Mmc Fourth Year Solutions

8/10/2019 Mmc Fourth Year Solutions

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MMC FOURTH YEAR SOLUTIONS

A.

Elimination Round

1)Subtract x4+ 2x3

3x2+ 5 from 4x4

5x3+ x

3.

4x4 5x3 + x 3 x4 + 2x3 3x2 + 5 = 3x4 7x3 + 3x2 + x 82)Which of the following is/are functions? [ A and D]

a) x2+ 4y2= 16 b) x + 3y + 2 = 9

c) y = 4x2+ x 2 d) x + y22y + 5 = 0Solution #1: Graph and vertical line test

In a), the graph is a circle, and it fails the vertical line test. So it is not a

function.

In b), the graph is a line. And all lines pass the vertical line test. So it is a

function.

In c), the graph is a parabola opening upwards. Parabolas opening upwards

and downwards pass the vertical line test. Otherwise, fails. So it is a function

In d), the graph is a parabola opening to the left. And as stated in c), it fails the

vertical line test. So it is not a function.

Therefore, A and D are not functions.

Solution #2: Term observation

Key point: observe the terms with ym(for m>1)

In a), there is a term with y2. Therefore, it is not a function

In b), there are terms with variables x and y only. Therefore, its a functionIn c), there are terms with more than degree it is in the variable x. and y has a

maximum degree of 1. So it is a function.


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In d), there is a term with y2in the equation. So it is not a function.

Therefore, A and D are not functions

3)

What is the domain of y =

2x

4

x2 4 ? [, 2 2,2 2, or \ 2, 2]Identity: Domain a set of real numbers that are possible to be an x-value of agiven function so that the value of f(x) will be a defined real number.

y =2x 4x2 4 = 2(x 2)x + 2(x 2)

It is known that when a denominator becomes zero, the number will beundefined. Therefore, x + 2x 2 0 ~ x 2Therefore, the domain is {2} or , 2 2,2 (2, )x 2 cannot be cancelled because when x = 2,

f2 = 2(2 2)

2 + 2

(2

2)

=2(0)

4(0)

Division by zero is not allowed.

4)If fx = 2x + 1 and gx = x2 4, what is f + g2 + fg2? [5]Identity: (f + g)(x) = f(x) + g(x), (fg)(x) = f(x) * g(x)

(f + g)(2) = f(2) + g(2) = 5 + 0 = 5

(fg)(2) = f(2) * g(2) = 5 * 0 = 0

(f + g)(2) + (fg)(2) = 5 + 0 = 5

5)Same with the given in 4), what is the domain of g o f?

[, 32

12

, ]


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Solution #1 Conventional

g fx = g fx = (2x + 1)2 4 = 4x2 + 4x 3To get the domain, the radicand (the number inside the radical symbols) of

even roots, like the square root, should be nonnegative

4x2 + 4x 3 0 ~ 2x 12 x + 3 0Critical intervals: , 3

2 , 3

2,

1

2 , and 1

2,

(, 32

] 32

,1

2 1

2,

2x

1 - - or 0 + or 0

2x+3 -

Or 0 + or 0 +2x 1(2x + 3) + or 0 - Or 0 + or 0The intervals that satisfy 4x2 + 4x 3 0 is

, 32

12

, Hence, the domain.

Solution #2 Sets

Denote {dom f} and {dom g} as the domains of g and f.

Identity: dom g o f = dom f x fxdom g}xfx dom g}means set of x such that f(x) is an element in the domain of g.

fx = 2x + 1, since it s a linear function, dom f = x fx dom g} = (fx)2 4

From there, (fx)2 4 ~ (2x + 1)2 4 ~ (2x + 1)2 4 0~ 2 x + 1 22 x + 1 + 2 = 2x 12 x + 3 0So, x fx dom g} = , 3

2 1

2,


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Therefore, dom g o f = , 32

12

, = , 32

12

,

6)What is the equation of the line that is perpendicular to the line x

y + 2

= 0 and passes through (1, 0)? [x + y 1 =0]The perpendicular line can be expressed as:-x y + k = 0.Since (1, 0) is in x y + k = 0, substitute the coordinates,

-(1) (0) + k = 0, k = 1,the equation will be

x

y + 1 = 0,

Therefore, the equation of the line is x + y 1 = 0.7)Find the distance between the two parallel lines x 3y + 2 = 0 and

2x 6y 3 = 0. [ 720

10 units]Identity: given the lines a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0 and given that

they are parallel, the distance between them is defined as:

d = | c1 c2|A2 + B2A = a1 = a2and B =b1= b2

First, we make sure that the coefficients of x and y of the two equations are the

same,

x

3y + 2 = 0 and 2x

6y

3 = 0 have different coefficients. So we multiply 2

on all sides of x 3y + 2 = 0 ~ 2x 6y + 4 = 0.So, d =

|4 3|22 + (6)2 = 740 = 720 10 units8)Four more than four times a number is the same as eight less than seven

times the number. Find the number. [4]


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4N + 4 = 7N 8 ~ N = 49)Write the equation y = 2x2 + 5x + 2 in the form y = a(x h)2 + k.

[ y = 2

x +

5

42

9

8

]

Key point: completing the square

y = 2 x2 + 5 x + 2 y = 2 x2 + 52

x + 2 y = 2 x2 + 5

2x +

25

16 25

16 + 2

y = 2 x2 + 52

x +25

16 25

8+ 2

y = 2 x + 542 98

10) With the same given in 9), what is the minimum value of y? [ 98]

Identity: y = a(x + h)2 + k is the vertex for of the equation. Where k is the

minimum value of y, x = -h is the axis of symmetry, and (-h, k) is the vertex ofits graph.

Therefore, the minimum value of the function is k = 98

11) With the same given in 9), what is the axis of symmetry?

[x = 5

4]

x +5

4= 0 x = 5

4


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12) Find a quadratic equation whose roots are 2 and -5.

[x2 + 3x 10 = 0]Using Vietas identity,

+ =-

3 and =-10.

the equation is x2 (3)x + 10 = 0 x2 + 3x 10 = 013) Find a quadratic equation with integral coefficients whose roots are

1

3(2 2). [9x2 12x + 2 = 0]

Using Vietas identity, + = 43

and = 29

x2 43

x +2

9= 0

9x2 12x + 2 = 014) If x + 2 is a factor of kx3+ kx2x + 2, what is k?

[k = 1]

Using the remainder theorem,

let Px = kx3 + kx2 x + 2x + 2 = 0, therefore, x = -2,

P2 = 0 = 8 k + 4 k + 2 + 2 k = 1

15) Find the remainder when 2x35x2+ 3x 2 is divided by x 2. [0]


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x 2 = 0, x = 2Remainder = 16 2 0 + 6 2 = 0

16) How many possible real roots can 2x3

5x2+ 3x

2 = 0 have? [1]

Key point: Descartes Rule of signsPx = 2x3 5x2 + 3x 2

In P(x), there are 3 sign variations. (From 2x3to -5x2, -5x2to 3x, and 3x to -2),

which means there are 3 or 3-2 = 1 possible positive real root/s.

Then try P(-x),

Px = 2x3 5x2 3x 2In P(-x), there are no variation of signs.

Therefore, there are 3 or 1 possible roots are all are positive.

By the Rational Root Theorem, these are the possible roots:22

= factors:1,2

1,2= 1, 2,

1

2

And only P(2) = 0. Therefore x 2 is a factor of 2x35x2+ 3x 2.2x35x2+ 3x 2 = (x 2)(2x2x + 1)Now, we try to find if 2x2x + 1 has real roots.Using its discriminant, D = (-1)24(1)(2) < 0Therefore, 2x2

x + 1 have imaginary roots. Therefore, there is only 1 real

root.

17) If 1 and 2 are the roots of x3 8x2 + 17x 10 = 0, what is the thirdroot? [5]


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Solution #1 Factor Theorem

Since 1 and 2 are the roots of the equation, it means that x 1(x 2) is afactor of x3 8x2 + 17x 10.

x3

8x2 + 17x

10 =

x

1

x

2

(x

5)

Therefore, the third root is 5

Solution #2 Vietas identityx x x = x3 + + x2 + + + x

+ + = 8, + + = 17, = 10And

= 1 , = 2

from the first equation, = 5, similarly, in equation 2, =5, and same withequation 3, = 5.Which clearly means that 5 is the third root

18) What is the range of y = 2x 2? [(0, )]Key point: The domain of the inverse of a function is the range of the function.

Finding the inverse of y = 2x 2,x = 2y 2 y = l nx + 2 f1(x)

Finding the domain of the inverse of f,

In logarithms, the one inside the logarithm operator should always be positive.

Therefore, dom f1 = range f = (0, )

19) Solve for x: 44x = 64x+1 . [x = 3]

44x = 43x+3

4 x = 3 x + 3

x = 3


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20) The population of a certain town is presently 100000. If the

population increases by 2% per year, what shall be the population after 3

years to the nearest thousand? [106000]

100000 2100 + 100000 year 1therefore, 100000 102

100x year x

Year 3 100000 102100

3 = (102)310

10600

21) If log2(3x + 4) = 4, what is x? [x = 4]

24 = 3 x + 4 16 = 3x + 4 x = 4and since 3x + 4 > 0 > 4

3, x = 4 satisfies

22) If f

x

= 4x

7, what is f1

x

? [f1

x

=

x+7

4]

y = 4 x 7 x = 4 y 7 y = f1 x = x + 74

23) If fx = 3 2x , what is f1 x? [f1 x = log2 x3]y = 3

2x

x = 3

2y

log2 x = log2

3

2y

log2 x = log2 3 + y l o g2 2 y = log2 x log2 3 y = f1 x = log2 x3

24) The radius of a circle is 5cm. What is the length of an arc, to the

nearest tenth of a cm, subtending an angle of 30 degrees? [2.6 cm]


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Identity: arc length =x

3602r

arc length =30

36025 cm = 5

6 5

63.14 2.6 cm

25) The minute hand of a clock is 10 cm long. How far does its tip travel

in 25 minutes? [25

3cm]

Identity: In the minute hand of a clock, it moves 6 degrees per minute.

25 minutes = (25)(6) = 150 degrees

distance covered = arc length =150

3602

10 cm

=

25

3 cm

26) On a unit circle, what are the coordinates of the terminal point of an

arc in standard position of length7

6? [ 3

2, 1

2]

Since7

6

terminates in the third quadrant, the sign of the coordinates will be

both negative.

