Answers Qqad 2006

ANSWERS FOR THE QQAD 2006 ----------------------------------------------------------- Quant Answer # 1 ------------------------------------------------------------ gcd(a,b) = gcd(a, b-a) when b > a. Thus gcd(n^2 + 11, (n+1)^2 + 11) = gcd(n^2 + 11, 2n+1) 2n+1 is always odd, thus gcd(2n+1, 2) = 1, gcd(n^2 + 11, 2n+1) = gcd(2*(n^2 + 11), 2n+1)) gcd(2*(n^2 + 11), 2n+1) = gcd(2*n^2 +n +22 -n, 2n+1) 2*n^2 + n is always divisible by 2n+1, thus 22-n must divide 2n+1, thus n = 22, gcd which is 2n+1 is maximum. Hence, choice (b) is the correct answer. Note that gcd(n^2 + 11, (n+1)^2 + 11) can attain 5 different values for n=1,2,4,7,22. And gcd(n^2 + 11, (n+1)^2 + 11) = 1 for n=3,5,6,... Thus in all, gcd(n^2 + 11, (n+1)^2 + 11) has 6 different values and 45 is max. among them. Note that 45 = (3^2)*(5) has in all (2+1)*(1+1) = 6 divisors Hence, choice (b) is the correct solution. ----------------------------------------------------------- Quant Answer # 2 ------------------------------------------------------------ Let PO = x = RO, and QO = y, then by stewart's theorem (http://planetmath.org/encyclopedia/ProofOfStewartsTheorem.html) 21*x^2 + 9*x^2 = 30*y^2 + 30*9*21 => x^2 - y^2 = (3^3)*(7) (x+y)(x-y) = (27*7)(1) = 63(3) = 27(7) = 21(9) We are required to find the sum of all 2x + 30. From, (x+y)(x-y) = (27*7)(1) = 63(3) = 27(7) = 21(9) (x,y) can take values as (95,94) (33,30), (17,10) (15,6) but (x,y) as (15,6) is not possible as QO + RO > QR. Hence, the required sum is 2(95 + 33 + 17) + 3*30 = 380

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Hence, choice (c) is the correct answer. Note that remembering the result from stewart's theorem can make you derive the length of median, angle bisector easily. CAT 2002 had a question on finding the length of angle bisector in a triangle. ----------------------------------------------------------- Quant Answer # 3 ----------------------------------------------------------- Since all the terms in B(x,y,z) are symmetric we assume without the loss of any generality that x >= y >= z Thus, B(x,y,z) = |x-|x-y|| + |y-|y-z|| + |z-|z-x|| = |y| + |z| + |2z-x| |x-y| + |y-z| + |z-x| = x-y + y-z + x-z = 2(x-z) putting x = 1, y=1, z= 1 we see that choice (a) is not true x² + 3y² - 4xy - 2yz + 2zx = (x - 2y + z)^2 -y^2 -z^2 + 2yz => A(x,y,z) = (x - 2y + z)^2 -(y-z)^2 => A(x,y,z) = (x - 3y + 2z)(x-y) Let x > y > z > 0 and x > 2z, then B(x,y,z) = x+y-z, if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get 2y(3x-y) + 2x(3y-2z) which is > 0 Let x > y > z > 0 and x < 2z, then B(x,y,z) = y+z+(2z-x) = 3z +y -x if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get (3z +y -x)^2 - (x - 3y + 2z)(x-y) > 0 => 9z^2 - 2y^2 +8yz +2xy -8zx > 0 => z(9z-8x+8y) + 2x(x-y) > 0, which we can see always is Hence, (b) is the right option put x=1, y = 0, z = -4 we get A(x, y, z) = -7 hence (c) is also not true ----------------------------------------------------------- Quant Answer # 4 ------------------------------------------------------------ [1/3] = 0 Note that 2/3 + 2^2/3 = 2 or 2â is of the form 3k-1, and 2^(a+1) is of the form 3k+1 when a is odd. [2/3] + [2^2/3] = [(3-1)/3] + [(3+1)/3] = 1 = (2 + 2^2)/3 - 1 [2^3/3] + [2^4/3] = [(9-1)/3] + [(15+1)/3] = 7 = (2^3 + 2^4)/3 - 1 and so on... Thus, [2/3] + [2^2/3] + [2^3/3] + [2^4/3] +...+ [2^100/3] = (2/3 + 2^2/3 -1) + (2^3/3 + 2^4/3 -1) +...+ (2^99 + 2^100/3 -1) =

2/3 + 2^2/3 + 2^3/3 + 2^4/3 +...+ 2^100/3 - (50) 1/3(2 + 2^2 + 2^3 + ...+ 2^100) - 50 = 2/3( 2^100 - 1) - 50 Hence, choice (c) is the correct solution ----------------------------------------------------------- Quant Answer # 5 ------------------------------------------------------------ Its same as asking we have 7 books in a shelf and we have to choose place for 4 more books so that no two of these 4 will be together. Answer 8C4 = 70. ----------------------------------------------------------- Quant Answer # 6 ------------------------------------------------------------ We can always have a system of equations with unique, infinite and zero solutions. Consider a system in 2 variables. ax + by = c, dx + ey = f. Let (m,n) be its solution then am + bn = c, dm + en = f Let (p,q) be another solution then ap + bq = c, dp + eq = f => a(m-p) = b(q-n) and d(m-p) = e(q-n); either m=p and q=n or a/b = d/e in which case we have infinite solutions, or 0 solutions depending on the value of variables c and f. Hence, choice (d) is the correct answer ----------------------------------------------------------- Quant Answer # 7 ------------------------------------------------------------ Minus the sequence from m to n at diagonal and you will get , 1^2/1 + (2^1 - 1^1)/3 + (3^2 - 2^2)/5 + ... + (500^2 - 499^2)/999 - 500^2/1001 = 500 - 500^2/1001 which is close to 250. Hence, choice (a) is the right answer. ----------------------------------------------------------- Quant Answer # 8 ------------------------------------------------------------ We have 2 quadratic equations (x^2 - 2ax -4(a^2 + 1)) = 0, (x^2 - 4x -2a(a^2 + 1)) = 0 which can be written as (x-a)^2 = 5a^2 + 4, and (x-2)^2 = 2(a^3 + a +2)

Now, we can have 3 different roots in the following scenarios (a) one of the equations has a double root while the other doesn't (b) both the equations have a common root Case 1: when (x-a)^2 = 5a^2 + 4 has double root Not possible as RHS can never be zero. Case 2: when (x-2)^2 = 2(a^3 + a +2) has double root Thus, 2(a^3 + a +2) = 0, (a+1)(a^2 - a + 2) = 0, thus a = -1 putting a = -1 quadratic equations (x^2 - 2ax -4(a^2 + 1)) = 0, (x^2 - 4x -2a(a^2 + 1)) we get the 1st equation roots as 2, -4 and 2nd eq. has double root as 2. Thus, in all we have 2 different roots. Hence, a = 1 is also ruled out. Case 3: both the equations have a common root Let p be the common root. If p is the root of A(x) and B(x), then it must be the root of A(x) - B(x). Hence p is the root of (x^2 - 2ax -4(a^2 + 1)) - (x^2 - 4x -2a(a^2 + 1)) = 0 hence, (p^2 - 2ap -4(a^2 + 1)) - (p^2 - 4p -2a(a^2 + 1)) = 0 solving, (2a - 4)p = (2a - 4)(a^2 + 1) so, a =2 or a^2 + 1 = p. putting a = 2 in our 2 quadratic equations we get (a-2)^2 = 24 for both, and hence we have only 2 different roots. putting p = a^2 + 1, we get (a^2 - 1)^2 = 2(a^3 + a + 2) which can be factored as (a+1)(a-3)(a^2 + 1) = 0, hence a = -1 or a = 3. But a = -1 is already ruled out. putting a = 3 we get the equations as x^2 - 6x - 40 = 0, and x^2 - 4x - 60 = 0. the roots of first eq. are 10, -4 and that of 2nd are 10, -6 and we have 3 different roots. For a = 3 only the pair of quadratics have 3 different roots. Hence, choice (d) is the correct answer. ----------------------------------------------------------- Quant Answer # 9 ------------------------------------------------------------ Note a very simple fact. Let us denote degree of a vertex by the number of edges passing through it. In a triangle the degree of each vertex is 2 and sum of the degrees of vertices in 6. In any graph, the sum of degrees of vertices is double the number of edges in that

graph. Let us consider each person as a vertex. 100th person(vertex) has 100 edges passing through it => an edge from 100th vertex meets 99th also. 99th person(vertex) has 98 new edges passing through it but 99th hasn't shaken hands with 1st person as 1st has already with 100th. => 99th has shaked hands with 98th person 98th person(vertex) has 96 new edges passing through it so on ... Thus, total number of edges is 100+98+96+ ... + 2 + 0 = 50*51 Let 101th person(vertex) has degree n i.e. the number of edges passing through it or the number of persons he has shaken hands with. Then, 2*50*51 = 1+2+3+ ... + 100 + n => 100*51 = 101*50 + n => n = 50 Hence, choice (d) is the correct answer ----------------------------------------------------------- Quant Answer # 10 ------------------------------------------------------------ Let us denote S(n) as the sum n of a, b, c, d the 4 elements from the set such that 1 <= a a+b+c+d < 201 => (101-p) + (101-q) + (101-r) + (101-s) < 201 => P+q+r+s > 203. **Note: We are subtracting 101 from each element because for each "a" belonging to set A, "101-a" also belongs to the set A. Thus, for each set {a,b,c,d} where S(n) < 201 we have an exact map where S(n) > 203. => the number of 4 element subset that have sum < 201 is same as number of 4 element subset that have sum > 203. But for S(n)=202, we have few extra sets of {a,b,c,d}. Hence, (1) is true. Consider the statement (2) S(n) > 203 => a+b+c+d > 203 => (101-p) + (101-q) + (101-r) + (101-s) > 203 => p+q+r+s < 201 Thus, for each set {a,b,c,d} where S(n) > 203 we have an exact map where S(n) < 201.

=> the number of 4 element subset that have sum > 203 is same as number of 4 element subset that have sum < 201. But for S(n)=202, we have few extra sets of {a,b,c,d}. Hence, (2) is false. Hence, choice (a) is the correct answer. ----------------------------------------------------------- Quant Answer # 11 ------------------------------------------------------------ 2f(x) + f(1-x) = 2*x^2 + 1, (1) 2f(1-x) + f(1-(1-x)) = 2f(1-x) + f(x) = 2*(1-x)^2 + 1, (2) subtracting 2nd from 1st we get, f(x) - f(x-1) = 4x -2, (3) adding 3rd with the first we get, 3f(x) = 2*x^2 + 4x - 1 thus f(x) = 1/3(2*x^2 + 4x - 1) = 1/3(2(x+1)^2 - 3) which assumes the min. value at x = - 1 and f(x) >= -1. Hence, choice (b) is the correct answer. ----------------------------------------------------------- Quant Answer # 12 ------------------------------------------------------------ Any number in base k is divisible by k+1 or its divisors if and only if the sum of the digits(alternating and sign changed) is divisible by k+1 or its divisors respectively. eg. in base 10, if a number is div. by 11 then sum of its digits(sign changed alternatively) is div. by 11 and vice-versa. In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b => 6 -(a +b +c) is divisible by 4 what values can a+b+c take? a+b+c = 2, (0,1,1), (0,0,2), 3+3 = 6 solutions a+b+c = 6, (0,0,6), (0,1,5), (0,2,4), (0,3,3), (1,1,4), (1,2,3), (2,2,2), 3+6+6+3+3+6+1 = 28 solutions a+b+c = 10, (1,3,6), (1,4,5), (2,2,6), (2,3,5), (2,4,4), (3,3,4), (4,0,6), (5,0,5), 6+6+3+6+3+3+6+3 = 36 solutions a+b+c = 14, (2,6,6), (3,5,6), (4,4,6), (4,5,5), 3+6+3+3 = 15 solutions a+b+c = 18, (6,6,6), 1 solution In all we have 86 ordered pairs of (a,b,c) Hence, choice (c) is the correct option.

----------------------------------------------------------- Quant Answer # 13 ------------------------------------------------------------ |x^2-1|/(x-2) = x => |x^2-1| = x(x-2). Since LHS >= 0 RHS = x(x-2) >= 0 => x >= 2, x <= 0. case when x >= 2: |x^2-1| = x(x-2) => x^2 - 1 = x^2 - 2x => x=1/2, rejected case when x <= 0: In this we should have 2 sub-cases as well since |x^2-1| = x^2-1 for x <= -1 and 1-x^2 for -1 <= x <= 0. when x <= -1 we have x^2 - 1 = x^2 - 2x => x=1/2, rejected when -1 <= x <= 0, we have 1- x^2 = x^2 -2x => 2x^2 -2x -1 = 0 solving we get (1-3^1/2)/2 and (1+3^1/2)/2. The later value is rejected as it's > 0. Hence, choice (c) is the right answer. ----------------------------------------------------------- Quant Answer # 14 ------------------------------------------------------------ Let x, y be the ages. Then 100x + y = a^2, also 100(x + 9) + b + 9 = b^2, subtracting we get b^2 - a^2 = 909 = 101*9 = 303*3 = 909*1 thus, (b,a) are (55, 46) (156, 150), (455, 454) rejecting the last 2 pairs as a <= 99, we get (b, a) as (55, 46) a^2 = 2116, thus the sum of the ages is 21 + 16 = 37 Hence, choice (a) is the right answer. ----------------------------------------------------------- Quant Answer # 15 ------------------------------------------------------------ We know that every prime > 3 can be expressed as 6k+/-1, where k is a natural number. Let p = 6k+/-1, then p^2 = 36k^2 +/-12k +1 now, 36k^2 +/-12k = 12(3k^2 +/-k) 3k^ +/-k is always even for odd and even k, thus every p^2 > 3^2 can be written as 24n + 1 for some natural number n. Thus f(n) is prime for infinite n. Hence, (a) is false We know that (24a + 1)^2 is always composite for infinite values of a.

also (24a + 1)^2 is of the form 24n + 1 where n is positive integer. Hence, (b) is also false Hence, choice (d) is the correct answer ----------------------------------------------------------- Quant Answer # 16 ------------------------------------------------------------ (f(x)) = x simplifies to (a+d)(cx^2 + (d-a)x - b) = 0, so d = -a. There is no loss of generality in taking c = 1. Then f(-1) = 0, f(1) = 1 give -a +b = 0, 2a + b = 1. Solving, a = 1/3, b = 1/3. The unattained value is a/c (Why?) Hence, choice (a) is the correct answer. Solution by IdiotR From f(-1)=0 implies b-a=0 and b=a from f(1)=1 we get a+b=c+d i.e a+b=c+d i.e 2a=c+d now given f(f(x)) = x and we know by property finverse(f(x))=x we can conclude f(x)=finverse(x) now finverse(x)=dx-b/a-cx hence ax+b/cx+d =dx-b/a-cx put x=0 implies ab+bd=0 hence a+d=0 now from all these eqns we can find the relatn a=b=-d=c/3 =k say now putting values in f(x) we get f(x)=x+1/3x-1 hence value of x not in range when denominator is zero i.e x=1/3 ----------------------------------------------------------- Quant Answer # 17 ------------------------------------------------------------ a^3 + b^3 + c^3 -3abc = (a+b+c)(a^2 + b^2 + c^2 -ab -bc -ca) x^6 + 5x^3 + 8 = 0, put a = x^2, b= -x, c= 2. =>x^6 + 5x^3 + 8 = (x^2 -x +2)*(A bi-quadratic) Hence, choice (a) is the correct answer. Solution by niceguy123 Rewrite the eq. as x^6 + 6x^3 - x^3 + 8 =0 as a^3+b^3 + c^3 - 3abc = 0 ==> a +b +c =0

x^2+ (-x) + 2=0 (here a=x^6, b=x^3 and c=8) x + 2/x =1 hence option (a) ----------------------------------------------------------- Quant Answer # 18 ------------------------------------------------------------ Let (< ADP) =x, (< DDC) = y, since (< PDQ) = 45 degrees, x+y = 45 degrees Also tanx = AP/1 = 1/2, tan(x+y) = (tanx + tany)/(1 - tanx*tany) solving we get tany = 1/3, thus CQ = 1/3, and BQ=2/3 area of traingle PBQ = 1/2*1/2*2/3 = 1/6 Choice (d) is the answer ----------------------------------------------------------- Quant Answer # 19 ----------------------------------------------------------- 6 straight lines in a plane, no 2 of which are parallel and no 3 of which pass through the same point intersect in 6C2 = 15 points. These 15 lines can intersect at 15C2 points amonst themselves. At each 6 point we have 5 lines passing through it. These 5 lines give 5C2 points of intersection amonst themselves, but no such points exist. Hence, new lines thus introduced is 15C2 - 6*(5C2) = 105 - 60 = 45 lines. Hence, choice (a) is the correct answer. Solution by tired_soul pts of intersection of six lines = 6c2 =15 now each line contains 5 pts on them (pt. of intersection with five other lines) so the no. of lines that can be generated by joining these 5 points on a line= 5c2=10 total no. of such lines = 10*(no. of lines)=10*6=60 no of lines obtained by joining the 15 pts of intersections =15c2=105 but 60 0f these lines were present before as explained above. so total no. of new lines = 105 - 60 =45. hence ans option (a) ----------------------------------------------------------- Quant Answer # 20 -----------------------------------------------------------

Let A be the area of the triangle and third altitude has length x then the 3 sides are 2A/3, 2A/5, 2A/x. Also the sum of any two sides in a triangle is greater than the third side checking with options x can not be 7.5 as 4A/15 + 2A/5 > 2A/3 is false. Choice (a) is the answer. Solution by vineet.nitd since area is constant 4 a triangle .... 1/2 *3 * a=1/2*5*b=1/2*p*c=k for a triangle sum of 2 sides must be gr8er than 3rd .... (2k/p)+b > 5/3 *b so, 2k/p shud be > 2/3 *b but 2k = 5b => 5(b/p)>(2/3) b =>p<7.5 hence option (a) ----------------------------------------------------------- Quant Answer # 21 ----------------------------------------------------------- (-1)â + (-1)^b + (-1)^c + (-1)^d = A, at most A can range from [-4, 4] We further see that A can not take odd values If A=-4, then all of a,b,c,d is odd which is not possible as the sum is 99. If A=-2, 3 terms negative, 1 positive which means 3 among a,b,c,d are odd and one is even which is possible If A=0, 2 among a,b,c,d are odd, 2 even - not possible If A=2, 3 among a,b,c,d are even one odd - possible If A=4, All 4 are even - not possible Thus, A can only be -2 or 2. Hence, choice (d) is the correct answer Sum of 4 numbers is odd if any one of the four or any three of the four are odd.

----------------------------------------------------------- Quant Answer # 22 ----------------------------------------------------------- If B is the first letter, there's 120 ways to arrange the remaining. The same goes for A and K. However, in this counting, we've overcounted the number of ways to arrange the letters such that B and A are in the first and third places respectively, etc. each once, and each of these combinations (BA, BK, AK) have 24 ways to arrange the remaining letters. So far we have 120*3-24*3=288. Note also, however, that we've counted the ways to arrange the letters such that B, A, and K are in the 1st, 3rd, and 5th places respectively (which is 6 ways) twice when we subtracted the 3 24s. Thus, we add 6 back again. So we have 288+6=294 ways These are the ways that we don't want to find, so since there's 720 ways to arrange the letters BLACKI, there are 720-294=426 ways to arrange the letter with the specified conditions. Hence, choice (b) is the correct answer. Solution by AnnaScott n(A U B U C) is the total number of ways in which 3 letters can occur in the three respective places n(A U B U C) = n(A) + n(B) + n(C) - n(A � B) - n(B� C) - n( A � C) + n(A� B � C) n(A) = n(B) = n(C) = Number of word s with one letter at fixed position = 5! = 120 n(A � B) = n(B�C) = n( A � C) = Number of word s with 2 letters at fixed position = 4! = 24 n(A� B � C) = ) = Number of word s with three letters fixed at their position(Arrange remaining three) = 3! = 6 n(A U B U C) = 120*3 – 3*24 + 6 =366 – 72 = 294 Ans = Total number of arrangements – n(A U B U C) = 720 – 294 = 426 ----------------------------------------------------------- Quant Answer # 23 -----------------------------------------------------------

Aditya can chose first lady in 9 ways. The other 2 can be chosen in 6C2 - 5 ways = 10 ways(why?) Since the two ladies can be arranged in 2 ways, the number of ways of selecting 3 contestants is 2*9*10. But the order is immaterial in our case, hence total ways = 1/3!(2*9*10) = 30 Hence, choice (d) is the correct answer. Solution by sidhesh Its good to be back with a solution. total ways of selecting 3 contestants of 9 = 9C3 =84 total ways in which all three contestants are neighbours = 9 total ways in which 2 contestants are neighbours = 9*5={total diff combos}*{no of ways in which 3rd one in team can be selected}=45 total ans = 84-54=30 Solution by niceguy123 ya my ans is also 30 but by a different way( probably a litle better if one understand the concept. this prob is same as in a polygon with n side how many triangles can be formed using the vertices such that no side of triangle is common to the side of polygon? ans (n-4)C2 * n/ 3( how? this one can be further generalise for any polygon in any polygon in this case put n = 9 ----------------------------------------------------------- Quant Answer # 24 ----------------------------------------------------------- Since, [x^1/2] = 10 => 100 <= x = 121 => 10000 <= 100x < 12100 Also, [(100x)^1/2] = 100 => 10000 <= 100x < 10201 Probability = (10201 - 10000)/(12100 - 10000) = 201/2100 = 67/700 Hence, choice (c) is the correct answer.

