2009 Metrobank-MTAP-DepEd Math ChallengeNational Finals, Fourth-Year Level4 April 2009Questions, Answers, and Solutions

Questions

Team Oral Competition

15-SECOND QUESTIONS

1. If 361−x = 6x, find x.

2. The larger angles of a rhombus are double the smaller angles. The length of the shorterdiagonal is 10 cm. What is the perimeter of the rhombus?

3. If the price of a diamond varies as the square of its weight, and a diamond weighing 2grams is worth P10,000, what is the value of a 10-gram diamond?

4. A bottle of Refreshing Monster Crush (RMC) makes enough drink to fill sixty glasseswhen it is diluted in the ratio 1 part RMC to 4 parts water. How many glasses of drinkwould a bottle of RMC make if they are diluted in the ratio 1 part RMC to 6 partswater?

5. If |x− 2| = a and x < 2, what is x− a?

6. Two circles lying in the same plane have the same center. The radius of the largercircle is twice the radius of the smaller. The area of the region between the circles is7 cm2. What is the area of the larger circle?

7. The angles of a triangle are in the ratio of 1 : 2 : 3. The shortest side is 6 cm long.Find the area (in terms of π) of the circle that circumscribes this triangle.

8. What is the sum of the digits of the decimal expansion of the product 22009 · 52013?

9. A boys’ club decides to build a cabin. The job can be done by 3 skilled workmen in 4days or by 5 of the boys in 6 days. How long will it take if all work together?

10. The sides of one triangle have lengths 3, 8, and x, while the sides of a second trianglehave lengths 3, 8, and y. What is the smallest positive number that can not be a valueof |x− y|?

30-SECOND QUESTIONS

1. Define an operation “?” as follows: a ? b = ab+2a+b+22a

. What is 29 ? 288?

2. In 4ABC, with AB = AC, let D be a point on side BC such that AD = BD. If∠DAC = 90◦, what is ∠BAD?

3. If x = 3, what is the average of 2x2 − 11x, 12x, and 2x3 − 2x2?

4. After working for the required number of hours, Mary Lou works 4 hours overtime,each at 150% of her regular hourly pay. Her total pay that day is equivalent to 12hours at her regular hourly salary. What is her required number of working hours eachday?

5. Let ABCD be a convex quadrilateral with AB = AD = 25 cm, CB = CD = 10√

13cm, and DB = 40 cm. How long is AC?

60-SECOND QUESTIONS

1. The parabola whose equation is 8y = x2 meets the parabola whose equation is x = y2

at two points. What is the distance between these two points?

2. If f(x) = (x+ 3)x+2(x+ 1)x, then what is f(0) + f(1) + f(2)?

3. Let ABCD be a parallelogram, and P be a point on segment AB such that AP : PB =2 : 3. Let DP cut AC at point Q. If AQ = 6 cm, what is AC?

4. A fair die is tossed three times. Given that the sum of the first two tosses equals thethird, what is the probability that at least one “1” is tossed?

5. The area of a rhombus is 300 in2, and the lengths of its diagonals are in the ratio of2 : 3. Compute the length of a side of the rhombus.

Individual Oral Competition

15-SECOND QUESTIONS

1. Mr. Lee borrowed P25,000 on March 10, 2008, and paid it back with interest at 5%on March 10, 2009. Find the amount he paid.

2. If 322a = 12, what is a?

3. If sin θ = 23, what is | cot θ|?

4. For |x| ≥ 2, what is the maximum value of y in y = 3x− x2?

5. Melanie eats x bars of chocolate every y days. Maintaining this rate, how many barsdoes she get through in xy weeks?

6. The ratio of the areas of two similar right triangles is 4 : 5. If the legs of the smallertriangle are 3 cm and 4 cm long, how long is the hypotenuse of the larger triangle?

7. Bill and Chris have the same favorite barbershop. Bill goes to this barbershop afterevery 8 days, while Chris after every 6 days. If they met one Tuesday afternoon, onwhat day will they possibly meet again?

