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2016 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS

1. Find the value of βπ₯ + β14 β π₯ when π₯ = β2. ββ2 + β14 β (β2) = ββ2 + β14 + 2 = β2 ANSWER: βπ 2. Find the area of the triangle formed by the coordinate axes and the line 3π₯ + 2π¦ = 6. The triangle has a base equal to the x-intercept of the graph of 3π₯ + 2π¦ = 6 and a height equal to the y-intercept. The area, therefore is half the product of the intercepts. 1 π΄ = πβ 2

1 π΄ = (2)(3) π΄ 2 = 3 π ππ’πππ π’πππ‘π ANSWER: 3 square units 3. A sequence is defined by ππ = 3(ππβ1 + 2) for π β₯ 2, where π1 = 1. What is π4 ? π1 = 1 ( ) π2 = 3 π1 + 2 π2 = 3(1 + 2) π2 = 9 π3 = 3(π2 + 2) π3 = 3(9 + 2) π2 = 33 π4 = 3(π3 + 2) π4 = 3(33 + 2) π2 = 105 ANSWER: πππ 4. Christa leaves Town A. After traveling 12 km, she reaches Town B at 2:00 P.M. Then she drove at a constant speed and passes Town C, 40 km from Town B, at 2:50 P.M. Find the function π(π‘) that models the distance (in km) she has traveled from Town A t minutes after 2:00 P.M. π1 = 12 ππ TOWN A

π2 = 40 ππ π‘ = 50 πππ

TOWN B 2:00 PM

π = π =

π π‘

TOWN C 2:50 PM

40 ππ = 0.8 πππ 50 πππ

π = π π‘ π = 0.8π‘ πππ πππ π‘ππππ πππ‘ππ πππ€π π΅ π(π‘) = 12 + 0.8π‘ ANSWER: π (π) = ππ + π. ππ 1|P ag e

Grade 10 Elimination Round

5. What is the largest negative integer that satisfies the inequality |3π₯ + 2| > 4? Two conditions will satisfy this inequality, (1) If 3π₯ + 2 > 4; and (2) If 3π₯ + 2 < β4 2 In case (1), π₯ > 3. In case (2), π₯ < β2. This is the solution set to the inequality. Considering case (2), the largest negative integer that satisfies this condition is -3. ANSWER: βπ 6. A person has two parents, four grandparents, eight grand-parents, and so on. How many ancestors does a person have 10 generations back? This being a geometric series of 2+4+8+β¦+π10 , 1βπ π ) Use the formula, ππ = π1 ( 1βπ

1 β 210 π10 = 2 ( ) 1β2 π10 = 2046 ANSWER: ππππ 7. In Ξπ΄π΅πΆ, β π΅ is twice β π΄, and β πΆ is three times as large as β π΅. Find β πΆ If πβ π΅ = 2 πβ π΄ πππ πβ πΆ = 3 πβ π΅, then πβ πΆ = 6 πβ π΄ πβ π΄ + πβ π΅ + πβ πΆ = 180Β° πβ π΄ + 2 πβ π΄ + 6 πβ π΄ = 180Β° 9 πβ π΄ = 180Β° πβ π΄ = 20Β° πβ πΆ = 6 πβ π΄ = 120Β° ANSWER: πππΒ° 8. If β3 β€ π₯ β€ 0, find the minimum value of π(π₯) = π₯ 2 + 4π₯. The graph of π(π₯) opens upward based on its leading coefficient. Therefore, the absolute minimum value of the function is the y-coordinate, k, of its vertex. π2 π=πβ 4π 42 π=0β 4(1) π = β4 ANSWER: -4

βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

9. The 9th and the 11th term of an arithmetic sequence are 28 and 45, respectively. What is the 12th term? π11 β π9 2π = π11 β π9 π= 2 45 β 28 17 π= = ππ 8.5 2 2 π12 = π11 + π π12 = 45 + 8.5 ANSWER: 53.5 10. Perform the indicated operation, and simplify: 2 3 4 + β 2 π₯ π₯β1 π₯ βπ₯ 2(π₯ β 1) + 3π₯ β 4 π₯2 β π₯ 5π₯ β 6 π₯2 β π₯ ππβπ ANSWER: π π βπ 11. Find the range of the function π (π₯) = |2π₯ + 1|. For any absolute value function of this form, π (π₯) = |π(π₯)| where π(π₯) is a linear function, the range is always {π¦|π¦ β β, π¦ β₯ 0}. ANSWER: π β₯ π 12. Find the value of [π₯ β (π₯ β π₯ β1 )β1 ]β1 when π₯ = 2. [2 β (2 β 2β1 )β1 ]β1 1 β1 [2 β (2 β ) ] 2 3 β1 [2 β ( ) ] 2 π ANSWER: π

β1

β1

2 β1 4 β1 3 = [2 β ] β ( ) = 3 3 4

13. Find the equation (in the form ππ₯ + ππ¦ = π) of the line through the point (5, 2) that is parallel to the line 4π₯ + 6π¦ + 5 = 0. 2 5 4π₯ + 6π¦ + 5 = 0 β π¦ = β π₯ β 3 6 2 Solve for π given that π = β 3 and (5, 2) π¦1 = ππ₯1 + π 2 16 2 = β (5) + π π = 3 3 2 16 π¦=β π₯+ β 2π₯ + 3π¦ = 16 3 3 ANSWER: ππ + ππ = ππ 2|P ag e

Note on an alternative: When asked for the equation of a line, ππ₯ + ππ¦ = π, parallel to a given linear equation of the form ππ π₯ + ππ π¦ = π1 and passing through the point, (ππ , ππ ), use the algorithm: ππ π + ππ π = ππ ππ + ππ ππ When asked for the equation of a line, ππ₯ + ππ¦ = π, perpendicular to a given linear equation of the form ππ π₯ + ππ π¦ = π1 and passing through the point, (ππ , ππ ), use the algorithm: ππ π β ππ π = βππ ππ + ππ ππ 14. In Ξπ΄π΅πΆ, β π΅ = 90Β°, β π΄πΆπ΅ = 28Β°, and D is the midpoint of AC. What is β π΅π·πΆ? A D

