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2018 mtap for g10 with answers

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MMC 2018 Elimination Round Grade 10
mbalvero
Page1
1. The average of a number and √7 +
√5 is √7–√5. Find the number.
Solut...
MMC 2018 Elimination Round Grade 10
mbalvero
Page2
n = 181
But there are some numbers that are
divisible by 2 are also div...
MMC 2018 Elimination Round Grade 10
mbalvero
Page3
Solution:
Notice that B contains all even
numbers up to 100 and C conta...
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2018 mtap for g10 with answers

  1. 1. MMC 2018 Elimination Round Grade 10 mbalvero Page1 1. The average of a number and √7 + √5 is √7–√5. Find the number. Solution: Let x be the missing number. x + (√7 + √5) 2 = √7–√5 x + √7 + √5=2√7– 2√5 x = 2√7–√7– 2√5– √5 x = √7– 3√5 2. Find n if 27.63% of 349 is 2.763% of n. Solution: (27.63%)(349) = (2.763%)(n) (27.63%)(349) 2.763% = n n = 10(349) = 3,490 3. Simplify: 101! 99! Solution: Using the definition n! = n(n – 1)!, 101! 99! = 101 × 100 × 99! 99! = 101 × 100 = 10,100 4. If 3n = 90, find the integer closet to n. Solution: The number 90 is between 81 = 34 and 243 = 35. But 90 seems to be closer to 81 than to 243, so n should be closest to 4. 5. Find the largest positive integer n such that (n – 18)4036 ≤ 992018. Solution: Taking the 2018th root both sides, we have (n – 18)2 ≤ 99 Since we are looking for the largest positive integer n, we observed that the largest value of (n – 18)2 less than 99 is 81 since 92 = 81. Hence, n – 18 = 9  n = 27. 6. How many integers between 60 and 600 (inclusive) are divisible by 2 or by 3? Solution: To get the number of multiples of a certain number, use the formula of the arithmetic sequence which is an = a1 + (n – 1)d. The number of multiples of 2 from 60 to 600 can be found as follows: a1 = 60, an = 600, d = 2 an = a1 + (n – 1)d 600 = 60 + (n – 1)(2) 540 = 2(n – 1) 270 = n – 1 n = 271 The number of multiples of 3 from 60 to 600 can be found as follows: a1 = 60, an = 600, d = 3 an = a1 + (n – 1)d 600 = 60 + (n – 1)(3) 540 = 3(n – 1) 180 = n – 1
  2. 2. MMC 2018 Elimination Round Grade 10 mbalvero Page2 n = 181 But there are some numbers that are divisible by 2 are also divisible by 3. Those are the numbers divisible by 6. The number of such numbers can be found as follows: a1 = 60, an = 600, d = 6 an = a1 + (n – 1)d 600 = 60 + (n – 1)(6) 540 = 6(n – 1) 90 = n – 1 n = 91 To get the number of elements in either sets (A ∪ B), use the Addition Rule formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) where n(A) is the number of elements in a set. Let n(A) and n(B) be the number of multiples of 2 and 3, respectively. n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 271 + 181 – 91 = 361 7. How many factors does the product of 3 different prime factors have? Solution: Let a, b and c be the different prime factors, and abc be their product. To get the number of factors of a number, take first the prime factorization in exponential form. Take all exponents of all prime factors, add all of these by 1 and multiply all of the results. The exponents of the three different prime factors are all 1. Adding it allby 1 gives 2 and thus, we have 2 × 2 × 2 = 8 factors. 8. A triangle has area 30 cm2, and two sides are 5 cm and 12 cm long. How long is the third side? Solution: Recall the area of a triangle A = bh 2 where b and h be the base and height, respectively. Assume that 12 cm which is the given be the base of a triangle. A = bh 2 30 = 12h 2 60 = 12h h = 5 cm The height of a triangle is found to be 5 cm, which is already given in the problem. Since the height and base lines are perpendicular, the triangle in the problem is a right triangle whose lengths of both of its legs are given. To find the third side, use the Pythagorean Theorem. a2 + b2 = c2 52 + 122 = c2 25 + 144 = 169 = c2 c2 = 13 cm 9. If A = {all prime numbers from 1 and 100}, B = {2, 4, 6, …, 98, 100}, and C = {3, 6, 9, …, 96, 99}, how many elements does (A ∪ B) ∩ C have?
