2017 mmc grade 10 eliminations questions with answers

2017 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS

1. Arrange the following from least to 3 greatest: 15%, 0.1, 25. Transform all quantities into the same numerical notation. 3 15% = 0.15, 0.1, = 0.12 25 𝟑 ANSWER: 𝟎. 𝟏, 𝟐𝟓 , 𝟏𝟓% 2. A box contains red, blue, and yellow balls. If the ratio of red to blue to yellow balls is 3:6:10, and there are 42 blue balls, how many are yellow? Assume that each part (or unit) in the ratio is x. 6𝑥 = 42 → 𝑥 = 7 10𝑥 = 10(7) = 70 ANSWER: 70 3. If 3% of a number is 18, what is 20% of twice the number? Let x be the number, 0.03𝑥 = 18 → 𝑥 = 600 0.2(2𝑥 ) = 0.2(2(600)) ANSWER: 240 2

4. If 𝑥 = −1, find the value of 𝑥 𝑥 + 𝑥 𝑥 . 2 (−1)−1 + (−1)(−1) −1 + (−1) ANSWER: -2 5. What is 𝑄(𝑥) in (𝑥 + 2)𝑄(𝑥) + (4𝑥 + 2) = 𝑥 3 + 5𝑥 2 + 9𝑥? 𝑄(𝑥) represents the quotient when 𝑥 3 + 5𝑥 2 + 9𝑥 is divided by 𝑥 + 2 and with a remainder of 4𝑥 + 2. (𝑥 + 2)𝑄(𝑥) + (4𝑥 + 2) = 𝑥 3 + 5𝑥 2 + 9𝑥 (𝑥 3 + 5𝑥 2 + 9𝑥) − (4𝑥 + 2) 𝑄(𝑥) = 𝑥+2 𝑥 3 + 5𝑥 2 + 5𝑥 − 2) 𝑄 (𝑥) = 𝑥+2 By synthetic division, -2 1 5 -2 1 3 𝟐 ANSWER: 𝒙 + 𝟑𝒙 − 𝟏

5 -6 -1

6. Which is greater, 2√5 or 3√3? 2√5 = √20, 3√3 = √27 ANSWER: 𝟑√𝟑 1|P ag e

-2 2 0

Grade 10 Elimination Round

7. If the point (-2,11) lies on the line 𝑦 = 𝑚𝑥 + 3, what is m? 11 = 𝑚 (−2) + 3 11 − 3 𝑚= −2 ANSWER: -4 8. The exterior angle of an equiangular polygon is 40°. How many sides does the polygon have? The sum of the exterior angles of a polygon is 360°. If the polygon is equiangular, it is regular and all its exterior angles are equal. Hence, divide 360 by 40. ANSWER: 9 9. What is the equation of the line (in slopeintercept form) that is parallel to 2𝑥 + 3𝑦 = 1 and passes through (1, -2)? 2

The slope of 2𝑥 + 3𝑦 = 1 is − . This is also 3 the slope of the line whose equation is being asked. Solve for its y-intercept using, 𝑦 = 𝑚𝑥 + 𝑏 2 4 −2 = − (1) + 𝑏 → 𝑏 = − 3 3 2 4 𝑦=− 𝑥− 3 3 𝟐 𝟒 ANSWER: 𝒚 = − 𝟑 𝒙 − 𝟑 10. If −11 − 6𝑤 ≥ −35, what is the largest value of 𝑤? −11 − 6𝑤 ≥ −35 −6𝑤 ≥ −35 + 11 1 1 − (−6𝑤 ≥ −35 + 11) − 6 6 𝑤≤4 ANSWER: 4 11. The average of two radical expressions is √3 − √2. If one of them is √3 + √2, what is the other radical expression? (√3 + √2) + 𝑥 = √3 − √2 2 𝑥 = 2(√3 − √2) − (√3 + √2) 𝑥 = 2√3 − 2√2 − √3 − √2 𝑥 = √3 − 3√2 ANSWER: 𝒙 = √𝟑 − 𝟑√𝟐 ●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