C

B

The median to the hypotenuse of a right triangle is half the measure of the hypotenuse. Since π΅π· = π·πΆ, ΞBDC is isosceles. It also follows that πβ πΆ = πβ π·π΅πΆ. πβ πΆ + πβ π·π΅πΆ + πβ π΅π·πΆ = 180Β° 28Β° + 28Β° + πβ π΅π·πΆ = 180Β° πβ π΅π·πΆ = 180Β° β 56Β° πβ π΅π·πΆ = 124Β° ANSWER: πππΒ° 15. Find all solutions of the system: 2π₯ + π¦ = 1 { 3π₯ + 4π¦ = 14 2π₯ + π¦ = 1 β π¦ = β2π₯ + 1 By substitution, 3π₯ + 4(β2π₯ + 1) = 14 3π₯ β 8π₯ + 4 = 14 β5π₯ = 10, π₯ = β2 π¦ = β2π₯ + 1 β π¦ = β2(β2) + 1 π¦=5 ANSWER: π = βπ, π = π 16. If π(2π₯ β 1) = π₯, what is π(2)? π(2π₯ β 1) = π (2) 3 2π₯ β 1 = 2 π₯ = 2 π ANSWER: π = π βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

17. What is the quotient when 3π₯ + 5π₯ β 4π₯ 3 + 7π₯ + 3 is divided by π₯ + 2? -2 3 5 -4 0 7 -6 2 4 -8 3 -1 -2 4 -1 ANSWER: πππ β ππ β πππ + ππ β π 5

4

3 2 5

2m

6m

18. A man is walking away from a lamppost with a light source 6 meters above the ground. The man is 2 meters tall. How long is his shadow when he is 10 meters from the lamppost?

x

10 m

2m

The two triangles are similar, based on the Triangle Proportionality Theorem.

6m

x

10 m + x

6 10 + π₯ = 2 π₯ 6π₯ = 20 + 2π₯ β π₯ = 5 ANSWER: π π

1,2,3 and 5 if repetition of digits is not allowed? 3 possible 2 possible 2 possible digits: digits: digits: 1 or 2

By the fundamental principle of counting, the product of 2, 3, and 2 is the answer. ANSWER: ππ 21. Factor completely the expression π₯ 3 β 7π₯ + 6. Use the rational zeroes theorem. Employ trial and error. 1 1 0 -7 6 1 1 -6 1 1 -6 0 Since 1 is a zero, π₯ 3 β 7π₯ + 6 = (π₯ β 1)(π₯ 2 + π₯ β 6) π₯ 2 + π₯ β 6 = (π₯ + 3)(π₯ β 2) ANSWER: (π β π)(π + π)(π β π) 22. An integer between 1 and 10000 inclusive is selected at random. What is the probability that it is a perfect square? The probability of selecting a perfect square is equal to the number of perfect squares within the interval divided by the number of integers in that interval. There are 100 perfect squares from 1 to 10 000. π

23. If a chord 24 cm long is 5 cm from the center of a circle, how long is a chord 10 cm from the center? Let π be the chord length, π be the perpendicular distance from the center to the chord; and π be the radius. c

x

20-x

B

20

By the Angle Bisector Theorem, 12 π₯ 2 π₯ = β = 18 20 β π₯ 3 20 β π₯ 40 β 2π₯ = 3π₯ β π₯ = 8 ANSWER: π ππ 20. How many different three-digit numbers less than 300 can be formed with the digits 3|P ag e

Whichever digits are left

ANSWER: πππ or 0.01

19. Find the length of the shorter segment made on side AB of Ξπ΄π΅πΆ by the bisector of β πΆ, if π΄π΅ C= 20; π΄πΆ = 12; and π΅πΆ = 18 ππ.

A

One from (1 or 2), 3, or 5

By Pythagorean theorem, 1 2 π 2 = ( π) + π2 2 Given that π = 24, π = 5 Compute for r. 2 1 2 π = ( (24)) + 52 2 π 2 = 122 + 52 β π 2 = 169 β π = 13 1 2 2 π = ( π) + π2 β π = 2βπ 2 β π 2 2 βVGChuaβEastern Samar Division d

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

π = 2β132 β 102 β π = 2β69 ANSWER: πβππ ππ 24. Pipes are being stored in a pile with 25 pipes in the first layer, 24 in the second, and so on. If there are 12 layers, how many pipes does the pile contain? The problem presents an arithmetic series with π1 = 25, π = β1, and π = 12. π ππ = π1 π + (π β 1)π 2 12 π12 = 25(12) + (12 β 1)(β1) β π12 2 = 234 ANSWER: πππ 25. Find the solution set of the inequality π₯ 2 < 5π₯ β 6. π₯ 2 < 5π₯ β 6 2 π₯ β 5π₯ + 6 < 0 (π₯ β 3)(π₯ β 2) < 0 Split into three intervals to determine the domain of the inequality: (ββ, 2), (2,3), (3, +β) Test at 5 π₯=

Test at π₯=1

Test at π₯=4

2

2

FALSE

3

TRUE

FALSE

ANSWER: π < π < π 26. Suppose an object is dropped from the roof of a very tall building. After t seconds, its height from the ground is given by β = β16π‘ 2 + 625, where h is measured in feet. How long does it take to reach ground level? 25 0 = β16π‘ 2 + 625 β π‘ = 4 ANSWER: π. ππ π

28. Find a polynomial π(π₯) of degree 3 that has zeroes of -2, 0, and 7, and the coefficient of π₯2 is 10. π(π₯) = ππ₯ 3 + 10π₯ 2 + ππ₯ + π ππ₯(π₯ + 2)(π₯ β 7) = ππ₯ 3 β 5ππ₯ 2 β 14ππ₯ β5ππ₯ 2 = 10π₯ 2 β5ππ₯2 10π₯2 = β π = β2 β5π₯2 β5π₯2 3 2 ππ₯ β 5ππ₯ β 14ππ₯ = β2π₯ 3 + 10π₯ 2 + 28π₯ ANSWER: π·(π) = βπππ + ππππ + πππ 29. Find the equation (in the form ππ₯ + ππ¦ = π) of the perpendicular bisector of the line segment joining the points (1, 4) and (7, -2). A perpendicular bisector of a segment is perpendicular to the segment at its midpoint. π¦2 β π¦1 β2 β 4 π= βπ= = β1 π₯2 β π₯1 7β1 The slope of the perpendicular bisector is 1. π₯1 + π₯2 π¦1 + π¦2 π=( , )βπ 2 2 1+7 4β2 =( , ) 2 2 The perpendicular bisector should pass through the point (4, 1). π¦1 = ππ₯1 + π β 1 = 1(4) + π β π = β3 π¦ = π₯β3 ANSWER: π β π = π 30. The sides of a polygon are 3, 4, 5, 6, and 7cm. Find the perimeter of a similar polygon, if the side corresponding to 5 cm is 9 cm long. By similarity, π2