And the coordinates are: cos,sin cos 76

, sin 76

= 32

, 12

27) If sin A = 1/3, what is cos A is A is in the first quadrant? [2

32]

Identity: sin2

A + cos2

A = 1 ~ the Pythagorean identity

13

2 + cosA = 1 since A is in quadrant I, cosA > 0 cos A = 23

2

28) In ABC, BC = 20 cm, sin A = 0.4, and sinB = 0.9. Find AC. [45 cm]


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Find a point D on AB such that CD will be perpendicular with AB. From there,

triangles BDC and ADC are right triangles with ADC and BDC are 90 degreesIn ADC, sin A = DC

AC DC = AC sin A

And in BDC,sin B =

DC

BC DC = BC sin B

From the two equations,

AC sin A = BC sin B AC = BCsinBsinA

=20 cm(0.9)

(0.4)= 45 cm

29) Two roads diverge from point A at an angle of 30 degrees. Pete

walks along one road at 8 km per hour. And Joe walks along the other

road at 6 km per hour. How far apart are they after two hours?[400 1923 km]

Identity: by cosine law, c2 = a2 + b2 2ab cos CAfter two hours, Pete already covered (8kph)(2hr) = 16km

while Joe covered (6kph)(2hr) = 12 km

Construct ABC such that A is the point A in the problem, so we can deducethat BAC is 30 degrees, AB = 16 km, and AC = 12 km. To get the distances of Pete and Joe, we will use the cosine law to get the length

of BC,

BC2 = AC2 + AB2 2ABAC cos 30BC2 = 12 km2 + 16 km2

216km12km

32

BC2 = 144 + 256 1923 km2BC = 400 1923 km, hence the final answer.

Because it cannot be simplified anymore


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30) A base angle of an isosceles triangle is 30 degrees. If the base is 24

cm, find the length of a leg of the triangle. [83 cm]Construct isosceles ABC such that AB = BC, and BAC = BCA = 30 degrees,and AC = 24 cm. The legs are AB and BC

Produce segment BD where D is on AC and BD is perpendicular to AC. By thetheorems, BD also bisects AC.

Therefore, AD = DC = 12 cm.

Now we have right triangles BDA and BDC which are congruent.cos A = cos C = cos 30 =

AD

AB=

CD

AC

AD = CD = AB cos 30 = AC cos 30 = 12 cm

hence, AB = AC =12

cos 30c m = 83 cm

31) A wheel has radius of 25 cm. Points A and B are on its rim. Find the

length, to the nearest tenth of a cm, of the arc between A and B if the arc

subtends and angle of 160 degrees. [69.8 cm]

arc length = 160360

225 cm = 2009 = 2009 3.14 = 69.8 cm

32) From a certain point on a plain, the elevation of a mountain peak is

45 degrees. From a point 300 m nearer, the elevation is 60 degrees. Find

the height of the peak to the nearest meter given that (tan 60 = 1.732).

[710 m]

Let x = CB.


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From the given, ABC = 45 degrees, ADC = 60 degrees. From from a point 300 m nearer,BD = 300 m, CD = x 300.

tan ABC = tan 45 , tan ADC = tan 60

tan 45 =AC

x AC = x tan 45

tan 60 =AC

x 300 AC = x 300 tan 60AC = x tan 45 = x 300 tan 60 tan 60 1x = 300 tan 60x =

300 tan 60

tan 60

1

=300(1.732)

1.732

1

=519.6

0.732= 709.8

710 m

33) Iflog2 0.3010, log 3 0.4771, and log 7 0.8451, find anapproximate value of log 42. [1.6232]

Identity: log AB = log A + log B

log 42 = log 2 + log 3 + log 7 = 0.3010 + 0.4771 + 0.8451 = 1.6232

34) The half-life of a substance is 15 hours, how long will it take for a

given amount of the substance to decay to 25% of its original amount?

[30 hours]

Identity: Amount = Pekt

where: P = initial amou

nt, e = 2.718281828,k = constant ,t = time

half life: Amount =1

2P = P e15k e15k = 1

2

25% decay:1

4P = P ekt ekt = 1

4

since 12

2 = 14

, (e15k)2 = e30k = ekt t = 30 hours


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35) Solve for x in the equation 3x+1 = 22x+1.

[log 3log 2log 4log 3 or log 3 2112 log 3 2]

Solution #1: Common logarithm

log3x+1 = log22x+1 x + 1log 3 = 2 x + 1log22log2x + log 2 = log3x + log 32log2 log3x = log 3 log 2x =

log3 log2log4

log3

Solution #2: Base 3

log3 3x+1 = log3 22x+1 x + 1 = 2 x + 1log3 2x =

log3 2 11

2log3 2

36) Find the intersection of the lines:

x y + 2 = 0 and 5x + 4y 8 = 0. [(0, 2)]Solve simultaneously for x and y:

x

y + 2 = 0

4x

4y + 8 = 0

#1

5 x + 4 y 8 = 0 5x+4y 8 = 0 #2#1+#2 9 x = 0 x = 0, then y = 2 (0, 2)37) For what values of x is x2 x 6 < 0?

[2, 3or 2 < < 3]x2

x

6 < 0

x

3

x + 2

< 0


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Critical intervals: , 2, 2, 3, and (3, ), 2 2, 3 (3, )x + 2 - + +x 3 - - +

x

3

x + 2

+ - +

In the interval (-2, 3), (x 3)(x + 2) is negative, that satisfies the inequality,thus, the answer. Or can be written as -2 < x < 3

38) What polynomial of lowest degree must be multiplied to

P

x

= 4x3

4x2

15x + 18 to make it a perfect square polynomial? [(x

+ 2)]Using factor theorem, remainder theorem, and the rational root theorem,

P(x) can be factored as:

Px = x + 2(2x 3)2since 2x 32is already a perfect square,x + 2 should be multiplied to a polynomial so that

P will be a perfect square polynomial.and the polynomial that will be multiplied is x + 2

39) Given the table, what is the defining quadratic equation?

[ y = 2x23x + 1]Table:

x -1 0 1 2 3 4

y 6 1 0 3 10 21

Let P(x) = ax2+ bx + c

In quadratic equations, 3 points will already be enough to define one.

Tip: choose the easiest points.Points: (-1, 6), (0, 1), and (1, 0)


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P(-1) = a b + c = 6, P(0) = c = 1, P(1) = a + b + c = 0From there, a = 2, b = -3, c = 1

P(x) = y = 2x23x + 140)

What is the domain of y = log5(x2 16)?[, 4 (4, )]

From the definition of logarithms,

x2 1 6 > 0 x + 4x 4 > 0Critical intervals:

,

4,

4, 4, and 4,

, 4 4, 4 4, x + 4 - + +x 4 - - +x + 4x 4 + - +From the table, , 4, and 4, satisfies the inequality.Uniting the intervals, , 4 (4, ) hence, the domain

41) What is the range in 40? [ or (, ) or all real nos.]y = log5(x

2 16) x2 = 5y + 16 x = 5y + 16To solve for the range, we must find ALL possible values of y. the technique is

the same like solving for the domain.

5y + 16 > 0 5y > 0, , 5y + 16 > 0 Range: y

42) Solve for x in the equation:1

3logx 8+lo gx 18 = 2.

[x = 6]


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Identity: logb ac = c logb a

1

3logx 8+lo gx 18 = 2 logx 2+lo gx 18 = 2 logx 3 6 = 2 x2 = 6 x = 6 (x > 0)

43) If a = log4 x, what is log4(16x1) in terms of a?

[2 a]log4(16x

1) = log4 16 + log4 x1 = 2 log4 x = 2 a44) If f(x) = x2+ 4x

5 and x -2, find f -1(x).

[y = 2 x + 9 ]Identity: a2 = a if a > 0a2 = a if a < 0

f

x

= y = x2 + 4x

5 and x

2

x = y2 + 4y 5 y2 + 4 y = x + 5 y2 + 4y + 4 = x + 9 y + 22 = x + 9 y 2, therefore, y + 22 = y + 2 y + 2 = x + 9 y = f1 x = 2 x + 9

45) A polygon had d diagonals. If the polygon had one more side, it

would have d+10 diagonals. How many sides does the polygon have? [11]

Identity: if there are n sides in a polygon, there aren(n 3)

2diagonals


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Let n= number of sides.

If it is an n-gon, there are d =n(n 3)

2diagonals.

If it will be an (n+1)-gon, there will be

d + 1 0 = n+1

(

[n+1

]3)

2 = n+1

(n

2)

2 diagonals

The equations will be:

d =n(n 3)

2and d + 10 =

n + 1(n 2)2

Substituting the first equation to the second,

n

n

3

2+ 1 0 =

n + 1

n

2

2

nn 3 + 2 0 = n + 1n 2 n2 3 n + 2 0 = n2 n 2 2n = 22 n = 1146) Find all values of , where 0 < 2, so that 3 3cos = 2sin2 .

[

=

1

3

,

5

3

, 2

]

3 3cos = 2sin2 31 cos = 21 cos2 31 cos = 21 cos 1+c o s 1 cos 3 [2 + 2 c o s ] = 0 1 cos 1 2cos = 0 cos = 1, 1

2

= 2

,1

3

,5

3

47) If sin A =5

13and cos B =

8

17, where A and B are both in the first

quadrant, find sin(A + B). [220

221]

Identity: sin2 + cos2 = 1, sin(A + B) = sin A cos B + cos A sin B


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cos A = 1 sin2A = 1 513

2 = 1213

cos A > 0sin B = 1

8

17

2

=15

17(sin B > 0)

sin(A + B) = sin A cos B + cos A sin B = 513

817

+ 1213

1517

= 220221

48) For what values of x is 3x2

7x

6

0?

[ , 23 [3, )]3x2 7x 6 0 3 x + 2x 3 0

Critical intervals: , 23

, 23

, 3 ,[3, )

,

2

3

2

3, 3

[3, )3 x + 2 - Or 0 + or 0 +x 3 - - or 0 + or 03 x + 2x 3 + or 0 - Or 0 + or 0

Therefore, , 23

[3, )

49) Divide 43 2 by 23 . [ 326 ]23 23 223 = 23 2 = 2

13

+12 = 2

56 = 326


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50) If 4x + 6 = 5 2x , what is x? [x = 1, log2 3]Let A = 2x

A2 + 6 = 5A A 3A 2 = 0 A = 2, 32x = 2, 3

x = 1, log2 3

B.

Division Finals

A.15 seconds

1)What is the slope of the line given by 2x + 3y 5 = 0? [-2/3]slope = a

b= 2

3

2)What is the domain of the function

x

=

x2

9 ?