12. If 𝑓(𝑥) = 1 + 5𝑥 − 2𝑥 2 , what is 𝑓(1 + ℎ ) − 𝑓(1)? 𝑓 (1 + ℎ ) − 𝑓 (1) = [1 + 5(1 + ℎ ) − 2(1 + ℎ )2 ] − [1 + 5(1) − 2(1)2 ] = 1 + 5 + 5ℎ − 2(1 + 2ℎ + ℎ 2 ) − (1 + 5 − 2) = −2ℎ 2 + ℎ ANSWER: 𝒉 − 𝟐𝒉𝟐 13. Find the least value of k for the equation 2𝑘 = 2(𝑥 − 1)2 + 5 to have a real root. Use the discriminant, transform to standard form: 2(𝑥 2 − 2𝑥 + 1) + 5 − 2𝑘 = 0 2𝑥 2 − 4𝑥 + (7 − 2𝑘) = 0 𝐷 = 𝑏2 − 4𝑎𝑐 𝐷 ≥ 0 for the equation to have real roots. (−4)2 − 4(2)(7 − 2𝑘) ≥ 0 16 − 56 + 16𝑘 ≥ 0 5 𝑘≥ 2 𝟓 ANSWER: 𝟐

14. A liquid mixture is 4% pesticide. Some amount of this mixture and some amount of water are mixed to make 500 L of new mixture. Find the amount of water needed so that the new mixture is 1% pesticide. Let x be the amount of the 4% pesticide mixture in the new mixture; y be the amount of pure water in the new mixture. 𝑥 + 𝑦 = 500 The 4% pesticide mixture is 96% water so the amount of water in the mixture is .96 times the volume. When the water content from this mixture is added to some amount of pure water and the sum is divided by the total volume of 500 L, we must get a 99% water content in the new mixture. Hence, 0.96𝑥 + 𝑦 = 0.99 500 By substitution, 0.96(500 − 𝑦) + 𝑦 = 0.99 500 480 + 0.04𝑦 495 − 480 = 0.99 → 𝑦 = 500 0.04 𝑦 = 375 𝐿 ANSWER: 375 L 2|P ag e

15. If 𝑎𝑛 = (−1)𝑛+1 − 2𝑛 + 2𝑛, what is 𝑎4 ? By substitution, 𝑎4 = (−1)4+1 − 24 + 2(4) 𝑎4 = (−1) − 16 + 8 𝑎4 = −9 ANSWER: -9 16. If the numbers is −15, 𝑎2 , 𝑎3 , 𝑎4 , 21 form an arithmetic sequence, what is 𝑎4 ? There are three arithmetic means between 15 and 21 and 4 times the common difference. 21 − (−15) 𝑑= =9 4 𝑎4 = 𝑎5 − 𝑑 = 21 − 9 ANSWER: 12 17. If the third and ninth terms of an arithmetic sequence are 13 and 37, respectively, what is the sum of the first six terms of the sequence? 𝑎𝑏 − 𝑎𝑎 𝑑= 𝑏−𝑎 37 − 13 24 𝑑= = =4 9−3 6 𝑎1 = 𝑎𝑛 − (𝑛 − 1)𝑑 𝑎1 = 𝑎3 − (3 − 1)𝑑 𝑎1 = 13 − (3 − 1)4 𝑎1 = 5 𝑛 𝑆𝑛 = 𝑎1 𝑛 + (𝑛 − 1)𝑑 2 6 𝑆6 = 5(6) + (6 − 1)4 2 𝑆6 = 90 ANSWER: 90 18. How many three-digit numbers from 100 to 500 are divisible by 12? The least multiple of 12 between the given interval is 108. The difference between 500 and 108 is 392. The quotient (plus 1 for 108) between 392 and 12 is the number of three-digit numbers divisible by 12 within the interval. ANSWER: 33 ●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS 1

1

19. If the numbers − 10 , 𝑎2 , 𝑎3 , 𝑎4 , 𝑎5 , 320 form a geometric sequence, what is 𝑎4 ? 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 1 1 = (− ) 𝑟 6−1 320 10 1 1 𝑟 5 = 320 = ( ) (−10) 1 320 − 10 1 𝑟5 = − 32 5