27. In Ξπ΄π΅πΆ, π΄π΅ β₯ π·πΈ, AC = 10 cm, CD = 7 cm, and CE = 9 cm. Find BC. In order to apply the Triangle C Proportionality Theorem, assume that D is on AC and E is on BC. πΆπΈ π΅πΆ = πΆπ· π΄πΆ D

E

9 A

B

ANSWER: 4|P ag e

ππ π

cm or 12.86 cm

7

=

π΅πΆ 10

9 ππ

25 ππ

π1

π

2 = 9 ππ β 5 ππ

π2

=

5 ππ 9 ππ ANSWER: ππ ππ

3+4+5+6+7 5 ππ

=

β π2 = 45 ππ

2

31. If tan π = and π is in Quadrant III, find 3

cos π. tan π =

π π

=

2 3

β π = β2 and π = β3 since π

is in Quadrant III. cos π =

π

π Solve for c, through the Pythagorean theorem, βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

π2 = π2 + π2 β π2 = (β2)2 + (β3)2 β π = β13 3 3β13 cos π = β =β 13 β13 ANSWER: β

πβππ ππ

32. The first term of a geometric sequence is 1 1536, and the common ratio is . Which 2

term of the sequence is 6? 1 π1 = 1536, π = , ππ = 6, π =? 2 ππ = π1 ππβ1 1 πβ1 6 = (1536) ( ) 2 6 1 πβ1 =( ) 1536 2 1 1 = πβ1 256 2 πβ1

2

πβ1

8

= 256 β 2 =2 πβ1= 8β π= 9 ANSWER: 9th term

33. In Ξπ΄π΅πΆ; β π΅ = 2β π΄. The bisectors of these angles meet at D, while BD extended meet AC at E. If β πΆπΈπ΅ = 70Β°, find β π΅π΄πΆ. β πΆπΈπ΅ and β π΄πΈπ΅ form a linear pair and are supplementary.

B

D

E

35. What is the remainder when π₯3 β π₯ + 1 is divided by 2π₯ β 1? By the remainder theorem, 1 1 3 1 1 5 π( ) = ( ) β +1 β π( ) = 2 2 2 2 8 π ANSWER: π

36. Triangle ABC has right angle at B. If β πΆ = 30Β° and BC = 12 cm, find AB. Ξπ΄π΅πΆ is a special right triangle. In a 30-60 right triangle, the side opposite the 60degree angle is β3 times the length of the side opposite the 30-degree angle. Thus, 12 = ABβ3 β π΄π΅ = 4β3 ANSWER: πβπ 37. What is the largest root of the equation 2π₯ 3 + π₯ 2 β 13π₯ + 6 = 0? Use the Rational zeroes theorem to try out possible zeroes.

70Β° C

2 π 2 π 1 1 β (3) 1 1 β (3) ) β ππ = ( ) ππ = ( 2 1 3 3 1β 3 3 2 π 2π ππ = 1 β ( ) β ππ = 1 β π 3 3 2π is a quantity that approaching zero as n 3π gets relatively smaller. ππ = 1 β 0 β ππ = 1 ANSWER: 1

A

2 πβ π΄πΈπ΅ = 110Β° Since πβ π΅π΄πΆ = πβ πΈπ΅π΄ and πβ π΄πΈπ΅ + πβ π΅π΄πΆ + πβ πΈπ΅π΄ = 180Β° 110Β° + 2 πβ π΅π΄πΆ = 180Β° πβ π΅π΄πΆ = 35Β° ANSWER: ππΒ° 34. Find the sum of the series: 1 2 22 23 + + + +β― 3 32 33 34 The infinite sequence described in the series is geometric in nature. 1 β ππ ππ = π1 ( ) 1βπ 5|P ag e

2 2

1 4 5

-13 10 -3

6 -6 0

2π₯ 3 + π₯ 2 β 13π₯ + 6 = 0 (π₯ β 2)(2π₯2 + 5π₯ β 3) = 0 (π₯ β 2)(2π₯ β 1)(π₯ + 3) = 0 π The zeroes of the equation are π, , βπ π ANSWER: π 38. Find the equation (in the form π₯2 + π¦2 + ππ₯ + ππ¦ = π) of the circle that has the points (1, 8) and (5,-6) as the endpoints of the diameter. βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

The center is the midpoint of the two given points. π₯1 + π₯2 π¦1 + π¦2 π=( , )βπ 2 2 1+5 8β6 =( , ) 2 2 π = (3,1) The radius is the distance between one of the endpoints and the midpoint. π = β(π₯2 β π₯1 )2 + (π¦2 β π¦1 )2 π = β(5 β 3)2 + (8 β 1)2 β π = β53 The equation of a circle whose center (β, π) is not at the origin may be obtained through (π₯ β β)2 + (π¦ β π)2 = π2 (π₯ β 3)2 + (π¦ β 1)2 = 53 π₯2 β 6π₯ + 9 + π¦2 β 2π¦ + 1 = 53 π₯2 + π¦2 β 6π₯ β 2π¦ = 43 ANSWER: ππ + ππ β ππ β ππ = ππ 39. The distance from the midpoint of a chord 10 cm long to the midpoint of its minor arc is 3 cm. What is the radius of the circle? Refer to item #23 for a similar example. π = 2βπ 2 β π2

10 = 2βπ2 β (π β 3)2 10 = 2βπ2 β (π 2 β 6π + 9) 10 = 2β6π β 9 5 = β6π β 9 25 = 6π β 9 β π = ππ

ANSWER:

π

17 3

cm

40. Find the equation (in the form ππ₯ + ππ¦ = π) of the line tangent to the circle π₯2 + π¦2 = 25 at the point (3,-4). The slope of the line containing the radius whose endpoints are the center and the given point must be the negative reciprocal of the slope of the tangent line.