8. On the clock’s face, what is the smaller angle between the segments joining the centerto the 2 o’clock and to the 7 o’clock marks?

9. If 4 ≤ x ≤ 6 and 2 ≤ y ≤ 3, what is the minimum value of (x− y)(x+ y)?

10. Ricky was told to add 4 to a certain number, and then divide the sum by 5. Instead,he first added 5, and then divided the sum by 4. He came up with the wrong answerof 9. What should the correct answer be?

30-SECOND QUESTIONS

1. Aling Rosa bought 12 dozen oranges at P45 per dozen. She found 18 bad oranges, andshe sold the rest of them at 6 oranges for P25. Did she lose or gain? By how much?

2. Two times the father’s age is 8 more than six times his son’s age. Ten years ago, thesum of their ages was 44. How old is his son now?

3. In 4ABC, AB = 18 cm, AC = 20 cm, and BC = 9 cm. Let D be a point on side BCsuch that AD bisects ∠A. Find DC.

4. Two negative numbers differ by 3 and their squares differ by 63. Find the largernumber.

5. Let E and F be points on sides AB and AC, respectively, of 4ABC such that EF isparallel to BC. If EF = 4 in, BC = 6 in, and the area of 4AEF is 10 in2, what isthe area of the quadrilateral BEFC?

60-SECOND QUESTIONS

1. Find all real numbers a such that a4 − 15a2 − 16 = 0 and a3 + 4a2 − 25a− 100 = 0.

2. Tangent PT and secant PAB are drawn to a circle froman external point P . The tangent is 14 cm long, and theratio of the lengths of the internal segment AB to theexternal segment PA is 3 : 1. Find PB.

A

T

P

B

3. Solve for x: log 6 + x log 4 = log 4 + log(32 + 4x).

4. An arithmetic progression and a geometric progression have the same first term, whichis 4. Their third terms are also equal, but the second term of the arithmetic progressionexceeds the second term of the geometric progression by 2. Find the second term ofthe geometric progression.

5. Ella has three-fourths as much money as Jake. If Jake gives one-half of his money toElla, and Ella gives Jake one-fifth of what she has then, then Jake will have P1 lessthan Ella. How much had each at first?

Individual Written Competition

PART I. Write your answer on the space before the corresponding number. Each correctanswer earns 2 points.

1. Simplify:(

23

+ 13· 3

4

)÷ 81

4.

2. The word “thirty” has six letters, and 6 is a factor of 30. How many of the numbersfrom one up to twenty have this curious property?

3. Solve for a: 41+a = (13)1−a.

4. Mark’s average score on three tests was 84. His score on the first test was 90. His scoreon the third test was 4 marks higher than his score on the second test. What was hisscore on the second test?

5. Solve the equation 3x3 + x2 − 15x− 5 = 0.

6. A diameter of a circle is 24 cm long and is divided into parts that are in the ratioof 1 : 2 : 3. Through the points of division, chords are drawn perpendicular to thediameter. Find the lengths of the chords.

7. A line x = k intersect both the graphs of y = log5 x and y = log5(x+4). If the distancebetween these points of intersection is 1

2, find the value of k.

8. If log2 x = a, what is log4 2x+ log8 x2 in terms of a?

9. A bag contains 3 white and 5 black balls. If two balls are drawn in succession withoutreplacement, what is the probability that both balls are black?

10. The area of a right triangle and the length of its hypotenuse are both numerically equalto 5. What are the lengths of its legs?

11. Solve for x: xx√

x = (x√x)

x.

12. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3, and P (4) = 5. Findthe value of P (6).

13. In rectangle ABCD, AD = 1, P is on side AB, and DB and DP trisect ∠ADC.Determine the perimeter of 4BDP .

14. A man calculates that if he continues at the present speed, to drive the remaining 100km of his trip, he will arrive 30 minutes late. In order to arrive on time, he must travelat an average rate of 10 kph faster. What is his present speed?