𝑟 = √−

1 1 =− 32 2

1 1 4−1 (− ) 10 2 1 𝑎4 = 80

𝑎4 = −

𝟏

ANSWER: 𝟖𝟎

𝟓

𝑝

20. If the second and fourth terms of a geometric sequence are 250 and 10, respectively, what is the sum of the first five terms of the sequence? 1 − 𝑟𝑛 𝑆𝑛 = 𝑎1 ( ) 1−𝑟 𝑎𝑛 10 = 𝑟 𝑚−𝑛 → = 𝑟 4−2 𝑎𝑚 250 𝑟=√

1 1 =± 25 5

1

Case 1: when 𝑟 = 5 250 𝑎1 = = 1250 1 5 1 5 1−( ) 5 ) 𝑆5 = 1250 ( 1 1−5 𝑆5 = 1562 1 Case 2: when 𝑟 = − 5 250 𝑎1 = = −1250 1 −5 1 5 1 − (− ) 5 ) 𝑆5 = −1250 ( 1 1 − (− 5) 𝑆5 = −1042 ANSWER: 1562 or -1042 3|P ag e

21. Find the sum of the infinite geometric series 3 3 3 − 8 + 32 − ⋯. 2 An infinite geometric series may be computed as 𝑎1 𝑆𝑛 = 1−𝑟 3 (− ) 8 𝑟= 3 ( ) 2 1 𝑟=− 4 3 2 𝑆𝑛 = 1 1 − (− 4) 6 𝑆𝑛 = 5 𝟔 ANSWER: ̅̅̅̅ in the form where p and q 22. Express 2. 24 𝑞 are positive integers with gcf(𝑝. 𝑞) = 1. Repeating, non-terminating decimals can all be expressed as a fraction by regarding it as an infinite geometric series. ̅̅̅ = 0.24 + 0.0024 + 0.000024 + ⋯ 0. ̅24 24 24 24 ̅̅̅ = 0. ̅24 + + +⋯ 100 10000 1000000 24 10000 𝑟= 24 100 1 𝑟= 100 24 100 𝑆𝑛 = 1 1− 100 8 𝑆𝑛 = 33 8 74 ̅̅̅̅ = 2 + 2. 24 = 33 33 𝟕𝟒 ANSWER: 𝟑𝟑 23. When the polynomial 𝑃(𝑥) is divided by 2𝑥 2 − 𝑥, the quotient is 𝑥 2 + 1 and the remainder is 𝑥 + 2. What is 𝑃(𝑥)? 𝑃 (𝑥) = 𝑄 (𝑥) ∙ 𝐷(𝑥) + 𝑅(𝑥) 𝑃(𝑥) = (𝑥 2 + 1)(2𝑥 2 − 𝑥) + (𝑥 + 2) 𝑃(𝑥) = 2𝑥 4 − 𝑥 3 + 2𝑥 2 + 2 ANSWER: 𝟐𝒙𝟒 − 𝒙𝟑 + 𝟐𝒙𝟐 + 𝟐 ●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

24. What is the remainder when 𝑃 (𝑥) = 2𝑥 4 − 7𝑥 2 − 4𝑥 + 2 is divided by 𝑥 + 2? By the remainder theorem, 𝑃(−2) = 2(−2)4 − 7(−2)2 + 4(−2) + 2 𝑃(−2) = −2 ANSWER: -2 (Answer Key: 14) 25. Find the value of k so that 𝑥 + 1 is a factor of 𝑃 (𝑥) = 2𝑘𝑥 4 + 4𝑥 3 + 𝑘𝑥 2 − 𝑥 − 3. By the factor theorem, 𝑃(−1) must be 0. 2𝑘(−1)4 + 4(−1)3 + 𝑘(−1)2 − (−1) − 3 =0 𝑘=2 ANSWER: 𝟐 26. Solve for x in the equation 3𝑥 3 + 𝑥 2 − 12𝑥 − 4 = 0 Use the Rational zeroes theorem to try out possible zeroes. 2 3 1 -12 -4 6 14 4 3 7 2 0 3𝑥 3 + 𝑥 2 − 12𝑥 − 4 = (𝑥 − 2)(3𝑥 2 + 7𝑥 + 2) Factor out 3𝑥 2 + 7𝑥 + 2. 3𝑥 3 + 𝑥 2 − 12𝑥 − 4 = (𝑥 − 2)(3𝑥 + 1)(𝑥 + 2) 𝟏 ANSWER: 𝒙 = 𝟐, − , −𝟐 𝟑