π¦2 β π¦1 β4 β 0 4 3 βπ= = β β ππ = π₯2 β π₯1 3β0 3 4 π¦ β π¦1 = π(π₯ β π₯1 ) 3 π¦ + 4 = (π₯ β 3) β 3π₯ β 4π¦ = 25 4 ANSWER: ππ β ππ = ππ

π=

41. How many triangles can be formed by 8 points, no three of which are collinear? 8! 3! (8 β 3)! = = 56 2 2 ANSWER: ππ

(πΆ83 )

42. Two tangents to a circle form an angle of 80Β°. How many degrees is the larger intercepted arc? The measure of the angle formed outside by two tangents to a circle is half the difference of the measures of the intercepted arcs. 1 Μ β ππ΅πΆ Μ) πβ π΄ = (ππ΅π·πΆ 2 Μ + ππ΅πΆ Μ = 360Β° Note however that ππ΅π·πΆ Μ = 360Β° β ππ΅π·πΆ Μ Therefore, ππ΅πΆ 1 Μ β 360Β° + ππ΅π·πΆ Μ) πβ π΄ = (ππ΅π·πΆ 2 1 Μ β 360Β°) 80Β° = (2 ππ΅π·πΆ 2 Μ β 360Β°) 160Β° = (2 ππ΅π·πΆ Μ 160Β° + 360Β° = 2 ππ΅π·πΆ Μ = 260Β° ππ΅π·πΆ ANSWER: πππΒ° 43. Find all real solutions of the equation 1

1

π₯3 + π₯6 β 2 = 0. Treat the equation like a quadratic trinomial, 1

2

1

1

(π₯6 ) + (π₯6 ) β 2 = 0 where π₯6 is the variable. Factor out as, 1

1

(π₯6 + 2) (π₯6 β 1) = 0 Solve for x. 1

π₯6 + 2 = 0 β π₯ = 64 6|P ag e

βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS 1

π₯6 β 1 = 0 β π₯ = 1 The equation does not hold true for π₯ = 64, this being an extraneous root. ANSWER: π

Grade 10 Elimination Round

47. Find the area (in terms of π) of the circle inscribed in the triangle enclosed by the line 3π₯ + 4π¦ = 24 and the coordinate axes.

44. The sides of a triangle are 6, 8, and 10, cm, respectively. What is the length of its shortest altitude?

6

a

x

10

The altitude to a side of a triangle divides the 8 triangle into two similar triangles. The shortest 10-x altitude is that to the longest side.

π΄π‘ π Where π is the radius of the circle, π΄π‘ is the area of the triangle, and π is the semi1 perimeter (π = 2 (π + π + π)) 1 π΄π = ππ 2 , π΄π‘ = πβ 2 2 1 πβ π΄π = π (2 ) π π=

π₯

6

π

By proportion, = = π 8 10βπ₯ 3π 3 π 24 π₯= , = βπ= ππ 4.8 3π 4 4 10 β 5 4 ππ ANSWER: ππ π. π ππ π

45. In a circle of radius 10 cm, arc AB measures 120Β°. Find the distance from B to the diameter through A. The triangle is special r=10

A

x 60

B

2

π΄π = π ( ) 1 (6 + 8 + 10) 2 24 2 π΄π = π ( ) β π΄π = 4π 12 ANSWER: ππ πππ

120

The side opposite the 30-degree angle is 5 cm because it is half the hypotenuse which happens to be a radius. Therefore, the length of x is 5β3 ANSWER: πβπ ππ 46. Find the remainder when π₯100 is divided by π₯ 2 β π₯. π₯100 π₯100 π₯ π₯ 99 = = β π₯ 2 β π₯ π₯(π₯ β 1) π₯ π₯ β 1 By the remainder theorem, π(1) = 199 = 1 Multiplying the remainder by x, the remainder becomes x. ANSWER: π

7|P ag e

1 (6)(8) 2

48. In how many different ways can 5 persons be seated in an automobile having places for 2 in the front seat and 3 in the back if only 2 of them can drive and one of the others insists on riding in the back? (2) Driverβs (1) The last seat person 2 possibilities (1) The (3) Any (2) Any person of the of the insisting three two left to sit on others the back By Fundamental Principle of Counting, 2 β 1 β 1 β 3 β 2 = 12 The three passengers at the back may be rearranged in three different ways, so, 12 β 3 = 36 ANSWER: ππ βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS

Grade 10 Elimination Round

49. Factor completely the expression π₯ 4 + 3π₯ 2 + 4. π₯ 4 + 3π₯ 2 + 4 β π₯ 4 + 4π₯ 2 + 4 β π₯ 2 (π₯ 4 + 4π₯ 2 + 4) β π₯ 2 (π₯ 2 β 4)2 β π₯ 2 2 [(π₯ β 4) β π₯][(π₯ 2 β 4) + π₯] (π₯ 2 β π₯ β 4)(π₯ 2 + π₯ β 4) ANSWER: (ππ β π β π)(ππ + π β π) 50. A bag is filled with red and blue balls. 1 Before drawing a ball, there is a 4 chance of drawing a blue ball. After drawing out a ball, 1 there is now a 5 chance of drawing a blue ball. How many red balls are originally in the bag? Since π(π2 ) < π(π1 ), the ball first drawn is blue. π 1 π (π1 ) = = π+π 4 πβ1 1 π (π1 ) = = πβ1+π 5 π 1 = β 4π = π + π β π = 3π π+π 4 πβ1 1 πβ1 1 = β = π β 1 + π 5 π β 1 + 3π 5 πβ1 1 = β 5π β 5 = 4π β 1 4π β 1 5 π = 4; π = 12 ANSWER: 12