----------------------------------------------------------- Quant Answer # 25 -----------------------------------------------------------

The sequence of numbers when written in base 3 have no digit as 2 in any of them. 50 = 110010 in base 2. In base 3 110010 is 327.

Hence, choice (c) is the correct answer.

My Solution Seen from the series 4th term of the series is 3^2=9 Similarly 8th term(2^3) is 3^3=27 Therefore, 32nd term = 3^5=243 To get the 50th term we need 18th term from the begining and add it to 243. To get the 18th term, i know 16th term is 81. So, 18th term is 81+3=84 Hence, 50th term is 243+84 = 327 ----------------------------------------------------------- Quant Answer # 26 ----------------------------------------------------------- By fermat's theorem 31^40 = 1 mod(41) But is 40 the smallest values such that 31^x = 1 mod(41)? clearly, x has to be the divisor of 40. 31^2 = 18 mod(41), 31^4 = 324 mod(41) => 31^4 = -4 mod(41) => 31^5 = -124 mod(41) => 31^5 = -1 mod(41) clearly such smallest x is 10 as 31^10 = (-1)^2 mod(41) => 31^30 = 1 mod(41), if 31^29 = a mod(41) => 31^30 = 31a mod(41) => 31a = 41K + 1, => a = 4. Thus, 31^29 = 41*Q + 4. => Q is always odd. If Q is divisible by 5 then 31^29 should leave remainder 4 on division by 5. but 31^29 = (30+1)^29, hence the remainder is 1 by 5. Similarly, 31^29 leaves 5 by 7 and not 4. Hence, choice (d) is the correct answer. ----------------------------------------------------------- Quant Answer # 27 ----------------------------------------------------------- Initially he won n out of 2n, then n+9 out of 2n+10 with n+9 >51/100*(2n+10), or thus, 100n + 900 > 102n + 510, 390 > 2n, thus n at max is 194. Total no. of matches played is 2n+10 = 398 Among the given options Hence, choice (a) is the correct answer.

----------------------------------------------------------- Quant Answer # 28 -----------------------------------------------------------

We are to find Min(x^2 + 2y^2 + z^2 + 2yz) = Min(x^2 + y^2 + (y+z)^2) subject to x + 2y + z = –6

if x =a, y=b, y+z=c, then a^2 + b^2 + c^2 is to be minimized when a+b+c is constant

Thus, a=b=c

x=y=y+z => x= -2, y= -2, z=0

Hence, Choice (c) is the correct answer

Solution by Amar_kashyap

x^2 + 2y^2 + z^2 + 2yz =x^2+y^2+(y+z)^2=a^2+b^2+c^2 (where a=x, b=y, c=y+z) Also x+2y+z=x+y+(y+z)=a+b+c=-6 (a^2+b^2+c^2)/3 >= {(a+b+c)/3}^2=4 Hence (a^2+b^2+c^2)>=12. henec the minimum value =12.

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Take X on QR with QX = QA. Then RX = PA and so RX/RA = PA/RA = PQ/QR = PR/QR. Then the triangles QPR and RXA have a common angle R and the ratio of the two sides containing the angle the same. Hence they are similar. So XR = XA. Hence <QXA = 2 <R. But <R = 90(degrees) - P/2 and <QXA = 90(degrees) - Q/4 (since QAX is isosceles) = 90(degrees) - (90(degrees) - P/2)/4. Solving gives <(P) = 100 degrees. Hence, choice (d) is the right answer.

----------------------------------------------------------- Quant Answer # 30 -----------------------------------------------------------

No candidate can have only 1 vote. If a candidate has just 2 votes, then 2/n <= 1/100, so n >= 200. If a candidate has 3 votes, then n >= 150. So in a minimal solution each candidate must have at least 4 votes. If all have at least 5, then n >= 135. If a candidate has 4, then 4/n <= 3/100, so n >= 134. This can be achieved: 1 candidate has 4 votes, the other 26 have 5 each. Then 5/134 = 3.7%, 4/134 = 2.99%.

Hence, choice (c) is the right answer.

----------------------------------------------------------- Quant Answer # 31 -----------------------------------------------------------

For x > 0, y^2 +2xy + 4|x| = 4 => y^2 +2xy + 4x = 4

=> (y+2)(y+2x-2) = 0

For x < 0, y^2 +2xy + 4|x| = 4 => y^2 +2xy - 4x = 4

=> (y-2)(y+2x+2) = 0

The region bounded by the lines y+2=0, y-2=0, y+2x-2=0, y+2x+2=0 is a parallelogram.

(Plot the co-ordinates and check yourself)

The area is height*base = 2*4 = 8

Hence, choice (a) is the right answer.

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The bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored 6 1/2 and the top player 7. But then the top player must have won all his games, and hence the number 2 player lost at least one game and could not have scored 6 1/2. Hence the number 2 player scored exactly 6, and the bottom 4 players lost all their games with the top 4 players. In particular, the number 3 player won against the number 7 player.


----------------------------------------------------------- Quant Answer # 33 -----------------------------------------------------------

AB is the common tangent to the 2 circles. The length of AB is 2*(4*9)^1/2 cm. Let the radius of the third circle be R => 2*(4*R)^1/2 + 2*(9*R)^1/2 = 2*(4*9)^1/2 => 1/R = 1/2 + 1/3 => R = 36/25 Hence, choice (b) is the right answer.

----------------------------------------------------------- Quant Answer # 34 -----------------------------------------------------------

I. => if the speed of the first contestant is s, then that of nth is s*n! and that of (n-1)th is s*(n-1)!, of (n+1)th s*(n+1)! Let d be the length of the track be d nth meets (n+1)th once in lcm(d/s*n!, d/s*(n+1)!) = d/s*n! times nth meets (n-1)th once in lcm(d/s*n!, d/s*(n-1)!) = d/s*(n-1)! times

=> I is true.

II. => The speeds are s, 2s, 3s, ..., ns 2nd and 4rth meet each other once in d/(4s-2s) = d/2s time 6th and 7th contestant meet once in d/(7s-6s) = d/s time

=> II is true


----------------------------------------------------------- Quant Answer # 35 -----------------------------------------------------------

For g(y) to be maximum, f(1/4,y) and f(3/4,y) has to be maximum when -1/4 <= y <= 3/4 f(1/4,y) = 1 when 1/4 <= y <= 5/4 f(3/4,y) = 1 => y lies in [1/4, 3/4] Hence, choice (d) is the correct answer.

----------------------------------------------------------- Quant Answer # 36 -----------------------------------------------------------

AM = (a+b)/2, GM = (ab)^1/2 => (a+b)/2 = (ab)^1/2 + 2 => ab = a^2/4 + b^2/4 + ab/2 -2a -2b + 4 => a+b = 2 + 2((a-b)/4)^2 => a-b is a multiple of 4, Let a-b = 4k => a+b = 2 + 2*k^2 => a = (k+1)^2, b = (k-1)^2 => k = 2, 3, ..., 30 => 29 values are possible Choice (c) is the answer

----------------------------------------------------------- Quant Answer # 37 -----------------------------------------------------------

Let q = p+d; The numbers are p, p+d, p+2d, (p+2d)^2/(p+d). So (p+2d)^2-a(p+d) = 30(p+d) or 3pd + 4d^2 = 30p + 30d. Hence d is multiple of 3. Also d(4d-30) = 3p(10-d). So 7.5 < d < 10. Hence d = 9. Hence p = 18. So numbers are 18, 27, 36, 48. Sum 129. Choice (a) is the answer

----------------------------------------------------------- Quant Answer # 38 -----------------------------------------------------------

I. Let (x+5)(x+2)/(x+1) = y = (x^2 + 7x + 10)/(x+1) = (x^2 + 2x + 1 + 5x + 5 + 4)/(x+1) => (x+5)(x+2)/(x+1) = x+1 + 4/(x+1) + 5 By AM-GM rule, x+1 + 4/(x+1) >= 4 => I is true II. (100^100)/100! = (100^99)/99! For n > 100, (100^n)/n! = ((100^100)/100!)*(100^x/(101*102*...xterms) where n = x + 100 But (100^x/(101*102*...xterms) < 1 For n < 100, (100^n)/n! < (100^100)/100! always as numerator/denominator > 1 as n increases from 1 to 99 Hence, II is false III. We have (x-2)^2 = 0, so x^2 = 4(x-1). Hence x/(x-1)^1/2 = 2. Now by AM/GM, x^2/(y - 1) + y^2/(x - 1) = 2xy/((x-1)(y-1))^1/2. But rhs = 2·2·2 => III is true Choice (d) is the answer

----------------------------------------------------------- Quant Answer # 39 -----------------------------------------------------------

S =1 + 3x + 4x^2 + 10x^3 + 18x^4 + ...

2xS= 2x + 6x^2 + 8x^3 + 20x^4 +.....

=> (1 - 2x)S= 1 + x - 2x^2(1 - x + x^2 - x^3 ......)

=> (1 - 2x)S= 1 + x - 2x^2/(1+x) [since 0<x <1]

=> S = 7/2 for x=1/3

Hence, choice (d) is the correct answer

----------------------------------------------------------- Quant Answer # 40 -----------------------------------------------------------

We have xyz = x^2 + y^2 + z^2 > y^2 + z^2 = 2yz. Hence x > 2. Hence x^3 > 4x. Similarly, y^3 > 4y and z^3 > 4z. But, x^3 + y^3 + z^3 = 4(x + y + z). Contradiction. So there are no solutions. Choice (b) is the answer

Solution by Chautauqua

x^2 + y^2 + z^2 = xyz ie xyz - z^2 = x^2 + y^2 >= 2xy ie xy(z-2)>=z^2 ie z>2 similarly for x and y ... we get x, y,z >2 but then clearly x^3>4x for x>2 ... which implies that the first equation cannot be satiafied given the above constraints... => no positive real solutions...

----------------------------------------------------------- Quant Answer # 41 -----------------------------------------------------------

|x(1)| + |x(2)| + ... + |x(n)| < n, so we need n >= 100 Taking x(i) = alternately +/(-99/100) works for n = 100 for the minimum n. Choice (b) is the answer

Alternate solution

LHS <n and 99<RHS< 99+n n can never be les than 100 or 2nd cond wiill ...fail hence n>=100 ...since n is an integer.. so (n)min =100

----------------------------------------------------------- Quant Answer # 42 -----------------------------------------------------------

Let the fares initially be 8x, 6x, 3x for first, second and the third class respectively. After reduction, fares become 20/3x, 11/2x, 3x Let the passengers for first, second and the third class be 9y, 12y, 26y respectively => 20/3x*9y + 11/2x*12y + 3x*26y = 1326 => 204xy = 1326 => 2xy = 13 First class passengers paid 20/3x*9y = 60xy = 13*30 = 390 Choice (d) is the answer

----------------------------------------------------------- Quant Answer # 43 -----------------------------------------------------------

I. Let a, b be two other sides( b > a). => either a^2 + b^2 = 324 or b^2 = a^2 + 324 € If a^2 + b^2 = 324 => either both a, b are odd or both a, b are even m^2 + n^2 = 81 has no solutions in positive integers. and 4m^2 + 4m + 4n^2 + 4n = 322 is not possible as RHS is not divisible by 4. If b^2 = a^2 + 324 => (b-a)*(b+a) = 324 b-a = 2, b+a = 162 gives one solution where a is maximum, area is 1/2*18*80 b-a = 6, b+a = 54 gives one solution where a is minimum, area is 1/2*18*24 Difference in areas is 576 sq. cm I is false II. Let ABC be the equilateral triangle, and PQRS is a square inside ABC of maximum possible area. P lies on AB, Q on AC and S, R on BC. Let D be a point on BC such that AD is perpendicular to BC. Triangles PRB and ADB are similar. Let each side of the square be a cm. => PR/BR = AD/BD, but AD = ((3^1/2)/2)*(2 + 3^1/2), and BR is (2 + 3^1/2)/2 - a/2 => 2a/((2 + 3^1/2) - a) = 3^1/2 => a = 3^1/2 => the circle inscribed within this square has diameter as 3^1/2 => the circle has area as 3*pi/4 sq. cm II is false Hence, choice (d) is the right answer.

Solution by niceguy

1)in rt tringle a^2 +b^2=c^2 assuming c=18 no soln exist if a=18

c^2 - b^2 = 18^2=(c-b)(c+b)=2*162=6*54 from this v get c=82, b=80 or c=30, b=24 max area possible 720 and min area possible 216 so diff <600 2)side of square wil b (3^1\2)/2=s so area of circle will be pi(s^2)/4= 3.14*3/16<2 hence my ans (d)

Funda

The radius of the inscribed circle in a n-regular polygon has radius (a/2)*cot(pi/n), where a is the length of each side of the polygon. The radius of the circumscribed circle of a n-regular polygon has radius (a/2)*cosec(pi/n), where a is the length of each side of the polygon.

----------------------------------------------------------- Quant Answer # 44 -----------------------------------------------------------

Let on X-Y axis my home be (0,0) => Anupams's house is (40, -12) => Shrikant;s house is (-13, 110) => Nilesh's house is (15, 14) Area bounded by the houses of my three friends is 836 sq. m The shortest distance taken by Anupam is the altitude from Anupam's house to line connecting my other two friends. Nilesh's and Shrikant's houses are ((28)^2 + (96)^2)^1/2 = 100 m apart => 1/2*100*(shortest distance taken by Anupam) = 836 => shortest distance taken by Anupam = 16.72 m Hence, choice (a) is the right answer

Alternate solution

A => (40,-12) N => (15,14) S => (-13,110) Now, to find equation of line SN (14-110)/(15+13) = (y-14)/(x-15) => -96/28 = (y-14)/(x-15) => -96x + 1440 = 28y - 392 => -96x -28y = -392 - 1440 => 96x + 28y = 1832

=> 24x + 7y = 458 Now to find distance between A(40,-12) and 24x + 7y -458 =0 (24.40 - 7.12 -45 /(24^2 + 7^2)^1/2 => 418/25 => 16.72 Hence the answer is a)

----------------------------------------------------------- Quant Answer # 45 -----------------------------------------------------------

I. (p-1)! + 1 = 0 mod(p) when p is prime => 30! + 1 = 0 mod(31) => 29! = 1 mod(31) Let 28! = x mod(31) => 29! = 29x mod(31) => 29! = -2x mod(31) => -2x by 31 leaves 1 as the remainder => x = -16 + 31 = 15 28! by 31 is 15, 28! by 2 is 0 => 28! by 62 is 46. I is true II. N = 9*10*11 - 5 = 985 N^3 + 2N^2 - N - 3 = (N-1)*(N+1)*(N+2) - 1 N-1 is divisible by 4 and leaves remainder as 0 => (N-1)*(N+1)*(N+2) - 1 leaves 3 by 4 N-1 is divisible by 4 and leaves remainder as 0 => (N-1)*(N+1)*(N+2) - 1 leaves 2 by 3 N^3 + 2N^2 - N - 3 leaves 2 by 5 as N is divisible by 5 so will be N^3 + 2N^2 - N II is true. Hence, choice (c) is the right answer.

----------------------------------------------------------- Quant Answer # 46 -----------------------------------------------------------

We can draw a triangle with 4 centers in it. Three of the centers are the vertices of the triangle, and lets call them A,B,C. radius(A) = radius(B) = 5, radius(C) = 8. Then ABC is an isoceles triangle, with AC=BC=13, AB=10. Drop a perpendicular height from C to AB and call the point D. AD = 12. The fourth center lies on AD. Call the fourth Center E. Then EB = 5+a, EC = 8+a, ED = 12-8-a = 4-a, BD = 5. => 5^2 + (4-k)^2 = (5+k)^2 Hence, choice (a) is the right answer.

----------------------------------------------------------- Quant Answer # 47 -----------------------------------------------------------

Each box will be touched by children whose number is the divisor of the box number. And the toffees taken overall from the box is the sum of the divisors for the box number. The sum of the divisors of a number of the form (a^x1)*(b^x2)*(c^x3)*... where a, b, c ... are all prime and x1, x2, x3 are positive integers is (1 + a + a^2 + ... + a^x1)*((1 + b + b^2 + ... + b^x2)*(1 + c + c^2 + ... + c^x3)*... e.g for the number 54 it's (1 + 3 + 3^2 + 3^3)*(1 + 2) I. The sum of the divisors of 96 is max. among first 100 natural numbers, hence I is false II. There are 25 primes less than 100, hence I is true III. Only 6 and 28 are perfect numbers less than 100, hence III is true Hence, choice (b) is the right answer

----------------------------------------------------------- Quant Answer # 48 -----------------------------------------------------------

a^4 + 2(a^3) + 2(a^2) +2a + 1 = (a+1)^2*(a^2 + 1) = b^2 (i) a^2 + 1 is a perfect square only for a=0, b = +/-1 (ii) when a = -1, b = 0 Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 49 -----------------------------------------------------------

(n^4 + 256 + 4n(n^2 + 16))/(n+4)^2 = (n^3 + 64)(n+4)/(n+4)^2 = n^2 - 4n + 16 4 <= n^2 <= 49 => -7 <= x <= -2 or 2 <= n <= 7

x = n^2 - 4n + 16 = (n-2)^2 + 12

x is max at n = -7, and min. at n=2

Hence, choice (d) is the right answer

----------------------------------------------------------- Quant Answer # 50 -----------------------------------------------------------

Let the manufacturers cost be Rs 100, then marked price is Rs 800. Let x be the discount given to the distributor by the manufacturer.

The cost price for distributor, wholesaler, retailer, customer is 800-x, 800-3/4x, 800-x/2, and 800-x/4 respectively.

Given, ((800-x/2) - (800-3/4x))/(800-3x/4) = 1/2 => x = 640. => the profit of manufacturer, distributor, wholesaler, retailer is Rs 60, Rs 160, Rs 160, Rs 160 respectively.

Hence, choice (a) is the correct answer

ANSWERS FOR THE QUESTION BANK QUESTIONS

(51-100)

----------------------------------------------------------- Quant Answer # 51 -----------------------------------------------------------

c > 2b, 3m > c, 3b > 4m. => c < 3m < 3b and c > 2b > 2m => c > b + m, but c + b + m = 42, => c > 21. Also, c < 3/2(b + m) => c < 26

Now, plugging the value of c in original equations we see that only c = 23 satisfies. c = 23, m = 8, b = 11


----------------------------------------------------------- Quant Answer # 52 -----------------------------------------------------------

x^4-x^3-x^2-1 = (x+1)(x^3-2x^2+x-1), so a = -1, b+c+d = 2, bc+cd+db = 1. Hence b^2+c^2+d^2 = 2^2-2 = 2. We have f(x) = (x^3-2x^2+x-1)(x^3+x^2+x+1)+x^2-x+1. Hence f(a) = 3, f(b) = b^2-b+1, f(c) = c^2-c+1, f(d) = d^2-d+1, so f(a)+f(b)+f(c)+f(d) = 6 + (b^2+c^2+d^2)-(b+c+d) = 6.

Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 53 -----------------------------------------------------------

Label equations given (1), (2), (3). Then (1) - 2 (2) + (3) = 2(a + ... + g), (3) - (2) = 5a + 7b + 9c + ... . Hence 7a + 9b + 11c + ... = (1) - 3 (2) + 2 (3) and required value = (1) - 3 (2) + 3(3).


Solution by IdiotR:

I hav done ths with the traditional way of doin ths let k1(a + 4b + 9c + 16d + 25e + 36f + 49g )+k2(4a + 9b + 16c + 25d + 36e + 49f + 64g )+k3(9a + 16b + 25c + 36d + 49e + 64f + 81g )=16a + 25b + 36c + 49d + 64e + 81f + 100g then

k1(-1)+k2(-2)+k3(3) is our answer equating co effs k1+4k2+9k3=16 4k1+9k2+16k3=25 9k1+16k2+25k3=36 solvin we get k1=1 k2=-3 k3=3 hence value=1(-1)+(-3)(-2)+3(3)=14

Logic of cosmic_glitch:

Take any of the 7 variables. It's coefficients in the 3 equations are of the form: k^2, (k+1)^2, (k+2)^2 We need to find it's value when the coefficient is (k+3)^2. So, we write (k+3)^2 as a linear combination of k^2, (k+1)^2 and (k+2)^2: (k+3)^2 = k^2 - 3 * (k+1)^2 + 3 * (k+2)^2 Formula : A(n+1) = 3*A(n) - 3* A (n-1) + A(n-2) . where A(n) = n*n

---------------------------------------------------------- Quant Answer # 54 -----------------------------------------------------------

Plot the graph of f(x) = x + | x^2 - 1 | It is clear from the graph that there are no roots for k < -1, and one root for k = -1 (namely x = -1). Then for k > -1 there are two roots except for a small interval [1, 1+h]. At k = 1, there are 3 roots (x = -2, 0, 1). The upper bound is at the local maximum between 0 and 1. For such x, y = x + 1 - x^2 = 5/4 - (x - 1/2)^2, so the local maximum is at 5/4. Thus there are 3 roots at k = 5/4 and 4 roots for k belonging to (1, 5/4).