27. Let 𝑃(𝑥) be a third-degree polynomial such that 𝑃(−1) = 𝑃(1) = 𝑃 (2) = 0 and 𝑃 (3) = −16. What is 𝑃(−2)? The factors of 𝑃(𝑥) are 𝑘(𝑥 + 1)(𝑥 − 1)(𝑥 − 2) 𝑃 (𝑥) = 𝑘(𝑥 3 − 2𝑥 2 − 𝑥 + 2) 𝑃 (3) = 𝑘(33 − 2(3)2 − 3 + 2) −16 = 8𝑘 → 𝑘 = −2 𝑃 (𝑥) = −2𝑥 3 + 4𝑥 2 + 2𝑥 − 4 𝑃 (−2) = −2(−2)3 + 4(−2) + 2(−2) − 4 𝑃(−2) = 24 ANSWER: 24

4|P ag e

28. In a circle, chord AB is bisected by chord CD at M. If 𝐶𝑀 = 8 𝑐𝑚 and 𝑀𝐷 = 18𝑐𝑚, find AB. If two chords intersect A in the interior of a circle, then the D product of the segment lengths of one chord is equal to the product of the segment lengths of the other chord.

M

C B

𝐶𝑀 ∙ 𝑀𝐷 = 𝐴𝑀 ∙ 𝑀𝐵 2 1 8 𝑐𝑚 ∙ 18 𝑐𝑚 = ( 𝐴𝐵) 2 𝐴𝐵 = √4(144𝑐𝑚 2 ) = 24𝑐𝑚 ANSWER: 24 cm. 29. Chords AB and CD intersect at W such that 𝐶𝑊: 𝑊𝐷 = 3: 2. If 𝐴𝑊 = 6 𝑐𝑚 and 𝑊𝐵 = 16 𝑐𝑚, what is the length of CD? With the same concept as that in item 28, 𝐶𝑊 ∙ 𝑊𝐷 = 𝐴𝑊 ∙ 𝑊𝐵 3𝑥 ∙ 2𝑥 = 6 𝑐𝑚 ∙ 16 𝑐𝑚 6𝑥 2 = 96 𝑐𝑚 2 𝑥 = 4 𝑐𝑚 𝐶𝐷 = 𝐶𝑊 + 𝑊𝐷 𝐶𝐷 = 3𝑥 + 2𝑥 𝐶𝐷 = 5(4 𝑐𝑚 ) ANSWER: 20 cm 30. Quadrilateral ABCD is inscribed in a circle. Given that ∠𝐴 = (2𝑥 + 12)°, ∠𝐶 = (3𝑥 + 18)° and ∠𝐷 = (2𝑥 − 15)°, what is ∠𝐷? In any quadrilateral inscribed in a circle, the sum of any two opposite angles of the quadrilateral is 180 degrees. 𝑚∠𝐴 + 𝑚∠𝐶 = 180° 2𝑥 + 12° + 3𝑥 + 18° = 180° 𝑥 = 30° 𝑚∠𝐷 = 2(30°) − 15° 𝑚∠𝐷 = 45° ANSWER: 𝟒𝟓°

●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

31. Find the center of the circle 4𝑥 2 + 4𝑦 2 + 24 = 8𝑥 + 24𝑦. 4𝑥 2 + 4𝑦 2 − 8𝑥 − 24𝑦 = −24 𝑥 2 + 𝑦 2 − 2𝑥 − 6𝑦 = −6 (𝑥 2 − 2𝑥) + (𝑦 2 − 6𝑦) = −6 (𝑥 2 − 2𝑥 + 1) + (𝑦 2 − 6𝑦 + 9) = −6 + 1 + 9 (𝑥 − 1)2 + (𝑦 − 3)2 = 4 ANSWER: (𝟏, 𝟑) 32. A circle with radius 4 cm is tangent to both x and y axes. How far is its center to the origin? The segment joining the center of the circle and the origin is the diagonal 4 cm of the square formed by the radii of the circle at the points of tangency and the axes. ANSWER: 𝟒√𝟐 cm