8|P ag e

βVGChuaβEastern Samar Division

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ANSWERS WITH SOLUTIONS

1. Find the value of βπ₯ + β14 β π₯ when π₯ = β2. ββ2 + β14 β (β2) = ββ2 + β14 + 2 = β2 ANSWER: βπ 2. Find the area of the triangle formed by the coordinate axes and the line 3π₯ + 2π¦ = 6. The triangle has a base equal to the x-intercept of the graph of 3π₯ + 2π¦ = 6 and a height equal to the y-intercept. The area, therefore is half the product of the intercepts. 1 π΄ = πβ 2

1 π΄ = (2)(3) π΄ 2 = 3 π ππ’πππ π’πππ‘π ANSWER: 3 square units 3. A sequence is defined by ππ = 3(ππβ1 + 2) for π β₯ 2, where π1 = 1. What is π4 ? π1 = 1 ( ) π2 = 3 π1 + 2 π2 = 3(1 + 2) π2 = 9 π3 = 3(π2 + 2) π3 = 3(9 + 2) π2 = 33 π4 = 3(π3 + 2) π4 = 3(33 + 2) π2 = 105 ANSWER: πππ 4. Christa leaves Town A. After traveling 12 km, she reaches Town B at 2:00 P.M. Then she drove at a constant speed and passes Town C, 40 km from Town B, at 2:50 P.M. Find the function π(π‘) that models the distance (in km) she has traveled from Town A t minutes after 2:00 P.M. π1 = 12 ππ TOWN A

π2 = 40 ππ π‘ = 50 πππ

TOWN B 2:00 PM

π = π =

π π‘

TOWN C 2:50 PM

40 ππ = 0.8 πππ 50 πππ

π = π π‘ π = 0.8π‘ πππ πππ π‘ππππ πππ‘ππ πππ€π π΅ π(π‘) = 12 + 0.8π‘ ANSWER: π (π) = ππ + π. ππ 1|P ag e

Grade 10 Elimination Round

5. What is the largest negative integer that satisfies the inequality |3π₯ + 2| > 4? Two conditions will satisfy this inequality, (1) If 3π₯ + 2 > 4; and (2) If 3π₯ + 2 < β4 2 In case (1), π₯ > 3. In case (2), π₯ < β2. This is the solution set to the inequality. Considering case (2), the largest negative integer that satisfies this condition is -3. ANSWER: βπ 6. A person has two parents, four grandparents, eight grand-parents, and so on. How many ancestors does a person have 10 generations back? This being a geometric series of 2+4+8+β¦+π10 , 1βπ π ) Use the formula, ππ = π1 ( 1βπ

1 β 210 π10 = 2 ( ) 1β2 π10 = 2046 ANSWER: ππππ 7. In Ξπ΄π΅πΆ, β π΅ is twice β π΄, and β πΆ is three times as large as β π΅. Find β πΆ If πβ π΅ = 2 πβ π΄ πππ πβ πΆ = 3 πβ π΅, then πβ πΆ = 6 πβ π΄ πβ π΄ + πβ π΅ + πβ πΆ = 180Β° πβ π΄ + 2 πβ π΄ + 6 πβ π΄ = 180Β° 9 πβ π΄ = 180Β° πβ π΄ = 20Β° πβ πΆ = 6 πβ π΄ = 120Β° ANSWER: πππΒ° 8. If β3 β€ π₯ β€ 0, find the minimum value of π(π₯) = π₯ 2 + 4π₯. The graph of π(π₯) opens upward based on its leading coefficient. Therefore, the absolute minimum value of the function is the y-coordinate, k, of its vertex. π2 π=πβ 4π 42 π=0β 4(1) π = β4 ANSWER: -4

βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

9. The 9th and the 11th term of an arithmetic sequence are 28 and 45, respectively. What is the 12th term? π11 β π9 2π = π11 β π9 π= 2 45 β 28 17 π= = ππ 8.5 2 2 π12 = π11 + π π12 = 45 + 8.5 ANSWER: 53.5 10. Perform the indicated operation, and simplify: 2 3 4 + β 2 π₯ π₯β1 π₯ βπ₯ 2(π₯ β 1) + 3π₯ β 4 π₯2 β π₯ 5π₯ β 6 π₯2 β π₯ ππβπ ANSWER: π π βπ 11. Find the range of the function π (π₯) = |2π₯ + 1|. For any absolute value function of this form, π (π₯) = |π(π₯)| where π(π₯) is a linear function, the range is always {π¦|π¦ β β, π¦ β₯ 0}. ANSWER: π β₯ π 12. Find the value of [π₯ β (π₯ β π₯ β1 )β1 ]β1 when π₯ = 2. [2 β (2 β 2β1 )β1 ]β1 1 β1 [2 β (2 β ) ] 2 3 β1 [2 β ( ) ] 2 π ANSWER: π

β1

β1

2 β1 4 β1 3 = [2 β ] β ( ) = 3 3 4

13. Find the equation (in the form ππ₯ + ππ¦ = π) of the line through the point (5, 2) that is parallel to the line 4π₯ + 6π¦ + 5 = 0. 2 5 4π₯ + 6π¦ + 5 = 0 β π¦ = β π₯ β 3 6 2 Solve for π given that π = β 3 and (5, 2) π¦1 = ππ₯1 + π 2 16 2 = β (5) + π π = 3 3 2 16 π¦=β π₯+ β 2π₯ + 3π¦ = 16 3 3 ANSWER: ππ + ππ = ππ 2|P ag e

Note on an alternative: When asked for the equation of a line, ππ₯ + ππ¦ = π, parallel to a given linear equation of the form ππ π₯ + ππ π¦ = π1 and passing through the point, (ππ , ππ ), use the algorithm: ππ π + ππ π = ππ ππ + ππ ππ When asked for the equation of a line, ππ₯ + ππ¦ = π, perpendicular to a given linear equation of the form ππ π₯ + ππ π¦ = π1 and passing through the point, (ππ , ππ ), use the algorithm: ππ π β ππ π = βππ ππ + ππ ππ 14. In Ξπ΄π΅πΆ, β π΅ = 90Β°, β π΄πΆπ΅ = 28Β°, and D is the midpoint of AC. What is β π΅π·πΆ? A D

