2009 MetrobankMTAPDepEd Math Challenge
National Finals, FourthYear Level 4 April 2009 Questions, Answers, and Solutions Questions Team Oral Competition 15SECOND QUESTIONS 1. If 361−x = 6x , ﬁnd x. 2. The larger angles of a rhombus are double the smaller angles. The length of the shorter
diagonal 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 2 grams is worth P10,000, what is the value of a 10gram diamond? 4. A bottle of Refreshing Monster Crush (RMC) makes enough drink to ﬁll sixty glasses
when it is diluted in the ratio 1 part RMC to 4 parts water. water. How many many glasses of drink would a bottle of RMC make if they are diluted in the ratio 1 part RMC to 6 parts water? 5. If x
  − 2 = a and x < 2, what is x − a?
6. Two circles circles lying in the same plane have have the same center center.. The radius radius of the larger larger
circle circle is twice twice the radius of the smaller. The area of the region between between the circles circles is 2 7 cm . What is the area of the larger circle? 7. The angles angles of a triangle triangle are in the ratio ratio of 1 : 2 : 3. The shorte shortest st side is 6 cm long. 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 2 2009 52013 ?
·
9. A boys’ club decides to build a cabin. The job can be done by 3 skilled workmen in 4
days 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 triangle have lengths 3, 8, and y . What is the smallest positive number that can not be a value of x y ?
  − 
30SECOND QUESTIONS 1. Deﬁne an operation “” as follows: a b =
ab+2a+b+2
2a
. What What is 29 288?
2. In ABC , 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 2 x2
3
2
− 11x, 12x, and 2x − 2x ?
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 12 hours at her regular hourly salary. What is her required number of working hours each day?
√
5. Let ABCD be a convex quadrilateral with AB = AD = 25 cm, CB = CD = 10 13 cm, and DB = 40 cm. How long is AC ? 60SECOND QUESTIONS 1. The parabola whose equation is 8y = x 2 meets the parabola whose equation is x = y 2
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 : P B = 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 the
third, what is the probability that at least one “1” is tossed? 5. The area of a rhombus is 300 in 2 , and the lengths of its diagonals are in the ratio of
2 : 3. Compute the length of a side of the rhombus.
Individual Oral Competition 15SECOND 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 = 21 , what is a? 3. If sin θ = 32 , what is cot θ ?

4. For x

2
  ≥ 2, what is the maximum value of y in y = 3x − x ?
5. Melanie eats x bars of chocolate every y days. Maintaining this rate, how many bars does 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 smaller
triangle 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 after
every 8 days, while Chris after every 6 days. If they met one Tuesday afternoon, on what day will they possibly meet again? 8. On the clock’s face, what is the smaller angle between the segments joining the center
to 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 answer of 9. What should the correct answer be? 30SECOND QUESTIONS 1. Aling Rosa bought 12 dozen oranges at P 45 per dozen. She found 18 bad oranges, and she 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, the
sum of their ages was 44. How old is his son now? 3. In
ABC , AB = 18 cm, AC = 20 cm, and B C = 9 cm. Let D be a point on side BC
such that AD bisects
∠A.
Find DC .
4. Two negative numbers diﬀer by 3 and their squares diﬀer by 63. Find the larger
number. 5. Let E and F be points on sides AB and AC , respectively, of ABC such that EF is parallel to BC . If EF = 4 in, BC = 6 in, and the area of AEF is 10 in2 , what is the area of the quadrilateral BEFC ?
60SECOND QUESTIONS 1. Find all real numbers a such that a4
2
− 15a − 16 = 0 and a
3
+ 4a2
− 25a − 100 = 0. T
2. Tangent P T and secant P AB are drawn to a circle from an external point P . The tangent is 14 cm long, and the ratio of the lengths of the internal segment AB to the external segment P A is 3 : 1. Find P B .
P A
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, which
is 4. Their third terms are also equal, but the second term of the arithmetic progression exceeds the second term of the geometric progression by 2. Find the second term of the geometric progression.
5. Ella has threefourths as much money as Jake. If Jake gives onehalf of his money to Ella, and Ella gives Jake oneﬁfth of what she has then, then Jake will have P1 less
than Ella. How much had each at ﬁrst?