  3. 3. MMC 2018 Elimination Round Grade 10 mbalvero Page3 Solution: Notice that B contains all even numbers up to 100 and C contains all multiples of 3 less than 100. A ∪ B must contain all prime OR even numbers less than or equal to 100. Let us look the elements of C. We identify each of them whether it is prime or even to include in (A ∪ B) ∩ C. Inspecting them one by one in C, we have the following results: - 3 is prime - 6 is even - 9 is neither prime nor even - 12 is even - 15 is neither prime nor even - 18 is even - …and so on According to the results, the set (A ∪ B) ∩ C must contain 3 and all multiples of 6 less than 100. In other words, (A ∪ B) ∩ C = {3, 6, 12, 18, 24, …, 90, 96} From 6 to 96, we have 96 ÷ 6 = 16 elements, adding it by 1 for the number 3 gives a total of 17 elements in (A ∪ B) ∩ C. 10. Find the perimeter of a regular hexagon inscribed inside a circle whose area is 49π cm2. Solution: Recall also the area of a circle A = πr2 to find the radius. A = πr2 49π = πr2 r2 = 49 r = 7 cm Note that the regular hexagon divides into 6 equilateral triangles that have three equal sides each. Therefore, the side of an equilateral triangle has length 7 cm. Since hexagon has 6 sides, the perimeter is 6(7) = 42 cm. 11. Sets A and B are subsets of a set having 20 elements. If set A has 13 elements, at least how many elements should set B have to guarantee that A and B are not disjoint? Solution: Two sets are disjoint if their intersection is empty set (Ø). There should be at least 1 element in intersection to make them non- disjoint sets. To guarantee that there’s always an intersection between two sets, the combined sets A and B, which isA∪ B, should have 20 elements. By letting n(A ∩ B) = 1 and using the Addition Rule, we can solve for n(B), the number of elements of set B. n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 20 = 13 + n(B) – 1 20 = 12 + n(B) n(B) = 8 elements
  4. 4. MMC 2018 Elimination Round Grade 10 mbalvero Page4 12. If the 2nd and 5th elements are -2 and 7, respectively, find the 101st term. Solution: Using the formula for arithmetic sequence, we can solve as follows. If n = 2 and an = -2, an = a1 + (n – 1)d -2 = a1 + (2 – 1)d -2 = a1 + d eq. 1 If n = 5 and an = 7, an = a1 + (n – 1)d 7 = a1 + (5 – 1)d 7 = a1 + 4d eq. 2 We form a system of equations in 2 variables that can be solved for them. -2 = a1 + d  eq. 1 -2 – d = a1 7 = a1 + 4d  eq. 2 7 = -2 – d + 4d 9 = 3d d = 3 a1 = -2 – d  eq. 1 a1 = -2 – 3 = -5 Therefore, if n = 101, an = a1 + (n – 1)d a101 = -5 + (101 – 1)3 a101 = -5 + (100)3 a101 = -5 + 300 = 295 13. Find the sum of the first 71 terms of the arithmetic sequence above. Solution: The sum of terms of an arithmetic sequence is defined as Sn = n 2 (2a1 + (n – 1)d). With results obtained in problem #12 which is a1 = -5, d = 3 and n = 71, we have Sn = n 2 (2a1 + (n – 1)d) S71 = 71 2 (2(-5) + (71 – 1)3) S71 = 71 2 (2(-5) + (70)3) S71 = 71 2 (-10 + 210) S71 = 71 2 (200) S71 = 71(100) = 7,100 14. Find k such that k – 2, 2k + 2, and 10k + 2 are consecutive terms of a geometric sequence. Solution: If a1, a2, a3 are consecutive terms of a geometric sentence, then a2 a1 = a3 a2 . By applying this, we have 2k + 2 k – 2 = 10k + 2 2k + 2 2k + 2 k – 2 = 2(5k + 1) 2(k + 1) 2k + 2 k – 2 = 5k + 1 k + 1 (2k + 2)(k + 1) = (5k + 1)(k – 2) 2k2 + 4k + 2 = 5k2 – 9k – 2 2k2 – 5k2 + 4k + 9k + 2 + 2 = 0 -3k2 + 13k + 4 = 0 3k2 – 13k – 4 = 0 By using quadratic formula x= -b±√b 2 –4ac 2a with a = 3, b = -13 and c = -4, we have