[, 3] [3, )]x2 9 0 x + 3x 3 0 , 3] [3, )3)If the perimeter of a rectangle is 16x + 8 cm and its width is 2x + 6 cm,

what is its length? [6x 3 cm]P = 2(L + W) L =

P

2 W =16x+8

2 2 x + 6 = 6x 3 cm4)For what values of x is |3x 9| = 6? [x = 1, 5]

3x 9 = 6 x 3 = 2


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x 3 = 2 x = 53 x = 2 x = 1 both satisfies, x = 1 , 5

5)What are the roots of the equation 2x2 + x 15 = 0?[x = 3,

5

2]

2x2 + x 15 = 2x 5x + 3 = 0 x = 3, 52

6)How many degrees is an angle that is7

6radians? [210]

Identity: radians = 18076

radians = 76

180 = 210

7)For what value(s) of k will (1, 0) be on the graph of

y = x2 + 4x + k ? [k = -5]

0 = (1)2 + 41 + k k = 58)What is the range of y = x2 3? [ [3, )]

x2 0 for x therefore, minimum x2 = 0, miny = 3range

y

= [

3,

)

9)If fx = 5x 3, what if f1?[ 1415

]

f1 = 51 3 = 15

3 = 1415


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10) What are the coordinates of the terminal point of an arc of a unit

circle of length5

6, which starts at (1, 0) and is directed

counterclockwise? [

32

,1

2

]

cos ,sin = cos 56

,sin 56

= 32

,1

2

B.30 seconds

1)If y units varies inversely as the square of x, and y = 3 when x = 4, find y

when x = 2. [y = 12]

y =k

x2 k = x2y = (4)2 3 = (2)2 y y = 12

2)

If 64x = 162x 1, what is x?[x = 2]43x = 44x 2 3x = 4x 2 x = 2

3)Bob stands 500 m from a point directly below a flying helicopter. Theangle of elevation of the helicopter from Bobs position is 30 degrees.How high is the helicopter flying? [

500

33 m]


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From the given, Bob = 30 degrees.Let h = distance from the helicopter to the ground

tan 30 =h

500 h = 500 mtan 30 = 500

33 m

4)What is the remainder when 3x3 9x2 + 8x 5 is divided by x 3? [19]Solution #1: Remainder theorem

x 3 = 0 x = 3Px = 3x3 9x2 + 8x 5 P3 = 19

5)For what values of x is 3x2 + 2x 5 > 0? [x > 1, x < -5/3]3x2 + 2x 5 > 0 3 x + 5x 1 > 0 > 1, < 5

3


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C.1 minute

1)Solve for x: log2(3x + 1) log2(x 3) = 3. [x = 5]log2 3 x + 1 log2 x 3 = 3 log2

3 x + 1

x 3 = 3 3 x + 1x 3 = 23 3 x + 1 = 8 x 24 x = 5

since 3x + 1 > 0 > 13

, and x 3 > 0 > 3and x = 5 satisfies the domain, the solution is x = 5

2)For what values of on 0 < < 2 is 3 3cos = 2sin2 ?[ = 1

3, 5

3]

3 3cos = 2sin2 31 cos = 21 cos 1 + cos 1 cos 1 2cos = 0 cos = 1, 1

2

cos

= 1 have no solutions in 0 0 d = 5

8s

5)If fx = 3x 1 and gx = 3x 5, what is the domain of g o f ? [89

, or x 89]

Solution #1 Composite functions

g o fx = 33x 1 5 = 9x 8dom g o f = 9x 8 0 x 8

9or [

8

9, )

Solution #2 Sets

dom

g o f

= dom f

x

g

x

dom g}

dom f = x since fxis linearxgx dom g} = gx 53

3x 1 53

x 89

dom g o f = x x 89

= 89

, or x 89

D.

Sectoral Finals NCR A

A.15 seconds

1)If f

x

= 3 + 4x, find the function g(x) such that f

g

x

= x. [g

x

=

x 34

]


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fgx = 3 + 4gx = x gx = x 34

2)

For what value(s) of k does kx3 5x2 3x leave a remainder of -6 whendivided by x 1. [k = 2]x 1 = 0 x = 1Px = kx3 5x2 3xRemainder = P1 = k 8 = 6 k = 2

3)For what value(s) of x is log(x2 9) a real number?[, 3 (3, )]

x2 9 > 0 x + 3x 3 > 0 > 3, < 3 , 3 3, 4)What is the sum of all the real zeroes of

y = 4x(4x 8)(4x + 64)? [ 23]4x 4x 84x + 64 = 0, since 4x > 0 ,4x and 4x + 64 0. So 4x 8 = 0 x = 2

3

5)What is the value of log8

(log4

16) ? [1

3]

log4 16 = 2 log8 2 = 13

6)In rolling two dice, what is the probability of getting a sum of 8? [5

36]


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The numbers given in a dice is from 1 to 6 only.

And 8 = 2 + 6, 3 + 5, and 4 + 4

So the combinations are: 2, 6 3, 5 and 4, 4The combinations can be permuted because in 2 dice, the first can show 1 then

the second can show 2, and vice versa.

So, the number of ways that two numbers can show a sum of 8 is:

2 ! + 2 ! +2!

2!= 5

if 4, 4is permuted, there are 2! permutations.but there are 2 identical numberss

so the total permutations is 2!2!

probability =possible ways

possible outcomes=

5

6 6 = 536Possible outcomes = 6 * 6 since there are 6 possible outcomes in the first dice

and same with the second.

7)If 42x 2 = 2x , what is x? [ x = 43]

24x 4 = 2x 4x 4 = x x = 43

8)

What is cos if the terminal side of the angle in standard position lieson the segment joining the origin and the point(4, -5)? [

4

4141 ]cos = abscissa

distance=

4(4)2 + (5)2 = 441 41


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9)The vertex of the parabola y = x2 + 4x 3 lie on the liney = k, what is k? [k = -7]

y = x2 + 4x

3 = (x + 2)2

7

vertex

2,

7

k =

7

10) Convert7

4radians into degrees. [315 degrees]

7

4 = 7

4180 = 315

B.

30 seconds

1)If 4x + 2x1 = 12, what is the value of x? [x = -1]

22x +2x

2=

1

2 2x = A A2 + A

2=

1

2 2A2 + A 1 = 0

2A 1A + 1 = 0 A = 1, 12 2x = 1, 12 but 2x > 0, 2x = 12

x = 1

2)If the polynomial f(x) has zeroes 5, -2, and 1, and a y-intercept of -15,

what is f(x) in factored form?

[y = 32 x 5x + 2(x 1)]The zeroes of f(x) are 5, -2, and 1.

Therefore, x = 5, x = -2, x = 1,

x 5 = 0, x + 2 = 0, x 1 = 0 x 5x + 2x 1 = 0


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We assume that fx = x 5x + 2(x 1)If the y-intercept is -15, f(x) should satisfy f(0) = -15

But, f0 = 0 50 + 20 1 = 10 15. This means that f(x) is NOTONLY (x

5)(x + 2)(x

1). There must be a multiplier to the factored

polynomial. So we let it as A.

fx = Ax 5x + 2x 1 f0 = 10A = 15 A = 32

fx = 32

x 5x + 2(x 1)

3)

If logx+11

16 = 2

3 , what is x? [x = 63]

x + 1 23 = 116

x + 123 = 16 x + 123 = 24 x + 12 = 212 x + 12 212 = 0 x + 1 + 26 x + 1 26 = 0 x + 6 5x 63 = 0 x = 63, 65 but x + 1 > 0 > 1 = 63

4)Find the vertex of the parabola y = 4x2 + 3x 1.[ 3

2, 25

16]

y = 4

x2 +

3

4

x +9

64 9

64 1 = 4

x2 +

3

4

x +9

64

9

16 1

= 4 x + 38

2 2516

vertex 38

, 2516


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5)If tan = 43, and the terminal side of the angle in standard position

lies on the second quadrant, find sin 2.[ 24

25]

Identity: sin 2

= 2 sin

cos

In the second quadrant, cos < 0, sin > 0In right triangles,

sin A =opposite

hypotenuse, cos A =

adjacent

hypotenuse, tan A =

sinA

cosA=

opp.

adj.

tan = sin cos

=

4

3

=4k

3k

sin = 4k, cos = 3kfrom sin2 + cos2 = 1, 4k2 + 3k2 = 1 k2 = 125 k = 1

5, k > 0

sin = 45

,cos = 35

fromsin 2 = 2 sin cos ,sin2

= 2

4

5 3

5=

24

25

C.1 minute

1)Find the value(s) of k for which the minimum value of y = 4x2 2kx + 7is equal to the minimum of

y = k 4x x2. [k = -6, 2]Identity: in y = ax2 + bx + c, the vertex is b

2a,

4ac b24a

the y-coordinate of the vertex is the minimum value of a quadratic equation.


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In y = 4x2 2kx + 7, the minimum value is447 (2k)2

4(4)=

112 4k216

= 7 k24

In y = k 4x x2or y = x2 4x + k, minimum value is41k (4)

2

4(1) = k + 4Equating the two minimum values,

k + 4 = 7 k24

k2 + 4k 12 = 0 k + 6k 2 = 0 k = 6,2

2)The area of a right triangle is 12 cm2. If the length of the hypotenuse is

45 cm, how long are its legs? [62 cm, 22 cm]Let b and h be the legs of the right triangle

bh

2= 12, b2 + h2 = 80

bh = 24, and b2 + h2 = 80

Identity: a

2

+ b

2

= (a + b)

2

2ab, (a b)2

= a

2

+ b

2

2abb + h2 = b2 + h2 + 2bh = 80 + 224 = 128 b + h = 82b h2 = b2 + h2 2bh = 80 224 = 32 b h = 42from the two equations, b = 62 cm, and h = 22 cm

3)For what values of x is

2 sin 2x = 2 cos x if 0

x < 360?

[x = 90, 270, 45, 135]

22 cos x sin x = 2 cos x 2 cos x sin x = cos x cosx2sinx 1 = 0 cosx = 0, sin x = 1

22

from cos x = 0, x = 90, 270 and from sin x = 12

2,


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x = 45, 135 x = 90, 270, 45, 1354)Two balls are drawn from an urn that contains 2 red, 3 green, and 5 black

balls. What is the probability that both balls are of the same color if the

first ball is not replaced? [1445]

Let P(X) = probability that event X will occur.

P(Same color) = P(2 red) + P(2 green) + P(2 black)

Solving for P(2 red):

In the first draw, there are 10 balls. 2 of which are red. The probability that the

first draw will be a red is

2

10 =

1

5.After the first draw, there will be 9 remaining balls. And if the first draw was

red, there will be 1 remaining red balls. So the probability will be1

9.

So the probability that the two will occur will be

P2 red = 210

19

=1

45

Similarly, P2 green =3

10 2

9 =1

15, and P2 black =5

10 4

9 =2

9

Therefore, Psame color = 145

+1

15+

2

9=

14

45

5)For what values of a and b is x3 ax2 7x + b divisible by x2 + 2x 3?[a = 0, b = 6]

x2 + 2x 3 = 0 x = 3,1Px = x3 ax2 7x+bFor P(x) be divisible by x2 + 2x 3, it should satisfy,P(1) = P(-3) = 0

P1 = 1 a 7 + b = 0 a b = 6P3 = 27 9 a + 2 1 + b = 0 9a b = 6


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from the two equations, a = 0, b = 6

D.