Solution by Anupam will return:

Since we have overlapped options here, let us consider each option. Considering option (a) k = 5/4. Now, we have 3 cases with us : 1)x >1 2)x<-1 3)-1 <x<1 Taking Case (1), we have x + x^2 - 1 = 6/5 or 5x^2 + 5x -11 = 0 . We get two real roots here. Taking Case (2), we have x -X^2 + 1 = 6/5 solving for x we get another set of 2 real roots. Taking Case (3), we have x + x^2 -1 = 6/5 (same as (1)) Hence the number of roots obtained is 4 making option (a) true. Take option (b) k = 5/4.Let's again consider cases 1,2 & 3. (1) we have x + x^2 -1=5/4 or 4x^2 + 4x -9 =0 .. solving we obtain two real roots. (2) we have x - x^2 + 1 =0 .. solving we get x = 2( only one root) (3) will again yield two roots which we obtained in (1). Hence the number of roots is 3 making option (b) true. At this point.. we need not proceed further..as we see multiple options satisfied.. we should mark option (d). But just incase you are interested you should try and find (c) true too.

----------------------------------------------------------- Quant Answer # 55 -----------------------------------------------------------

Applying cosine rule to velocity triangle, we get that Amar's velocity relative to Vineet's is k where 7^2 = 8^2 + k^2 - 8k. So k = 3 or 5. Hence largest possible k is 5. Time 100/5 = 20s.

Hence, choice (a) is the right answer

----------------------------------------------------------- Quant Answer # 56 -----------------------------------------------------------

Setting x = 1, and x = 8 gives us f(2) = 0, f(8) = 0 respectively. since f(8) = 0, f(4) = 0. If the degree of f(x) is k => f(2x)/f(x) = 8(x-1)/(x-8) for large x, f(2x)/f(x) -> 8 => 2^k = 8 => k = 3


----------------------------------------------------------- Quant Answer # 57 -----------------------------------------------------------

x^2 + y^2 = 7 => (x+y)^2 - 2xy = 7 x^3 + y^3 = 10 => (x+y)(7 - xy) = 10 => (x+y)(7 - ((x+y)^2 - 7)/2) = 10 Solving cubic for x+y, roots are -5, 1, 4. Hence, choice (a) is the right answer

Solution by amar_kashyap:

x^2+y^2=7 and x^3+y^3=10 From first eqaution xy= {(x+y)^2-7}/2 From second equation (x+y)^3-3xy(x+y)=10 (x+y)^3-3(x+y){(x+y)^2-7}/2=10 21(x+y)/2- {(x+y)^3}/2= 10 (x+y){21-(x+y)^2}=20 Now let's say x+y= a+ib So we have (a+ib){21-(a+ib)^2}=20 (a^2+b^2)(21-a^2+b^2-2iab)=20a-20ib Eqauting the real and imaginary parts on both sides we have, ab(a^2+b^2)=10b....1) and (a^2+b^2)(21-a^2+b^2)=20a...2) from equation 1) a^2+b^2=10/a.. case 1. For this case a cannot be negative or b=0 case 2. Proceeding with case 1. From 2nd equation: Sustituting for a^2+b^2=10/a in eqquation2) 21-a^2+b^2=2a^2 ie 3a^2-b^2=21. Again substituting for b^2=10/a-a^2 in this equation we get 3a^2-10/a+a^2=21 4a^3-10-21a=0 (a+2)(2a+1)(2a-5)=0 Now since a cannot be negative a=5/2 which when put into equation a^2+b^2=10/a results into imaginary value of b, so even a= 5/2 is rejected. Case 2. when b=0

From equation..2) a^2(21-a^2)=20a => a^4-21a^2+20a=0 => a^3(a-1)+a^2(a-1)-20a(a-1)=0 =>a.(a-1)(a^2+a-20)=0 =>a.(a-1)(a-4)(a+5)=0 So a can be 0, 1, 4 or -5 Hence the minimum value of a ie the real part of x+y=-5. I have taken all the cases..hope the above solution is general..

----------------------------------------------------------- Quant Answer # 58 -----------------------------------------------------------

Let AB = L => Anupam travells 8L by the time he crosses C for the second time. Also OA = L. Let the speeds of Anupam and Shruti be a and b respectively. Check yourself that Shruti doesn't overtake Anupam on AB in the first round. She does so in the second round.

When the distance covered by Anupam is 3L, 5L, 7L repectively then the min. distance covered by Shruti is 5L, 8L, 11L respectively. When the distance covered by Anupam is L, 2L, 4L, 6L repectively then the max. distance covered by Shruti is 2L, 4L, 7L, 10L respectively.

=> b:a is 5:3 (WHY?)


----------------------------------------------------------- Quant Answer # 59 -----------------------------------------------------------

Let x be the certainity of Varun to try first option before the second. Suppose he calls Amar and gets his number then he makes x calls or he fails in which case he makes (1-x)*3 + (1-x) calls. WHY? To opt for second option would take him 3 calls. => x + 4(1-x) <= 3 => x >= 1/3


Alternate Solution:

Option A: Call Amar first Chance of success = P with 1 call Chance of success = (100-P) with 4 calls Option B: Call Vineet first Chance of success = 100 with 3 calls

Compunding chance with number of calls: P*1 + (100-P)*4 <= 100*3 P >= 100/3

----------------------------------------------------------- Quant Answer # 60 -----------------------------------------------------------

Let AP = a, BP = b, CR = c. Since, AP=AR, PB=BQ, CQ=CR => (a+b)^2 + (b+c)^2 = (a+c)^2 => ab + b^2 + bc = ac Let a/b = x an integer => c/b = (x+1)/(x-1) c/b is integer only when x = 2 or 3

Hence, choice (b) is the right answer

----------------------------------------------------------- Quant Answer # 61 -----------------------------------------------------------

Plot the co-ordinates and check yourself.

Let A = (3, 4) and C = (5, 6) B = f(3, 4) will lie on y=4, and D = g(5, 6) will lie on x=5. A polygon is convex if all the angles are less than 180 degrees. if b > 1 => (5, 6b) is one point => the point on y=4 is such that x < 5 => a < 5/3

Note here that though ABCD is not convex for all a < 5/3 but for some a. However, when a > 5/3 ABCD can never be convex but we are given ABCD is convex.

choice (b), (c) doesn't give convex polygon


----------------------------------------------------------- Quant Answer # 62 -----------------------------------------------------------

Let OA = 1 km and AB = x km Let V km/min be the velocity of the bird when there is no wind resistance. Let U km/min be the wind velocity from A to B. Given that, 1/V + x/(V+U) = 12, 1/V + x/(V-U) = 14, 1/(V+U) + x/(V+U) = 11 solving we get, x = 5/6, U=1/42, V=1/7

Time taken by the bird to travel from B to O with wind resistance throughout = (1+x)/(V-U) = 15.4 minutes


----------------------------------------------------------- Quant Answer # 63 -----------------------------------------------------------

When only the mid-points of ABC are joined, we have triangles ABC, PQR, APQ, BPR, and CQR whose sum is 2*6 = 12 sq. cm, area(PQR) = 3/2 sq. cm In the next sequence of triangle formation where P', Q', R' are mid-points of sides of triangle PQR, sum of new three triangles is 2*area(PQR) - area(PQR) as triangle PQR has already been considered. Continuing with the process,

sum of areas of all the triangles formed = 12 + 3/2 + 3/8 + 3/32 + ... = 14 sq. cm


----------------------------------------------------------- Quant Answer # 64 -----------------------------------------------------------

For the number of those who do not play one or more games to be maximum, the students who do not play any given game should be as distinct as possible from those not playing an other game.

=> Maximum = 9 + (16 - 9) + (21 - 9) + (11 - 9) + (15 - 9) = 36


----------------------------------------------------------- Quant Answer # 65 -----------------------------------------------------------

A cone can be formed from the sector of a circle. Since C is the vertex, the angle of sector can be maximum 90 degrees. The radius of this sector is maximum when its radius is equal to the length of the prependicular from C to AB which is 6*8/10 = 4.8 cm => Slant height of this cone is 4.8 cm and radius of the cone is 1/4*(4.8) = 1.2 cm Volume of the cone is 1/3*pi*((1.2)^2)*(4.8^2 - 1.2^2)^1/2 =~ 7


----------------------------------------------------------- Quant Answer # 66 -----------------------------------------------------------

Let f(x) = [x] + [2x] + [3x] + [4x] and g(x) = f(x) + [5x/3]. Since [y+n] = n + [y] for any integer n and real y, we have that f(x+1) = f(x) + 10. So for f it is sufficient to look at the half-open interval [0, 1). f is obviously monotonic increasing and its value jumps at x = 0, 1/4, 1/3, 1/2, 2/3, 3/4. Thus f(x) takes 6 different values on [0, 1).

g(x+3) = g(x), so for g we need to look at the half-open interval [0, 3). g jumps at the points at which f jumps plus 4 additional points: 3/5, 1 1/5, 1 4/5, 2 2/5. So on [0, 3), g(x) takes 3 x 6 + 4 = 22 different values. Hence on [0, 99), g(x) takes 33 x 22 = 726 different values. Then on [99, 100] it takes a further 6 + 1 + 1 (namely g(99), g(99 1/4), g(99 1/3), g(99 1/2), g(99 3/5), g(99 2/3), g(99 3/4), g(100) ). Thus in total g takes 726 + 8 = 734 different values


Solution by cosmic_glitch

Previously, I was thinking that number of unique values generated by the expressions depends only on the fractional part. But in fact because of the existence of [5x/3] in the expression, the pattern repeats not for every integer but for every 3 integers: [x] changes value at 0, 1, 2, 3 [2x] changes values at 0/2, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2. [3x] changes values at 0/3, 1/3, 2/3 ... 9/3 [5x/3] changes values at 3/5, 2*3/5, 3*3/5, 4*3/5, 5*3/5 Total number of intervals from 0-3 is 22. Till 99 there are 22*(99/3)=726 unique values. Then from 99 to 100 there are 8 more. So, answer is 726+8=734

----------------------------------------------------------- Quant Answer # 67 -----------------------------------------------------------

Let AB be the wall and BC the floor. BDEF is the one surface of the cube, D is on AB and F is on BC, and E on AC. AEC is the ladder. Let AD = a, FC = b. Triangles AED and CFE are similar => a/12 = 12/b. Also AC^2 = AB^2 + BC^2 => 35^2 = (a+12)^2 + (b+12)^2, solving we get a = 16, b = 9 => ladder touches the wall at a height 16+12 = 28 ft


----------------------------------------------------------- Quant Answer # 68 -----------------------------------------------------------

We need to factorize 100 in such a way that following 2 constraints are met

1. “Sum of factors – number of factors” is least 2. Number of factors = Number of dimensions in which the cut can be made

100 = 5*5*4

Hence, minimum number of cuts = (5-1)+(5-1)+(4-1) =11 cuts.


----------------------------------------------------------- Quant Answer # 69 -----------------------------------------------------------

Consider the set S` = {166, 167, ... , 999}. The smallest possible value for x + 2y + 3z, for distinct x, y, z in S` is 168 + 2.167 + 3.166 = 1000. So we cannot find distinct x, y, z, k in S` with x + 2y + 3z = k. So the smallest n > 834.

Now suppose S` is any subset of 835 elements which satisfies the condition. Take it elements to be m = x1 < x2 < ... < x835 = M. Obviously M = m + 834 = 835, so -3m = 3*834 - 3M and hence M - 3m = 2502 - 2M = 2502 - 2*999 = 504. Put p = M - 3m.

There are at least 167 disjoint pairs (x, y) of numbers taken from {1, 2, ... , 999} with x + 2y = p, namely (p - 2, 1)

(p - 4, 2)

(p - 6, 3)

...

(p-334, 167) - note that in the extreme case p = 504 this is (170, 167)

At least one number from each pair must either (1) be M or m or (2) not belong to S` - or otherwise we would have x + 2y + 3m = M for distinct elements a, b, m and M in S`. None of the numbers can be M and at most one of them can be m, so we have at least 166 numbers which are not in S`. That means S` contains at most 999 - 166 = 833 numbers. Contradiction. So S` cannot have 835 elements. Nor can it have more than 835 elements.


----------------------------------------------------------- Quant Answer # 70 -----------------------------------------------------------

Let the length of each side of the cube be a cm => the length of the diagonal of one of the faces of the cube is 2^1/2*a and it's parallel to the base of the cone. => (2 - a)/2 = (length of the diagonal)/(Base length) = a/4 => a = 4/3 Volume remaining is 1/3*pi*8*2 - 64/27 =~ 13.5 cubic cm


Solution by vineet

for max volume ..if l be the length of cube cut , then 2r / l = h / h-l =>l = 4*root(2)/[2*root(2)+1] volume remaining = 16pi/3 - 64 *2root2/(2root2 + 1)^3 =16.75 - 3.25 =13.5 (approx)

----------------------------------------------------------- Quant Answer # 71 -----------------------------------------------------------

Let v be the volume of the tank. When the tank was full to half the height then the volume of water was v/8. WHY? Hence, v/2 was expected but it was v/8 => 3v/8 of the water was missing. => leak can empty (3v/8)*2 water in the time the tap alone can fill the tank = 3v/4


----------------------------------------------------------- Quant Answer # 72 -----------------------------------------------------------

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, draw the graphs of y = |1 - |x|| and y = (1.01)^(1.01x) and see they intersects at 3 points.


Rule

Let f(x) = 0, then if f(a).f(b) < 0, then there are odd number of roots in the interval (a, b) f(a).f(b) > 0, then there are even number of roots in the interval (a, b) You can check for exact number by further inspecting the behaviour(increasing/decreasing) of f(x) in (a, b)

----------------------------------------------------------- Quant Answer # 73 -----------------------------------------------------------

y^2 - xy = (y - x/2)^2 - x^2/4, so y^2 - xy has a minimum at y = x/2. It is decreasing for y < x/2 and increasing for y > x/2.

Thus the largest value of y^2 - xy must be at y = 0, x/2 or 1. The values there are 0, x^2/4, |1-x|. So for x outside the interval (0,1), f(x) >= 1/4. For x in (0,1) we have f(x) = max(x^2/4, 1-x). The quadratic x^2 + 4x - 4 = 0 has roots -2 ± 2*2^1/2, which are approx and -4.83 and 0.83. Hence f(x) = 1-x for x <= 2*2^1/2 - 2, and x^2/4 for x <= 2*2^1/2 - 2, and the minimum value of f(x) in the interval (0,1) is f(2*2^1/2 - 2) = 3 - 2*2^1/2, which is approx 0.18 and < 1/4. This is also the minimum value of f(x) for all x


----------------------------------------------------------- Quant Answer # 74 -----------------------------------------------------------

Let each side has length as 4 => AD = 8. Area of ABCDEF = ((3*3^1/2)/2)*16 = 24*(3^1/2) sq. units

area(BADC) = area(BCQ) + area(BQP) + area(PQA) + area(QAE) If area(BQP) = x => area(PQA) = 3x as AP/BP = 3. area(BCQ) = 1/2*4*3*sin(120 degrees) = 3*3^1/2 sq. units area(QAE) = 1/2*1*8*sin(60 degrees) = 2*3^1/2 sq. units

=> 12*(3^1/2) = 4x + 5*(3^1/2) => x = 7/4*(3^1/2)

area(BPQC) = x + 3*3^1/2 = 3^1/2(19/4)


2nd solution given by Rajsher and better than the above.

considering the length of hexagon as 4. just extend AB and DC to meet at G. Now area of BPQC = area(GPQ) - area(GBC) = 1/2*5*7*sin60 - 4sqrt(3) =19/4*sqrt(3) area of hexagon = 6*4sqrt(3) and hence the ratio 19:96

Solution by lemma

RED IS THE REQUIRED AREA. BLUE IS THE AREA OF TRIANGLE. 'h' is total height of trapezium. '3h/4' is height of smaller trapezium. 'h/2' is height of triangle. 'a' and 'a+a+3+/4' are sides of smaller trapezium.

'a+3a/4' is base of triangle.

area of BPQC is 1/2(3h/4)(a+a+3a/4)-1/2(h/2)(a+3a/4)=1/2(h)(a)(33/16-7/8 )=1/2(h)(a)(19/16)=19/32(h)(a) area of ABCDEF is (h)(a+a+a)=3(h)(a) hence ration of areas = (19/32)/(3) = 19/96

----------------------------------------------------------- Quant Answer # 75 -----------------------------------------------------------

The minutes hand travels a full circle in 120 minutes => 720 degrees in 120 minutes => 6 degrees in 1 minute Hour hand travels 720/36 = 20 degrees in 1 hour = 1/6 degrees in 1 minute

at 11:24 the hour hands will be at angle 11*20 + 24/6 = 224 degrees and the minute hand at 24*6 = 144 degrees


----------------------------------------------------------- Quant Answer # 76 -----------------------------------------------------------

Let the four digits be a, a+d, a+2d, a+3d => (4a+6d)/4 is an integer => d is even and also a+3d <= 9 Check yourself that d=2 and a <=3 is the only possibility for all even d. Hence, four digits can be from {0, 2, 4, 6}, {1, 3, 5, 7}, {2, 4, 6, 8}, {3, 5, 7, 9} The first set gives 18 numbers while next three give 24 each. Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 77 -----------------------------------------------------------

f(x+4) = 1/2 + (f(x+2) - (f(x+2))^2)^1/2.

But, f(x+2) = 1/2 + (f(x) - (f(x))^2)^1/2, subsituting and solving we get f(x+4) = 1/2 + ((1/2 - f(x))^2)^1/2, but since f(x) > 1/2, 1/2 + ((1/2 - f(x))^2)^1/2 = 1/2 + f(x) -1/2 => f(x+4) = f(x)


Solution by vineet

[f(x+2)- 1/2]^2 = 1/4 - [f(x) - 1/2]^2 putting x=x+2 [f(x+4)- 1/2]^2 = 1/4 - 1/4 +[f(x) - 1/2]^2 = [f(x) - 1/2]^2 => f(x+4) = f(x) periodicity is 4

----------------------------------------------------------- Quant Answer # 78 -----------------------------------------------------------

Divisors of powers of 3 never add up: 1 + 3 + 9 = 14 < 27, etc. So 3^b isn't enough. Try 3^b * 5^c: n = 3*5 = 15: 1 + 3 + 5 = 9 < 15; 9/15 = 0.6 n = 45: 1 + 3 + 5 + 9 + 15 = 33 < 45; 33/45 :=: 0.733 (For n = 3^6 * 5^4 = 455,625: sum of proper div = 398,008 ; ratio :=: 0.874) This seems to fall short of our goal of 1, so we bring in powers of 7. n = 3^2 * 5 * 7 = 315; sum of all div = (1+3+9)(1+5)(1+7) = 624, sum of proper div = 624 - 315 = 309, just missed! n = 3^3 * 5 * 7 = 945; sum of all div = (1+3+9+27)(1+5)(1+7) = 1920, sum of proper div = 1920 - 945 = 975 , it's abundant!

The smallest odd abundant number is 945.


----------------------------------------------------------- Quant Answer # 79 -----------------------------------------------------------

Total distance Shrikant can cover = 10/60*(15 + 15/2 + 15/4 + ...) = 5 km Total distance Sachin can cover = 10/60*(25 + 25/2 + 25/4 + ...) = 25/3 km

First time they cover 750 m and subsequently they cover a distance of 1500 m to meet. The total distance they cover together is 40/3 km. Number of meetings possible is 1 + (40000/3 - 750)/1500 = 9.4


----------------------------------------------------------- Quant Answer # 80 -----------------------------------------------------------

In 13 hours exactly 780 pizzas are made out of which we need all those numbers which are neither multiples of 5 nor 7 nor 8.

A U B U C = A + B + C - A/B - A/C - B/C + A/B/C (A/B here denotes A intersection B)

Multiples of 5 are 780/5 = 156 Multiples of 7 are [780/7] = 111 Multiples of 8 are [780/8] = 97 Multiples of 5 and 7 are [780/35] = 22 Multiples of 5 and 8 are [780/40] = 19 Multiples of 7 and 8 are [780/56] = 13 Multiples of 5, 7 and 8 are [780/280] = 2

Since, numbers are counted from 1st pizza

=> A U B U C = 156 + 112 + 98 - 23 - 20 - 14 + 3 = 312


----------------------------------------------------------- Quant Answer # 81 -----------------------------------------------------------

Let the triangle be ABC with <A obtuse and b > c. We have b sin C = c sin 2C and c^2 = a^2 + b^2 - 2ab*(sin C). Hence, b/2c = sin 2C/sin C = cos C = (a^2 + b^2 - c^2)/2ab. So a*b^2 = a^2*c + b^2*c - c^3. Hence b^2*(a - c) = c*(a^2 - c^2). => b^2 = c(a + c).