C

B

The median to the hypotenuse of a right triangle is half the measure of the hypotenuse. Since π΅π· = π·πΆ, ΞBDC is isosceles. It also follows that πβ πΆ = πβ π·π΅πΆ. πβ πΆ + πβ π·π΅πΆ + πβ π΅π·πΆ = 180Β° 28Β° + 28Β° + πβ π΅π·πΆ = 180Β° πβ π΅π·πΆ = 180Β° β 56Β° πβ π΅π·πΆ = 124Β° ANSWER: πππΒ° 15. Find all solutions of the system: 2π₯ + π¦ = 1 { 3π₯ + 4π¦ = 14 2π₯ + π¦ = 1 β π¦ = β2π₯ + 1 By substitution, 3π₯ + 4(β2π₯ + 1) = 14 3π₯ β 8π₯ + 4 = 14 β5π₯ = 10, π₯ = β2 π¦ = β2π₯ + 1 β π¦ = β2(β2) + 1 π¦=5 ANSWER: π = βπ, π = π 16. If π(2π₯ β 1) = π₯, what is π(2)? π(2π₯ β 1) = π (2) 3 2π₯ β 1 = 2 π₯ = 2 π ANSWER: π = π βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

17. What is the quotient when 3π₯ + 5π₯ β 4π₯ 3 + 7π₯ + 3 is divided by π₯ + 2? -2 3 5 -4 0 7 -6 2 4 -8 3 -1 -2 4 -1 ANSWER: πππ β ππ β πππ + ππ β π 5

4

3 2 5

2m

6m

18. A man is walking away from a lamppost with a light source 6 meters above the ground. The man is 2 meters tall. How long is his shadow when he is 10 meters from the lamppost?

x

10 m

2m

The two triangles are similar, based on the Triangle Proportionality Theorem.

6m

x

10 m + x

6 10 + π₯ = 2 π₯ 6π₯ = 20 + 2π₯ β π₯ = 5 ANSWER: π π

1,2,3 and 5 if repetition of digits is not allowed? 3 possible 2 possible 2 possible digits: digits: digits: 1 or 2

By the fundamental principle of counting, the product of 2, 3, and 2 is the answer. ANSWER: ππ 21. Factor completely the expression π₯ 3 β 7π₯ + 6. Use the rational zeroes theorem. Employ trial and error. 1 1 0 -7 6 1 1 -6 1 1 -6 0 Since 1 is a zero, π₯ 3 β 7π₯ + 6 = (π₯ β 1)(π₯ 2 + π₯ β 6) π₯ 2 + π₯ β 6 = (π₯ + 3)(π₯ β 2) ANSWER: (π β π)(π + π)(π β π) 22. An integer between 1 and 10000 inclusive is selected at random. What is the probability that it is a perfect square? The probability of selecting a perfect square is equal to the number of perfect squares within the interval divided by the number of integers in that interval. There are 100 perfect squares from 1 to 10 000. π

23. If a chord 24 cm long is 5 cm from the center of a circle, how long is a chord 10 cm from the center? Let π be the chord length, π be the perpendicular distance from the center to the chord; and π be the radius. c

x

20-x

B

20

By the Angle Bisector Theorem, 12 π₯ 2 π₯ = β = 18 20 β π₯ 3 20 β π₯ 40 β 2π₯ = 3π₯ β π₯ = 8 ANSWER: π ππ 20. How many different three-digit numbers less than 300 can be formed with the digits 3|P ag e

Whichever digits are left

ANSWER: πππ or 0.01

19. Find the length of the shorter segment made on side AB of Ξπ΄π΅πΆ by the bisector of β πΆ, if π΄π΅ C= 20; π΄πΆ = 12; and π΅πΆ = 18 ππ.

A

One from (1 or 2), 3, or 5

By Pythagorean theorem, 1 2 π 2 = ( π) + π2 2 Given that π = 24, π = 5 Compute for r. 2 1 2 π = ( (24)) + 52 2 π 2 = 122 + 52 β π 2 = 169 β π = 13 1 2 2 π = ( π) + π2 β π = 2βπ 2 β π 2 2 βVGChuaβEastern Samar Division d

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

π = 2β132 β 102 β π = 2β69 ANSWER: πβππ ππ 24. Pipes are being stored in a pile with 25 pipes in the first layer, 24 in the second, and so on. If there are 12 layers, how many pipes does the pile contain? The problem presents an arithmetic series with π1 = 25, π = β1, and π = 12. π ππ = π1 π + (π β 1)π 2 12 π12 = 25(12) + (12 β 1)(β1) β π12 2 = 234 ANSWER: πππ 25. Find the solution set of the inequality π₯ 2 < 5π₯ β 6. π₯ 2 < 5π₯ β 6 2 π₯ β 5π₯ + 6 < 0 (π₯ β 3)(π₯ β 2) < 0 Split into three intervals to determine the domain of the inequality: (ββ, 2), (2,3), (3, +β) Test at 5 π₯=

Test at π₯=1

Test at π₯=4

2

2

FALSE

3

TRUE

FALSE

ANSWER: π < π < π 26. Suppose an object is dropped from the roof of a very tall building. After t seconds, its height from the ground is given by β = β16π‘ 2 + 625, where h is measured in feet. How long does it take to reach ground level? 25 0 = β16π‘ 2 + 625 β π‘ = 4 ANSWER: π. ππ π

28. Find a polynomial π(π₯) of degree 3 that has zeroes of -2, 0, and 7, and the coefficient of π₯2 is 10. π(π₯) = ππ₯ 3 + 10π₯ 2 + ππ₯ + π ππ₯(π₯ + 2)(π₯ β 7) = ππ₯ 3 β 5ππ₯ 2 β 14ππ₯ β5ππ₯ 2 = 10π₯ 2 β5ππ₯2 10π₯2 = β π = β2 β5π₯2 β5π₯2 3 2 ππ₯ β 5ππ₯ β 14ππ₯ = β2π₯ 3 + 10π₯ 2 + 28π₯ ANSWER: π·(π) = βπππ + ππππ + πππ 29. Find the equation (in the form ππ₯ + ππ¦ = π) of the perpendicular bisector of the line segment joining the points (1, 4) and (7, -2). A perpendicular bisector of a segment is perpendicular to the segment at its midpoint. π¦2 β π¦1 β2 β 4 π= βπ= = β1 π₯2 β π₯1 7β1 The slope of the perpendicular bisector is 1. π₯1 + π₯2 π¦1 + π¦2 π=( , )βπ 2 2 1+7 4β2 =( , ) 2 2 The perpendicular bisector should pass through the point (4, 1). π¦1 = ππ₯1 + π β 1 = 1(4) + π β π = β3 π¦ = π₯β3 ANSWER: π β π = π 30. The sides of a polygon are 3, 4, 5, 6, and 7cm. Find the perimeter of a similar polygon, if the side corresponding to 5 cm is 9 cm long. By similarity, π2