Sectoral Finals NCR B

A.15 seconds

1)If fx = 2 7x, find g(x) such that f gx = x.[gx = 2x

7]

gx = f1 x = 2 x7

2)For what value of k does kx3 3x2 5x leave a remainder of -6 whendivided by x + 1? [k = 8]

x + 1 = 0

x =

1,

Px = kx3

3x2

5x P1 = k 3 + 5 = 6 k = 8

3)For what values of x is log |x2 9| a real number? [x 3 or /{3}]x2 9 > 0 0but in the inequality, it is STRICTLY > 0.therefore, x2 9 0 x 3 or x /{3}


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4)What is the sum of all the real zeroes of

y = 1 6x 16x 8(16x + 64)? [ 34]

16x 8 = 0 24x = 23 4x = 3 x = 34

5)What is the value of log8 log16 2? [ 23]log16 2 =

1

4 log8 1

4= 2

3

6)

If the polynomial P(x) has zeroes -1, and 2 and y-intercept 4, what isP(x)? [ Px = 2x2 +2x+4]

Px = ax + 1x 2 P0 = 2 a = 4 a = 2Px = 2x + 1x 2 = 2x2 x 2 = 2x2 + 2 x + 4

7)

If 4

92x

1

3

2x

= 0, what is x? [x =

2

5]

49

2x 1 = 32

x 23

4x 2 = 23

x 4x 2 = x x = 25

8)For what angle in the third quadrant is sin 53

equals cos ?[ =

7

6 ]Solution #1 Brute Force

sin 53

= 12

3 = cos in third quadrant = 76


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Solution #2 Identities

Identity: sin(x +2

) = cos x

sin 53

= cos sin 53

= sin x + 2

x + 2

=5

3

x = 76 and it is in the third quadrant

9)What is the maximum value of the function y = 8 + 2 x x2.[3]

Key point: Since the radicand is a quadratic polynomial, get its maximum value

and then do the square root operation

8 + 2 x x2 = 9 x 12 maximum is 9maximum of y = 8 + 2 x x2is 9 = 3

10) What is the probability that a card drawn from a standard deck of

playing cards is a diamond or a jack? [ 413]

Identity: if P(x) is the probability that the event x will occur,

PA B = PA + PB P(A B)PDiamond OR Jack = PDiamond JackPDiamond Jack = PDiamond + PJack P(Diamond Jack)

PDiamond =n

Diamonds

nCards =13

52 =1

4

PJack = 452

=1

13 PDiamond Jack = nJack of diamonds

ncards = 152PDiamond Jack = 1

4+

1

13 1

52=

4

13


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B.30 seconds

1)If 4x + 2x 1 = 12, what is the value of 9x 1? [ 8

9]

22x + 2x

2= 1

2 A = 2x A2 + A2 = 12 2A2 + A = 1 2A2 + A 1 = 0 2A 1A + 1 = 0 2x = 12

x = 1 91 1 = 19

1 = 89

2)

For what values of x is cos 2x = cos x if 0 x < 360? [x =0, 120, 240]Identity: cos 2x = 2cos2x 1 = 1 2sin2x

2cos2x 1 = cos x 2cos2x cosx 1 = 0 2 c o s x + 1cosx 1 = 0 cos x = 1, 1

2 x = 0, 120, 240

3)If the polynomial Px = ax5 + bx3 + cx + 1 is divided byx 3, the remainder is 5, What is the remainder when P(x) is divided by x

+ 3? [-3]

x

3 = 0,

x = 3

P3 = 243a + 27b + 3c + 1 = 5 243a + 27b + 3c = 4x + 3 , x = 3P3 = 243a 27b 3 c + 1 243a + 27b + 3c + 1= 4 + 1 = 3 P3 = 3


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4)From a top of a 60 ft building, a man observes a car moving towards the

building. If the angle of depression of the car changes from 45 degrees to

60 degrees during the observation, how far did the car travel?

[60 203 ft]5)In a high school graduating class of 100 students,64 studied mathematics,

36 studied physics, and 2 studied neither subject. What is the probability

that a student selected at random from this class took only mathematics?

[13

25or 0.52 or 52%]

Let n(E) = the graduating class = 100,

n(M) = math = 64, n(P) = physics = 36,

Now, the number of students who did not take both subject is:

n

M

P

= 12

So the total number of students who took any of the two subjects is

100 12 = 88 = n(M P)


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Now, from the Venn Diagram, nM P = nM only + nP only + n(M P),nM only = 64 x, and nP only = 36 x

88 = 64 x + 36 x + x x = nM P = 12Therefore, nM only = 64 12 = 52, and nP only = 36 12 = 24Finally, PM only =

n(M only )

n(E) =

52

100 =

13

25 or 0.52 or 52%

C.1 minute

1)If xlog x = 1000x2, what are the values of x? [x = 1000,1

10]

xlog x = 1000x2 logxlog x = log1000x2 logxlogx = log 1000 + 2 log x logx2 = 3 + 2 log x Let A = log x A2 = 3 + 2 A A2 2A 3 = 0 A 3A + 1 = 0 A = 3, 1 log x = 3, 1 x = 103 , 101 x = 1000, 1

10

2)A committee of 4 people is chosen at random from among 6 men and 4

women. What is the probability that the committee consists of more

women than men? [5

42]

Identity: Cn, rmeans number of ways of n objects taking r at a time such thattheir arrangement is not a factor. And repetition is not allowed. And

Cn, r = n!r!n r!

Identity: n! = 1 2 3 4 n 2 n 1 n


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Possible ways to choose 4 people such that n(women) > n(men):

(4Women, 0 Men), (3Women, 1Man)

(4Women, 0Men) = C(4,4)*C(6,0) = 1 way ~ 4 people are chosen from the 4

women and 0 people are chosen from the 6 men.

C4,4 =4!

4 4! 4! =4!

0!4! = 1, and C6,0 =6!

6 0! 0! = 1 0! = 1(3Women, 1Men)= C(4,3)*C(6,1) = 24 ways

C4,3 = 4!4 3! 3! = 4, C6,1 = 6!6 1! 1! = 6Total required outcomes: 1 + 24 = 25

Total Possible outcomes in choosing 4 people from 10 people:C10,4 = 10!10 4! 4! = 10!6!4! = 7 8 9 102 3 4 = 210

Probability =25

210=

5

42

3)Identical squares are cut off from each corner of a rectangular piece of

cardboard measuring 7cm by 12 cm. The sides are then folded up to make

a box with an open top. If the volume of the box is 33 cm3, what is the

largest possible length of each side of the square? [9 152 cm]Key point: Equation manipulation

Let x be the length of the sides of the squares.

Each corner is cut by x cm. So each dimension of the rectangle will be cut by 2x

cm. And if the sides will be folded up to make a box, the height will be x cm.


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So the dimensions of the rectangular box are 12 2x cm, 7- 2x cm, and x cm.The volume can be expressed as:

12 2x7 2xx = 3 3 = 1 1 3 1Now, as seen at the left hand side of the equation

, all factors except x has a 2xin it. So for easier manipulation, we let t = 2x,12 t7 t t2

= 12 t7 tt 12

= 33

12 t7 tt = 6 6 = 1 1 6 1Now, we can assume that: 12 t = 11, 7 t = 6, and t = 1 and solving for thevalue of t, the three equations will come up with the same value of t which is

t = 1,t = 2x = 1 x = 1

2

This means that 2x 1 = 0 and 2x 1 is a factor of 12 2x7 2xx 33(from the equation for the volume)

Now, 12 2x7 2xx 33 = 4x3 38x2 + 84x 334x3

38x2 + 84x

33 =

2x

1

2x2

18x + 33

= 0

Solving for ALL values of x in the equation,

x =1

2and 2x2 18x + 33 = 0 x = 9 15

2

Now from the problem, the required answer is the largest value of the side of

the squares that were cut.

From the values of x,9+

15

2 is the largest value. But geometrically speaking, The

side of the square should be less than the dimension of the rectangle which is

7cm by 12cm.

And since9 15

2and

1

2are less than the dimensions


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and9 15

2>

1

2, the largest possible length of each side of the square

is x =9 15

2cm

4)Find the inverse of the function =10x 10x10x +10x . [f1 x = 12 log x+11x]

x =10y 10y10y + 10y 10

y

10y=

102y 1102y + 1

x102y + x = 102y 1 x102y 102y = 1 x 102y x 1 = 1 x

102y =

1 xx 1

=x + 1

1 x log 102y = log

x + 1

1 x 2y = log

x + 1

1 x

y = f1 x = 12

log x + 11 x

5)The sum of all the terms of an infinite geometric sequence is 12 while the

sum of the squares of all its terms is 72. What is the first term of the

sequence? [8]

Let a=first term, r=common ratio.

Let S1= sum of the geometric sequence = a + ar + ar2 + ar3 + = 12Since the sum of the infinite geometric sequence is convergent, the sequence is

also convergent. Therefore, we will use this identity:

Identity: Sum of infinite geometric sequence =a

1

r, where a is the first term

and r is the common ratio.

S1 = a + ar + ar2 + ar3 + = a

1 r = 12Let S2= sum of squares,


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S2 = (a)2 + (ar)2 + (ar2)2 + = a2 + a2r2 + a2r4 + = a2

1 r2 = 72From the equations, a = 121 rand a2 = 721 + r(1 r)Squaring the first equation, a2 = 144(1

r)2. Now, equating a2 = a2,

1441 r2 = 721 + r1 r since r 1 from r < 1,1441 r = 721 + r r = 1

3. And from a = 121 r,

a = 1 2 1 13

= 8

E.

Regional Finals

I. Individual Written Competition (Nationwide)

A.Part I

1)What is the quotient if x6 + y6is divided by x2 + y2? [x4 x2y2 + y4]Key point: Factoring

Identity: a3 + b3 = a + b(a2 a b + b2)x6 + y6 =

x2

3 +

y2

3 =

x2 + y2

(x4

x2y2 + y4)

x6 + y6

x2 + y2= x2 + y2 x4 x2y2 + y4 x2 + y2 = x4 x2y2 + y4

2)Solve for x in the equation 2x 2x 1 = 32. [x =6]


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2x 2x2

= 32 2x2

= 32 2x = 64 x = 6

3)What is the smallest number, which when multiplied by 120, gives a

perfect cube? [225]

120 = 23 3 5 23is already a cube. 3 5 should be multipliedto a numberto make 3 5 a perfect cube which is 32 52 = 225

4)Find the remainder when x50 2x49 + 3x48 4x47 + 10x41whendivided by x + 1. [55]

x + 1 = 0 x = 1Px = x50 2x49 + 3x48 4x47 + 10x41

= x41 x9 2x8 + 3x7 10P1 = 141 1 2 3 4 10 = 155 = 55

5)How many different scalene triangles are possible if the measures of the

three sides are selected from the lengths 2, 3, 4, and 5? [3]

Identity: Triangle inequality a b < < + , where a, b, c are sides of atriangle and it runs through all sides of a triangle.

Sides to be tested: (2,3,4), (2,3,5), (2,4,5), (3, 4, 5)

In (2, 3, 4), 4

2 < 3 < 4 + 2, 4

3 < 2 < 4 + 3, 3

2 < 4 < 3 + 2 the

triangle inequality is satisfied, so 2, 3, 4 can form a triangle

Similarly, (3,4,5) and (2,4,5) can also form triangles. Except for (2,3,5) because

2+3 = 5 but in the triangle inequality, the inequality symbols are strictly < or

>.