Now the triangle with smallest perimeter will have a, b, c relatively prime. => c must be a square. For if c and a+c have a common factor, then so do a and c and hence a, b and c, which means they cannot be the minimal set. Clearly c is not 1, c = 4 gives a = 5, which is too short, or a >= 21, which is too long. c = 9 gives a = 7 (too short), 18 (3 divides a, b, c), or >= 40 (too long). c = 16 gives a = 20 (too short) or a = 33 which works. Hence, sides have lengths as 16, 28 and 33.


Solution by cosmic_glitch

Let ABC be the triangle obtuse angled at C. Angle A = x Angle B = 2x Angle C = 180-3x Also, let AB=r, BC=q, AC=p Now, p/sin(2x)=q/sin(x)=r/sin(3x) Since p, q and r are integers: sin(2x)/sin(x) and sin(3x)/sin(x) should be rational numbers. => 2cos(x) and 4cos^2(x)-1 are rational numbers Also, we know that: 0 < x < 30 (otherwise C will not be obtuse). => cos(30) < cos(x) < 1 => sqrt(3)/2 < cos(x) < 1 => .86 < cos(x) < 1 If cos(x) has to be a rational number m/n, such that both m and n are as small as possible and m/n is less than 1: cos(x) = m/(m+1) > 0.86 => m > 6 => m = 7 So, 2cos(x) = 7/4. From that, 4*cos^2(x)-1 = 33/16 Finally: r/q = 33/16 p/q = 7/4 = 28/16 So, min values of p, q, r are: 28, 16, 33. => Perimeter is 77

----------------------------------------------------------- Quant Answer # 82 -----------------------------------------------------------

f(p) = p3 - 3p2 + 5p - 17, f(3).f(4) < 0, and the first derivative of f(p) > 0 for all p => f(p) = 0 has only one real root. Similarly f(q) = q3 - 3q2 + 5q + 11 = 0 has one real root. Putting p = 2-q in we get p3 - 3p2 + 5p - 17 we get -(q3 - 3q2 + 5q + 11 ) => the two real roots satisfy p+q = 2.


Solution by chautauqua

putting p=x+1 and q=y+1: we get the following equations: x^3+2x=14 ... (i) and y^3+2y=-14 ... (ii) It is obvious that if 'k' is a root of (i), then '-k' is a root of (ii) => x+y=0 => p+q = 2 Further it can also be easily shown that (i) and (ii) have only one real root each... The derivative of f(x) = x^3 + 2x -14 or the original eqn is always positive and f(infinty) is positive while f(-infinity) is negative ... => f(x) has only one real root Similarly for the second equation.

----------------------------------------------------------- Quant Answer # 83 -----------------------------------------------------------

Let the number of snickers, eclairs, and pan-pasand be a, b and c respectively. => a + b + c = 100, 2a + 0.5b + 0.1c = 100 => 19a + 4b = 900 => a is a multiple of 4 lying between 27 and 47. checking, we see only a = 36, 40 and 44 satisfy.


----------------------------------------------------------- Quant Answer # 84 -----------------------------------------------------------

We have 3 choices for the first letter, and 2 for each subsequent letter, so 3*(2^5).


----------------------------------------------------------- Quant Answer # 85 -----------------------------------------------------------

Aarav’s comments

Sorry folks, mistake at my end. The ratio is 6/5 indeed. Simpler way to do is by drawing a circle passing through D and E and AB as diameter. Use secant rule: CD*CB = CE*AC.

Solution by vineet.nitd

16 -(5-CD)^2 = 36 -CD^2 =>CD =4.5 25 - CE^2= 16 - (6-CE)^2 = >CE =3.75

in question ratio of CD /CE is asked and not CE /CD .... Hence CD / CE = 6/5

----------------------------------------------------------- Quant Answer # 86 -----------------------------------------------------------

Let d 1st term is a1 n second a2 The term of the series repeat themselves after every 6th term.. S(1492)=a1+a2+(a2-a1)-a1-a2+(a1-a2)+.....upto 1492th term=2a2-a1=1985 => 2a2-a1=1985..............(1) S(1985)=a1+a2+(a2-a1)-a1-a2+(a1-a2)+.....upto 1985th term=a2-a1=1492 =>a2-a1=1492................(2) Solving 1 n 2 v get a1=-999 a2=493 S(2001)=a1+a2+(a2-a1)-a1-a2+(a1-a2)+......(upto 2001th term) = 2a2=986


----------------------------------------------------------- Quant Answer # 87 -----------------------------------------------------------

Relative speed of Manish wrt to Harish is 18 + 12 = 30 km/ h. They would have travelled 7 km by the time they meet for the first time. When they cross each other for the second time they would have covered 21 km, third time 35 km, fourth time 49 km, and fifth time 63 km. Time taken to meet for the fifth time is 2 hour and 6 minutes. Distance from Kothrud is 37.8 km. => 37.8 - 35 from Kothrud colony = 2.8 km.


----------------------------------------------------------- Quant Answer # 88 -----------------------------------------------------------

From any point 3 lines can be drawn which pass through only two quadrants. Four points A, B, C and D in 4 quadrants will give 12 such lines, of which 4 are vertical, 4 are horizontal, and the remaining 4 are inclined and pass through the origin. To get maximum number of points of intersection we assume that each of the 12 lines is distinct and not more than 2 lines intersect at a point, except for A, B, C and D and the origin. 4 vertical lines can intersect 4 horizontal lines in 16 points(which includes A, B, C and D). The inclined line through A intersects each of the 3 vertical lines through B, C, D and the 3 horizontal lines through B, C, D at 6 new points. Similarly, for inlined lines through B, C, D we get 18 lines. In addition 4 inclined lines intersect at origin.

=> points of intersection are 16+24+1 = 41


----------------------------------------------------------- Quant Answer # 89 -----------------------------------------------------------

Since f(x) = 2 we have 1 + (1-x) + (1-x)^2 + (1-x)^3 = 2*(1-x)^4; putting 1-x = t => 0 < t < 1 1 + t + t^2 + t^3 = 2t^4 put 1/t = p we get p> 1 and p^4 + p^3 + p^2 + p = 2

Clearly no such p > 1 exits.


----------------------------------------------------------- Quant Answer # 90 -----------------------------------------------------------

Let AB = x cm. Since BP = PC => <PBC = <BCP = < APB => Triangles BAP and BCP are isoscles triangles. Also triangles BAP and BCP are similar => AD = BC = (x+5) cm and x/6 = 6/(x+5) => x = 4 cm.


----------------------------------------------------------- Quant Answer # 91 -----------------------------------------------------------

Each person works for 10 days with someone else + 1 day more = 11 days. Let a(i) be the capacity(days in which work is completed) of each worker => (1/a(1) + 1/a(2) + ... + 1/a(11)) of the work in a single day = w => 11w = 1 => All working together take 11 days to finish the work


----------------------------------------------------------- Quant Answer # 92 -----------------------------------------------------------

The highest power of 2 that divides N! is always greater than the highest power of 5 that divides N! => The quotient will contain atleast one factor as 2. => The quotient is always even Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 93 -----------------------------------------------------------

Let f(x)= A(x^2) + Bx + C and let p , q be the roots of f(x). f(x)=a(x-p)(x-q) f(0)=C=A.p.q>0 and f(1)=A+B+C=A(1-p)(1-q)>0 f(0)f(1)>0 so f(0)f(1) >= 1 ,or A^2.p.(1-p).q.(1-q) >= 1 (*) we know that for 0<x<1, x(1-x) <= 1/4 and equality holds iff x=1/2 since p and q are different we have p.(1-p).q.(1-q) < 1/16 from (*) ;A^2 > 16 => A > 4 => A >= 5 For a=5, 5(x^2) -5x +1 satisfies the conditions. Hence, choice (c) is the right answer

Solution by mansoor316

The sum of roots = b/a < 2; b < 2a ....(i) Since roots are distinct, b²-4ac>0 => c < b²/4a For a=2, b=1,2,3 ......from (i) Since c is +integer and c < b²/4a = b²/8, only possibility=(a,b,c)=(2,3,1) [Since, b² has to be > 8] For a=3, b=1,2,3,4,5......from (i) c < b²/4a = b²/12, possibilities=(3,4,1),(3,5,1)(3,5,2) In ax²-bx+c=0, for (2,3,1)(3,4,1) and (3,5,2) one root is 1 For (3,5,1): 3x²-5x+1=0 has one root in (1,2) For a=5, b=1..9; c < b²/4a = b²/20. Consider first b=5, we get c=1; 5x²-5x+1 has both roots less than 1 Hence, a=5: option (c)

----------------------------------------------------------- Quant Answer # 94 -----------------------------------------------------------

The problem must hold true for a square. Consider a square of side 2 with the above coordinates. eqn of AF: x=2y eqn of DE: 2x+y=2 Solving, AF, DE intersect at y=2/5 Area of shaded triangle with base AE= 1/2*2/5*1 = 1/5 Area of all shaded triangles = 8/5 Required ratio = (8/5)/4 = 2/5 Hence, choice (b)

----------------------------------------------------------- Quant Answer # 95 -----------------------------------------------------------

(a^2 + b^2)*(c^2 + d^2) >= (ac + bd)^2 9*25 = ((x-3)^2 + (y+2)^2)*(3^2 + 4^2) >= (3(x-3) + 4(y+2))^2 => 9*25 >= (3(x-3) + 4(y+2))^2 => 3x + 4y <= 16 Hence, choice (d) is the correct answer

This was based on cauchy-schwarz’s inequality.

Solution by vineet

We have with us a circle with centre(3,-2) and radius 3. so for 3x+4y to be max , straight line 3x + 4y =c must be a tangent to this circle ...This will give me a max value for c. hence , |3*3 - 4*2 -c|/5 = 3 =>c(max) =16

----------------------------------------------------------- Quant Answer # 96 -----------------------------------------------------------

Let small cone have height h. Then vol cone/vol frustrum = h^3/(64-h^3). If we unroll large cone, it forms part of circle with radius 5, arc length 6p, so area (6p/10p)*25p = 15p. Base has area 9p. So painted area of small cone is h^2*15p/16, painted area of frustrum = 9p + (16-h^2)15p/16. Thus k = h^3/(64-h^3) = 5h^2/(128-5h^2). So h = 5/2, k = 125/387 Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 97 -----------------------------------------------------------

Suppose the 2-digit sum is d. We must have 1 = d = 18. For d = 9, there are d+1 choices for the 2nd pair and one less for the 1st pair (no. cannot have leading 0). For d > 9, there are 19-d choices for each pair.

Hence 1·2 + 2·3 + ... + 9·10 + 9^2 + 8^2 + ... + 1^2 = 615 Hence, choice (a) is the right answer

Solution by vineet.nitd

for sum as 1 , 1 possibility at the two left most digits and 2 possibilities at two right most digits , for sum as 2, 2 possibilities at the two left most digits and 3 possibilities at two right most digits And so on ......upto 9 , giving u 1*2 +2*3 +....9*10 = 330 for sum as 10 , u will have 9 possibilities at each place , for sum as 11 u will have 8 and so on ...til u reach sum as 18 which will have 1 posibilit at each pair ... 9^2+8^2 +....+1 =285 hence total = 285 + 330

----------------------------------------------------------- Quant Answer # 98 -----------------------------------------------------------

Let T be time Aarav takes to make 25 steps. Then Aarav takes 3T to make 75, and Rahul takes 2T to make 150. Suppose the escalator has N steps visible and moves n steps in time T. Then Rahul covers N + 2n = 150, N - 3n = 75. Hence N = 120, n = 15. Hence, choice (d) is the right answer

----------------------------------------------------------- Quant Answer # 99 -----------------------------------------------------------

If n = rs, with 1 < r < s, then r < s < n, and hence rsn = n^2 divides n!. Similarly, if n = r^2 with r > 2, then r < 2r < n, and hence n^2 divides n!. This covers all possibilities except n = 4 or n = prime, and it is easy to see that in these cases n^2 does not divide n!. There are 25 primes in first 100 naturals. Hence, choice (c) is the right answer

----------------------------------------------------------- Quant Answer # 100 -----------------------------------------------------------

We find that f(f(f(x))) = (m + n2 + mnx)/(n3+2mn + (m2+mn2)x). Now f(f(f(x))) = x is equivalent to (m^2+mn^2)x^2 + (n^3+mn)x - (m+n^2) = 0. But this quadratic has three distinct solutions p, q, r so it must vanish identically. The constant term vanishes if m + n2 = 0, but then the other coefficients also vanish. So we must have m = -n2. If m = -n2, then it is easy to check that f(x) = 1/(n - n2x), f(f(x)) = (nx-1)/n2x, f(f(f(x))) = x, so p, q, r certainly exist.



(101-150) ----------------------------------------------------------- Quant Answer # 101 -----------------------------------------------------------

Let Shirish's speed be a km/hr and the speed of the current be b km/hr => 18/(a-b) - 18/(a+b) = 9 and 18/(2a-b) - 18/(2a+b) = 1 Let k = a/b => 1/(k-1) + 1/(k+1) = v/2, and 1/(2k-1) - 1/(2k+1) = v/18 solving for k we get v = 20/3


----------------------------------------------------------- Quant Answer # 102 -----------------------------------------------------------

Cauchy-Schwarz's inequality: (a1^2 + a2^2 + a3^3 +...)*(b1^2 + b2^2 + b3^3 +...) >= (a1*b1 + a2*b2 + a3*b3 + ...)^2

(a^2 + b^2 + c^2 + d^2)*(2^2 + 3^2 + 6^2 + 24^2) >= (2a + 3b + 6c + 24d)^2

100*625 >= (2a + 3b + 6c + 24d)^2


----------------------------------------------------------- Quant Answer # 103 -----------------------------------------------------------

Let the number of students who arrive early, on time and late be a, b, c respectively. Let the number of students who depart early, on time and late be d, e, f respectively. => d = 1/3*b + 1/4*a, e = 2/3*b + 1/2*a + 1/2*c, f = 1/4*a + 1/2*c, d = 1/2*a, e = 2*b - 4, f = c - 1 solving we get a = 20, b = 15, c = 12, d = 10, e = 26, f = 11


Solution by ashish banaya:

Let the no. of people who arrived early be 4x, arrived on time be 3y arrived late be 2z ...................Early........ On time........... Late Arrival ...........4x............. 3y................. 2z

dep. ............x+y.......... 2y+2x+z ..........x+z The 3 eqns. read like this x+y=2x............(1) 2y+2x+z=6y-4...(2) y+z=2z-1............(3) Equate & we get x=y=5 & z=6.........So ans. is (C) 47

----------------------------------------------------------- Quant Answer # 104 -----------------------------------------------------------

The overlap is a hexagon, and there are six congruent triangles outside the hexagon, each similar to the original triangle and 1/3 the (linear) size (because the centroid divides the median in the ratio 2:1). So the area of the union is 4/3 times the area of the triangle. We have the semiperimeter s = 21, so by Heron's formula the triangle has area (21·8·7·6)^1/2 = 84.


----------------------------------------------------------- Quant Answer # 105 -----------------------------------------------------------

Put QP = 2a, take perpendicular distance A, B from QP be b, c. Take line parallel to QP through B meeting perpendicular bisector of QP at D. Considering triangle ADB gives 4a^2 + (b-c)^2 = 144. Also b^2 = 64 - a^2, c^2 = 36 - a^2. Hence 4a^2 = 130.


----------------------------------------------------------- Quant Answer # 106 ----------------------------------------------------------- Solution by raul_v20: even i am getting {1,2,3,5,8} u can proceed like this 1,2 1+2 = 3 u can take 3 since sum >3 3+2=5 thus 4 can't be used 5+3=8 thus 6,7 cant be used 8+5=13 which is out of the set thus B-5 is d correct answer to me Concept – Fibinacci series

----------------------------------------------------------- Quant Answer # 107 -----------------------------------------------------------

Put n = a + k, where a = [n], and 0 ≤ k < 1. Then the equation is 2ak = [2ak + k2]. The rhs is an integer, so 2ak must be an integer, and that is evidently a sufficient condition. So k = 0, 1/2a, 2/2a, ... , (2a-1)/2a. In other words, there are 2 solutions for a = 1 (namely 1 and 1 1/2), 4 solutions for a = 2 (namely 2, 2 1/4, 2 1/2, 2 3/4), 6 solutions for a = 3, ... , 2(m-1) solutions for a = m-1. Hence there are 2(1 + 2 + ... + m-1) + 1 = m(m-1) + 1 solutions in the range 1 ≤ x ≤ m.


Solution by vaibhav_mba: If we take decimal part as x and number part as a then n=a+x so the eq will change to : a^2 +x^2 +2ax-[a^2+x^2+2ax]=x^2 bcoz (n-[n])^2= (a+x-a)^2=x^2 this will be possible only when a^2+2ax will form a natural no. and x^2 will be decimal part bcoz then [a^2+x^2+2ax]= a^2+2ax also a^2 will always be natural no. so we have to find cases where 2ax will be natural no. for a=1 x=1/2 1 case a=2 x=1/4,2/4,3/4 3 cases a=3 x=1/6,2/6 .......5/6 5 cases a=4 x=1/8,2/8........7/8 7 cases total 16 cases also x=0 will suffice for a = 1 to 5 5 more cases so total 16 + 5 = 21 cases

----------------------------------------------------------- Quant Answer # 108 -----------------------------------------------------------

Let P(a) = ab + b -a -1, then numerical value of P(1)*P(2)*P(3)*...*P(100) where b = (a^2 + 3a +1)/(a^2 + 2a +1) is P(a) = ab + b -a -1 = (a+1)*(b-1) also b = 1 + a/(a+1)^2, thus P(a) = a/(a+1) product is 1/2*2/3*3/4*...*100/101


----------------------------------------------------------- Quant Answer # 109 -----------------------------------------------------------

Let AC and BD be the 2 flagposts of heights 10*(3^1/2) and 20*(3^1/2) respectively. Let EF be on AB such that EF = x. Since angle of elevation is >= 60 degrees => AF = 5, BE = 10 => EF = 5 => chance = EF/AB = 1/2


----------------------------------------------------------- Quant Answer # 110 -----------------------------------------------------------

Sum of such rationals < 1 is (1+7+11+13+17+19+23+29)/30 = 4. The rationals between 1 and 2 are all 1 larger, so sum 12. Similarly, those between 2 and 3 have sum 20, so total sum = 4 + 12 + 20 + ... + 76 = 400. Hence, choice (b) is the right answer

Solution by warrior:

in 1 to 30 only 1,7,11,13,17,19,23, 29 can't be reduced whenb divided by 30. rest all other ( 2,3,5,.... are multiple of 30 so cancel all no <=30 which are divisible by them) also if we add them 1,7,11,13,17,19,23, 29 we get 120 so 120/30=4 in the next series of 31 to 60 we need to add 30 to 1,7,11,13,17,19,23, 29 so sum for thsi series is 4+8 ( there are 8 no in total). so same will follow for other 30 no groups so we have AP with 1st term =4 and d=8 and n=10 so sum of ap = 10/2(8+72)=400

Solution by mansoor:

p can take all values which are coprime to 300. Sum of coprimes to 300 = (300/2) * Totient Function (300) = 300/2 * 80 sum of p/30's = 400 So, Choice (b).

----------------------------------------------------------- Quant Answer # 111 -----------------------------------------------------------

Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so B and E are the same perpendicular distance from CD, so BE is parallel to CD. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at P. Then ABPE is a parallelogram, so area BPE = area EAB = 1. Also area CDP + area EDP = area CDP + area BCP = 1. Put area EDP = a. Then DP/PB = area EDP/area BPE = a/1 and also = area CDP/area

BCP = (1-a)/a. So a^2 + a - 1 = 0, a = (5^1/2 - 1)/2 . Hence area ABCDE = 3 + a = (5^1/2 + 5)/2.


----------------------------------------------------------- Quant Answer # 112 -----------------------------------------------------------

200 shirts work is same as 150 pair of trousers work. => Total work to be done is for 310 pair of trousers. On first day man works for 3 hours while woman works for 2 hours => work on 24 pairs of trousers get finished Every subsequent day work on 54 trousers get finished. 24 + 5*54 = 294 => work will get over on the 7th day itself Check yourself that it is the washerman who washed the last trouser.


----------------------------------------------------------- Quant Answer # 113 -----------------------------------------------------------

1411 = 83*17 or else why would a weird figure as 1411 appear :-)

Remember the rule: a^f(n) leaves remainder of 1 on division by n when a and n are co-prime, f(n) is the number of co-primes that less than n.