27. In Ξπ΄π΅πΆ, π΄π΅ β₯ π·πΈ, AC = 10 cm, CD = 7 cm, and CE = 9 cm. Find BC. In order to apply the Triangle C Proportionality Theorem, assume that D is on AC and E is on BC. πΆπΈ π΅πΆ = πΆπ· π΄πΆ D

E

9 A

B

ANSWER: 4|P ag e

ππ π

cm or 12.86 cm

7

=

π΅πΆ 10

9 ππ

25 ππ

π1

π

2 = 9 ππ β 5 ππ

π2

=

5 ππ 9 ππ ANSWER: ππ ππ

3+4+5+6+7 5 ππ

=

β π2 = 45 ππ

2

31. If tan π = and π is in Quadrant III, find 3

cos π. tan π =

π π

=

2 3

β π = β2 and π = β3 since π

is in Quadrant III. cos π =

π

π Solve for c, through the Pythagorean theorem, βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

π2 = π2 + π2 β π2 = (β2)2 + (β3)2 β π = β13 3 3β13 cos π = β =β 13 β13 ANSWER: β

πβππ ππ

32. The first term of a geometric sequence is 1 1536, and the common ratio is . Which 2

term of the sequence is 6? 1 π1 = 1536, π = , ππ = 6, π =? 2 ππ = π1 ππβ1 1 πβ1 6 = (1536) ( ) 2 6 1 πβ1 =( ) 1536 2 1 1 = πβ1 256 2 πβ1

2

πβ1

8

= 256 β 2 =2 πβ1= 8β π= 9 ANSWER: 9th term

33. In Ξπ΄π΅πΆ; β π΅ = 2β π΄. The bisectors of these angles meet at D, while BD extended meet AC at E. If β πΆπΈπ΅ = 70Β°, find β π΅π΄πΆ. β πΆπΈπ΅ and β π΄πΈπ΅ form a linear pair and are supplementary.

B

D

E

35. What is the remainder when π₯3 β π₯ + 1 is divided by 2π₯ β 1? By the remainder theorem, 1 1 3 1 1 5 π( ) = ( ) β +1 β π( ) = 2 2 2 2 8 π ANSWER: π

36. Triangle ABC has right angle at B. If β πΆ = 30Β° and BC = 12 cm, find AB. Ξπ΄π΅πΆ is a special right triangle. In a 30-60 right triangle, the side opposite the 60degree angle is β3 times the length of the side opposite the 30-degree angle. Thus, 12 = ABβ3 β π΄π΅ = 4β3 ANSWER: πβπ 37. What is the largest root of the equation 2π₯ 3 + π₯ 2 β 13π₯ + 6 = 0? Use the Rational zeroes theorem to try out possible zeroes.

70Β° C

2 π 2 π 1 1 β (3) 1 1 β (3) ) β ππ = ( ) ππ = ( 2 1 3 3 1β 3 3 2 π 2π ππ = 1 β ( ) β ππ = 1 β π 3 3 2π is a quantity that approaching zero as n 3π gets relatively smaller. ππ = 1 β 0 β ππ = 1 ANSWER: 1

A

2 πβ π΄πΈπ΅ = 110Β° Since πβ π΅π΄πΆ = πβ πΈπ΅π΄ and πβ π΄πΈπ΅ + πβ π΅π΄πΆ + πβ πΈπ΅π΄ = 180Β° 110Β° + 2 πβ π΅π΄πΆ = 180Β° πβ π΅π΄πΆ = 35Β° ANSWER: ππΒ° 34. Find the sum of the series: 1 2 22 23 + + + +β― 3 32 33 34 The infinite sequence described in the series is geometric in nature. 1 β ππ ππ = π1 ( ) 1βπ 5|P ag e

2 2

1 4 5

-13 10 -3

6 -6 0

2π₯ 3 + π₯ 2 β 13π₯ + 6 = 0 (π₯ β 2)(2π₯2 + 5π₯ β 3) = 0 (π₯ β 2)(2π₯ β 1)(π₯ + 3) = 0 π The zeroes of the equation are π, , βπ π ANSWER: π 38. Find the equation (in the form π₯2 + π¦2 + ππ₯ + ππ¦ = π) of the circle that has the points (1, 8) and (5,-6) as the endpoints of the diameter. βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

The center is the midpoint of the two given points. π₯1 + π₯2 π¦1 + π¦2 π=( , )βπ 2 2 1+5 8β6 =( , ) 2 2 π = (3,1) The radius is the distance between one of the endpoints and the midpoint. π = β(π₯2 β π₯1 )2 + (π¦2 β π¦1 )2 π = β(5 β 3)2 + (8 β 1)2 β π = β53 The equation of a circle whose center (β, π) is not at the origin may be obtained through (π₯ β β)2 + (π¦ β π)2 = π2 (π₯ β 3)2 + (π¦ β 1)2 = 53 π₯2 β 6π₯ + 9 + π¦2 β 2π¦ + 1 = 53 π₯2 + π¦2 β 6π₯ β 2π¦ = 43 ANSWER: ππ + ππ β ππ β ππ = ππ 39. The distance from the midpoint of a chord 10 cm long to the midpoint of its minor arc is 3 cm. What is the radius of the circle? Refer to item #23 for a similar example. π = 2βπ 2 β π2

10 = 2βπ2 β (π β 3)2 10 = 2βπ2 β (π 2 β 6π + 9) 10 = 2β6π β 9 5 = β6π β 9 25 = 6π β 9 β π = ππ

ANSWER:

π

17 3

cm

40. Find the equation (in the form ππ₯ + ππ¦ = π) of the line tangent to the circle π₯2 + π¦2 = 25 at the point (3,-4). The slope of the line containing the radius whose endpoints are the center and the given point must be the negative reciprocal of the slope of the tangent line.