35. Find all values of 𝜃 in the interval [0°, 360°] that satisfy the equation sin 𝜃 = tan 𝜃. sin 𝜃 tan 𝜃 = cos 𝜃 We want to find values for which sin 𝜃 sin 𝜃 = cos 𝜃 Or cos 𝜃 = ±1 within the interval [0°, 360°]. ANSWER: 𝟎°, 𝟏𝟖𝟎°, 𝟑𝟔𝟎° 5

36. If 0° < 𝜃 < 180° and cos 𝜃 = − 13, what is tan 𝜃? 𝑥 cos 𝜃 = ; 𝑥 = −5; 𝑟 = 13 𝑟 𝜃 is in Quadrant 2, 90° < 𝜃 < 180° 𝑦 = √𝑟 2 − 𝑥 2 𝑦 = √132 − (−5)2 𝑦 = 12 𝑦 tan 𝜃 = 𝑥 12 tan 𝜃 = − 5 𝟏𝟐 ANSWER: − 𝟓

33. For what value of k will the line through (2𝑘, −7) and (8,2)contain the origin? Since the line must pass through the origin, find the slope of the line passing through (0,0) and (8,2). 𝑦2 − 𝑦1 2 − 0 1 𝑚= = = 𝑥2 − 𝑥1 8 − 0 4 The slope must be the same for (2𝑘, −7) and (8,2). 1 2 − (−7) = 4 8 − 2𝑘 1 9 = 4 8 − 2𝑘 8 − 2𝑘 = 36 𝑘 = −14 ANSWER: -14

(Answer Key: −

𝟓

𝟏𝟐 2𝑥−5

)

37. If 642𝑥 = 16

, what is 42𝑥+10 ? 642𝑥 = 162𝑥−5 (43 )2𝑥 = (42 )2𝑥−5 46𝑥 = 44𝑥−10 6𝑥 = 4𝑥 − 10; 𝑥 = −5 42𝑥+10 = 42(−5)+10 ANSWER: 1 38. If 𝑥 + 𝑥 −1 = 5, what is 𝑥 2 + 𝑥 −2 ? 𝒙 + 𝒙−𝟏 = 

𝒙𝟐 +𝟏 𝒙

𝑥

=𝟓

𝑥4 + 1 𝒙 +𝒙 = 𝑥2 4 𝑥 +1 = 𝑥2 + 2 − 2 Add 2 (later expressed 𝟐



𝑥 2+1

−𝟐

as

34. Find k so that the line 3𝑥 + 2𝑦 = 7 is perpendicular to 𝑘𝑦 = 2𝑥 + 2. 3 7 3𝑥 + 2𝑦 = 7 → 𝑦 = − 𝑥 + 2 2 2 𝑚𝑝 = 3 2 2 𝑘𝑦 = 2𝑥 + 2 → 𝑦 = 𝑥 + 𝑘 𝑘 2 2 = ; 𝑘=3 𝑘 3 ANSWER: 𝟑 5|P ag e

4

𝟐

𝑥 + 1 𝟐𝒙 + 𝟐 −2 𝑥2 𝒙 4 𝑥 + 2𝑥 2 + 1 = −2 𝑥2 (𝑥 2 + 1)2 = −2 𝑥2 2 𝒙𝟐 + 𝟏 =( ) −2 𝒙 = 𝟓2 − 2 = 23 ANSWER: 23 =

2𝑥 2 𝑥2

) to make the

numerator a square of a binomial but subtract 2 to maintain the original value of the expression.