15. Four children are arguing over a broken toy. Alex says Barbara broke it. Barbara saysClaire broke it. Claire and David say they do not know who broke it. Only the guiltychild was lying. Who broke the toy?

PART II. Give a complete and neat solution. Each correct and complete solution earns 3points.

1. Solve the equation 1 + 68x−4 = 21x−2.

2. Let x and y be two 2-digit integers, where y is the number obtained by reversing thedigits of x. If x2 − y2 = 495, find x and y.

3. In rectangle ABCD, ∠C is trisected by segments CF and CE, where E is on AB, Fon AD, BE = 6, and AF = 2. Find the area of the rectangle.

4. The sum of the lengths of the three sides of a right triangle is 18. The sum of thesquares of the lengths of the three sides is 128. Find the area of the triangle.

5. Let ABCD be a trapezoid with bases AB and CD, where CD > AB. The medianMN cuts the diagonal AC at P and the diagonal BD at Q. Express PQ in terms ofAB and CD.

PART III. Give a complete and neat solution. Each correct and complete solution earns 5points.

1. Determine the value(s) of k such that the system of equationsx+ y = 1x2 + y2 = 2x3 + y3 = k

has at least one solution. For such k, determine the solutions.

2. Suppose that the roots of x3 + 3x2 + 4x− 11 = 0 are a, b, and c, and that the roots ofx3 + rx2 + sx+ t = 0 are a+ b, b+ c, and c+ a. Find the value of t.

3. In 4ABC, AB =√

30, AC =√

6, and BC =√

15. Let D be a point such that thesegment AD bisects side BC and that ∠ADB is right. If E is the midpoint of BC,determine the ratio AE : AD.

Answers and Solutions

Team Oral Competition

15-SECOND QUESTIONS

1. 23

2. 40 cm

3. P250,000

4. 84

5. 2− 2a

6. 283

cm2

7. 36π cm2

8. 13

9. 125

days

10. 6

30-SECOND QUESTIONS

1. 150

2. 30◦

3. 19

4. 6

5. 45 cm

60-SECOND QUESTIONS

1. 2√

5

2. 5762

3. 21 cm

4. 35

5. 5√

13 in

Individual Oral Competition

15-SECOND QUESTIONS

1. P26,250

2. − 110

3.√

52

4. 2

5. 7x2

6. 5√

52

cm

7. Friday

8. 150◦

9. 7

10. 7

30-SECOND QUESTIONS

1. Aling Rosa lost P15.

2. 15 years old

3. 9019

cm

4. −9

5. 12.5 in2

60-SECOND QUESTIONS

1. −4

2. 28 cm

3. 3

4. 8

5. Jake has P4, while Ella has P3.

Individual Written Competition

PART I

1. 19

2. 3

3. ln 4+ln 3ln 3−ln 4

4. 79

5. −13,√

5,−√

5

6. 8√

5 cm and 24 cm

7. 1 +√

5

8. 12

+ 76a

9. 514

10.√

5 and 2√

5

11. 94

and 1

12. 16

13. 2 + 4√

33

14. 40 kph

15. Barbara

PART II

1. x = ±2,±√

17

Let a = x−2. Then the given equation is equivalent to

1 + 68a2 = 21a,

which has roots 14

and 117

. Thus, since a = x−2, the roots of the original equation are

2, −2,√

17, and −√

17.

2. x = 32 and y = 23

Let x = 10a+ b and y = 10b+a, where a, b ∈ {1, 2, . . . , 9}. Then x2−y2 = (x−y)(x+y) = 495 implies that

9(a− b) · 11(a+ b) = 495 =⇒ (a− b)(a+ b) = 5.

It follows that {a− b = 1a+ b = 5

=⇒ a = 3 and b = 2

Thus, we have x = 32 and y = 23.

3. 108√

3− 36

Since ∠BCE = 30◦, using right triangle BCE, we get

BC = 6 cot 30◦ = 6√

3.

It follows that FD = 6√

3−2. Since ∠DCF = 30◦, usingright triangle DCF , we also get

CD =(

6√

3− 2)

cot 30◦ =(

6√

3− 2)√

3 = 18− 2√

3.