Individual Written Competition PART I. Write your answer on the space before the corresponding number. Each correct
answer earns 2 points. 1. Simplify:
2 3
1 3
3 4
1 4
+ · ÷ 8 .
2. The word “thirty” has six letters, and 6 is a factor of 30. How many of the numbers
from 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 score
on the third test was 4 marks higher than his score on the second test. What was his score on the second test? 5. Solve the equation 3 x3 + x2
− 15x − 5 = 0.
6. A diameter of a circle is 24 cm long and is divided into parts that are in the ratio
of 1 : 2 : 3. Through the points of division, chords are drawn perpendicular to the diameter. 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 distance between these points of intersection is 21 , ﬁnd the value of k. 8. If log2 x = a , what is log 4 2x + log8 x2 in terms of a? 9. A bag contains 3 white and 5 black balls. If two balls are drawn in succession without
replacement, 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 equal
to 5. What are the lengths of its legs? √ x
11. Solve for x: xx
√
x
= ( x x) .
12. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3, and P (4) = 5. Find the value of P (6). 13. In rectangle ABCD, AD = 1, P is on side AB , and DB and DP trisect Determine the perimeter of BDP .
∠ADC .
14. A man calculates that if he continues at the present speed, to drive the remaining 100
km of his trip, he will arrive 30 minutes late. In order to arrive on time, he must travel at 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 says
Claire broke it. Claire and David say they do not know who broke it. Only the guilty child was lying. Who broke the toy? PART II. Give a complete and neat solution. Each correct and complete solution earns 3
points. 1. Solve the equation 1 + 68x−4 = 21x−2 . 2. Let x and y be two 2digit integers, where y is the number obtained by reversing the digits of x. If x2 y 2 = 495, ﬁnd x and y.
−
3. In rectangle ABCD, ∠C is trisected by segments CF and CE , where E is on AB , F on 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 the
squares 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 median MN cuts the diagonal AC at P and the diagonal BD at Q. Express P Q in terms of AB and CD. PART III. Give a complete and neat solution. Each correct and complete solution earns 5
points. 1. Determine the value(s) of k such that the system of equations
x + y = 1 x2 + y 2 = 2 x3 + y 3 = 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 of x3 + rx2 + sx + t = 0 are a + b, b + c, and c + a. Find the value of t.
−
3. In
ABC , AB = √ 30, AC = √ 6, and BC = √ 15.
Let D be a point such that the segment 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 15SECOND QUESTIONS 1.
2 3
2. 40 cm
3. P250,000
5. 2
4. 84
6.
7. 36π cm2
− 2a
28 3
cm2
30SECOND QUESTIONS
8. 13
9.
12 5
days
10. 6
60SECOND QUESTIONS
√
1. 150
1. 2 5
2. 30◦
2. 5762
3. 19
3. 21 cm
4. 6
4.
5. 45 cm
5. 5 13 in
3 5
√
Individual Oral Competition 15SECOND QUESTIONS 1. P26,250 2.
−
1 10
3.
√
5 2
4. 2
30SECOND QUESTIONS
5. 7x2 6.
5
√
5
2
7. Friday
cm
8. 150◦
1.
2. 15 years old
2. 28 cm
90 19
4.
−9
cm
5. 12.5 in2
10. 7
60SECOND QUESTIONS
1. Aling Rosa lost P15.
3.
9. 7
−4
3. 3 4. 8 5. Jake has P4, while Ella has P3.
Individual Written Competition PART I 1.
√
1 9
6. 8 5 cm and 24 cm
9 4
11.
√ 5
and 1
2. 3
7. 1 +
ln4+ln 3 3. ln 3−ln 4
8.
1 2
4. 79
9.
5 14
14. 40 kph
√ 5 and 2√ 5
15. Barbara
5.
− , √ 5, −√ 5 1 3
10.
12. 16
+ 67 a
13. 2 +
√
4
3
3
PART II 1. x =
±2, ±√ 17
Let a = x −2 . Then the given equation is equivalent to 1 + 68a2 = 21a, 1 which has roots 41 and 17 . 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 y ) = 495 implies that 9(a It follows that
− b) · 11(a + b) = 495
a b = 1 a + b = 5
−
2
∈ {1, 2, . . . , 9}. Then x − y =
(a
⇒
=
⇒
2
= (x
− y)(x +
− b)(a + b) = 5 .
a = 3 and b = 2
Thus, we have x = 32 and y = 23.