  5. 5. MMC 2018 Elimination Round Grade 10 mbalvero Page5 k= -b±√b 2 – 4ac 2a k = -(-13) ± √(-13) 2 – 4(3)(-4) 2(3) k = 13 ± √169 + 48 6 = 13 ± √217 6 15. Three numbers form an arithmetic sequence with common difference 15. If the first is doubled, the second is unchanged, and the third is increased by 21, a geometric sequence is formed. Find the first number of the geometric sequence. Solution: In arithmetic sequence, if a1 = x, then a2 = x + 15 and a3 = x + 30. The a1 of a geometric sequence is2x, a2 is still x + 15, and a3 is x + 30 + 21 or x + 51. Applying the same method as problem #14, we have x + 15 2x = x + 51 x + 15 (x + 15)(x + 15) = (2x)(x + 51) x2 + 30x + 225 = 2x2 + 102x x2 – 2x2 + 30x – 102x + 225 = 0 -x2 – 72x + 225 = 0 x2 + 72x – 225 = 0 (x – 3)(x + 75) = 0 x – 3 = 0 x = 3 x + 75 = 0 The first number in a geometric sequence is 2x. If x = 3, 2x = 2(3) = 6 and if x = -75, 2x = 2(-75) = -150. 16. The sum of an infinite geometric sequence have sum 14. If their squares have sum 196 3 , find the sum of their cubes. Solution: The sum of infinite geometric sequence is defined as S = a 1 – r where |r| < 1. S = a 1 – r 14 = a 1 – r 14(1 – r) = a 14 – 14r = a  eq. 1 The squares of terms have a2 as its first term and r2 its ratio. 196 3 = a2 1 –r2 196 3 = ( a 1 – r ) ( a 1 + r ) But a 1 – r =14, so use substitution. 196 3 = 14 ( a 1 + r ) 14 3 = a 1 + r 14(1 + r) = 3a 14 + 14r = 3a  eq. 2 We have a system of equations formed. Solving for a and r, we have 14 – 14r = a  eq. 1 14 + 14r = 3a  eq. 2 14 + 14r = 3(14 – 14r) 14 + 14r = 42 – 42r 14r + 42r = 42 – 14 56r = 28
  6. 6. MMC 2018 Elimination Round Grade 10 mbalvero Page6 r = 28 56 = 1 2 a = 14 + 14( 1 2 ) eq. 1 a = 14 – 7 = 7 Therefore, the sum of the cubes is S = a3 1 – r3 S = 7 3 1 – ( 1 2 ) 3 S = 343 1 – 1 8 = 343 7 8 S =343 ( 8 7 ) = 392 17. A sequence {an} has two terms a1 = 3 and a2 = 2. For every n ≥ 3, an is the sum of all the preceding terms of the sequence. Find a12. Solution: Using the rule stated in the problem, we have the terms obtained below: a1 = 3 a2 = 2 a3 = 3 + 2 = 5 a4 = 3 + 2 + 5 = 10 a5 = 3 + 2 + 5 + 10 = 20 a6 = 3 + 2 + 5 + 10 + 20 = 40 …and so on Notice that the terms starting a3form a geometric sequence. The common ratio is 2, the first term a3 = 5 so a12 should be the 10th term (n = 10). Therefore, a12 = 5(2)10 – 1 = 5(2)9 a12= 5(512) = 2,560 18. Let r and s be the roots of x2 – 9x + 7 = 0. Find (r + 1)(s + 1). Solution: Sum of roots: r + s =-(-9) = 9 Product of roots: rs = 7 (r + 1)(s + 1) = rs + r + s + 1 = 7 + (-9) + 1 = -1 19. If P(x + 2) = 3x2 + 5x + 4, find P(x). Solution: To get P(x), take the inverse of x + 2 to “undo” the expression and change to x. The inverse of x + 2 is x – 2. Replacing all x to x – 2, we have P(x + 2) = 3x2 + 5x + 4 P((x – 2) + 2) = 3(x – 2)2 + 5(x – 2) + 4 P(x – 2 + 2) = 3(x2 – 4x + 4) + 5(x – 2) + 4 P(x) = 3x2 – 12x + 12 + 5x – 10 + 4 P(x) = 3x2 – 7x + 6 20. Find the remainder if P(x) = 27x3 + 81x2 + 5x + 20 is divided by x + 3. Solution: We use the Remainder Theorem to find the remainder by finding P(c) if P(x) is divided by x – c. Thus we find P(-3) for this problem because of the divisor x – (-3). P(-3) = 27(-3)3 + 81(-3)2 + 5(-3) + 20 = 27(-27) + 81(9) + 5(-3) + 20