=> 17^82 == 1 (83) => 17^83 == 17 (83*17)

also 83^16 == 1 (17) => 83^17 == 83 (83*17)


----------------------------------------------------------- Quant Answer # 114 -----------------------------------------------------------

No. of combinations with 4 Alike (Identical) Results = 6C1 = 6 No. of combinations with 3 Alike and 1 Non Alike = 6C1*5C1 = 30 No. of combinations with 2 Alike and 2 Non Alike = 6C1*5C2 = 60 No. of combinations with 2 Alike of 1 Kind and 2 Alike of another kind i.e.1122, 2233 etc = 6C2 = 15 No. of combinations with all 4 different = 6C4 = 15

In general if we have N dices which are r-faced, distinct enumerations such as what we want in our question would be (N+r-1)C(r-1).


----------------------------------------------------------- Quant Answer # 115 -----------------------------------------------------------

We start of with 2*3*...*11*12*23 because we need to use all the smallest integers because every integer greater than 3 can be replaced with a sum of two others whose product is greater or equal than that integer .i.e. 4 = 2+2 <=2*2 5 = 2+3 < 2*3 = 6 so this means we should just decrease the number 23 untill it gets to 14 and increase all the other numbers by 1. Then we get the solution 2*3*...*10*12*13*14 or shortly 14!/4


----------------------------------------------------------- Quant Answer # 116 -----------------------------------------------------------

Let f(n) = 4n^2 - 40[n] + 51. For n <= 1 f(n) > 51, so there are no solutions with x < 1. f(1) = 15, f(2-) = 16 - 40 + 51 = 27, so there are no solutions in the interval [1, 2). f(2) = -13, f(3-) = 7, so we expect a solution in [2, 3). For all n in this interval we have f(n) = 4n^2 - 29, so the unique solution is 2n = 29^1/2. f(3) = -33, f(4-) = -5, so there are no solutions in [3, 4). f(4) = -45, f(5-) = -9, so there are no solutions in [4, 5). f(5) = -49, f(6-) = -5, so there are no solutions in [5, 6). f(6) = -45, f(7-) = 7, so we expect a solution in [6, 7). For all n in this interval we have f(n) = 4n^2 - 189, so the unique solution is 2n = 189^1/2 f(7) = -33, f(8-) = 27. For all n in the interval [7, 8) we have f(n) = 4n^2 - 229, so the unique solution is 2n = 229^1/2 f(8) = -13, f(9-) = 55. For all n in the interval [8, 9) we have f(n) = 4n^2 - 269, so the unique solution is 2n = 269^1/2 f(9) = 15. f(10-) = 91, so there are no solutions in [9, 10). For n >= 10, we have 4n^2 - 40[n] + 51 = 4n^2 - 40n + 51 = 4n(n - 10) + 51 > 0, so there are no solutions.


----------------------------------------------------------- Quant Answer # 117 -----------------------------------------------------------

Let A, B, M, O, N, C be the points in the same order and AB = x miles. Let the speeds of X, Y, Z be a, b and c respectively. => ON = 100 - 2x Ratio of speeds of X and Y = a/b = AM/BM = (20+x)/20 Ratio of speeds of X and Z = a/c = (MO + ON)/NC = (120 - 2x)/x Ratio of speeds of Y and Z is b/c. => ((120-2x)/x)/((20+x)/20) = MO/OC = 20/(100-x) => x = 40 miles as x < AC and x=300 is ruled out from the quadratic x^2 - 340x + 12000 = 0 Hence, choice (d) is the right answer

----------------------------------------------------------- Quant Answer # 118 -----------------------------------------------------------

We need to understand the symmetry in this problem, let, XY=WZ=b and AX=BY=CZ=DW=a hence,b=2a+19 Also area of ABCD =4 area(XPQY) =>19(2a+b)=4*1/2*19/2 *(87+b) = > 2a=87 henc b=87+19=106 hence AB=2a+b=87+106=193 units.


----------------------------------------------------------- Quant Answer # 119 -----------------------------------------------------------

The figure is a kite. The area is 1/2*(product of the diagonals).

The co-ordinates of the vertices are (48,0),(60,15),(80,0),(60,-15).


----------------------------------------------------------- Quant Answer # 120 -----------------------------------------------------------

Let N be a positive integer satisfying the condition and let n be the largest integer not exceeding its cube root. If n = 7, then 3·4·5·7 = 420 must divide N. But N cannot exceed 8^3 - 1 = 511, so the largest such N is 420. If n = 8, then 3·8·5·7 = 840 divides N, so N > 729 = 9^3. Hence 9 divides N, and hence 3·840 = 2520 divides N. But we show that no N > 2000 can satisfy the condition.

Note that 2(x - 1)^3 > x^3 for any x > 4. Hence [x]^3 > x^3/2 for x > 4. So certainly if N > 2000, we have n^3 > N/2. Now let p(k) be the highest power of k which does not exceed n. Then p(k) > n/k. Hence p(2)p(3)p(5) > n^3/30 > N/60. But since N > 2000, we have 7 < 11 < n and hence p^2, p^3, p^5, 7, 11 are all <= n. But 77 p^2p^3p^5 > N, so N cannot satisfy the condition.

420 = 4*3*5*7


----------------------------------------------------------- Quant Answer # 121 -----------------------------------------------------------

The key is to cut the cone along VC and unroll it to give the sector of a circle. If the sector is bounded by VC and VC' and the arc CC', then evidently the particle travels along the straight line CC'. It is closest to V when it is at the midpoint M. The arc CC' has length 2c, so the <CVC' is 2c/3. Hence <VCC' = p/6, so VM = 3/2.


----------------------------------------------------------- Quant Answer # 122 -----------------------------------------------------------

Sanjog's ratio was 8/15 on day 1, 7/10 on day 2, and 8/15 < 7/10. We want Rahul's ratio to be weighted towards day 2, so we take 1/2 on day 1 and 348/498 on day 2 (we must have a total of 500 attempted), giving 349/500 overall. Hence, choice (b) is the right answer

Solution by ankur483:

Here the problem boils down to these things :- 1. Maximize the marks obtained by rahul under the given constraints. a. Total questions attempted should be 500 b. sucess rate for the first day should be less than 53.33 % c. sucess rate for the second day should be less than 78 % d. Both days he get atleast 1 question right (+ive integer) Now from here if we want to obtain max marks we have to attemp the Max number of questions on the 2nd day as the sucess rate is 70%. So we keep the least number of questions that should be right on 1st day as 1 . now sucess rate should be less than 53.33 so questions attemted are 2 (one correct n one wrong) now he attemted 498 question on the 2nd day . 70 % of 498 = 348.6 now we can keep 348 so that sucess rate for the 2nd day is less than 70%. thus we get max total of 349 for 2 days......

----------------------------------------------------------- Quant Answer # 123 -----------------------------------------------------------

lhs >= 0, so n^2 + n >= 30, so n >= 5 or <= -6. Hence n^2-n >= 20, so we do not need the absolute value signs and lhs = n^2-n-15. Hence n = 15/2


----------------------------------------------------------- Quant Answer # 124 -----------------------------------------------------------

ABC15 gives us 300 numbers. AB15C and A15BC gives us 60 each. 15ABC gives us 33 for c=5 and 34 for c=0. From this we subtract cases in which there are more than one block of 15 i.e. 8 cases.


My solution :

Here is the explaination -: Before that just see that 15=5*3. Hence we need to look for divisibility by 5 and 3. Case 1: - - - 1 5 For the first 3 digits, thier sum should be multiple of 3. 300 such nos. are possible. (from 102 to 999) Case 2: - - 1 5 - a) - - 1 5 0 Sum of first 2 digits should be multiple of 3. 30 such nos. are there. b) - - 1 5 5 Sum of first 2 digits + 5 should be multiple of 3 31 such nos. are there So, total nos. for case 2 are 61 Case 3: - 1 5 - - a) - 1 5 - 0 Sum of the remaining 2 digits should be multiple of 3 30 such nos. are there b) - 1 5 - 5 Sum of the remaining 2 digits should be multiple of 3 30 such nos. are there but out of which 3 are already counted in Case 1 which are 31515,61515 and 91515. Hence, only 27 nos. are for this case So, total nos. for case 3 are 57

Case 4: 1 5 - - - a) 1 5 - - 0 Sum of the remaining 2 digits should be multiple of 3. 34 such nos are there bcoz we can use the digit 0 in both the places but out of these nos. one number has been counted in Case 2a which is 15150. Thus, 33 possible nos. are there. b) 1 5 - - 5 Sum of the remaining 2 digits + 5 should be multiple of 3. 28 such nos. are there Hence, total nos. for case 4 are 61. Sum of all the cases comes out to be 479.

----------------------------------------------------------- Quant Answer # 125 -----------------------------------------------------------

200 = 2^3*5^2 => sum of divisors of 200 is (1+2+2^2+2^3)*(1+5+5^2) = 15*31

We want all positive integers less than 200 such that their number of divisors is either 1, 3, 5 or 15

1 has only 1 divisor. All perfect squares of primes have number of divisors as 3 => 6 such numbers excluding 1. Only numbers(less than 200) 1+2+2^2+2^3+2^4+2^5 = 63 or 121(1+3+3^2+3^3+3^4+3^5) have 5 divisors.

144 has 15 divisors.

=> In all 10 such numbers satisfy the criteria.


----------------------------------------------------------- Quant Answer # 126 -----------------------------------------------------------

By AM-GM rule, x^8 + y^8 + 1 + 1 + 1 + 1 + 1 + 1 >= 8xy and the equality occurs when x=y.

=> x^8 = 4x^2 - 3, put x^2 = p we get p^4 - 4p + 3 = (p-1)^2*(a quadratic with complex roots) => x^2=1.


----------------------------------------------------------- Quant Answer # 127 -----------------------------------------------------------

Let the number of pages in the novel be "n". Now 1+2+....+n=n(n+1)/2 >15,000. So n > 172 (after manipulations). The sum of numbers of the pages torn attain a maximum value of (n-1) + n = 2n-1. So n(n+1)/2 -(2n-1) <=15,000. So n < 176 after manipulations. so n =173 or 174 or 175. For n=173 ,n(n+1)/2 = 15,051. So sum of pages on torn leaf = 15,051 -15,000 =51= 2n-1. So n =26. For n = 174, n(n+1)/2 = 15,225. Here sum of pages on torn leaf = 15,225 -15,000=225 = 2n-1.So n = 112. For n = 175, n(n+1)/2 = 15,400 So sum of pages on torn leaf = 15,400 - 15,000 = 400 = 2n-1 implying n = 200.5 which is not possible. Therefore the permissible values of the page numbers on the leaf are 25,26 and 112,113.


----------------------------------------------------------- Quant Answer # 128 -----------------------------------------------------------

Put <R = 5x, <PRC = 2x, since AB=AR => <ABR = 2x...since AR=CR => <RAC = <RCA = 90 - x. <Q/2 = 90 -4x, <PAB = 4x.

Now AB/PA = AR/PA = QR/PQ (angle bisector theorem) but QR/PQ = sinP/sinR, AB/AP = sinP/sinPBA => <PBA = 5x.

In triangle PAB 3x + 4x + 5x = 180 => x = 15 degrees.

Hence, choice (b) is the right answer ----------------------------------------------------------- Quant Answer # 129 -----------------------------------------------------------

Not available

----------------------------------------------------------- Quant Answer # 130 ----------------------------------------------------------- For the first course number of admitted students = x/2 For the seconfiltered= 2x/3 -1/2(x/2) = 5x/12

For the third course number of admitted students = 3x/4 -1/2(2x/3) = 5x/12 For the fourth course number of admitted students = (x-1) -1/2(3x/4) = 5x/8 -1 Overall, PagalGuy trained 47x/24 - 1 students => x = 24 Hence, choice (a) is the right answer. ----------------------------------------------------------- Quant Answer # 131 ----------------------------------------------------------- Each pair of 0, 5, 12 differ by 5, 7 or 12, so f(0), f(5), f(12) must all be different, sonfiltered= 3. We can exhibit an f with n = 4. Define f(m) = 1 for m = 1, 3, 5, 7, 9, 11 (mod 24), f(m) = 2 for m = 2, 4, 6, 8, 10, 12 (mod 24), f(m) = 3 for m = 13, 15, 17, 19, 21, 23 (mod 24), f(m) = 4 for m = 14, 16, 18, 20, 22, 0 (mod 24). Hence, choice (a) is the right answer ----------------------------------------------------------- Quant Answer # 132 -----------------------------------------------------------

http://www.cut-the-knot.org/Generalization/Euler.shtml => 22^24 = 1 mod(45), if 22^23 = x mod(45) => 22x = 1 mod(45) => x = -2+45

10^2 = 10 mod(45) => 10^35 = 10^18 mod(45)...continuing 10^35 = 10 mod(45)


----------------------------------------------------------- Quant Answer # 133 ----------------------------------------------------------- A student can answer in following ways from each section (A, B, C) - (6,7,4), (4,7,6), (5,7,5) Numbers of ways of making selection is 8C6*7C7*6C4 + 8C4*7C7*6C6 + 8C5*7C7*6C5 = 28*15 + 70 + 56*6 = 826 ways Hence, choice (d) is the right answer ----------------------------------------------------------- Quant Answer # 134 ----------------------------------------------------------- Call stories with an odd number of pages odd stories and stories with an even number of pages even stories. There are 15 odd stories and 15 even stories.

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The odd stories change the parity of the starting page (in the sense that the following story starts on a page of opposite parity), whereas the even stories do not. So the odd stories must start alternately on odd and even pages. Hence 8 of them must start on odd pages and 7 on even pages (irrespective of how the stories are arranged). We can, however, control the even stories. In particular, if we put each of them after an even number of odd stories, then they will all begin on odd pages. For example, we could put them all first (before any of the odd stories). Hence, choice (b) is the right answer

----------------------------------------------------------- Quant Answer # 135 -----------------------------------------------------------

Let initially milk and water be 4x, 5x. Let the rate of evaporation be 2y and 3y respectively per minute. After 90/7 minutes the ratio of milk and water is reveresed => y/x = 1/10 Now it's 1/2 after n minutes => (4x - 2xn/10)/(5x - 3xn/10) = 1 => n = 10 Hence, choice (b) is the right answer Solution by sidhesh: My answer Option B My eqns ... let initial ratio be 4x:5x ... and reqd time be t and r o evap be r (4x-rt)/(5x-1.5rt)=1/1 (4x-r*90/7)/(5x-1.5*r*90/7)=5/4 solving t=10 ... so addtnal time =20/7 ----------------------------------------------------------- Quant Answer # 136 ----------------------------------------------------------- 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 => they meet in the mid-way as they have same speed. ----------------------------------------------------------- Quant Answer # 137 ----------------------------------------------------------- Let N be the smallest number of calls such that after they have been made at least one person knows all the news. Then N=7, because each of the other 7 people must make at least one call, otherwise no one but them knows their news. After 7 calls only one person can know all the news, because otherwise at least one person would

have known all the news before the 7th call and 7 would not be minimal. So at least a further 7 calls are needed, one to each of the other 7 people. So, at least 14 calls are needed in all. Hence, choice (b) is the right answer. ----------------------------------------------------------- Quant Answer # 138 -----------------------------------------------------------

Let Chetna has N marbles and Apple has 3N marbles, and n be the no. of bags

=> N = nQ1 + 31, where Q1 is the number of marbles per bag

=> 4N = 4nQ1 + 124 = nQ2 + 16 = n(4Q1 + 108/n) + 16, where Q2 is the number of marbles per bag

=> n = 36, 54 or 108 as n > 31.

=> Number of marbles with Apple are 3*(36*8 + 31) = 957

957 = 3*11*29


----------------------------------------------------------- Quant Answer # 139 -----------------------------------------------------------

n is divisible by 8 => n = 8k => k is odd

25n/8 - 211 > 0 also 25n/8 - 211 < n => k=9, 11 are the only possibility.


----------------------------------------------------------- Quant Answer # 140 -----------------------------------------------------------

P must be the opposite side of AB to C (or we could increase CP, whilst keeping AP and BP the same, by reflecting in AB). Similarly it must be on the same side of AC as B, and on the same side of BC as A. For any P in this region the quadrilateral APBC is convex and hence satisfies Ptolemy's inequality CP·AB = AP·BC + BP·AC, with equality iff APBC is cyclic. But AB = BC = CA, so we have CP = AP + BP = 5 with equality iff P lies on the arc AB of the circle ABC. Note that there is just one such point, because the locus of P such that BP=1.5 AP is a circle which cuts the arc just once. Hence, choice (c) is the right answer.

----------------------------------------------------------- Quant Answer # 141 ----------------------------------------------------------- First read multiples of 2 and 5 (up to 400). Then since 399 = 3·7·19, read remaining multiples of 3, 7 and 19 up to 399. Then read 397 (a prime). Then read remaining multiples of 17 and 23 up to 391 (=17·23). Now all primes up to 23 have been covered except 11 and 13. Now the largest non-prime number not yet read must be divisible by 11 or 13 (since all multiples of smaller primes have already been read and 292 > 400), so the candidates are 377 = 13·29 and 341 = 11·31. So we keep reading primes until we get down to 377 and then read all remaining multiples of 13. We then read primes until we get down to 341, and then read all remaining multiples of 11. At this point the only remaining numbers are primes > 31. So we read them in reverse order, the last being 37.


----------------------------------------------------------- Quant Answer # 142 -----------------------------------------------------------

AE*AF = AB^2, AD*AC = (AB/2)*(2AB) = AB^2 => AE*AF = AD*AC => triangles ADE and AFC are similar. Thus, <(AED) = <(ACF) => DECF is cyclic. Since M is the perpendicular bisector of DE and CF, it must be the circumcentre of DEFC.=> M lies on the perpendicular bisector of CD. => AM/MC = 5/3 Hence, choice (c) is the right answer. Solution by santosh_s: let AC = 4x then AB = 2x, AD = x we know that AE*AF = (AB)^2 => AE*AF = 4*x^2 -------- (1) now, AD*AC = x*4x = 4*x^2 ----------- (2) so, from the above two expressions, we have AE*AF = AD*AC => AE/AC = AD/AF in triangles AED and ACF, angle A is common and the sides containing A are in proportion. So, the triangles AED and ACF are similar. looking at quadrilateral EDCF, (angle E + angle C) = 180 degrees and (angle D + angle F) = 180 degrees. hence its a cyclic quad. given that perpendicular bisectors of ED and CF intersect in M, hence M is mid point of DC. we have, DC = 3x so, MC = 3x/2 and AM = 4x- 3x/2 = 5x/2 hence we have the result, AM/MC = 5/3

----------------------------------------------------------- Quant Answer # 143 -----------------------------------------------------------

Let t = abc. Then a, b, c are the roots (= zeros) of f(x) = x^3 - 6x^2 + 9x - t = 0. We have f'(x) = 3(x - 1)(x - 3) so f(3) < 0 < f(1). Hence a < 1 0, t = abc < 4, and since f(3) = -t < 0, abc > 0. As b, c are positive, a > 0. Also, f(x) is strictly increasing on (3, +oo) and f(4) = f(1) = 4 - p > 0 so c < 4. Hence, choice (d) is the right answer. Solution by twanger: Firstly, b+c = 6-a ,and bc = 9-a(b+c) = 9-a(6-a) on applying to (b-c)^2 > 0 gives a > 0 Now since we know all the numbers are +ve , we can apply AM > GM to get abc < 8 in which case (1/a) + (1/b) + (1/c) > 1 so that at least one of them is less than one - which means the smallest number a certainly has to be < 1 Now abc < 8 and bc = (3-a) ^ 2 implies 8 < bc < 9 and 5 < b+c < 6. b+c < 6, b< c implies b < 3,c>3 (b-c)^2 from the above two eqns is < 4 so that c-b > 2 so 1 3) Both options are true, so only (d) can be the possible answer. ----------------------------------------------------------- Quant Answer # 144 ----------------------------------------------------------- a+b+c+d = 8-e, also a^2 + b^2 + c^2 + d^2 = 16 - e^2 (a^2 + b^2 + c^2 + d^2)(1+1+1+1) >= (a+b+c+d)^2 => 4*(16 - e^2) >= (8-e)^2 => e <= 16/5Hence, choice (a) is the right answer.