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6)What is the value of x if 10 + 10 + 10 + + 10 100 terms = 10x5?[x = 8]

1 0 + 1 0 + 1 0 + + 10 100 terms = 10100 = 1000 = 103 = 10x 5 x 5 = 3 x = 87)Find x if log3(logx 125) = 1. [x = 5]

3log 3 log x 125 = 31 logx 125 = 3 x3 = 125 x = 58)Find all possible values of x that satisfy the equation

2x + 1 = x 3 + 2. [x = 4, 12]2 x + 1 = x 3 + 2 2 x + 12 = x 3 + 22 2 x + 1 = x 3 + 4x 3 + 4 x = 4x 3 x2 = 16x 3 x2 16x + 48 = 0 x 4x 12 = 0 x = 4,12Substituting the values of x in the original equations,

x = 4 24 + 1 = 4 3 + 2 3 = 3x = 1 2 212 + 1 = 12 3 + 2 5 = 5

Since they both satisfy the equations, x = 4,12

9)What is the domain of the function fx = x2 1? [, 1] [1, )]x2 1 0 x + 1x 1 0 x 1, x 1 , 1] [1, )


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10) A chord of a circle of radius 10 cm is 5cm from the center. Find the

length of the chord. [103 cm]Let x = length of the chord.

Identity: If the radius of a circle intersects a chord in a right angle, it bisects the

chord.

Since the 5cm distance of the center from the chord is its perpendicular

distance, the chord is already bisected. Therefore, the chord is divided into two

parts with lengthx

2cm each.

Now, connect one endpoint of the chord and the center of the circle. Since the

Endpoint of the chord is in the circumference of the circle, the segment joining

the center and the endpoint of the chord is a radius of the circle which is 10cm.

Notice that there is a right triangle formed by the perpendicular distance from

the chord to the center, one bisected part of the chord, and the formed radius.

Moreover, the formed radius is opposite the right angle, therefore, it is the

hypotenuse and the legs are the bisected part of the chord and the

perpendicular distance.

So by the Pythagorean Theorem,

x2

2 + 5 cm2 = 10 cm2 x24

+ 25 = 100 x = 103 cm

11) For what values of x is x2 x < 12? [3,4or 3 < < 4]x2

x

12 < 0

x

4

x + 3

< 0

3 0 in logarithms, therefore, x 0

10) Solve for x in 3x+1 = 4x 7. [x = 1+7 log 3 4log 3 41 ]

3x+1 = 4x 7 x + 1 = x 7 log3 4 x = 1 + 7 l o g3 4log3 4 1

11) If 42x+y = 256 and 4x+2y = 1024, solve for x and y. [x = 1, y = 2]

42x+y = 256 256 = 44 2 x + y = 44x+2y = 1024 1024 = 45 x + 2 y = 5adding the two equations, 3x + 3y = 9 x + y = 3from 2x + y = 4 and x + y = 3, x = 1

from x + 2y = 5 and x + y = 3, y = 2

12) Find two pairs of two consecutive integers between which the graph

of Px = 2x4 + 5x3 3x2 + x + 2 crosses the x-axis. [1, 0 and 4, 3]By the rational root theorem, these are the possible roots:

2

2=

1, 2

1, 2= 1, 2,

1

2

But since the problem requires integers, 1

2will be disregarded.

By trial and error,

P1 = 7 which is quite far from 0, if we try a high x-value, x =2, P2 = 64.This means, as the x-value starts that starts from 1, the value of P(x) will also

increase. Therefore, it goes farther from the x-axis. So trying a lower x value,

try x = 0, P(0)=2, it goes nearer to the x-axis. Another lower x-value, x = -1,

P

1

=

5. So from x = 0 to x = -1, it went from P

0

> 0

1

< 0.


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Which means the graph already crossed the x-axis in the interval 1 < < 0.So one pair is -1 and 0

Before trying another possible root, we already knew that x-values greater

than 1 will already give values of P(x) greater than 0 which means, P(x) will

not cross the x-axis on those values. So the negative values will be tried.

Trying values smaller than -1, x = -2, P2 = 20, thats too far from the x-axis. Try x = -3, P(-3) = -1. So it went nearer to the x-axis. Another smaller x-

value, x = -4, P 4 = 142. From there, in the interval -4 < x


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AP

PB=

2

3=

2k

3k AP = 2k, PB = 3k and AQ

QB=

3

4=

3k

4k AQ = 3k, QB = 4k

From the given line, by line segment addition/subtraction:

AQ = AP + PQ and QB = PB PQSo, substituting to the ratio,

AQ

QB=

A P+PQ

PB PQ = 2 k + 23k 2 = 34 k = 14Solving for the length of AB,

A B = A P + P B = 2 k + 3 k = 5 k = 514 = 70 units

B.

Part II

1)Find the equation of the line that passes through the point of intersection

of the lines 2x = 3y + 2 and 5x + 3y = 26, and parallel to the line

x + y = 4. [x + y 6 = 0]Solution #1: Conventional

Solving for the point of intersection of the two lines, and using the slope of the

parallel line, the equation will be x + y 6 = 0.Solution #2: Existence of a real number k

The required equation can be expressed as:

2x 3y 2 + k5x+3y 26 = 0 where k This was derived from the two equations, 2x=3y+2 ~ 2x - 3y - 2=0 and5x + 3y = 26 ~ 5x+3y - 26 =0

Rewriting the expression of the required equation,5 k + 2x + 3k 3y 2 26k = 0


58/113

Now, expressing its slope,

m = 5 k + 23k 3

But the line is parallel to x + y = 4, so they have equal slopes.

5 k + 2

3k 3 = 1 k = 5

2

Substituting to the expression of the required equation,

2x 3y 2 52

5x+3y 26 = 0 x + y 6 = 0

2)The vertices of the base of an isosceles triangle are at (1, 2) and (4, -1).

Find the y-coordinate of the third vertex if its x-coordinate is 6. [4]

Let A(1, 2), B(4, -1) and C(6, y)

Since ABC is isosceles with base AB, the congruent legs are AC and BC.

AC = BC AC2 = BC2Expressing the squares of their respective distances,

AC2 =

6

1

2 +

y

2

2 =

y

2

2 + 25

BC2 = 6 42 + y 1 2 = y + 12 + 4y 22 + 2 5 = y + 12 + 4 2 1 = 2 y + 1 + 4 y 4 y = 43)In counting n colored balls, some red and some black, it was found that 49

of the first 50 counted were red. Thereafter, 7 out of every 8 counted

were red. If, in all, 90% or more of the balls counted were red, find the

maximum value of n. [210]

Key point: In terms of the number of red balls.

Finding the total number of red balls:

After the first 50 balls, there are 49 red balls. After it, there are n 50 balls areleft. And it is given that

7

8of the remaining balls are also red. So the total


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number of red balls is:

49 +7

8(n 50)

It is also given that all in all, the total number of red balls is 90% or more of the

total number of balls. So it can be expressed as:

9

10n or more

To solve for the maximum value of n, the two expressions can be written as an

inequality because as said in the given, it is not exactly 90% of the total is red.

It is 90% or more. So the inequality is:

49 +7

8

n

50

9

10n

49

40

50

35

n

n

210

Therefore, the maximum number of n is 210.

4)If 1 is placed after the last digit of a 30digit number, the resulting four-

digit number is three times the four-digit number formed when 2 is

placed before the first digit of the original number. [857]

To know how to express in algebraic expression the placing of digits to the firstor last digit of a number, we will try to look at these examples:

If 4 is placed in the number 567 after its last digit, the resulting number will be

5674. It is like multiplying 567 by 10 to make it 5670 then add 4 to have the

same result, 5674.

Now, instead of placing 4 after the last digit, we will try to place 4 before the

first digit. So the resulting number will be 4567. It is like simply adding 4000 ti

567 to make it 4567.

Applying the examples given, let N be the 3-digit number.

If 1 is placed after the last digit, the number will be 10N+1.

and If 2 is placed before the first digit of N, the resulting number would be

2000+N. and 10N+1 is thrice 2000+N, as cited in the given.

Solving for N,


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1 0 N + 1 = 32000 + N N = 8575)Prove that any polynomial equation Px = 0 that contains only odd

powers of x and positive coefficients has no nonzero real roots.

Let Px = a1xn + a2xn+2 + a3xn+4 + + am xm The exponents are odd so there are no constant terms.

Pxcan be rewritten as Px = xm Q(x) by factoring.Since P(x) have positive coefficients Q(x) also has positive coefficients. Thus,

Qx > 0 for x . Thus Q(x) has no real root. Which means that x = 0 is theonly real root of P(x). Therefore, there are no nonzero real roots of anypolynomial P(x) with positive coefficients and odd exponents. QED

C.Part III

1)Two ships leave the same pier at the same time. One ship sails on a course

of 110 at 34 km/hr, while the other ship sails on a course of 230 at

40 km/hr. Find the distance between them after 3 hours. [4221 km]After 3 hours, the first ship will have covered 3(34) = 102 km while the other

ship will have covered 3(40) = 120 km.

The first ship is travelling 110 degrees with the pier and the other at 230

degrees along the same pier. The angle formed by the track made by the two

ships is 230 110 = 120 degrees.Using the cosine law, and letting d be the distance between them after 3 hours,

d2 = 1022 + 1202 2102120 cos 120 = 37044 d = 4221 km


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2)Find the coordinates of the centroid of the triangle ABC, given that

A(-3, -3), B(7, 1), and C(5, 9). [3, 73

]Definition: Centroid = the point of intersection of the medians of a triangle

Solution #1: Concurrency of the medians

Since there exist a centroid, the medians of a triangle are concurrent. So by

simply finding the equations of any two medians of a triangle and getting their

point of intersection, the point is the centroid which is 3, 73

.In writing the equation of the medians, connect the midpoint of a certain side

of a triangle and its opposite vertex and then get its equation.

Solution #2: If theres half, theres third.Generalizing solution #1, we get the identity to immediately solve for the

centroid/center of mass of a triangle.

Identity: The center of mass = x1+x2+x33

,y1+y2+y3

3

By applying the identity, The center of mass or the centroid is:

3 + 7 + 5

3 , 3 + 1 + 9

3 = 3,7

3


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3)Point F is taken on the extension of side AD of parallelogram ABCD, where

D is between A and F. The line FB intersects segments AC and CD at E and

G, respectively. If FG = 24 cm, GE = 8cm, find BE. [16 cm]

As seen from the figure, we can see that there are similar triangles.

FDG ~

FAB

FG

FB=

DG

ABand

GEC ~

BEA

GC

AB=

GE

BE

From FDG ~ FAB, FG = 24, and FB = FG+GE+EB = 32+BE,FG

FB=

DG

AB 24

BE + 32=

DG

AB DG = 24AB

BE +32

Now, we see that GC = DC DG. But AB = DC, So GC = AB DG.So from GC = AB DG, GE = 8, and GEC ~ BEA,

GC

AB =

GE

BE AB

DG

AB =

8

BE 8AB = AB DGBEbut DG =

24AB

BE + 32, 8AB = AB 24AB

BE + 32 BE

8 = 1 24BE + 32

BE 8BE + 256 = BE2 + 8BE BE = 16 cm


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III. Team Orals (Nationwide)

A.15 seconds

1)What is the slope of the line 3x 2y + 7 = 0? [ 32]slope = A

B= 32 = 32

2)What is the difference between the roots of x2

12x + 35 = 0. [2]

From a b2 = a + b2 4abLet A and B be the roots of the equation.