  7. 7. MMC 2018 Elimination Round Grade 10 mbalvero Page7 = -729 + 729 – 15 + 20 = 5 21. Find the quotient of the quotient when p(x) = 6x4 – 8x3 + x2 + 9x + 7 is divided by x + 2 3 . Solution: Using synthetic division, we have - 2 3 6 -8 1 9 7 -4 8 6 10 6 -12 9 15 17 The quotient is 6x3 – 12x2 + 9x + 15 and hence, its constant term is 15. 22. Find the constant k if x + 3 is a factor of f(x) = 2x4 + 8x3 + (k2+ 1)x2 + kx + 15. Solution: Since x + 3 is a factor of f(x), f(-3) = 0 by Remainder Theorem. With x = -3 and f(-3) = 0, we have f(x) = 2x4 + 8x3 + (k2 – 1)x2 + kx + 15 0 = 2(-3)4 + 8(-3)3 + (k2 + 1)(-3)2 + k(-3) + 15 0 = 2(81) + 8(-27) + 9(k2 + 1) + k(-3) + 15 0 = 162 – 216 + 9k2 + 9 – 3k + 15 0 = 9k2 – 3k – 30 0 = 3k2 – k – 10 0 = (k + 3)(3k – 10) 0 = k + 3 k = -3 0 = 3k – 10 10 = 3k k = 10 3 23. Find the largest real root of 2x3 – 5x2 – 2x + 2 = 0. Solution: We use Rational Roots Theorem to find all possible rational roots of a polynomial equation. Let p be the factor of a constant term, q be the factor of a leading coefficient (the term with the highest power) and p q be the possible rational root. Applying this, we have p = factors of 2 = ±1, ±2 q = factors of 2 = ±1, ±2 p q = ± 1 2 , ±1, ±2 Testing somepossible values of p q , we have 1 2 as a root because 1 2 2 -5 -2 2 1 -2 -2 2 -4 -4 0 We have depressed equation 2x2 – 4x – 4 = 0 which is equivalent to x2 – 2x – 2 = 0. Solving it by quadratic formula, we have x= -b±√b 2 – 4ac 2a x = -(-2) ± √(-2) 2 – 4(1)(-2) 2(1) x = 2 ± √4 + 8 2(1) = 2 ± √12 2 x = 2 + 2√3 2 , 2 – 2√3 2 x = 1 + 1√3, 1 – 1√3
  8. 8. MMC 2018 Elimination Round Grade 10 mbalvero Page8 The roots of 2x3 – 5x2 – 2x + 2 = 0 are 1 2 , 1 + 1√3 and 1 – 1√3. The value of the root1 + 1√3 is greater than 1, which is greater than the other root 1 2 . Therefore, the largest root is 1 + 1√3. 24. If p(x) is a 3rd degree polynomial, p ( 5 3 ) = p(4) = p(-5) = 0, and p(0) = - 200, find p(2). Solution: If p(c) = 0, then we can say that c is a zero of p(x) and x – c is a factor of p(x). In the problem above, we say that 5 3 , 4 and -5 are all zeros of p(x), and p(x)= a (x – 5 3 ) (x – 4)(x + 5) where a is the constant. With x = 0 and p(x) = - 200, we can solve for a. p(x)= a (x – 5 3 ) (x – 4)(x + 5) -200 = a (0 – 5 3 ) (0 – 4)(0 + 5) -200 = a (- 5 3 ) (-4)(5) -200 = 100 3 a -600 = 100a a = -6 Thus, p(x)= -6 (x – 5 3 ) (x – 4)(x + 5) p(2)= -6 (2 – 5 3 ) (2 – 4)(2 + 5) p(2)= -6 ( 1 3 ) (-3)(7) p(2) = -6(-1)(7) = 42 25. The midpoint of P(1, 2) and A is Q; the midpoint of P and C is S; the midpoint of Q and S is R(5, 1.5). Find the coordinates of B, the midpoint of A and C. Solution: The midpoint of two points (x1, y1) and (x2, y2) is defined as ( x1 + x2 2 , y1 + y2 2 ). This is known as Midpoint Formula. Let (a, b) be the coordinates of Q. If the midpoint of P(1, 2) and A(xA, yA) is Q(a, b), then A can be computed as follows. ( xP + xA 2 , yP + yA 2 ) = (xQ, yQ) ( 1 + xA 2 , 2 + yA 2 ) = (a, b) (1 + xA, 2 + yA) = (2a, 2b) (xA, yA) = (2a – 1, 2b – 2) The midpoint of Q(a, b) and S(xS, yS) is R(5, 1.5). S is computed as follows. ( xQ + xS 2 , yQ + yS 2 ) = (xR, yR) ( a + xS 2 , b + yS 2 ) = (5, 1.5) (a + xS, b + yS) = (10, 3) (xS, yS) = (10 – a, 3 – b) The midpoint of P(1, 2) and C(xC, yC) is S(10 – a, 3 – b). C is computed as follows. ( xP + xC 2 , yP + yC 2 ) = (xS, yS) ( 1 + xC 2 , 2 + yC 2 ) = (10 – a, 3 – b) (1 + xC, 2 + yC) = (20 – 2a, 6 – 2b) (xC, yC) = (19 – 2a, 4 – 2b)