√ − 36
3. 108 3
Since
∠BC E =
30◦ , using right triangle BC E , we get
√
BC = 6 cot30◦ = 6 3.
A
√ It follows that F D = 6 3 − 2. Since ∠DCF = 30 ◦, using
E
6
B
2
F
right triangle DCF , we also get CD
√ √ √ √ − − − = 6 3 2 cot30 = 6 3 2 3 = 18 2 3
.
◦
Thus, the area of ABCD is
·
BC CD
D
√ √ √ = 6 3 · 18 − 2 3 = 108 3 − 36
.
C
4. 9 square units
Let a and b be the lengths of the legs and c the length of the hypotenuse of the right triangle. Then (1) a + b + c = 18 a2 + b2 + c2 = 128 (2) 2 2 2 (3) a + b = c .
Equations (2) and (3) give c = 8. Thus, the above system of equations reduces to
a + b = 10 a2 + b2 = 64,
− √
(4) (5)
√
√ − √ − √
which has two solutions: (a, b) = 5 7, 5 + 7) and (a, b) = (5 + 7, 5 7). In 1 both solutions, the area of the right triangle is 2 5 + 7 5 7 = 9 square units. 5. P Q = 21 (CD
√
− AB)
A
By the Midsegment Theorem for Trapezoids, we know that M N = 21 (AB + CD). Moreover, since MN AB , P and Q are the midpoints of AC and B D, respectively. It follows, by the Midsegment Theorem for Triangles, that MQ = N P = 21 AB . Thus, we get
B
O M
N Q
P
D
C
− M Q − N P = 12 (AB + CD) − 2 · 12 AB = 12 (CD − AB).
P Q = M N PART III 1. k = 25 , solutions:
1 2
−
1 2
√ 3,
1 2
1 2
+
√ 3 and
1 2
+
1 2
From the ﬁrst two equations, we have
√ 3, − √ 3 1 2
1 2
1 = ( x + y )2 = x 2 + 2xy + y 2 = 2 + 2 xy, which implies that xy =
−
1 2
.
On the other hand, in order that the system admits at least one solution, we must also have k = x 3 + y 3 = (x + y )(x2 xy + y 2 ) = (1)(2 xy ), and so xy = 2
−
− k.
−
Combining the two expressions for xy above, we have 2 k = 25 .
−k = −
1 2
, which gives us
To solve for the solutions of the system, we use the ﬁrst equation and xy = x(1
− x) = − 12
Thus, we get ( x, y) =
1 2
−
1 2
=
x =
⇒
√ 3,
1 2
+
1 2
1 2
− 12
√
−
√
1 1 3 or x = + 3 2 2
√ 3 and (x, y) =
1 2
+
1 2
√ 3, − √ 3. 1 2
1 2
1 2
.
2. 23
By Factor Theorem, we know that x3 + 3x2 + 4x
3
2
− 11 = (x − a)(x − b)(x − c) = x − (a + b + c)x + (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) −[ab + ac + ba + bc + ca + cb + 2abc] −[(a + b + c)(ab + bc + ac) − 3abc + 2abc] −[(−3)(4) − 11]
= = = = 23.
2
2
2
2
2
2
3. 19 : 27 C
We ﬁrst note that A, E , and D are collinear. The main tool in the following argument is the Law of Cosines.
E A
The Law of Cosines applied to
ABC gives √ 30 + 15 − 6 13 2 AB + BC − AC . cos ∠ABC = = √ √ = 2AB · BC 20 2 15 30 Again, when the Law of Cosines is applied to ABE , we get AE = AB + BE − 2AB · BE cos ∠ABC √ √ 15 13√ 2 15 = 30 + − 2 30 · 2 · 20 4 2
2
2
= which gives AE =
1 2
2
2
2
57 , 4
√ 57, and so
−√ = 27√ 190 . · 30 380
57 + 30 AE 2 + AB 2 BE 2 cos ∠BAE = = 4 √ 57 2AE AB 2 2
·
−
·
It follows that AD = AB cos ∠BAE =
Thus, we obtain AE : AD = 19 : 27.
√
√
15 4
√
27 190 27 57 . 30 = 380 38
·
D
B