π¦2 β π¦1 β4 β 0 4 3 βπ= = β β ππ = π₯2 β π₯1 3β0 3 4 π¦ β π¦1 = π(π₯ β π₯1 ) 3 π¦ + 4 = (π₯ β 3) β 3π₯ β 4π¦ = 25 4 ANSWER: ππ β ππ = ππ

π=

41. How many triangles can be formed by 8 points, no three of which are collinear? 8! 3! (8 β 3)! = = 56 2 2 ANSWER: ππ

(πΆ83 )

42. Two tangents to a circle form an angle of 80Β°. How many degrees is the larger intercepted arc? The measure of the angle formed outside by two tangents to a circle is half the difference of the measures of the intercepted arcs. 1 Μ β ππ΅πΆ Μ) πβ π΄ = (ππ΅π·πΆ 2 Μ + ππ΅πΆ Μ = 360Β° Note however that ππ΅π·πΆ Μ = 360Β° β ππ΅π·πΆ Μ Therefore, ππ΅πΆ 1 Μ β 360Β° + ππ΅π·πΆ Μ) πβ π΄ = (ππ΅π·πΆ 2 1 Μ β 360Β°) 80Β° = (2 ππ΅π·πΆ 2 Μ β 360Β°) 160Β° = (2 ππ΅π·πΆ Μ 160Β° + 360Β° = 2 ππ΅π·πΆ Μ = 260Β° ππ΅π·πΆ ANSWER: πππΒ° 43. Find all real solutions of the equation 1

1

π₯3 + π₯6 β 2 = 0. Treat the equation like a quadratic trinomial, 1

2

1

1

(π₯6 ) + (π₯6 ) β 2 = 0 where π₯6 is the variable. Factor out as, 1

1

(π₯6 + 2) (π₯6 β 1) = 0 Solve for x. 1

π₯6 + 2 = 0 β π₯ = 64 6|P ag e

βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS 1

π₯6 β 1 = 0 β π₯ = 1 The equation does not hold true for π₯ = 64, this being an extraneous root. ANSWER: π

Grade 10 Elimination Round

47. Find the area (in terms of π) of the circle inscribed in the triangle enclosed by the line 3π₯ + 4π¦ = 24 and the coordinate axes.

44. The sides of a triangle are 6, 8, and 10, cm, respectively. What is the length of its shortest altitude?

6

a

x

10

The altitude to a side of a triangle divides the 8 triangle into two similar triangles. The shortest 10-x altitude is that to the longest side.

π΄π‘ π Where π is the radius of the circle, π΄π‘ is the area of the triangle, and π is the semi1 perimeter (π = 2 (π + π + π)) 1 π΄π = ππ 2 , π΄π‘ = πβ 2 2 1 πβ π΄π = π (2 ) π π=

π₯

6

π

By proportion, = = π 8 10βπ₯ 3π 3 π 24 π₯= , = βπ= ππ 4.8 3π 4 4 10 β 5 4 ππ ANSWER: ππ π. π ππ π

45. In a circle of radius 10 cm, arc AB measures 120Β°. Find the distance from B to the diameter through A. The triangle is special r=10

A

x 60

B

2

π΄π = π ( ) 1 (6 + 8 + 10) 2 24 2 π΄π = π ( ) β π΄π = 4π 12 ANSWER: ππ πππ

120

The side opposite the 30-degree angle is 5 cm because it is half the hypotenuse which happens to be a radius. Therefore, the length of x is 5β3 ANSWER: πβπ ππ 46. Find the remainder when π₯100 is divided by π₯ 2 β π₯. π₯100 π₯100 π₯ π₯ 99 = = β π₯ 2 β π₯ π₯(π₯ β 1) π₯ π₯ β 1 By the remainder theorem, π(1) = 199 = 1 Multiplying the remainder by x, the remainder becomes x. ANSWER: π

7|P ag e

1 (6)(8) 2

48. In how many different ways can 5 persons be seated in an automobile having places for 2 in the front seat and 3 in the back if only 2 of them can drive and one of the others insists on riding in the back? (2) Driverβs (1) The last seat person 2 possibilities (1) The (3) Any (2) Any person of the of the insisting three two left to sit on others the back By Fundamental Principle of Counting, 2 β 1 β 1 β 3 β 2 = 12 The three passengers at the back may be rearranged in three different ways, so, 12 β 3 = 36 ANSWER: ππ βVGChuaβEastern Samar Division

2016 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS

Grade 10 Elimination Round

49. Factor completely the expression π₯ 4 + 3π₯ 2 + 4. π₯ 4 + 3π₯ 2 + 4 β π₯ 4 + 4π₯ 2 + 4 β π₯ 2 (π₯ 4 + 4π₯ 2 + 4) β π₯ 2 (π₯ 2 β 4)2 β π₯ 2 2 [(π₯ β 4) β π₯][(π₯ 2 β 4) + π₯] (π₯ 2 β π₯ β 4)(π₯ 2 + π₯ β 4) ANSWER: (ππ β π β π)(ππ + π β π) 50. A bag is filled with red and blue balls. 1 Before drawing a ball, there is a 4 chance of drawing a blue ball. After drawing out a ball, 1 there is now a 5 chance of drawing a blue ball. How many red balls are originally in the bag? Since π(π2 ) < π(π1 ), the ball first drawn is blue. π 1 π (π1 ) = = π+π 4 πβ1 1 π (π1 ) = = πβ1+π 5 π 1 = β 4π = π + π β π = 3π π+π 4 πβ1 1 πβ1 1 = β = π β 1 + π 5 π β 1 + 3π 5 πβ1 1 = β 5π β 5 = 4π β 1 4π β 1 5 π = 4; π = 12 ANSWER: 12

8|P ag e

βVGChuaβEastern Samar Division

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