----------------------------------------------------------- Quant Answer # 145 -----------------------------------------------------------

We want to find the intersection between the area defined by this inequality and the line x+y = -1. It will be easier to do this if we rotate the equations by 45 degrees, and the length of the line segment will remain the same. We find x^2 - y^2 >= 1/3=> x = -(1/2)^1/2, the line segment has end opints (-(1/2)^1/2, +/-(1/6)^1/2)Hence, choice (b) is the right answer. Solution by prade: x^3 - 3xy^2 >= 3x^2*y - y^3 (x+y) (x^2-4xy+y^2) >=0 (x^2-4xy+y^2) <=0

put x = -1-y 6x^2 + 6x + 1 <= 0 (solve this to get corner values x1 = (-3+sqrt(3))/6 and x2 = (-3-sqrt(3))/6 x1-x2 = 1/sqrt(3) y1-y2 = -(1+x1) + (1+x2) = -1/sqrt(3) length of the line segment = sqrt((x1-x2)^2 + (y1-y2)^2) now A = x1-x2 = 1/sqrt(3) where x1 and x2 are the roots and B= y1-y2 = 1/sqrt(3) length of the line segment= sqrt(A^2 + B^2) = sqrt(2/3) Solution by nagi_govi3: x^3+y^3+3xy(x+y) = (x+y)^3 so given equations are (x+y)^3-6xy(x+y) >=0 and x+y = -1 6xy>=1 and x+y = -1 y= -1-x 6.x(-1-x) >=1 6x(1+x)<=-1 6x^2+6x+1<=0 so x= (-3+sqrt3)/6 and (-3-sqrt3)/6 y is (2-sqrt3)/6 and (2+sqrt3)/6 distance will become 2sqt6/6 = sqrt(2/3)

----------------------------------------------------------- Quant Answer # 146 -----------------------------------------------------------

Let the speed of the child who delivers the message be x m/min. When the child travels from first to last he is moving in opposite direction of the group, and when he is coming back he is coming along with the group. => 300/(x+10) + 300/(x-10) = 1080/x => x = 15Hence, choice (c) is the right answer.

----------------------------------------------------------- Quant Answer # 147 -----------------------------------------------------------

Let the line parallel to BE through D meet AC at G. Then DCG, BCE are similar and BC = 2 DC, so BE = 2 DG. Hence AD = DG, so <(DGA) = <(DAG) = 60o. FE is parallel to DG, so <(FEA) = 60o.

Hence, choice (b) is the right answer. Solution by amar_kashyap: Draw DM parallel to BE from D to M on AC. So DM= BE/2=AD. IN triangle ADM , AD=DM and <FAE=60 so ,<MDA=<AMD=60. Since BE is parallel to DM,<FEA= <AMD=60 Hence 60 degree is the answer..

----------------------------------------------------------- Quant Answer # 148 -----------------------------------------------------------

It is convenient to put X = x-2, Y = y-2, Z = z-2, so Y = 2 - X^2, Z = 2 - Y^2, X = 2 - Z^2. Substituting gives X = 2 - Z^2 = -2 + 4Y^2 - Y^4 = -2 + 4(2 - X^2)^2 - (2 - X^2)^4 = -(X^8 - 8X^6 + 20X^4 - 16X^2 + 2). Taking out the factors X = 1, -2 already noticed we get: (X - 1)(X + 2)(X^6 - X^5 - 5X^4 + 3X^3 + 7X^2 - X - 1) = 0. we get (X^3 - 3X - 1)(X^3 - X^2 - 2X + 1).Note that X, Y, Z satisfy the same equation. So we hope that if X satisfies X^3 - 3X - 1 = 0, then Q and R satisfy the same cubic (rather than the other one). Substituting X = 2 - Z^2 gives Z^6 - 6Z^4 + 9Z^2 - 1 = 0, which factorises as (Z^3 - 3Z - 1)(Z^3 - 3Z + 1) = 0, so if Z does not satisfy the same cubic as X, then it satisfies Z^3 - 3Z + 1 = 0. But then it also satisfies the cubic Z^3 - Z^2 - 2Z + 1 = 0, so subtracting it satisfies Z^2 - Z = 0, so Z = 0 or 1. Those do not satisfy Z^3 - 3Z + 1 = 0. Hence Z does satisfy the same cubic as X. Similarly so does Y. Note that we cannot have X = Y because then X^2 + X - 2 = 0, so X = 1 or -2 and it is easy to check that they do not satisfy X^3 - 3X - 1 = 0. Similarly for the Y = Z or Z = X. Thus the roots of X^3 - 3X - 1 = 0 are all distinct.We can check this more directly by noting that at -2 it is negative, at -1 positive, at 0 negative, at 2 positive. Hence there is a root in each of the intervals (-2,-1), (-1,0) and (0,2). We have X + Y + Z = 0. Hence x+y+z = 6.The three roots of the other equation must form another possible X, Y, Z family. Again it is easy to check directly that the cubic is negative at -2, positive at -1, positive at 0, negative at 1 and positive at 2, so it has a root in each of (-2,-1), (0,1) and (1,2). X+Y+Z = 1, so x + y + z = 7.Hence, choice (b) and (c) is the right answer.

----------------------------------------------------------- Quant Answer # 149 -----------------------------------------------------------

loga + logb + logc = 6log6 is equivalent to abc = 6^6. a, b, c GP is equivalent to b^2 = ac, so b^3 = 6^6, and b = 36. b-a is square, so a = 27. Hence c = 48 Hence, choice (c) is the right answer.

----------------------------------------------------------- Quant Answer # 150 -----------------------------------------------------------

Let the numbers be a(1), a(2), a(3), ..., a(100), a(101), a(102), ... where a(101) = a(1) and a(102) = a(2).

a(n) + a(n+1) + ... + a(n+7) = a(n+1) + a(n+2) + ... + a(n+8) => a(n) = a(n+8) => a(1) = a(9) = a(105) = a(5) => a(n) = a(n+4) => 2*(a(1) + a(2) + a(3) + a(4)) = 25/2 => a(50) = a(2) = 4 Hence, choice (a) is the right answer.


(151-200)

----------------------------------------------------------- Quant Answer # 151 -----------------------------------------------------------

Let the roots be m and n and the leading coefficient k, then the quadratic is k(x^2 - (m + n)x + mn). The sum of its coefficients is k(m - 1)(n - 1). We are told this is prime, hence k = 1 and m = 2. So the quadratic is (x - 2)(x - n) with n > 2 an integer. We have (x - 2)(x - n) = -55 for some x. But the only factors of 55 are 1, 5, 11, 55, so x = 3, n = 58, or x = 7, n = 18. But n -1 is prime, so on filtered= 18. Hence, choice (a) is the right answer. Solution by amar_kashyap: Let the roots be a and b So the quadratic is (x-a)(x-b) We have 1-a-b+ab=P(a prime number) b(a-1)-(a-1)=P (b-1)(a-1)=P a-1=-1 or 1 ie a=2 or 0 But roots are postive inetgers so none of a or b can be =0 So one of a or b=2 and other root= P+1 So the eqaution is (x-2)(x-p-1) which is equal to -55 for one of the values of x for p=17 and x=13 we have 11. (-5)=-55. So among given options the sum of the roots is 20

----------------------------------------------------------- Quant Answer # 152 -----------------------------------------------------------

The first derivative of f(x) is positive for all x => f(x) is strictly increasing function. It's easy to see that if f(a) = 16 then a > 14. Similarly when f(b) = 20 b > 14.Hence, choice (d) is the right answer.

----------------------------------------------------------- Quant Answer # 153 -----------------------------------------------------------

x is mutiple of y say x=ky we have ky. y(k+1).y(k+2).....y(k+n-1)=y^n.k.(k+1).(k+2)...(k+n-1)

product of any n consecutive numbers is divisible by n! so x is amutip-le of y..


----------------------------------------------------------- Quant Answer # 154 -----------------------------------------------------------

Cube the given _expression x and we get 10 - 9x = x^3, onfiltered= 1 is the only real solution.


----------------------------------------------------------- Quant Answer # 155 -----------------------------------------------------------

16^3 < 4321 < 67^2 => the base of the 3 digit number is greater than 16 and less than 67. Now let the base be b and the number be mn(21-m-n) => m*(b^2-1) + n(b-1) = 4300 = 43*4*25, => b-1 divides 4300, we get b=26 and 44. b=51 doesn't give us any desired number. For b = 26 we have 27m+n = 172 => m = 6, n = 10. The number in base 26 is 6105, similarly in base 44 it's 2109.

Hence, choice (b) is the right answer.

----------------------------------------------------------- Quant Answer # 156 -----------------------------------------------------------

There are 625 points (m,n) such that 1 <= m,n <= 25. Out of these 2m < n and 2n < m have to be excluded. The number of points satisfying 2n < m is 0 for m=1,2; 1 for m=3,4;...;11 for m=23, 24; 12 for m=25

=> total points are 2(1+2+3+ ... + 11) + 12 = 144 By symmetry 2m < n also gives 144 points.

=> in all 625 - 2*144 = 337 pairs satisfy m <= 2n <= 50, n <= 2m <= 50.

----------------------------------------------------------- Quant Answer # 157 -----------------------------------------------------------

Consider all points A' such that FA' = 3. They lie on a circle center F. The area of A'PQ is PQ/2 times the distance of A' from PQ. That distance is maximal for A'F perpendicular to PQ (because the distance is the distance of F from PQ + FA' sin(x), where x is the angle between A'F and PQ). Hence FP must be perpendicular to PQ. Similarly QF must be perpendicular to PR, so F must be the orthocenter.


----------------------------------------------------------- Quant Answer # 158 -----------------------------------------------------------

Harry's chance of winning is [area ( 25 <= x+y <= 35)/ area ( 20 <= x+y <= 50)]. Probability of Harry winning is (35^2 - 25^2)/(50^2 - 20^2) = 2/7 => Probability of milkman winning is 5/7.


----------------------------------------------------------- Quant Answer # 159 -----------------------------------------------------------

AC.BD=AD.BC+AB.CD (Ptolemy's theroem)

15^1/2. (16-x^2)^1/2= x+4 where CD=x, (BD = 15^1/2 as ABD is right angled at B)

16x^2+8x-224=0 => (2x-7)(x+4)=0


----------------------------------------------------------- Quant Answer # 160 -----------------------------------------------------------

5 similar balls can be placed in 3 cells in (5+3-1)C(3-1) ways. 4 similar balls can be placed in 3 cells in (4+3-1)C(3-1) ways. Hence, total 7C2*6C2 ways are possible.


----------------------------------------------------------- Quant Answer # 161 -----------------------------------------------------------

Let p, q be the roots of 1st equation and r, s of the 2nd.

=> (pq+rs)/2 = a+b-3 = p+q+r+s-3

=> (p-2)(q-2) + (r-2)(s-2) = 2 (This transformation is the key!!!)

and we have pq=rs. Thus, the solution set consists of (p, q, r, s) = (1,1,1,1), (3,3,3,3), (2,6,3,4).

Thus (a, b) is (2, 2) (8, 7 ) and (6, 6). In all 4 ordered pairs.


----------------------------------------------------------- Quant Answer # 162 -----------------------------------------------------------

John can drive along the axis and then head across the desert in a straight line. If he starts straight across the desert it can reach anywhere in a circle radius 1.4. If it drives 5 along the road, the circle has radius 0. For points in between the circle is scaled proportionately. So the envelope is the four-pointed star. This is tough one but try to visualise.

The part in the first quadrant is the difference between an isosceles right-angled triangle hypoteneuse 50^1/2 and an obtuse-angled triangle with the same base.

The height of the second triangle is 5/2^1/2 *( tan(45deg - x)), where tan x = 7/24 (the right-angled triangle has sides

5, 1.4 and hence 4.8 - it is a 7, 24, 25 triangle). So tan(45deg - x) = 17/31. Hence total area = 700/31.


----------------------------------------------------------- Quant Answer # 163 -----------------------------------------------------------

In CAT the questions that appear lengthy to read actually turn out be sitter most of the times.

Choice IV can be seen from the question itself, checking for the rest 3 options we conclude that all the statements are true.

Hence, choice (d) is the right answer.

----------------------------------------------------------- Quant Answer # 164 -----------------------------------------------------------

1,2,3,4 1,5,6,7 1,8,9,10 1,11,12,13 2,5,8,11 2,6,9,12 3,5,9,11 3,6,8,12 4,5,10,12 4,6,11,9


----------------------------------------------------------- Quant Answer # 165 -----------------------------------------------------------

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 + (2p-4+2q-4)5 + (p-2)(q-2)8)/2 = (12 + 10p + 10q - 40 + 8pq - 16p - 16q + 32)/2 = (16pq - 12p - 12q + 8)/4 = (4p - 3)(4q - 3)/4 - 1/4 = 327, so (4p-3)(4q-3) = 1309 = 7·11·17. Now 7 = 3 mod 4, 11 = 3 mod 4 and 17 = 1 mod 4, so we must have 4p-3 = 77, 4q-3 = 17 (or vice versa) and hence N = pq = 20·5 = 100.


----------------------------------------------------------- Quant Answer # 166 -----------------------------------------------------------

f(x) + f(1-x) =1, and f(1/2) = 1/2, f(1) = 3/4 => the required sum is 201/4.


----------------------------------------------------------- Quant Answer # 167 -----------------------------------------------------------

The pair (x, 11-x) for x = 1, 2, 3, 4, 5 has not to be there in the subset. Either of the 2 elements in the pair can not be in the subset or both can not be there. In all 3 possibilities and hence total subsets are 3^5.


----------------------------------------------------------- Quant Answer # 168 -----------------------------------------------------------

fig2 blue part is a rhombus. fig3 green part area = 1/2 (blue parallelogram) fig4 is 1/2 of the star...

area of star=6x+2y=1 where there 6 triangles of equal area and 2 other of equal area Area of required region= 3x+y=1/2..

Let AD and CE meet at R, and AD and BE meet at S. Then PQDR is a parallelogram, so PQD and QRD are both half of the parallelogram and hence have equal areas. Hence area PQD = area ERS. So area APQD = area APQD - area PQD + area ERS = area APDRES = (area star)/2 = 1/2.


----------------------------------------------------------- Quant Answer # 169 -----------------------------------------------------------

There are 24 engineers above 25 of which 12 are females. Out of 19 married above 25, 7 are males.


----------------------------------------------------------- Quant Answer # 170 -----------------------------------------------------------

Let the groups be 1, 2, 3, 4, 5, 6, A, B, C, D, E

1 2 3 A B C watch 4 5 6

1 4 5 A B watch 2 3 6 C

2 4 6 A C watch 1 3 5 B

3 5 6 B C watch 1 2 4 A

1 2 3 4 5 6 watch A B C


----------------------------------------------------------- Quant Answer # 171 -----------------------------------------------------------

Among 100 students there are 70+62+84+82 = 298 passes. 37 have 4 passes => remaining 63 students got 298 - 37*4 = 150 passes

These 150 passes can be divided among 50 students getting 3 each. => 13 are failing in all 4 subjects.


My Own Solution:

37 passed in all 4. Therefore 63 faliled in atleast one. 70 passed in physics=> 30 failed in physics 62 passed in Maths => 38 failed in Maths 84 passed in English=> 16 failed in English 82 passed in Chem => 18 faliled in Chem Suppose n fail all 4. To maximise n, the only common fails will be for all the four. (30-n)+(38-n)+(16-n)+(18-n)+n = 63 102-3n = 63 3n=39 n=13 Hence (d)

----------------------------------------------------------- Quant Answer # 172 -----------------------------------------------------------

There are 4.3 = 12 numbers with a given digit of n in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is 12.111 (a + b + c + d + e), where n = abcde. So 8668a = 332b + 1232c + 1322d + 1331e. We can easily see that a = 1 is too small and a = 4 is too big, so a = 2 or 3. Obviously e must be even. 0 is too small, so e = 2, 4, 6 or 8. Working mod 11, we see that 0 = 2b + 2d, so b + d = 11. Working mod 7, we see that 2a = 3b + 6d + e. Using the mod 11 result, b = 2, d = 9 or b = 3, d = 8 or b = 4, d = 7 or b = 5, d = 6 or b = 6, d = 5 or b = 7, d = 4 or b = 8, d = 3 or b = 9, d = 2. Putting each of these into the mod 7 result gives 2a - e = 4, 1, 5, 2, 6, 3, 0, 4 mod 7. So putting a = 2 and remembering that e must be 2, 4, 6, 8 and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for c. Similarly putting a = 4, gives a, b, d, e = 3, 4, 7, 8 or 3, 5, 6, 4 as the only possibilities. Checking, we find the solution above and no others. Thus, the number is 35964.


My Own Solution:

Let the digits of number N be a,b,c,d,e Now sum of all the distinct 3 digit nos. formed from the above digits = (a+b+c+d+e)*4*3*111 = 1332*(a+b+c+d+e) ------ (1) From (1), we can say that 15<=(a+b+c+d+e)<30

Between these numbers, 16,25,27 and 28 are numbers which are either perfect cube,square or perfect nos. We can reject 25 as it will give a 5 digit number which will have 0 in it. We can multiply rest of the numbers with 1332 and see whether we get distinct 5 digits number and whose sum is equal to the number multiplied. By this, we get 27 as the answer which is a perfect cube. 1332*27=35964 and 3+5+9+6+4=27

----------------------------------------------------------- Quant Answer # 173 -----------------------------------------------------------

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.


----------------------------------------------------------- Quant Answer # 174 -----------------------------------------------------------

The diagonals of the rectangle bisect each other, so AC is a median of the triangle BCD, BI is another median, so M is the centroid. AB = (8)^1/2, so CO = (2)^1/2. Hence BO = (6)^1/2 and BM = (2/3)*BO = (8/3)^1/2. Now AC = (12)^1/2, so CM = (2/3)AC/2 = (4/3)^1/2. Hence BC^2 = BM^2 + CM^2, so <AMB = 90 degrees.

area AEB = 2, area AMB = BM·AM/2 = (32)^1/2/3.


----------------------------------------------------------- Quant Answer # 175 -----------------------------------------------------------

Let the CP be = X. Then the SP for Amirchand = 5x/4 (since he gets a profit of 25% for a payment by credit card). And the marked price MP = 5x/4 * 3/2 = 15x/8 (since the discount is 33.33%). Now Garibchand robs Amirchand 20% of SP, so Garibchand pays 5x/4*4/5 = x (Since he pays by Credit Card so the SP is 5x/4)

Now Garibchand sells it at 40% discount of marked price implies, new SP = 15x/8*3/5 = 9x/8

Now, since Garibchand gets a profit of Rs. 20/- that means, 9x/8 - x = 20, so x = 160.


----------------------------------------------------------- Quant Answer # 176 -----------------------------------------------------------

6t^4 - t^2 + 5 = 6*(t^2 + 1)^2 - 13*(t^2+1) + 12 => f(x) = 6*x^2 - 13x + 12

Please see that the whole idea is to express 6t^4 - t^2 + 5 as a function of t^2 + 1 and hence we chose 6 as out first cofficient and 13 as our next.


----------------------------------------------------------- Quant Answer # 177 -----------------------------------------------------------

Let A be the moves towards A and B be the moves towards B => A+2B = 0 (mod 3).

When A+B = 8 we have (A, B) = (1, 7), (4, 4), (7, 1) => probability that ant comes back to the original position is (8C1 + 8C4 + 8C7)/2^8.

Similarly we have A+B = 10 and A+B = 12. Note that A+2B = 0 (mod 3).

B = (10C2 + 10C5 + 10C8)/2^10, C = (12C3 + 12C6 + 12C9)/2^12.


----------------------------------------------------------- Quant Answer # 178 -----------------------------------------------------------

Required area is Surface area of the frustum = pi*12*20- pi*6*10 = 240*pi-60*pi= 180*pi


Surface Area of a Conical Frustum = pi * (R1+R2) * sqrt[(R1-R2)2 + H2] ----------------------------------------------------------- Quant Answer # 179 -----------------------------------------------------------

Let m denote the minimum, and let a; b; c; d be the numbers written on a faces with a + b + c + d = m. Without loss of generality, we may assume that a = 10. Because 2 + 3 + 4 = 9, it follows that c >= 5, and so d >= 6. Hence m = (a + b + c) + d >= 10 + 6 >= 16


----------------------------------------------------------- Quant Answer # 180 -----------------------------------------------------------

Let x, y be 2 numbers => (x+y)/xy = 1003.We need to maximize (x+y).

y = 1003*x/(x - 1003). Max occurs when x = 1004 and y = 1003*1004.


----------------------------------------------------------- Quant Answer # 181 -----------------------------------------------------------

The small triangles can be painted in the "checkerboard" pattern so that any two adjacent triangles have different colors. There are n (n+1)/2 triangles (painted red) and n (n-1)/2 (orange) triangles. Every time one counts a new triangle, the count moves from a triangle of one color to a triangle of a different color. Thus one cannot count more than twice the number of orange triangles plus 1, provided the first triangle is red. This gives, as a maximum, n(n-1)/2+ n(n+1)/2+1= n^2-n+1. Hence 144-12+1=133 is the answer.


----------------------------------------------------------- Quant Answer # 182 -----------------------------------------------------------

For any convex n-gon, the sum of the interior angles is (n – 2)180°. If an n-gon has exactly three obtuse interior angles, then the remaining n – 3 angles have measure of at most 90° each, and the obtuse angles have measure less than 3 *180° together. Thus, (n – 2)180 < (n – 3)90 + 3 *180, 2(n – 2) < n – 3 + 6, n < 7.


----------------------------------------------------------- Quant Answer # 183 -----------------------------------------------------------

Let [x] denotes the greatest integer less than or equal to x then the highest power of a prime p in n! is the sum [n/pî] where i ranges from 1 to [logn/logp].