●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

ANSWERS WITH SOLUTIONS 2

39. If 0° < 𝜃 < 180° and 4sin 𝜃 = 2, find 𝜃. 2 (22 )sin 𝜃 = 2 1 √2 2sin2 𝜃 = 1; sin2 𝜃 = ; sin 𝜃 = 2 2 ANSWER: 𝟒𝟓°, 𝟏𝟑𝟓° 40. A group with 20 members plans to elect a president, a secretary, and a treasurer. If each member is eligible for each position, in how many ways can the group choose their officers? We consider permutation since electing three members is a different case as electing the same three members but in different positions. 20! 20 ∙ 19 ∙ 18 ∙ 17! 𝑃320 = = (20 − 3)! 17! 20 𝑃3 = 20 ∙ 19 ∙ 18 = 6840 ANSWER: 6840 41. In how many ways can the letters of the word ISLAND be arranged on a line so that the vowels are together? Taking A and I as one, 𝑃55 = 5! = 120 A and I can be arranged in 2 ways so, 𝑃 = 120 ∙ 2! = 240 ANSWER: 𝟐𝟒𝟎 42. A fair coin is tossed five times. What is the probability of getting exactly three heads? 5! 5 ∙ 4 ∙ 3! 𝐶35 = = = 10 (5 − 3)! 3! 2! ∙ 3! There are ten ways of getting exactly three heads in tossing a fair coin five times. Since there are 25 possible results in tossing a fair coin five times, 10 5 𝑃 (3 ℎ𝑒𝑎𝑑𝑠) = 𝑜𝑟 32 16 𝟓 ANSWER: 𝟏𝟔 43. It is found that if a hectare of land is planted with 42 mango trees, the annual yield of each tree will be 480 mangoes. For every additional tree planted on the land, the yield per tree will decrease by 8 mangoes. How many trees should be planted to this hectare of land to maximize the annual yield? 6|P ag e

Grade 10 Elimination Round

Let x be the number of mango trees planted in the hectare of land; and f(x) be the annual yield. f(x) is a piecewise function composed of two pieces: (1) if the number of trees planted is less or equal to 42, each tree yields 480, thus 480𝑥; (2) if the number of trees is more than 42, the yield of each tree is 8 times every mango tree in excess of the 42 less than 480. As a function, 480𝑥, 𝑥 ≤ 42 𝑓(𝑥) = { 480𝑥 − 8(𝑥 − 42)𝑥, 𝑥 > 42 or 480𝑥, 𝑥 ≤ 42 𝑓(𝑥) = { 816𝑥 − 8𝑥 2 , 𝑥 > 42 The maximum yield is the x-coordinate of the vertex of the parabola described in the second piece of the function, 816𝑥 − 8𝑥 2 𝑏 ℎ=− 𝑜𝑓 − 8𝑥 2 + 816𝑥 2𝑎 816 ℎ=− 2(−8) ℎ = 51 ANSWER: 51 44. Three regular polygons fit exactly together around a point on a plane surface. One is a square and another is a hexagon. How many sides does the third polygon have? Around a point would mean that the sum of the angles around this common vertex (one per regular polygon) is 360°. One angle of the square is 90°; one angle of the regular hexagon is 120°. The third regular polygon must therefore have an interior angle of 360° − (90° + 120°) or 150°. The interior angle(A) of a regular polygon is computed as 180°(𝑛 − 2) 𝐴= 𝑛 Solving for n, 360° 𝑛=− 𝐴 − 180° 360° 𝑛=− = 12 150° − 180° ANSWER: 12 ●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

45. How far is the line 3𝑥 − 4𝑦 + 15 = 0 from the origin? Look for the perpendicular distance. Identify the point on the line that when connected to the origin will form part of another line that is perpendicular to the given line. 3 15 3𝑥 − 4𝑦 + 15 = 0 → 𝑦 = 𝑥 + 4 4 3 4 𝑚= 𝑚𝑝 = − 4 3 Using the point-slope form, 𝑦 − 𝑦1 = 𝑚 (𝑥 − 𝑥1 ) 4 𝑦 − 0 = − (𝑥 − 0) 3 4 𝑦=− 𝑥 3 3 15 Find the intersection of 𝑦 = 4 𝑥 + 4 and 𝑦 = 4

− 3 𝑥 through substitution in systems of linear equations, 4 3 15 − 𝑥= 𝑥+ 3 4 4 9 12 𝑥 = − ;𝑦 = 5 5 Use the distance formula, 𝑑 = √(𝑦2 − 𝑦1 )2 + (𝑥2 − 𝑥1 )2 2 2 12 9 − 0) + (− − 0) 5 5 𝑑=3 ANSWER: 3 units

