p)/y is the length of the line. To make it an integer we can make x to be a multiple of 6. For q = x+1, we want to get an integer for y=1, 2, 3. More generally, for x(x+1), x(x+2)/2, and x(x+3)/3 to be integers we require x to be a multiple of 6.On Friday, there are 36k^2+6k people in the queue, on saturday 18k^2+6k and on Sunday 12k^2+6k.Hence, choice (e) is the right option

(Part B)Let the asked sum be x => 1+(1/2+1/4+1/8+...)+(1/3+1/9+1/27+...)+1/6(x) = x => 1+1+1/2+1/6(x) = x => x = 3


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