The desired _expression converts to 2*(1 + 2 + 2^2+...+ 2^9-1) + 2^2*(1 + 2 + 2^2+...+ 2^8 - 1) + ... + 2^9*1 = 2^18 + 2^17 + ... + 2^10 - 9*2^9 = 518656


----------------------------------------------------------- Quant Answer # 184 -----------------------------------------------------------

Both Prade and warrior knew 4 of the 5 digits. Both of them could have tried out all the 10 options for the digit which was unknown to him. If the unknown digit is 0,1, or 2 it can be 7,8 or 9 correspondingly, but if the unknown digit is 3,4,5 or 6 it cannot be anything else. As Prade could not determine the digit unknown to him, while Warrior could, the 1000th digit was 0,1,2,7,8,or 9 and the 100th digit was 3,4, 5 or 6.


----------------------------------------------------------- Quant Answer # 185 -----------------------------------------------------------

Let S be the required summation: We have S = 1/ 4^2+ 1/ 4^3 +2/ 4^4 + 3/ 4^5 + 5/ 4^6+….. 4S = 1/ 4^1 + 1/ 4^2 + 2/ 4^3 +3 /4^4 +5/ 4^5 + 8/ 4^6 16S = 1/ 4^0+ 1/ 4^1 + 2/ 4^2 + 3/ 4^3 + 5/ 4^4 + 8/ 4^5+………

So we see, S+4S= 16S-1


----------------------------------------------------------- Quant Answer # 186 -----------------------------------------------------------

Given a, b, c =/ 0 mod 7, we can always find d =/ 0 mod 7 such that ad = bc mod 7. However, it is possible that d = 6 mod 7. If bc = 1 mod 7, then a*6 =/ 1 mod 7, so d is not 6 for any choice of a. onfiltered=/ 1 mod 7, then if a = -bc mod 7 we will get d = 6 mod 7. Thus we may choose b arbitrarily (5 choices). Then for c = 1/b (one choice), we have 5 choices for a, and d is then determined. For c =/ 1/b (4 choices), we have 4 choices for a, and d is then determined. Thus in total there are 5*(5 + 4^2) = 105 possibilities.


----------------------------------------------------------- Quant Answer # 187 -----------------------------------------------------------

The problem boils down to finding out the sum of all the numbers co-prime to and less than 210 (LCM of (2,3,5,7) ). Which is given by= ½ n. P(n) Where n= 210 And P(n)= All the numbers co prime to and less than n= n. (1-1/m)(1-1/k)…. Where m, k….etc are prime factors of n

So in our case P(n)= 210 (1-1/2)(1-/3)(1-1/5)(1-1/7)=48 Required answer= ½ .210. 48= 48.105= 5040


----------------------------------------------------------- Quant Answer # 188 -----------------------------------------------------------

If n = 3k where k is a positive integer then we show that the given polynomial is not divisible by x^2 + x + 1. If x^2 + x + 1 = 0 => x + 1 = -x^2 => x^2 + x = -x^3 => x^3 = 1.

x^(2n) + 1 + (x+1)^(2n) = (x^3)^2k + 1 + (x^3)^4k = 1+1+1 = 3.

Note that when n = 3k+1 or 3k+2 then x^(2n) + 1 + (x+1)^(2n) = x^2 + x +1 = 0.


----------------------------------------------------------- Quant Answer # 189 -----------------------------------------------------------

Let p and q be the radii of circles P and Q, respectively. By symmetry, P and Q must lie on diagonal BD, and the tangency point of circles P and Q (call it T) must also lie

on diagonal BD, since it must lie on the same line as P and Q. Then BD = = BQ

+ QT + TP + PD = so and

Now let H be the foot of the perpendicular from P to Let the line through O1

perpendicular to intersect at X and at Z, and let Y be the foot of the perpendicular from O2 to this line, as illustrated in the diagram below.

We have XZ = XO1 + O1Y + YZ, and we will express term in this equation in terms of p and r (recall that r is the radius of O1 and O2). Note that XZ = p, XH = YZ = r. By

symmetry, O1O2Y is an isosceles right triangle, so Also (XO1)2 =

(O1P)2 - (PX)2 = (p + r)2 - (p - r)2 = 4pr. Thus Subtract and p from each side, square each side of the resulting equation, and substitute in the value of p to obtain a quadratic equation in r The larger solution is extraneous since it exceeds p. So the value of r, is 0.29.

Option (a) is the answer.

----------------------------------------------------------- Quant Answer # 190 -----------------------------------------------------------

Let there be P pagals. So number of lines connecting the pagals to each other is PC2= P(P-1)/2. Also each pagal point is connected to other 10-P points. Thus we get P(P-1)/2+ P(10-P) lines =<1000. The largest P=<100 satisfying this confiltered= 10


----------------------------------------------------------- Quant Answer # 191 -----------------------------------------------------------

We now have the sequence a, b, (b+1)/a, (b+1+a)/ab, (a+1)/b, a, b…. . Since we are given b already, we only consider four possible distinct sequences, since anything after the fifth term is repeating, and a different sequence is determined by where the second term is in the sequence. We see that we only have four distinct terms to choose 5 to be, so there are only four distinct sequences containing 5.


----------------------------------------------------------- Quant Answer # 192 -----------------------------------------------------------

Let the digits be a; b; c; d; e: Then abcde = 180 = 2^2. 3^2. 5: We observe that there are 6 ways to factor 180 into digits a; b; c; d; e (ignoring differences in ordering): 180 =1 . 1 . 4 . 5 . 9 = 1 . 1 . 5 . 6 . 6 = 1 . 2. 2 .5 . 9 = 1 . 2 .3 . 5 . 6 = 1 . 3 . 3 . 4 . 5 = 2 . 2 . 3 . 3 . 5:There are (respectively) 60, 30, 60, 120, 60, and 30 permutations of these breakdowns, for a total of 360 numbers.


----------------------------------------------------------- Quant Answer # 193 -----------------------------------------------------------

f(40) = 40^2 - f(39); f(39) = 39^2 - f(38); ... ; f(12) = 12^2 - 40 => f(40) = (79 + 75 + 71 + ... + 27) + 104 = 846.

846 when divided by 3 leaves remainder as 0. 846 when divided by 17 leaves remainder as 13 => we want a such that 13^11 = a (mod 17) => a = 4.

Remainder by 3 is 0, by 17 is 4 => by 51 is 21.


My Solution:

f(40) - f(12) = 2(40+38+36+....+14) - 14 = 742 f(40) = 104 + 742 = 846 Now (846^11)mod 51 = (30^11) mod 51 = (900^5)*30 mod 51 = (33^5)*30 mod 51 = (1089^2)*990 mod 51 = 324*990 mod 51 = 18*21 mod 51 = 21

----------------------------------------------------------- Quant Answer # 194 -----------------------------------------------------------

House n ends up red if and only if the largest odd divisor of n is of the form 4k+1. We have 25 values of n = 4k + 1; 13 values of n = 2(4k + 1) (given by k = 0, 1, 2, 3…..12); 7 values of n = 4(4k + 1) (k = 0, 1, 2,..6); 3 values of n = 8(4k + 1) (k = 0, 1, 2); 2 of the form n = 16(4k +1) (for k = 0, 1); 1 of the form n = 32(4k +1); and 1 of the form n = 64(4k + 1). Thus we have a total of 25 + 13 + 7 + 3 + 2 + 1 + 1 = 52 red houses.


----------------------------------------------------------- Quant Answer # 195 -----------------------------------------------------------

The lengths of AB and AC are irrelevant. Because the figure is symmetric about AD, lines BC” and B”C meet if and only if they meet at a point on line AD. So, if they never meet, they must be parallel to AD. Because AD and BC” are parallel, triangles ABD and ADC” have the same area. Then ABD and ADC also have the same area. Hence, BD and CD must have the same length, so BD = BC/2 = 5/2.


----------------------------------------------------------- Quant Answer # 196 -----------------------------------------------------------

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 mean 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.

Hence, choice (b) and choice (d) are the right answer.

----------------------------------------------------------- Quant Answer # 197 -----------------------------------------------------------

Let the point of tangency be T. So RT = RQ=2 Also say EP=ET=x In Triangle ESR, 2^2+ (2-x)^2= (2+x)^2..get x= ½ So required length= 2+1/2=5/2


----------------------------------------------------------- Quant Answer # 198 -----------------------------------------------------------

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.


----------------------------------------------------------- Quant Answer # 199 -----------------------------------------------------------

(1-p)/(1+p) + (1-q)/(1+q) + (1-r)/(1+r) = 2*(1/(1+p) + 1/(1+q) + 1/(1+r)) -3.

If P(x) = x^3 -x -1 has roots p,q,r then P(x-1) has roots as p+1, q+1, r+1 P(x-1) = x^3 - 3*x^2 +2x -1

In a cubic x^3 + px^2 + qx + r = 0, product of the roots is -r, sum taken 2 at a time is q, and the sum of the roots is -p.

thus 1/(1+p) + 1/(1+q) + 1/(1+r) = sum taken 2 at a time/product = 2/1 = 2.


----------------------------------------------------------- Quant Answer # 200 -----------------------------------------------------------

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 = 4x Also, (a+b+d) + (c+b+d) - (b+d) = 300 => x = 60 => a+c = 180



(201-end)

----------------------------------------------------------- Quant Answer # 201 -----------------------------------------------------------

The expression (x-2y+1)^2 + (x+y-1)^2 = 2*(x-1/2y)^2 + 1/2*(3y-2)^2. The _expression becomes minimum when y = 2/3 and x= 1/2y = 1/3. The min value is 0.

The maximum value is achieved y = 0 and x = 1. The max value is 4.


----------------------------------------------------------- Quant Answer # 202 -----------------------------------------------------------

Arrange 6 black balls and 5 white balls in a row. Then the 5 white balls divide the line into 6 parts corresponding to ai = 1, 2, ... , 6. The number of black balls in each part gives the number of ai with that value. Thus we have in all (6+6-1)C6 such possibilities.


----------------------------------------------------------- Quant Answer # 203 -----------------------------------------------------------

The figure formed by joining the mid-points of the sides of a rhombus is a rectangle and the figure formed by joining the mid-points of the sides of a rectangle is a rhombus.

Let x be the length of each side of the given rhombus => it's area is (1/2*3^1/2)*x^2. The diagonals of this rhombus have lengths x and (3^1/2)*x => length and breadth of the rectangle is x/2 and (1/2*3^1/2)*x respectively. Similarly we go on...

Area of all the rhombuses is 1/2*(3^1/2)*x^2/(1-1/4). Area of all the rectangles is 1/4*(3^1/2)*x^2/(1-1/4) => x = 8

Sum of the perimeter of all the rhombuses is 8*8. Sum of the perimeter of all the rectangles is 16*(1+ 3^1/2).


----------------------------------------------------------- Quant Answer # 204 -----------------------------------------------------------

rpms are in ratio 6:4:3 In 5 minutes G2 makes 300 revolutions => 75 beeps heard. In next 1 minute 1+2+3+...+15 = 120 beeps are heard.


----------------------------------------------------------- Quant Answer # 205 -----------------------------------------------------------

From x^3 - y^3 - z^3 = 3xyz we see that x > y and x > z as x, y, z are positive. From x^2 = 2(x + y + z) we see that x^2 < 6x and x is also even.

=> x can be either 4 or 2. When x = 2, y = z = 0, hence no solution.

When x = 4 we have 3 solutions. Please check this yourself.


----------------------------------------------------------- Quant Answer # 206 -----------------------------------------------------------

One way to do this problem is by looking at the answer choices. Take the case when there are 2 teams with 1 men each and the third team with just 1 women and no other team. All the conditions in the question are satisfied.

The formal proof is as follows.

Let there be n B-schools. Suppose the ith B-school sends Bi men and Gi women. Let B = sigma(Bi) and G = sigma (Gi). We are given that |B - G| = 1.

The number of same gender matches is 1/2 sigma( Bi(B - Bi)) + 1/2 sigma( Gi(G - Gi)) = (B^2 - sigma( Bi^2) + G^2 - sigma(Gi^2)). The number of opposite gender matches is sigma(Bi(G - Gi)) = BG - sigma(BiGi). Thus we are given that B^2 - sigma(Bi^2) + G^2 - sigma(Gi^2) - 2BG + 2 sigma(BiGi) = 0 or ±2. Hence (B - G)^2 - sigma(Bi - Gi)^2 = 0 or ±2. But (B - G)^2 = 1, so sigma(Bi - Gi)^2 = -1, 1 or 3. It cannot be negative, so it must be 1 or 3. Hence Bi = Gi except for 1 or 3 values of i, where |Bi - Gi| = 1. Thus the largest number of B-schools that can have Bi + Gi odd is 3.


----------------------------------------------------------- Quant Answer # 207 -----------------------------------------------------------

We need to find the largest possible number of subsets of {1, 2, 3, 4, 5, 6, 7} such that no 2 subsets are disjoint. Fix one element from the set to be presented in each subset and we can have 2^6 such possibilities.


----------------------------------------------------------- Quant Answer # 208 -----------------------------------------------------------

Solving the quadratic we get y^2 <= 4x

Option (B) and (D) are ruled out on account of the co-ordinates given (-1,-2)... since they don't lie on the curve 4x = y^2

For options (A) and (C) drawing the graph of the functions... y^2 - 4x = 0;

y=x+1; and y=-x-1; it can be clearly observed that only the left region of the curve y^2 = 4x satisfies all the condition.


----------------------------------------------------------- Quant Answer # 209 -----------------------------------------------------------

We have b = (2^n - 1)/a, hence 2^n - 2 + a - b = 2^n - 2 + a - (2^n - 1)/a = (2^na - 2a + a^2 - 2^n + 1)/a = (2^n + a - 1)(a - 1)/a. Now a divides 2^n - 1, so it must be odd. Let 2^m be the highest power of 2 dividing a - 1. Then m < n, so 2^m is also the highest power of 2 dividing 2^n + (a - 1). Hence k = 2m.

k is always even.


----------------------------------------------------------- Quant Answer # 210 -----------------------------------------------------------

Let PC be the hypotenuse. Let P' be the image of A through 60 degrees clockwise rotation around A. Now <(BPP') = 90 degrees if and only if <BPA = 30 degrees b'coz <(BPA) = <(BPP') - <(APP') = <(BPP') - 60 degrees.

So the locus is a circle with centre outside the triangle ABC in which AB determines an arc of 60 degrees.


----------------------------------------------------------- Quant Answer # 211 -----------------------------------------------------------

BE + EH + HK ≥ 510 and AK = 560, so AB ≤ 50, so AD = AB + BD ≤ 50 + 120 = 170. But AD ≥ 170. Hence AB = 50, BD = 120, BE = 170. Hence DE = 50. By symmetry, we also have GH = 50, HJ = 120, JK = 50. Hence GK = 50+120+50 = 220.


----------------------------------------------------------- Quant Answer # 212 -----------------------------------------------------------

Put p1 = h-k, q1 = h, so p2 = h+k, q2 = h+2k. Then h^2-k^2 = c/a, 2h = -b/a, h^2+2hk = a/c, 2h+2k = -b/c.

So h = -b/2a, k = b/2a - b/2c and b^2/2ac - b^2/4c^2 = c/a, b^2/2ac - b^2/4a^2 = a/c. Subtracting, (b^2/4)(1/a^2 - 1/c^2) = c/a - a/c, so (c^2-a^2)(b^2/4 - ac)/(a^2*c^2) = 0. Hence a = c or a + c = 0 or b^2 = 4ac. If b^2 = 4ac, then p1 = p2, whereas we are given that p1, p2, q1, q2 are all distinct. Similarly, if a = c, then {p1,p2} = {q1,q2}. Hence a + c = 0.


----------------------------------------------------------- Quant Answer # 213 -----------------------------------------------------------

Let the perpendiculars from X to the lines AB, AC meet them at Z, Y respectively. Triangles XBZ, XYC are congruent because XB = XC (X lies on angle bisector), XZ = XY (X lies on perpendicular bisector) and <(BZX) = <(CYX) = 90 degrees. Hence BZ = CY. Also AZ = AY. By Ceva's theorem, (AZ/ZB) (BD/DC) (CY/YA) = 1. Hence BD/DC = 1.


----------------------------------------------------------- Quant Answer # 214 -----------------------------------------------------------

Let the original number in base n Priyanka was to receive is pq. => 2*(pn+q) = qn + p. => p/q = (n-2)/(2n-1) => x = 5 and y = 8.

The 4 digit number abab in base z when converted to decimal system is az^3 + bz^2 + az + b = (az+b)*(b^2+1) is a perfect cube. Checking for values z = 2, 3, ..., we see that 2626 is 1000 in base z = 7.


----------------------------------------------------------- Quant Answer # 215 -----------------------------------------------------------

abbabb = 1001*(a 3 digit number of the form abb). 233 is a prime.

233233 = 7*11*13*233


----------------------------------------------------------- Quant Answer # 216 -----------------------------------------------------------

Put x = 5 - 22^1/2 => E = 3^1/2 + (10-x)^1/2 + x^1/2 => (E-3^1/2)^2 = 10 + 2*3^1/2.


----------------------------------------------------------- Quant Answer # 217 -----------------------------------------------------------

The total length of the track is 3L and let the speeds of A, B and C be v, 2v, and 3v respectively. A and B meet for the 1st time after t = L/3v and they meet for the nth time at L/3v + (3L/3V)*(n-1)

Similarly, B and C meet for the mth time at L/V + (3L/v)*(m-1) If all 3 were to meet => L/3v + (3L/3V)*(n-1) = L/V + (3L/v)*(m-1) => 1 + 3*(n-1) = 3 + 9*(m-1), which is not possible as RHS is divisible by 3 and LHS is not.


----------------------------------------------------------- Quant Answer # 218 -----------------------------------------------------------

Let 2 points be isolated and not even joined to each other. Let the rest 10 points be a part of another group such that 9 edges (from P1 to P2 to P3 ... to P10) are sufficient so that it is possible to reach any point from any other point in this group. Thus we have 9 edges in all => S = 18 as each edge contributes 2 in S.

Let 2 points be isolated but joined to each other => S = 2. For the group containing 10 points we can have max 10C2 = 45 edges => 90 more in S => total S = 92.


----------------------------------------------------------- Quant Answer # 219 -----------------------------------------------------------

Consider first the number of 5-digit numbers divisible by 3. The smallest is 10002 = 3334·3, the largest is 99999 = 33333·3, so there are 29999.

Now consider the number that do not contain the digit 6. There are 8 choices for the 1st digit, 9 choices for each of the 2nd, 3rd and 4th digits. If the total of the first 4 digits is 0 mod 3, then the last digit must be 0, 3 or 9. If it is 1 mod 3, then the last digit must be 2, 5 or 8. If it is 2 mod 3, then the last digit must be 1, 4 or 7. So in all cases there are 3 choices for the last digit. Hence the total number is 8*9^3*3 = 17496. So the total no. which do contain a 6 is 12503.


----------------------------------------------------------- Quant Answer # 220 -----------------------------------------------------------

Draw a line through B parallel to the rectangle sides length 5 cm. Suppose it meets the side through A at X. Then <(AXB) = 90 degrees and BX = 5 cm. So if AB <= 5 then there is no rectangle. If AB > 5, then X must lie on the circle diameter AB and on the circle center B radius 5. There are two such points and hence two possible rectangles.


----------------------------------------------------------- Quant Answer # 221 -----------------------------------------------------------

The total of the scores, 39, must equal the number of rounds times the total of the cards. But 39 has no factors except 1, 3, 13 and 39, the total of the cards must be at least 1 + 2 + 3 = 6, and the number of rounds is at least 2. Hence there were 3 rounds and the cards total 13.

The highest score was 20, so the highest card is at least 7. The score of 10 included at least one highest card, so the highest card is at most 8. The lowest card is at most 2, because if it was higher then the highest card would be at most 13 - 3 - 4 = 6, whereas we know it is at least 7. Thus the possibilities for the cards are: 2, 3, 8; 2, 4, 7; 1, 4, 8; 1, 5, 7. But the only one of these that allows a score of 20 is 1, 4, 8. Thus the scores were made up: 8 + 8 + 4 = 20, 8 + 1 + 1 = 10, 4 + 4 + 1 = 9. The last round must have been 4 to Veeru, 8 to Basanti and 1 to Gabbar. Hence on each of the other two rounds the cards must have been 8 to Veeru, 1 to Basanti and 4 to Gabbar.


----------------------------------------------------------- Quant Answer # 222 -----------------------------------------------------------

r is an integer as both the roots are integers and D = r^2 - 24r has to be a perfect square.

r(r-24) is a perfect square => (r-12)^2 = p^2 + 144, a perfect square. If r > 0, put r-12 as k => (k-p)*(k+p) = 144 = 72*2 = 36*4 = 24*6 = 18*8 = 12*12 Hence, choice (c) is the right answer.

=> k = 37, 20, 15, 13, 12 => r=49, 32, 27, 25, 24.

When r is non-positive, Let -r = t => t(t+24) is a perfect square. We repeat the same process (t+12)^2 is a perfect square, put t+12 as k => t = 25, 8, 3, 1, 0 => r = -25, -8, -3, -1, 0. Hence, 10 solutions in all.