A + B = 12, and AB = 35A B2 = A + B2 4AB = 122 435 = 144 140 = 4A B = 2

3)If a + b = 4 and a b = 2, what is 3a? [9]Adding the two equations, 2a = 6 a = 3 3a = 9

4)

If 8

x

= 16, what is x? [

4

3]

23x = 24 3x = 4 x = 43

5)For what values of x is the graph of y = 2x 6 below the x-axis? [x < 3]


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2x 6 < 0 < 3

6)What is the value of cos 120? [ 12]

120 is obtuse, so cosine is negative. cos180 x = cosx cos120 =cos 60 = 12

7)If the third term of a geometric progression is 12 and the first term is 3,

what is the second term? [6]

Identity: an

of a GP = arn

1

a3 = ar2 = 12, and a = 3 r2 = 4 r = 2 a2 = ar = 6

8)If the remainder is 3 when divided by 7, what is the remainder when 9n is

divided by 7? [6]

Solution #1: High School way

n = 7Q + 3 9n = 63Q + 27 9n7 = 277 = remainder 6Solution #2: Modular Arithmetic

n 3 mod7 9n 27 mod7 1 mod7 = 6 (mod7)

9)

What is the value of k if x + 2 is a factor of 2x2

+ 3x + k? [k=-2]

x + 2 = 0 x = 2222 + 3 2 + k = 0 k = 2

10) For what values of , where 0 2, is cos < 0? [ 2

< < 32

]


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A trivial question: cos < 0 Therefore,

2

< < 32

B.

30 seconds

1)For what values of x is x2 + 3x + 2 > 0? [x < 2, > 1]x2 + 3x + 2 > 0 x + 1x + 2 > 0 > 1, < 2

2)The terminal point of an arc of length in standard position lies on thesegment joining the origin and the point (-3, 4). Determine sin . [ 4

5]

sin = yd

d = distance of the pointsthe points are the origin and (-3, 4). Their distance is 5. Therefore,

sin

=

y

d=

4

5

3)A wheel of radius 8 cm made 7 revolutions. Find the distance traveled by

a point on the rim of the wheel in terms of . [112cm]1 revolution = circumference = 2r = 28cm = 16 7 revolutions = 716 = 112cm

4)If 5x+9 = 25x , find x. [x = 3]5x+9 = 25x 5x+9 = 52x x + 9 = 2x x = 3

5)Find the inverse of the function y =3x 5x+4

. [y =4x+5

3

x

]


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y = f(x) =3x 5x + 4

x = 3y 5y + 4

xy + 4x = 3y 5 yx 3 = 5 4x y = f1 x = 5 4x

x 3 = 4 x + 53 x

C.1 minute

1)If 3y 2x = 6 and 3 x 9, what is the range of values of y?[0 y 8]

Since the equation is linear, using the extreme values of x,

x = 3, y = 0 and x = 9, y = 8 0 y 82)If gx = x2 9, what is the domain of g(x)? [, 3] [3, )]

x2 9 0 x + 3x 3 0 x > 3, < 3 , 3] [3, )

3)A rectangle is divided into four smaller rectangles by drawing two

perpendicular lines. The areas of three of the rectangles are 6, 15, and 25

square units. What is the area of the fourth rectangle? [10 units2]


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By arranging the rectangles, the only arrangement is the one in the figure. The

arrangement is done by factoring.

25 can be factored as 5*5, 15=5*3, and 6=3*2. 25 and 15 can be adjacent

rectangles because the common side can be 5 units long. Then 6 and 15 can

have a common side with length 3 units. Now the fourth rectangle also has

common sides with the rectangle with area 25 and 6, respectively. X and the

25-area rectangle have a common side with length 5 units. While X and 6 have

2 since the factor 3 is already the common side of 6 and 15. Therefore, the

dimensions of the fourth rectangle are 5 units and 6 units. So the area of the

fourth rectangle is 10 square units.

4)What is the remainder when 2x4 3x3 + 4x2 x + 4 is divided byx + 1? [14]

x + 1 = 0 x = 1 214 313 + 412 1 + 4 = 14

5)

Find an equation of the line through the points (-4, 3) that is parallel tothe line 2y + 3x 7 = 0. [3x + 2y + 6 = 0]

2y +3x 7 = 0 3x+2y 7 = 0The equation of the line is:

3 x + 2 y + k = 0 34 + 23 + k = 0 k = 6 3x + 2y + 6 = 0


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IV. Team Orals (NCR-A)

A.Part I

1)What are the coordinates of the vertex of the parabola y = x2 + 6x?[(3,9)]

Solution #1: Completing the square

y = x2 + 6x y = x 32 + 9 (3, 9)Solution #2: Midway of zeroes

The zeroes of y = x2 + 6x is x = 0, 6. Since a parabola is symmetric to avertical axis, therefore, any two values of x, say A and B, such that PA = P(B),A+B

2is the x-coordinate of the vertex of that parabola.

The x-coordinate of the vertex is0+6

2= 3, and substituting to the equation, y=9

therefore, the vertex is (3, 9)

2)What is the period of the graph of y = 2 + 3 sin x3

? [6]period =

213

= 6

3)The circumference of a circle is numerically equal to its area. How long is

its diameter? [4]

2r = r2 r = radius = 2 diameter = 2r = 44)If 4x + 41+x = 80, find x. [x=2]


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4x + 41+x = 80 4x + 4 4x = 80 5 4x = 80 4x = 16 x = 2

5)What is the exact value of sin 75 cos 45 + cos 75 sin 45? [1

23 ]

Identity: sin(A + B) = sin A cos B + cos A sin B

sin 75 cos 45 + cos 75 sin 45 = sin(75 + 45) = sin(120) =1

23

6)Solve for x if logx 64 = 2. [x = 8]

x2 = 64

x > 0

= 8

7)If x 1 is a factor of fx = x3 + kx2 + 1, what is the value of k? [k = -2]x 1 = 0 x = 1 f1 = 13 + k12 + 1 = 0 k = 2

8)How many integers are there between log2 3 and log2 100? [5]

The expression, log2 x will give an integer value if and only if x is a power of 2.

Therefore, to count the number of integers between log2 3 and log2 100, we

will find the number of powers of 2 between 3 and 100. The powers are: 4, 8,

16, 32, and 64. Therefore, there are 5 integers.

9)If3

2 < < 2, find the radian measure of for which cos = sin 2

3.

[ = 116

]


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sin2

3 = 1

23 = cos = 11

6

10) A card is drawn from a standard deck of playing cards. What is the

probability that the card drawn is either a queen or a spade? [4

13]

PQueen Spade = PQueen + PSpade PQueen SpadePQueen = 4

52=

1

13 PSpade = 13

52=

1

4 PQueen Spade = 1

52

PQueen Spade = 113

+1

4 1

52=

4

13

B.30 seconds

1)What is the inverse of y = 4 + log3 x? [y = 3x 4]

y = 4 + l o g3 x x = 4 + l o g3 y log3 y = x 4 y = 3x 42)Let p and q be the highest and smallest roots of the equation

x3 3x2 x = 0, respectively. What is the value of p q? [13]Clearly, x = 0 is neither the highest nor the smallest root. Therefore, we will

consider the equation x2 3x 1 = 0.Now, p + q = 3, and pq = -1.

p q2 = p + q2 4pq = 32 4 1 = 13 p q = 133)For what values of x is xx

3+3x2+2x = 1? [x = -1, -2, 1]

Consider this: AB = 1 is satisfied if:


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Case 1: A=1, B x = 1 , x3 + 3x2 + 2x satisfied

Case 2: A 0, B = 0x3 + 3x2 + 2x = 0, x

0

x2 + 3 x + 2 = 0 ,

x =

1,

2

Case 3: A = 1, B = evenx = 1, now we will prove that x3 + 3x2 + 2x = even for x = 1.

x3 + 3x2 + 2x 13 + 312 + 21 = 0 evenTherefore, the values of x are x = 1, 2,1

4)

A 5x5x5 cube is built out of 1x1x1 cubes. If the outside surface of the large

cube is painted, how many small cubes will have paint on at least one

face? [98 cubes]

Key point: To count it fast, subtract the number of 1x1x1 cubes to the number

of cubes that are not painted of any face.

The number of 1x1x1 cubes is 5x5x5=125

The number of 1x1x1 cubes that is not painted on any face:

In an edge of the cube, there are 5 little cubes. The two ends are then painted.

Therefore, 5-2 = 3 cubes are not painted which means, there are 3x3x3 = 27

cubes that are not painted.

Therefore the required number cube is: 125 27 =98 cubes.5)

Find the exact value of sin 2x if sin x = 513, and x is in the second quadrant?

[ 120169

]

From sin2 x+c o s2 x = 1, and sin x =5

13, cos x = 12

13since it is in the second

quadrant.


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sin 2x = 2 sin x cos x = 2 513

1213

= 120169

C.Part III

1)

How many three-digit numbers are there for which the sum of its digits is

5? [15]

5 = 0 + 0 + 5, 0 + 1 + 4, 0 + 2 + 3, 1 + 1 + 3, 1 + 2 + 2

In 0,0,5 , only 500 is the accepted permutation.

In 0,1,4 , 0 cannot be the first digit. So there are 3! 2! = 4 numbers. The 3! isthe number of permutations of 0,1,4. The we will subtract it by 2! because 2! Is

the number of permutations of 0,1,4 such that 0 is in the first digit.

Similarly, 0,2,3, there are 3! - 2! =4 numbers.

Now in 1,1,3, there are3!

2!= 3 numbers. We divided it by 2! because there are 2

identical numbers.

Similarly, 1,2,2 also have3!

2!

= 3 numbers.

Therefore, there are 1 + 4 + 4 + 3 + 3 = 15 numbers

2)Find all ordered pairs (x, y) that satisfy the equations x = yandx2 + y2 = 1.

Since

x

=

y

, x2 = y2

substituting to x2 + y2 = 1, 2x2 = 1 x = y = 22 Therefore, the ordered pairs are: 2

2,

22


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3)In how many ways can the letters of the word PENCIL be arranged in a

row if the consonants are in alphabetical order? [30 ways]

Since the vowels will be in alphabetical order, there should only have 1

permutation of the consonants, which means we can treat the vowels as

identical.

Permutations =(number of letters in PENCIL)!number of consonants! = 6!4! = 30

4)Suppose that the angle between the minute hand and the hour hand of a

clock is 120 degrees. If the minute hand is 21 cm long and the hour hand

is 3 cm long, what is the distance between the tips of the hands of theclock? [37 cm]

Using the cosine law,

d2 = 212 + 32 2213 cos 120 = 37 d = 37 cm

5)Three positive numbers form an arithmetic sequence, the common

difference being 1. If the first number is decreased by 6, the second is

decreased by 1, and the third is doubled, the resulting numbers now forma geometric sequence. What is the biggest among the three numbers in

the original sequence? [36]

The arithmetic progression is expressed as: a, a + d, a + 2d and since the

common difference is 11, it will be a, a + 11, a + 22.