Thus, the area of ABCD is

BC · CD = 6√

3 ·(

18− 2√

3)

= 108√

3− 36.

2

6

F

EA

D C

B

4. 9 square units

Let a and b be the lengths of the legs and c the length of the hypotenuse of the righttriangle. Then

a+ b+ c = 18 (1)a2 + b2 + c2 = 128 (2)a2 + b2 = c2. (3)

Equations (2) and (3) give c = 8. Thus, the above system of equations reduces to{a+ b = 10 (4)a2 + b2 = 64, (5)

which has two solutions: (a, b) = 5 −√

7, 5 +√

7) and (a, b) = (5 +√

7, 5 −√

7). Inboth solutions, the area of the right triangle is 1

2

(5 +√

7) (

5−√

7)

= 9 square units.

5. PQ = 12(CD − AB)

By the Midsegment Theorem for Trapezoids, we knowthat MN = 1

2(AB + CD). Moreover, since MN‖AB, P

and Q are the midpoints of AC and BD, respectively. Itfollows, by the Midsegment Theorem for Triangles, thatMQ = NP = 1

2AB. Thus, we get

O

PQNM

D C

BA

PQ = MN −MQ−NP =1

2(AB + CD)− 2 · 1

2AB =

1

2(CD − AB).

PART III

1. k = 52, solutions:

(12− 1

2

√3, 1

2+ 1

2

√3)

and(

12

+ 12

√3, 1

2− 1

2

√3)

From the first two equations, we have

1 = (x+ y)2 = x2 + 2xy + y2 = 2 + 2xy,

which implies that xy = −12.

On the other hand, in order that the system admits at least one solution, we must alsohave

k = x3 + y3 = (x+ y)(x2 − xy + y2) = (1)(2− xy),

and so xy = 2− k.

Combining the two expressions for xy above, we have 2 − k = −12, which gives us

k = 52.

To solve for the solutions of the system, we use the first equation and xy = −12.

x(1− x) = −1

2=⇒ x =

1

2− 1

2

√3 or x =

1

2+

1

2

√3

Thus, we get (x, y) =(

12− 1

2

√3, 1

2+ 1

2

√3)

and (x, y) =(

12

+ 12

√3, 1

2− 1

2

√3).

2. 23

By Factor Theorem, we know that

x3 + 3x2 + 4x− 11 = (x− a)(x− b)(x− c) = x3− (a+ b+ c)x2 + (ab+ bc+ ac)x− abc.

It follows that a+ b+ c = −3, ab+ bc+ ac = 4, and abc = 11.

On the other hand, we also have

x3 + rx2 + sx+ t = [x− (a+ b)][x− (b+ c)][x− (a+ c)].

It follows that

t = −(a+ b)(b+ c)(a+ c)

= −[ab2 + ac2 + ba2 + bc2 + ca2 + cb2 + 2abc]

= −[(a+ b+ c)(ab+ bc+ ac)− 3abc+ 2abc]

= −[(−3)(4)− 11]

= 23.

3. 19 : 27

We first note that A, E, and D are collinear. The maintool in the following argument is the Law of Cosines.

DE

C

A BThe Law of Cosines applied to 4ABC gives

cos ∠ABC =AB2 +BC2 − AC2

2AB ·BC=

30 + 15− 6

2√

15√

30=

13√

2

20.

Again, when the Law of Cosines is applied to 4ABE, we get

AE2 = AB2 +BE2 − 2AB ·BE cos ∠ABC

= 30 +15

4− 2√

30 ·√

15

2· 13√

2

20

=57

4,

which gives AE = 12

√57, and so

cos ∠BAE =AE2 + AB2 −BE2

2AE · AB=

574

+ 30− 154

2 ·√

572·√

30=

27√

190

380.

It follows that

AD = AB cos ∠BAE =√

30 · 27√

190

380=

27√

57

38.

Thus, we obtain AE : AD = 19 : 27.