----------------------------------------------------------- Quant Answer # 223 -----------------------------------------------------------

1/2 = (x^4+y^4)/(x^4-y^4), so 1/2 + 2/1 = 2(x^8+y^8)/(x^8-y^8). Hence desired expresson = (1+4)/4 + 4/(1+4) = 41/20.


----------------------------------------------------------- Quant Answer # 224 -----------------------------------------------------------

There are (n+2)(n+1)/2 matches, so the total score is (n+2)(n+1)/2. Let the other players score k each. Then 8 + nk = (n+2)(n+1)/2, so n2 - (2k-3) - 14 = 0. We know this equation has one root which is a positive integer. The product of the roots is -14, so the possibilities for the roots are: 1, -14; 2, -7; 7, -2; 14, -1. Hence the sum of the roots is -13, -5, 5, or 13 (respecively). Hence k = -5, -1, 4 or 8 (respectively). But k must be non-negative, so n = 7 or 14 is a necessary condition.

We need to check that these values can be achieved. Take n = 7, so there are 9 players in total. If every match is a draw, then every player draws 8 matches and scores 4, which satisfies the conditions. Take n = 14, so there are 16 players in total. Suppose one player loses to everyone, and all the other games end in a draw. Then the first player scores 0 and all the other players score 1 + 14/2 = 8. That also satisfies the conditions.

----------------------------------------------------------- Quant Answer # 225 -----------------------------------------------------------

If 1 is green, take n to be blue (we are told n exists). Then n = 1·n is green. Contradiction. So 1 is blue. Suppose 4 is green, then 1+4 is blue. Contradiction, so 4 is blue. 1+5= 6 is blue. 4+5 = 9 is blue. If 2 is green, then 3=1+2 is blue, so 5=2+3

is blue. Contradiction, so 2 is blue. Hence 10=2·5 is green. If 8 is green, then 2+8 is blue. Contradiction, so 8 is blue. If 3 is green, then 2+3 is blue. Contradiction, so 3 is blue. If 7 is green, then 3+7 is blue. Contradiction. So 7 is blue.


----------------------------------------------------------- Quant Answer # 226 -----------------------------------------------------------

By symmetry we see that PQ passes through the midpoints of BC and DA. Let PQ meet BC at R and DA at S. RS = 1/2(7+5) = 6. PS=QR=3/2 (WHY?).


----------------------------------------------------------- Quant Answer # 227 -----------------------------------------------------------

The first 14 groups have 1+2+3+...+14 = 105 terms, so the first term of the 15th group is 1/2*15*14 + 1 and the last term of the 15th group is 1/2*15*16. The sum of the 15th group is 15/2[1st term + Last term].


----------------------------------------------------------- Quant Answer # 228 -----------------------------------------------------------

area BDE ≤ area DEA implies that the distance of A from the line DE is no smaller than the distance of B, so if the lines AB and DE intersect, then they do so on the B, D side. But area EAB ≤ area ABD implies that the distance of D from the line AB is no smaller than the distance of E, so if the lines AB and DE intersect, then they do so on the A, E side. Hence they must be parallel. But ABDE is cyclic (<ADB = <AEB = 90 degrees), so it must be an isosceles trapezoid and hence <A = <B.


----------------------------------------------------------- Quant Answer # 229 -----------------------------------------------------------

Let the numbers be x(1) >= x(2) >= x(3) >= ... >= x(15)

Then S equals 14x(1) + 12x(2) + 10x(3) + .... - 14x(15) => we can max S by taking each of x(i) with positive cofficients as 1 as with negative cofficients as 0.


----------------------------------------------------------- Quant Answer # 230 -----------------------------------------------------------

The centers of the 4 spheres form a regular tetrahedron side 2. The center of the fifth sphere must obviously be at the center of the tetrahedron. An altitude of the tetrahedron will have one end at the centroid of a face, so its length is (2^2 - ((2/3)*3^1/2)^2)^1/2 = (4 - 4/3)^1/2 = 2*(2/3)^1/2. So the distance from the center of the fifth sphere to one of the other centers is (3/4)2*(2/3)^1/2 = (3/2)^1/2. Hence the radius of the fifth sphere is (3/2)^1/2 + 1.


----------------------------------------------------------- Quant Answer # 231 -----------------------------------------------------------

Let f(n) = [n/3] - [n/5] - [n/7] + [n/35]. We are looking for the largest n with f(n) = 0. Now [n/5] + [n/7} <= [n/5 + n/7] = [12n/35] = [n/3 + n/105]. So for [n/5] + [n/7] to exceed [n/3] we certainly require n/105 = 1/3 or n = 35. Hence f(n) = 0 for n = 35. But f(n+35) = [n/3 + 11 + 2/3] - [n/5 + 7] - [n/7 + 5] + [n/35 + 1] = [n/3 + 2/3] - [n/5] - [n/7] + [n/35] = f(n) (*). Hence f(n) = 0 for all n. But f(n+105) = [n/3 + 35] - [n/5 + 21] - [n/7 + 15] + [n/35 + 3] = f(n) + 2. Hence f(n) = 2 for all n = 105.

Referring back to (*) we see that f(n+35) > f(n), and hence f(n+35) > 0, unless n is a multiple of 3. But if n is a multiple of 3, then n + 35 is not and hence f(n+70) > f(n+35) > 0. So f(n) > 0 for all n = 70.

f(70) = 1. So f(69) = 2 (we have lost 70, a multiple of 7). So f(68) = f(67) = f(66) = 1 (we have lost 69, a multiple of 3). Hence f(65) = 0 (we have lost 66, a multiple of 3).


--------------------------------------------------------- Quant Answer # 232 ---------------------------------------------------------

Let t be the time taken by the hands to meet when they are running in the same direction = 3*(when running in opposite) => t = 360 degrees/(m+h), where m and h are the speeds of the minute and hour hand) => 360/(m-h) = 3*(360/(m+h)) => m = 2h


--------------------------------------------------------- Quant Answer # 233 ---------------------------------------------------------

We will prove that the triangle MCD is equilateral. Let M' be a point inside ABCDE such that M'CD is equilateral. Then triangles CM'B and DM'E are both isoscles having 2 sides equal. <(M'CB) = <(DCB) - <(DCM') = 48 degrees, and by symmetry <M'DE = 48 degrees => <(M'BC) = <(M'EA) = 66 degrees. It follows <(M'BA) = <(M'EA) = 42 degrees => M and M' coincide.


--------------------------------------------------------- Quant Answer # 234 ---------------------------------------------------------

If a+b+c is x and ab+ac+bc = y then x^2 = 1 + 2y and xy = 0 => x = 0, 1, -1.


--------------------------------------------------------- Quant Answer # 235 ---------------------------------------------------------

Let S have k+1 elements with sum N. Then k divides N-1 and N-99 and hence also 98 = 7*7*2. If m is any element of S, then k divides N-m and hence m-1. In other words m = 1 mod k. So the largest element is at least 1+k^2. Since 99 is the largest element, k < 10. Thus the largest possible k is 7.


--------------------------------------------------------- Quant Answer # 236 ---------------------------------------------------------

x=4 gives a solution. Let x be positive and > 4. LHS > RHS for x = 19/2 where RHS is 1 and LHS is little less than 11/2. At x = 9 LHS is little less than 5 and RHS is 40. There is a change of sign in the interval [9, 19/2]. For x > 19/2 LHS is little less than x-4 while RHS is always less than 1 so no further change in signs. => 2nd positive solution is in (9, 19/2).

For x < 4 we will have to venture into complex numbers. Leave it, the question got framed wrongly due to missing pair of brackets as shown in colours.


--------------------------------------------------------- Quant Answer # 237 ---------------------------------------------------------

2x + y + 4z = 2, x + z = -3, x + 2y + Mz = 13 Multiplying 1st eq. by 2 and 2nd by -3 and adding both we get x + 2y + 5z = 13 => when M = 5 we have infinite solutions. => I is true

If z = 0, the system will always yield a unique solution. => II is false


--------------------------------------------------------- Quant Answer # 238 ---------------------------------------------------------

x + 20r1 = 200, x + 10r1 = 150 => x = 100, r1 = 5 => 100 + 5p = 25p => p = 5 days.

(n(n+1)/2 - k)/(n-1) = 15 for k <= n => n(n-29) = 2k - 30 => (n, k) = (30, 30), (28, 1), (29, 15) => q = 3.

Since each biscuit costs whole number of rupees it is Rs 1 as overall SP with profit is Rs 144 < Rs 200. 10r biscuits at 20% profit and (100 - 10r) at 50% profit => 10r*1.2 + (100 - 10r)*1.5 = 144 => r = 2.

RC = 6R + 8C + 15 => (R-8)(C-6) = 63. (R, C) takes 6 different values => s = 6.


--------------------------------------------------------- Quant Answer # 239 ---------------------------------------------------------

<AQB = <ACB + <CBQ = 90o + <CBQ = 90o + 1/2<COD = 90o + 1/4<AQB. Hence <AQB = 120o, and <COD = 60o. So OP = 1/cos(30o) = 2/(3^1/2).


--------------------------------------------------------- Quant Answer # 240 ----------------------------------------------------------

p = 15

n! + (n+1)! + (n+3)! = n!*(n+2)^3 => q = 28


--------------------------------------------------------- Quant Answer # 241 ---------------------------------------------------------

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.


--------------------------------------------------------- Quant Answer # 242 ---------------------------------------------------------

p = 1/20.

1/2*s1*a1 = area = 1/2*s2*a2 = 1/2*s3*a3. Inradius = area/semi-perimeter => 1/inradius = (1/4 + 1/5 + 1/10) => q = 1/11.


--------------------------------------------------------- Quant Answer # 243 ---------------------------------------------------------

2q/r + (1+r)/2p + 2p^2/q = 2q/r + 1/2p + r/2p + 2p^2/q >= 4*(2q/r*1/2p*r/2p* 2p^2/q) >= 4 => I is true

x^4 - x + 1/2 = (x^2 - 1/2)^2 + (x - 1/2)^2 = 0. We could only have equality if x^2 = x = 1/2, which is impossible, so the inequality is strict. => II is true


--------------------------------------------------------- Quant Answer # 244 ---------------------------------------------------------

8 has been typed, so 1, 2, ... , 7 have already been put into the tray. Any left must be typed in decreasing order. 9 can come anywhere in the order or not at all (if already typed). So for a subset of {1, 2, ... , 7} with k elements, there are k+2 possibilities for 9. Hence 1·2 + 7·3 + 21·4 + 35·5 + 35·6 + 21·7 + 7·8 + 1·9 = 2 + 21 + 84 + 175 + 210 + 147 + 56 + 9 = 704.


--------------------------------------------------------- Quant Answer # 245 ---------------------------------------------------------

Suppose that no point is joined to all the others. Then given any point X we can find Y not joined to X. So take arbitrary A and C. Then take B not joined to A and D not joined to C. Then the four points A, B, C, D do not meet the required condition. Contradiction.

So find X1 joined to all the other 99 points. Now repeat the argument for the other 99 points, that gives a point X2 joined to the other 98. But it is also joined to X1, so it is joined to all other 99 points. Now repeat for the other 98 points and so on. The last time we can repeat is when we have already found X1, X2, ... , X96 leaving four points. We can now take X97 joined to the other three and hence to all other 99. Thus we can get at least 97 points each joined to all points except itself.

That is best possible, because we can take the graph with 100 points including A, B, C and all edges except AB, BC and CA. That clearly has at most 97 points each joined to all points except itself, but it obviously satisfies the condition.


--------------------------------------------------------- Quant Answer # 246 ---------------------------------------------------------

Suppose x bricks are oriented to add 4 to the height, y to add 19 and z to add 10. Since 5·10 = 3·4 + 2·19 we can take z = 0, 1, 2, 3, or 4. Also we must have y = 94-x-z, so height = 4x+10z+19(94-x-z) = 1786-15x-9z. If z=0, height = 1786-15x = 1 mod 5; if z=1, height = 1777-15x = 2 mod 5; if z=2, height = 1768-15x = 3 mod 5; if z=3, height = 1759-15x = 4 mod 5; if z=4, height = 1750-15x = 0 mod 5. So these values of x,z all give different heights. We can check that 1750 > 15·94 = 1510, so for z=0 there are 95 possible values of x (0, 1, ... , 94), for z=1, 94 etc. Hence in total 95+94+93+92+91 = 465 possible heights.


--------------------------------------------------------- Quant Answer # 247 ---------------------------------------------------------

Let N have largest proper factor m < N. We show that N + m cannot be 2002. Suppose N + m = 2002. Put N = mp. Then p must be a prime (or N would have a larger proper factor than m). So 2002 = m(p+1). Also p <= m. Hence p < 44. So k = p+1 is a factor of 2002 smaller than 45 which is 1 greater than a prime. It is easy to check that the only possibility is k = 14. So N = 11·13^2. But this has largest factor 13^2, not 11·13. Contradiction.

=> A = 2002 is the only possibility.


--------------------------------------------------------- Quant Answer # 248 ---------------------------------------------------------

figure AW is also (1-a)now , BZ=AB , gives 2a^2 = 1 + (1-a)^2 solve for a, get then find root2 * a thats ur answer ...


--------------------------------------------------------- Quant Answer # 249 ---------------------------------------------------------

Adding the three equations we get, 2(x + y + z) = 9.2 => x + y + z = 4.6 subtracting first eq. from the above eq. we get {y} + [z] = 0.8 => {y} = 0.8 and [z] = 0, subtracting second eq. from the above eq. {x} + [y] = 1.4 => {x} = 0.4 and [y] = 1, subtracting third eq. from the above eq. [x] + {z} = 2.4 => [x] = 2, and {z} = 0.4 => x = 2.4, y = 1.8 and z = 0.4


--------------------------------------------------------- Quant Answer # 250 ---------------------------------------------------------

We must cut the longest edges, so the similar piece has dimensions L x 120 x k for some 1 ≤ k < H. The shortest edge of this piece cannot be L, so it must be k. Thus L x 120 x H and k x L x 120 are similar. Hence H = 120^2/L, k = L^2/120. Now 120 = 2^3·3·5, so 120^2 has 63 factors, of which (63-1)/2 = 31 are < 120

The answer (31) is not in the choices.

--------------------------------------------------------- Quant Answer # 251 ---------------------------------------------------------

At most 4 competitors can receive a rank 1. For a competitor with a rank 1 can only receive ranks 1, 2, 3 or 4. There are only 36 such ranks available and each competitor with a rank 1 needs 9 of them.

If only one competitor receives a rank 1, then his score is 9. If only 2 competitors receive a rank 1, then one of them must receive at least five rank 1s. His maximum score is then 5.1 + 4.4 = 21. If 4 competitors receive a rank 1, then they must use all the 36 ranks 1, 2, 3, and 4. The total score available is thus 9(1 + 2 + 3 + 4) = 90, so at least one competitor must receive 22 or less. Thus the winner's maximum score is at most 22. If 3 competitors receive a rank 1, then the winner's score is maximised by giving all three competitors the same score and letting them share the 27 ranks 1, 3 and 4. That gives a winner's score of 9(1 + 3 + 4)/3 = 24. That can be achieved in several ways, for example: each competitor gets 3 1s, 3 3s and 3 4s, or one competitor gets 4 1s and 5 4s, another gets 3 1s, 3 3s and 3 4s, another gets 2 1s 6 3s and one 4. Note that it is trivial to arrange ranks for the remaining 17 competitors. For example: give one 5 2s and 4 5s total 30, one 4 2s and 5 5s total 33, and then one 9 6s, one 9 7s and so on.

Thus the answer is 24, with three joint winners. If there is required to be a single winner, then the answer is 23.


--------------------------------------------------------- Quant Answer # 252 ---------------------------------------------------------

We use Ceva's theorem. Since AD, BM, CH are concurrent, we have (BD/DC).(CM/MA).(AH/BH) = 1. But CM = MA and since AD is the angle bisector BD/DC = AB/AC, so (AB/AC).(AH/BH) = 1. Hence AH/AC = BH/AB < 1. So angle HAC > angle HCA. But angle AHC = 90 deg, so angle A > 45 deg.


--------------------------------------------------------- Quant Answer # 253 ---------------------------------------------------------

I) Each meeting involves 10.9/2 = 45 pairs. So after 40 meetings, there have been 1800 pairs. We are told that these are all distinct. But if there are N people on the committee, then there are only N(N-1)/2 pairs available. For N=60, this is only 1770.

=> I is true

II) A subcommittee of 5 has 5.4/2 = 10 pairs. So 31 subcommittees have 310 pairs, and these are all distinct, since no two people are on more than one subcommittee. But a committee of 25 only has 25.24/2 = 300 pairs available.

=> II is true


--------------------------------------------------------- Quant Answer # 254 ---------------------------------------------------------

By similar triangles the total shadow (including the part under the cube) is a square with side (1+x)/x. So (1+x)2/x2 = 36, x = 1/5.

Plot the square ABCD on the x-y axis such that A(0, 0), B(1, 0), C(1, 1), D(0, 1). We get P(2, 0), Q(1, 2), R(-1, 1) and S(0, -1). Area(PQRS) = 5 => y = 1/5.


--------------------------------------------------------- Quant Answer # 255 ---------------------------------------------------------

(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.


--------------------------------------------------------- Quant Answer # 256 ---------------------------------------------------------

Reflect A in the line BP to get A'. Let Z be the intersection of BP and AA'. Let BA' meet AC at X. Since �ABX = 2 �ABP = 40o, and �BAX = 50o, we have �BXA = 90o. Now �PAA' = �BAA' - �PAB = �BAA' - 10o. But �BAA' = 90o - �PBA = 70o, so �PAA' = 60o.

Let BX meet CP at Y. �PYX = �YXC + �PCX = 90o + 30o = 120o = 180o - angle PAA', so PAA'Y is cyclic, so �A'YA = �A'PA = 2 �ZPA = 2(�PBA + �PAB) = 60o.

But �XYC = 90 - �PCA = 60o, so C is the reflection of A in BX. Hence BC = BA, so �ACB = �BAC = 50o. Hence �ABC = 80o and �PBC = 80o - �PBA = 60o.


--------------------------------------------------------- Quant Answer # 257 ---------------------------------------------------------

a + ak + ak2 + ... = a/(1-k). So a = 1-k and |k| < 1. If we have another sequence (1-h), (1-h)h, (1-h)h^2 ... with same second term, then (1-k)k^2 = (1-h)h^2 implies h-k=0 or h+k=1. h=k gives same sequence, so must have h = 1-k. Must have k > ½ or that also gives same. Third term = 1/8 gives 8k^3 - 8k + 1 = 0,

or (2k-1)(4k^2-2k-1) = 0. Hence 4k^2-2k-1=0. So k = (1+v5)/4 or (1-v5)/4. Corresponding h=1-k are (3-v5)/4 and (3+v5)/4. Latter is > 1. So k = (1+v5)/4 and second term (v5-1)/8.


--------------------------------------------------------- Quant Answer # 258 ---------------------------------------------------------

log20 = (logx - 1)*(logy - 1) 1 = (logy - 1)*(logz - 1) log20 = (logz - 1)*(logx - 1)

Multiplying 3 equations and taking square root we get +/-log20 = (logx - 1)*(logy - 1)*(logz - 1) => logz = 1 +/-1, logx = 1 +/-log20, logy = 1 +/-1 Thus, (x, y, z) = (200, 100, 100) or (1/2, 1, 1)


--------------------------------------------------------- Quant Answer # 259 ---------------------------------------------------------

Put n = (10^m)a+b, where 1 ≤ a ≤ 9. Then 7n/2 = 10b + a, so 13b = a(7·10^m-2). Obviously 13 cannot divide a, so it must divide 7·10^m-2. Hence 10^m = 4 mod 13. We have 10^1 = 10, 10^2 = 9, 10^3 = -1, 10^4 = 3, 10^5 = 4 mod 13. So the smallest possible m is 5. Obviously the smallest possible a is 1, and we get b = 53846.


--------------------------------------------------------- Quant Answer # 260 ---------------------------------------------------------

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.


--------------------------------------------------------- Quant Answer # 261 ---------------------------------------------------------

x^4 + 1/4 = (x^2 + x + 1/2)*(x^2 - x + 1/2)

Since, x^2 - x + 1/2 = (x-1)^2 + (x-1) + 1/2, the terms in the numerator and the denominaotor cancel outexcept for 1^1 - 1 + 1/2 in the numerator and (20)^2 + 20 + 1/2 in the denominator.


--------------------------------------------------------- Quant Answer # 262 ---------------------------------------------------------

Let a solve just 261, b solve just 262, c solve just 263, and d solve 262 and 263 but not 261. Then 25 - a - b - c - d solve 261 and at least one of 262 or 263. 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 = 26. But d = b - 2c ≥ 0, so b = 6, c = 2.


--------------------------------------------------------- Quant Answer # 263 ---------------------------------------------------------

Let the traingle be DEF and AB be || to EF height be h and inradius be r then AB/EF=h-2r/h Also r.p=h.EF thus AB=(rp-2rEF)EF/rp =(p-2EF)EF/p differentiating we get, its max when EF=p/4 and then AB=p/8 Hence, choice (d) is the right answer

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Answers Qqad 2006