  9. 9. MMC 2018 Elimination Round Grade 10 mbalvero Page9 Finally, the midpoint of A(2a – 1, 2b – 2) and C(19 – 2a, 4 – 2b) is B that we are looking for. ( xA + xC 2 , yA + yC 2 ) = (xB, yB) = ( 2a – 1 + 19 – 2a 2 , 2b – 2 + 4 – 2b 2 ) = ( 18 2 , 2 2 ) = (9, 1) 26. Find the radius of the circle with equation x2 + y2 + 4x – 3y + 5 = 0. Solution: The equation of the circle in center- radius form is (x – h)2 + (y – k)2 = r2 where (h, k) be the coordinates of the center and r the length of the radius. Rewrite the given equation into center-radius form. x2 + y2 + 4x – 3y + 5 = 0 x2 + 4x + y2 – 3y = -5 x2 + 4x + 4 + y2 – 3y + 9 4 = -5 + 4 + 9 4 (x + 2)2 + (y – 3 2 ) 2 = 5 4 = ( √5 2 ) 2 The radius is √5 2 . 27. Give the equation (in center-radius form) of the circle having as a diameter the segment with endpoints (-2, 10) and (4, 2). Solution: The endpoints of the diameter have given coordinates, so their midpoint is the center of the circle. Using Midpoint Formula, the center has an order pair of ( x1 + x2 2 , y1 + y2 2 ) = ( -2 + 4 2 , 10 + 2 2 ) = ( 2 2 , 12 2 ) = (1, 6) To compute for the radius, find the distance between the center and either of the points (-2, 10) or (4, 2) by using Distance Formula d = √(x2– x1) 2 + (y2 – y1 ) 2 . We use the center and (4, 2) for our purpose. d = √(x2– x1) 2 + (y2 – y1 ) 2 r = √(4 – (-2)) 2 + (2 – 10) 2 r = √(4 + 2) 2 + (2 – 10) 2 = √6 2 + (-8) 2 r = √36+ 64 = √100 r = 10 Using the center and (-2, 10) yields also the same result which is r = 10. Therefore, the equation of the circle is (x – 1)2 + (y – 6)2 = 102 or (x – 1)2 + (y – 6)2 = 100 in center-radius form. 28. The center of a circle is on the x-axis. If the circle passes through (0, 5) and (6, 4), find the coordinates of its center. Solution: Since the center lies on x-axis, the coordinates must be (x, 0). We know that the distance between the
  10. 10. MMC 2018 Elimination Round Grade 10 mbalvero Page10 center and any point on a circle is its radius. So, the coordinate (0, 5) has the same distance from the center (x, 0) as (6, 4). Recall also the distance formula d = √(x2– x1) 2 + (y2 – y1 ) 2 which is the distance between two points. Distance between (x, 0) and (0, 5) = Distance between (x, 0) and (6, 4) √(0 – x) 2 + (5 – 0) 2 = √(6 – x) 2 + (4 – 0) 2 (-x)2 + 52 = (6 – x)2 + 42 x2 + 25 = 36 – 12x + x2 + 16 25 = 52 – 12x -12x = -27 x = -27 -12 = 9 4 Therefore, the center has coordinates ( 9 4 , 0). 29. If θ is an angle in a triangle, and sec θ = 7 5 , find tan θ. Solution: By using trigonometric identity sec θ = 1 cos θ , we have sec θ = 1 cos θ = 7 5 5 = 7 cos θ cos θ = 5 7 Recall the memory aid SOH-CAH- TOA that is used to remember the ratios of sine, cosine and tangent of a certain angle. Since we have cos θ, we use CAH. CAH cos θ = adjacent hypotenuse = 5 7 To get the opposite side of a right triangle, we use Pythagorean Theorem. (opposite)2 + (adjacent)2 = (hypotenuse)2 (opp)2 + 52 = 72 (opp)2 + 25 = 49 (opp)2 = 24 opp = √24=2√6 To find tan θ, use again the memory aid above. TOA  tan θ = opposite adjacent Therefore, tan θ = 2√6 5 . 30. In ΔABC, BC = 6, CA = 7, and AB = 8. Find sin B+ sin C sin A . Solution: In solving triangles (sides and angles), use the formula for Law of Sines. BC sin A = AC sin B = AB sin C Substituting the given values, we have 6 sin A = 7 sin B = 8 sin C To make ratios equal, we assign the values of sin A, sin B and sin C. To reduce all 3 fractions to 1 n (where n ≠ 0) for our simplification purposes, we