If the first is decreased by 6, ~ a

6


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If the second is decreased by 1, ~ a + 1 1 1 = a + 1 0If the third is doubled, ~ 2a + 2 2 = 2a + 24Now they would form a geometric sequence:

2a +24

a + 1 0=

a + 1 0

a 6 2a2 + 32a

264 = a2 + 20a + 100

a = 14 (a > 0)

The original sequence is 14, 25, and 36. Therefore, the largest is 36.

V. Team Orals (NCR-A)

A.

15 seconds

1)How many integers are there between 3 and 2010? [43]Since 3 < 4, the least integer is 2

and 2010 > 44, 44 .Therefore, there are 44 2 + 1 = 43 integers.

2)If the parabola y = x2 6x + 7 is tangent to y = k, what is k? [k=-2]x2 6x + 7 = k should have only one root. So we can deduce that x2 6x +7

k = y is tangent to the x-axis, therefore, x2

6 x + 7

k should be a perfect

square trinomial. So 7 k = 9 ~ k=-2

3)For what values of x is 9x < 243? [x 0 2 > 0~ > 2 = 4

8)

Find the area of the triangle formed by the coordinate axes and the line

2x + 3y = 16. [64

3units2]

Key point: This line forms a right triangle because the coordinate axes are

perpendicular. The distance of each intercept to the origin is the length of the

leg of that triangle.

The intercepts are:

0,

16

3 and (8, 0). The respective lengths of the legs

therefore are:16

3units and 8 units area = 12 bh = 12 163 8 = 643 units2

9)What is the largest integer value of n for which 20! Is divisible by 3n .

[n=8]

Since 20! = 2

3

4

5

19

20, when divided by 3n where n satisfies the

problem, will give a quotient that does not have any prime factor of 3. So to

make it possible, we will do this algorithm:

20! = product of positive integers from 1 to 20, so there are20

3 6 integers

that has a prime factor of 3. And in that 6 integers, there are6

3= 2 integers that

has a factor of 9 which has a prime factor of 3. Therefore, the power of 3 if 20!

is prime factorized is 9. Therefore n = 9.

10) If the area of the square is increased from 32 cm2to 72 cm2, by how

much is the length of its diagonal increased? [4 cm]

The side of the square with area 32 cm2is 3 2 = 42 cm, and the length of thediagonal is 42 2 = 8 cm.


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Also, the square with area 72 cm2is 62 cm and has diagonal 12 cm.The change in the length of its diagonal is 12 cm 8 cm = 4 cm.

B.30 seconds

1)For what values of k is y =x 2x+k

its own inverse? [k = -1]

y = fx = x 2x + k

x = y 2y + k

y = f1 x = k x + 21 x

x

2

x + k=

k x + 2

1 x kx2

+ 2 x + k 2

x + 2 k = x2

+ 3x 2 kx2 + k2x + 2 k = x2 + x 2Now comparing the coefficient of the variables,

k = 1, k2 = 1, 2k = 2 all satisfy the same value of k = 12)Find the value of sin 2

if sin

< 0 and the terminal side of the standard

angle lies on the line 4x + 5y = 0. [ 4041]The graph of the line is both in the first and fourth quadrant. So cos > 0.Identity: slope = tan Applying the identity,

tan

=

4

5

sin

=

4

42

+ 52

=

4

41and cos

=

5

41

sin2 = 2 sin cos = 2 441 541 = 4041

3)Solve for x if log25 2x 1x+1 = 12. [x = 2, 23]


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log25 2x 1x + 1

= 12

log25 2x 1x + 1

= 12

and log25 2x 1x + 1

= 12

2x 1x + 1

= 2512 = 5 and

2x 1x + 1

= 251

2 =1

5

x = 2,2

3 and2x

1

x + 1 > 0 2x 1x + 1 > 0 < 1, >1

2 ,

x = 2, 23

4)One base of a triangle measures 30 degrees, and the side opposite this

angle is 2 in. What is its perimeter? [4 + 23 in]The two base angles of the triangle measure 30 degrees and the vertex angle is

60 degrees. The other leg of the isosceles triangle also has length 2 inches.

Identity: Sine law asin A

=b

sin B=

c

sin Cwhere the small letters are the length of

the sides of the triangles and the capital letters correspond to their respective

opposite angles. Applying it,

2

sin 30=

x

sin 60

x = 2

3 in

Therefore the perimeter of the triangle is

2 + 2 + 23 = 4 + 23 inches

5)A number is selected at random from the set {1, 2, 3,

, 100}. What is the

probability that the number selected is neither divisible by 3 nor 5? [53

100]

Let nE = 1,2,3, ,100 = 100, Px be the probability function andPx = n x

n ENote that:

n

not divisible by 3 or 5

= n

E

n

divisible by 3 OR 5


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ndivisible by 3 OR 5 = ndivi. by 3 + ndivi. by 5 nboth 3 and 5nboth 3 and 5 = ndivisible by 15 since LCM3, 5 = 15

Now, finding nboth 3 and 5:15, 30, , 15 6 = 6 elementsSimilarly, n

divi. by 3

and n(divi. by 5):

3,6,9, , 3 33 = 33 elements, and 5, 10, 15, , 5 20 = 20 elementsFinding ndivisible by 3 OR 5,

ndivisible by 3 OR 5 = 33 + 20 6 = 47 elementsTo solve for nnot divisible by 3 or 5

nnot divisible by 3 or 5 = 100 47 = 53 elementsFinally,

Pnot divi. by 3 OR 5 = 53100

C.1 minute

1)Suppose that x + 1 is a factor of fx = ax4 + 2x3 13x2 + bx + 4. If f(x)leaves a remainder of 12 when divided by x 1, what will be theremainder when f(x) is divided by x + 2? [168]

From the given, f 1 = 0, f 1 = 12, and f2 = unknownf1 = a 2 13 b + 4 = 0 a b = 11f1 = a + 2 1 3 + b + 4 = 1 2 a + b = 1 9a = 15, b = 4f2 = 16a 16 52 2b + 4 = 16a 2b 64 1615 24 64

= 168


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2)If 3x = 52x+1, then the value of x can be written in the formlog a

log b, where a

and b are positive rational numbers. Find ab. [3

5]

3x = 52x+1 x = 2 x + 1log3 5 xlog3 25 + log3 5 = x xlog3 25 1 = log3 5 x =

log3 5

1 log3 25 log3 5

log3 3 log3 25=

log3 5

log33

25

=log 5

log3

25

a = 5, b = 325

ab = 35

3)In how many ways can the letters of the word COSECANT be arranged in

a row so that the vowels are in alphabetical order? [3360]

The number of ways to jumble COSECANT can INITIALLY be counted as 8!Now, notice that there are 2 identical letter Cs. So 8! Will be divided by 2! Sothe running total now is

8!

2!.

Lastly, treat the vowels as identical letters so that it will only have 1

permutation since alphabetically arranged letters only have 1 permutation.

There are 3 vowels so the running total will be divided by 3!Therefore, the total ways is

8!

2!3!= 3360

4)

If 16cos2

x + 2 16sin2

x = 12, what are the possible values of cos2 x? [3

4 ,1

2]

16cos2 x + 2 16sin2 x = 12 16cos 2 x + 2 161cos 2 x = 12

16cos 2 x + 2 1616cos

2 x= 12 Let A = 16cos 2 x A + 32

A= 12

A2 12A + 32 = 0 A 8A 4 = 0 A = 4, 8


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16cos2 x = 4 cos2 x = 1

2, 16cos

2 x = 8 cos2 x = 34

cos2 x =3

4,1

2

5)A farmer wishes to fence a rectangular field along a river. No fencing is

needed for the side adjacent to the river. The fencing costs Php 30 per

foot for the side parallel to the river and Php 20 per foot for the other two

sides. If the farmer plans to spend a total of Php 2400 for fencing, what

dimensions will maximize the area of the field? [30ft by 40ft]

Note that the side adjacent to the river is not fenced; therefore, only three sides

of the field will be fenced.

Let x be the length of the side parallel to the river. And y be the other two sides.

From the given, x is Php 30 per feet. And y is Php 20 per feet.

The total fencing can be expressed as:

3 0 x + 220y = 2400 3x + 4y = 240Solving for y,

y =240 3x4 = 60 34 x

Multiplying by x to make it xy so that it can express the area of the field,

xy = 60x 34

x2 Max is at x = b2a

= 602 3

4 = 40 ft

and y = 30 ft by solving for y. Therefore the dimensions are 40 ft by 50 ft.

VI. National Finals

I. Individual Written Round


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A.Part I

1)Find the digit x in 18! = 6402370572x00. [x = 8]

Note that 10! =

800

8 is the last nonzero digit

Algorithm:

18

10 1 81 = 8 18! ~ 8 8! = 80 8 is the last nonzero digit

In 8 8!, 8 is from 10!, and 8! is from the algorithmTherefore, x = 8

2)If 2 is a root of the quadratic equation 2x2 + bx = 6, what is the other

root? [ 32]

2x2 + bx = 6 2x2 + bx 6 = 0 222 + b2 6 = 0 b = 1The quadratic equation is,

2x2

x

6 = 0

2 x + 3

x

2

= 0

other root is

3

2

3)If a two-digit number is divided by the sum of its digits, the quotient and

the remainder are 2. If the number is multiplied by the sum of its digits,

the product is 112. What is the number? [16]

Let 10x + y be the two-digit number. So the sum of its digits is x + y.

If the number is divided by the sum of its digits,10x+y

x + y= 2 +

2

x + y 1 0 x + y = 2x + y + 2 8x y = 2 y = 8 x 2

If the number is multiplied to the sum of its digits,1 0 x + yx + y = 112By substitution,


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10x + 8x 2x + 8x 2 = 112 18x 29x 2 = 1129x 19x 2 = 56 9x2 39x + 2 56 = 09x2 39x 5 4 = 0 9x 99 x + 6 = 0Since x is a digit of a number, 9x 9 = 0 x = 1, and y = 6

Therefore,10x + y = 101 + 6 = 16

4)In a given arithmetic sequence, the first term is 2, the last term is 29, and

the sum of all the terms is 155. What is the common difference? [3]

Let a = first term, l = last term, S = the sum, d = common difference

And n = number of terms

Expressing the sum,

S =na + l

2= 155

And expressing the terms,

a = 2, l = a + n 1d = 29Substituting,

S =n2 + 2 9

2= 155 31n

2= 155 n

2= 5 n = 10

Now to solve for the common difference,

a + n 1d = 29 2 + 10 1d = 29 d = 3

5)In ABC, the exterior angle at C is 60 degrees. If A : B = 1: 3, find themeasure of the smallest angle of ABC. [ 30]A: B = 1: 3 = k: 3k A = k, B = 3k

Since the sum of the interior angles of a triangle is 180 degrees,


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A + B + C = 180 k + 3k + 60 = 180 k = 30A = 30, B = 90Therefore, the smallest angle of the triangle is 30.

6)

The sequence an is defined as follows: a1 = 1, a2 = 3, andan = an 1 an 2for n 3. What is the sum of the first one thousandterms of the sequence? [5]

The sequence goes:

a3 = a2 a1 = 3 1 = 2 , a4 = 2 3 = 1, a5 = 3, a6 = 2, a7 = 1 OR

1,3,2, 1, 3, 2,1,3,2, The sequence only repeats the numbers 1, 3, 2, -1, -3, -2.