𝑑 = √(

46. A circle is inscribed in a triangle with perimeter 10 cm. if the radius of the circle is 4 cm, find the area of the triangle. 𝐴𝑡 = 𝑟𝑠 Where 𝐴𝑡 is the area of the triangle, 𝑟 is the radius of the circle, and 𝑠 is the semi1 perimeter (𝑠 = 2 (𝑎 + 𝑏 + 𝑐)) 𝐴𝑡 = (4 𝑐𝑚 )(5 𝑐𝑚 ) 𝐴𝑡 = 20 𝑐𝑚 2 ANSWER: 𝟐𝟎 𝒄𝒎𝟐 47. Find the quadratic equation whose leading coefficient is 2 and whose roots are the square of the sum and the square of the difference of the roots of the equation 2𝑥 2 + 4𝑥 + 10 = 0 2𝑥 2 + 4𝑥 + 10 = 0

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−4 ± √42 − 2(2)(10) 2(2) 1 𝑥 = −1 ± √−6 2 2 1 1 ((−1 + √−6) + (−1 − √−6)) = 4 2 2 2 1 1 ((−1 + √−6) − (−1 − √−6)) = −6 2 2 2(𝑥 − 4)(𝑥 + 6) = 0 2𝑥 2 + 4𝑥 − 48 = 0 ANSWER: 𝟐𝒙𝟐 + 𝟒𝒙 − 𝟒𝟖 = 𝟎 (Answer Key: 2𝑥 2 + 24𝑥 − 128 = 0) 𝑥=

48. The center of a circle lies on the line 2𝑥 + 5𝑦 = 42. If the line 𝑦 = 𝑥 + 4 is tangent to the circle at the point (4,8), find the equation of the circle in center-radius form.

The line that contains the center of the circle and the point of tangency is perpendicular to 𝑦 = 𝑥 + 4. 𝑚𝑝 = −1 Using the point-slope form, 𝑦 − 𝑦1 = 𝑚 (𝑥 − 𝑥1 ) 𝑦 − 8 = −1(𝑥 − 4) 𝑦 = −𝑥 + 12 The intersection between 𝑦 = −𝑥 + 12 and 2𝑥 + 5𝑦 = 42 is the center of the circle. By substitution, 2𝑥 + 5(−𝑥 + 12) = 42 𝑥 = 6; 𝑦=6 ℎ = 6; 𝑘=6 The radius of the circle is the distance between (4,8) and the center (6,6). Use the distance formula, 𝑟 = √(𝑦2 − 𝑦1 )2 + (𝑥2 − 𝑥1 )2 𝑟 = √(6 − 8)2 + (6 − 4)2 𝑟 = 2√2 𝟐 ANSWER: (𝒙 − 𝟔)𝟐 + (𝒚 − 𝟔)𝟐 = (𝟐√𝟐) ●VGChua●Eastern Samar Division