The sequence repeats1000

6= 166 times in the first 166 6 = 996 terms.

Finding the last four terms:

Since the last term in the first six terms is -2, it is also the last term in the first

996 terms. The next term or the 997thterm now is 1, then 3, 2 and the 1000thterm which is the end of the desired number of terms to be summed up is -1.

To sum up the 1000 terms, first notice that the first 6 terms has a sum of zero.

And since the 6 terms will be repeated 166 times, the whole 996 terms will

already have a sum of zero. The only terms that are not included in the sum is

the 997thto the 1000thterm. Which is:

a997 + a998 + a999 + a1000 = 1 + 3 + 2 +

1

= 5

Therefore, 5 is the sum of the first 1000 terms

7)Which positive integers x satisfy the equation log3 xlogx 5 = log3 5?[x 1 or /{1}]

log3 xlogx 5 = log3 5 logxlog3

log5logx

= log3 5


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So it looks like all real numbers are the solutions of the equation. But x 1because the quantities will have a zero denominator in the equation and 0

numerator and denominator will never cancel each other. Negative numbers

and zero are not included in the condition because the problem already said

that the values of x are positive integers.

Therefore, x 1 and /{1}8)In ABC, let D, E and F be the midpoints of AB, BD, and BC, respectively.The area of ABC is 96 in2. What is the area of AEF? [36 in 2]

Key Point: Area subtraction and Medians

Using the area subtraction, AEF = ABC BEF AFC

Since F is the midpoint of the side BC, the segment AF is the median of ABC tothe side BC. And by definition, AFC = AFB =96

2 = 48 in2

Now, in AFB, D is the midpoint of AB. So constructing a segment FD, that willbe the median to the side AB. So FDA = FDB = 48

2= 24 in2 .

Moreover, in BFD, FE is the median so DEF = BEF = 242

= 12 in2.


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So the areas of AFC and BEF are already known, AEF = ABC BEF AFC = 96 in2 12 in2 48 in2 = 36 in2

9)The bases of a trapezoid are 5cm and 8cm, respectively. The legs are 4cm

and 6cm, respectively, and they are produced to meet. How long has each

leg been produced? [10 cm,20

3cm]

From the figure, we let ABDC be the trapezoid. The bases are AB and CD with

lengths 5cm and 8cm, respectively. And the legs are AC and BD with lengths

6cm and 4cm, respectively. Now the legs are produced to meet, so the legs are

extended and they will meet at point E.

Now, C E D A E B, and AB||CD E A B E C D, and E B A E D CFrom the three angle congruencies, we can conclude that CED ~ AEB.(Triangle CED is similar to triangle AEB)

So if the two triangles are similar,


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CD

AB=

EC

EA=

ED

EB

And since,

ED = EB + BD and EC = EA + AC

CDAB

= ECEA

= ECEC AC 85 = ECEC 6 EC = 16cmCD

AB=

ED

EB=

ED

ED 4 85 = EDED 4 ED = 323 cmTherefore, each leg has been produced by:

EA = EC AC = 16cm 6cm = 10cmand EB = ED

BD =

32

3

cm

4cm =

20

3

cm

10) Each diagonal of a rectangle is 10cm. The area of a similar rectangle

is 6cm2, and one of its sides is 2cm. Find the area of the first rectangle.

[600

13cm2]

The similar rectangle has an area of 6cm2and a side of 2cm. So the other side

has a length of 3cm. So the length of the diagonal is 13 cm by the PythagoreanTheorem.By similarity,

side1side2

2 = diagonal1diagonal2

2 = area1area2

10cm13 cm2

=R

6 cm2 R = 600

13cm2

11) If f2 3x = x2 x, what is f(7)? [6]To get f(-7), 2 3x should be -7. So solving for x, 2 3x = 7 x = 3


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f7 = 32 3 = 6

12) Simplify: log9 54 log9 2. [ 34]log9 54 log9 2 = log9 542 = log9 542 = log9 27 = 12 log9 27

=1

2 log3 27

log3 9=

1

2 3

2=

3

4

13) The angles of a polygon are in arithmetic progression. If the smallest

angle is 100 degrees and the largest angle is 140 degrees, how many sides

does the polygon have? [6]

Key point: Sums of interior angles and Arithmetic progression

Identities: Sum of A. P =n

2(a + l) where n = number of terms, a = first term,

and l = last term.

Sum of interior angles of a polygon =

n

2

(180) where n =number of sides

The smallest angle is the first term of the progression, while the largest is the

last term. So expressing the sums and equating the necessary expressions,n

2100 + 140 = n 2180 n = 6 sides

14) Let be and acute angle such that sin2 = a. Express sin + cos in terms of a. [1+a ]

Derivation of an identity:

sin

+ cos

2 = sin2

+ cos2

+ 2 sin

cos

sin2

+ cos2

= 1


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sin + cos 2 = 1 + 2 sin cos sin2 = 2 sin cos sin + cos 2 = 1 + sin 2 sin + cos = 1 + s i n 2 , but sin 2 = a sin + cos = 1 + a15)

How many sets X satisfy the relationship a, b X a,b,c,d,e.[8]The relationship a, b X a,b,c,d,emeans the set {a, b} is a subset of Xand X is a subset of the set {a, b, c, d, e}.

So to X be a set that satisfies the relation, X should have the elements a and b.

So in counting the number of sets X, we should count the number of subsets the

set {c, d, e} has. The elements {a, b} does not count because the two elementsshould not be separated. So the number of subsets {c, d, e} has is 23 = 8. Since

null set is also a subset of {c, d, e}.

B.Part II

1)A circle of radius 8 cm is divided into four arcs: AB = 60 BC = 90,

CD = 120, and DA = 90.What is the area of ABCD? [64 + 323 cm2]Key point: Area addition

the points A, B, C, and D form a quadrilateral. Let O be the center. The measure

of the arcs is the same as the measure of the central angles.

Definition: Central Angles An angle formed by two points in thecircumference of a circle and its vertex is the center of the circleNow, the quadrilateral will be divided into four triangles AOB, BOC, COD, and

DOA. Where AOB = 60, BOC = DOA = 90 and COD = 120.


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So the area of ABCD can be expressed as:[ABCD] = [AOB] + [BOC] + [COD] + [DOA]Where [ABCD] means the area of ABCD.

Identity: Area of a triangle =1

2ab sin C. Where a and b are two sides of a

triangle and C is the included angle.

Now since the four points, A, B, C, and D are on the circumference of the circles

and O is the center, OA, OB, OC, and OD are all radii with length 8 cm.

Now, calculating the areas of the four triangles,

[AOB] = 12

OAOB sin 60 = 12

8 cm8 cmsin 60 = 163 cm2Similarly, [BOC] = 32 cm2 , [COD] = 163 cm2 , and [DOA] = 32 cm2Therefore, [ABCD] = 163 + 3 2 + 1 63 + 32 = 64 + 323 cm2


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2)Three equal circles are externally tangent to each other so that each circle

touches the other two. A belt that is placed around these circles has a

length of 50 feet. What is the length of the radius of each circle? [25 +3 ft]

The belt formed is the perimeter of the figure MNOPQR where it is divided

by the line segments MN, OP, and QR and the arcs NO, PQ, and RM.

Now, to find the measure of the arcs:

Since the circles A, C, and E are congruent, they also have congruent radii.

Connecting their respective centers, the triangle ACE is equilateral. Denote R

as the radius of each circle. So the length of each side of the equilateral

triangle ACE is 2R. So the measure of each interior angle of ACE is 60

degrees. So in circle A, EAC = 60 degrees.MN and RQ are segments tangent to the circle A. So AM MN and ARRQ,which implies that AMN = ARQ = 90 degrees. And also, in circle C and E,CNM = EQR = 90 degrees. We can conclude that the quadrilaterals AEQR,CEPO, and CAMN are rectangles.


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Going back to circle A, since CAMN and AEQR are rectangles, CAM = EAR= 90 degrees.

Knowing that circle A is 360 degrees,CAM + EAR + EAC + MAR = 360 9 0 + 9 0 + 6 0 + MAR = 360Therefore,

MAR= 120

So the angle MAR is the same as the measure of arc RM. Similarly, the arcs

PO and NQ are also 120 degrees each, which implies that the three arcs RM,

PO, and NQ are congruent.

From the three rectangles, it also implies that RQ = AE, OP = CE, MN = AC.

But ACE is equilateral and AC = CE = AE. So RQ = OP = MN = 2R.

Finding the perimeter of the figure MNOPQR,

MNOPQR = MN + arc NO + OP + arc PQ + QR + arc RM = 50

= 3MN + 3arc NO = 32R + 3 120360

2R = 32R + 3 23

R = 506 R + 2R = 5 0 R6 + 2 = 50 + 3R = 25 R = 25 + 3 feet

3)Five points O, A, B, C, D are taken in order (left to right) on a horizontal

line with distances OA = a, OB = b, OC = c, OD = d. Let P be a point on BC

such that AP: PD = BP: PC. Assuming that a + c b + d, express OP interms of a, b, c, and d. [ OP =

ac bda+c b d]

The given says that,

AP: PD = BP: PC APPD

=BP

PC

Expressing AP, PD, BP, and PC by segment subtraction,

AP = OP a, PD = d OP, BP = OP b, and PC = c OPSubstituting to the ratio,


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OP ad OP = OP bc OP OP ac OP = OP bd OP OP2 ac + OPa + OPc = OP2 bd + OPb + OPd OPa + c OPb + d = ac bd OPa + c b d = ac bd

OP =

ac

bd

a + c b d

4)Prove that the sum of the three medians of a triangle is greater than

three-fourths of the perimeter of the triangle.

Statement: ma + mb + mc >3

4( a + b + c )

Proof:

Denote G as the centroid of the triangle.

Let a = BC, b = AC, c = AB, ma = AF, mb = BE, and mc = CD

Identity: ma , mb , and mc are the medians of the triangle with centroid G,

CG =2

3mc , BG =

2

3mb , and AG =

2

3ma

Using the triangle inequality to the triangles BGC, CGA, and AGB,

BG + CG > , + > , + >,


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2

3mb +

2

3mc > , 2

3mc +

2

3ma > , 2

3ma +

2

3mb >

For easier calculations, modifying the inequalities,

mb + mc >3

2

a, mc + ma >3

2

b, and ma + mb >3

2

c

Adding the three inequalities,

2ma + mb + mc > 32

a + b + c ma + mb + mc > 34

( a + b + c )

This satisfies the statement. The left hand side of the inequality denotes the

sum of the medians while the right hand side of the inequality denotes the

perimeter of the triangle. So the sum of the medians is three-fourths theperimeter of the triangle. QED

5)The perimeter of a triangle is 60 inches. The bisector of an angle divides

the opposite side into two segments of 9 and 12 inches. Find the lengths

of the other two sides of the triangle. [117

7in and

156

7in ]

The angle bisector of the triangle divides the triangle into two similar triangle

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Mmc Fourth Year Solutions