2017 Metrobank-MTAP-DepEd Math Challenge

Grade 10 Elimination Round

ANSWERS WITH SOLUTIONS

49. When a polynomial 𝑃(𝑥) is divided by 2𝑥 − 4, the remainder is −3. If 𝑃(−1) = −6, what is the remainder when 𝑃(𝑥) is divided by (2𝑥 − 4)(𝑥 + 1)? Assume that 𝑃(𝑥) is a third-degree polynomial of the form, 𝑃 (𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 𝑃 (−1) = 𝑎(−1)3 + 𝑏(−1)2 + 𝑐 (−1) + 𝑑 −𝟔 = −𝒂 + 𝒃 − 𝒄 + 𝒅 𝑃(𝑥) 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 2𝑥 − 4 2(𝑥 − 2) 3 ( ) 𝑃 𝑥 𝑎𝑥 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 2𝑥 − 4 2𝑥 − 4 𝑃(2) = 𝑎(2)3 + 𝑏(2)2 + 𝑐 (2) + 𝑑 𝑃(2) = 8𝑎 + 4𝑏 + 2𝑐 + 𝑑 −𝟑 = 𝟖𝒂 + 𝟒𝒃 + 𝟐𝒄 + 𝒅 −𝑎 + 𝑏 − 𝑐 + 𝑑 = −6 8𝑎 + 4𝑏 + 2𝑐 + 𝑑 = −3 By elimination, −9𝑎 − 3𝑏 − 3𝑐 = −3 𝟑𝒂 + 𝒃 + 𝒄 = 𝟏 (2𝑥 − 4)(𝑥 + 1) = 2𝑥 2 − 2𝑥 − 4 𝑃(𝑥) 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 2𝑥 2 − 2𝑥 − 4 2𝑥 2 − 2𝑥 − 4 By synthetic division, 1 2 a b c d 2a 2a+2b a a+b a a+b 3a+b+c 2a+2b+d The remainder is (3𝑎 + 𝑏 + 𝑐)𝑥 + 2𝑎 + 2𝑏 + 𝑑 Solve for 2𝑎 + 2𝑏 + 𝑑 Given that −𝑎 + 𝑏 − 𝑐 + 𝑑 = −6, 𝑑 = 𝑎−𝑏+𝑐−6 2𝑎 + 2𝑏 + 𝑑 = 2𝑎 + 2𝑏 + 𝑎 − 𝑏 + 𝑐 − 6 = 3𝑎 + 𝑏 + 𝑐 − 6 = 1 − 6 = −5 The remainder is 𝒙 − 𝟓 ANSWER: 𝒙 − 𝟓

One revolution is 360 degrees. There are 360° or 30° in between 9 and 10. With this, 12

50. It is now between 9 and 10 o’clock. In 4 minutes, the hour hand of the clock will be directly opposite the position occupied by the minute hand 3 minutes ago. What time is it now?

2 𝑚𝑖𝑛/𝑑𝑒𝑔𝑟𝑒𝑒 1 23 min

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30°

the hour hand moves at a rate of 60 𝑚𝑖𝑛 or 1 degree every 2 minutes. On the contrary, the minute hand moves 360° or 6 degrees per minute. 60 𝑚𝑖𝑛

𝑡 − 3 𝑚𝑖𝑛 Hour (x)

𝑃(𝑥−3 ) 3 =𝑥− ° 2 𝑃(𝑦−3 ) = 𝑦 − 18°

Minute (y)

𝑡

𝑡 + 4 𝑚𝑖𝑛

𝑃(𝑥)

𝑃(𝑥4 ) = 𝑥 + 2°

𝑃(𝑦)

𝑃(𝑦4 ) = 𝑦 + 24°

The position of 𝑃(𝑥4 ) is directly opposite 𝑃(𝑦−3 ). 𝑃(𝑥4 ) − 𝑃 (𝑦−3 ) = 180° 𝑥 + 2° − (𝑦 − 18°) = 180° 𝑥 − 𝑦 = 160° Assign 0 degrees to 12 in the clock. 9 on the clock would signify 270 degrees, 10 represents 300 degrees. 𝑃 (𝑥) is between 270 and 300 degrees. 𝑃 (𝑦) must be between 110 and 140 degrees. If 𝑃(𝑦) is in the interval [110,140] degrees, then this also limits 𝑃 (𝑥) to the interval 1 2 [279 , 281 ] based on the following 6 3

computations: 110° 1 140° 1 = 18 𝑚𝑖𝑛; = 23 𝑚𝑖𝑛 6°/𝑚𝑖𝑛 3 6°/𝑚𝑖𝑛 3 This means that the minute hand shows 1 1 time between 183 𝑎𝑛𝑑 23 3 𝑚𝑖𝑛. The position of the hour hand within this interval must be 1 3

18 min

1

3

2 𝑚𝑖𝑛/𝑑𝑒𝑔𝑟𝑒𝑒

1

= 9 6 ° after 270 or 279 6 ° 2

2

= 11 3 ° after 270 or 281 3 °

From the interval, if we assume that 𝑃 (𝑥) = 280°, then 𝑃(𝑦) = 120°. Reverting to time measurement, 280 is between 270 and 300 so it must be sometime at 9. 120 degrees points to 4, translated into minutes, that is 20. Therefore, the exact time is 9:20. ANSWER: 9:20 ●VGChua●Eastern Samar Division