  11. 11. MMC 2018 Elimination Round Grade 10 mbalvero Page11 assign sin A = 6n, sin B = 7n and sin C = 8n. Substituting the values into sin B+ sin C sin A , we have 7n + 8n 6n = 15n 6n = 5 2 31. In ΔABC, AB = 3, AC = 2√3, and ∠A = 30°. Find BC. Solution: Notice that ∠A is between AB and AC, so we use Law of Cosines. BC2 = AB2 + AC2 – 2(AB)(AC)cos A Substituting the given values, we have BC2 = 32 + (2√3)2 – 2(3)(2√3)cos 30° BC2 = 32 + (2√3)2 – 2(3)(2√3)( √3 2 ) BC2 = 9 + 12 – 6(3) BC2 = 9 + 12 – 18 = 3 BC = √3 32. In a circle, chord PQ is bisected by chord RS at T. If RT = 3 and ST = 6, find PQ. Solution: Use the formula (TP)(TQ) = (TR)(TS) for two intersecting chords. Since RS bisects PQ at T, TP = TQ. Substituting the given values, we have (TP)(TP) = (3)(6) TP2 = 18 TP = √18 = 3√2 = TQ Note that TP + TQ = PQ. Therefore, PQ = 3√2+ 3√2= 6√2 See the figure above. 33. Rectangle ABCD is inscribed in a circle, BC = 2√3, and CD = 2. A point P is chosen on arc AB. Find ∠APB. Solution:
  12. 12. MMC 2018 Elimination Round Grade 10 mbalvero Page12 Showing the figure above, we have a right triangle BCD with right angle C. To get the degree measure of ∠BDC, we use tangent function with opposite side BC. Thus, we have tan θ = opposite adjacent = 2√3 2 = √3. Solving for θ, θ = ∠BDC = 60°. If ∠BDC = 60°, ∠BDA must be 30° because ∠BDC and ∠BDA are complementary angles. Notice that APBD is an inscribed quadrilateral whose opposite angles are supplementary. Therefore, the sum of ∠BDA and ∠APB is 180°. If ∠BDA =30°, therefore, ∠APB should be 150°. 34. Pentagon ABCDE is inscribed in a circle, BC = CD = DE, and AB = AE. If ∠BAE = 96°, find ∠BDA. Solution: If AB = AE, then their intercepted arc AB and arc AE are equal. The measure of an inscribed angle is one-half of its intercepted arc. If ∠BAE = 96°, then the major arc BCE = 2(96) = 192°. Remember that the degree measure of a whole circle is 360°. Therefore, if arc BCE = 192°, then arc BAE = 360 – 192 = 168°. But arc BAE = arc AB + arc AE and arc AB = arc AE. Therefore, arc AB = 168 ÷ 2 = 84°. The intercept of ∠BDA is arc AB and hence, ∠BDA = 84 ÷ 2 = 42°. 35. A line through a point A outside a circle is tangent to the circle at D. Another line through A intersects the circle at points B (closer to A) and C. If BC/AB = 2 and AD = 6, find AC. Solution:
  13. 13. MMC 2018 Elimination Round Grade 10 mbalvero Page13 If a tangent and a secant lines are drawn to a circle with their common point outside the circle, then we use the formula AD2 = (AB)(AC) in this case. The ratio of BC to AB is 2, so BC = 2x and AB = x for some number x. Note that AC = AB + BC. Substituting into the formula, we have AD2 = (AB)(AC) AD2 = (AB)(AB + BC) 62 = x(x + 2x) 36 = x(3x) 36 = 3x2 12 = x2 x = √12 = 2√3 AB = 2√3 BC = 2(2√3) = 4√3 Therefore, 2√3 + 4√3 = 6√3 36. In a circle, chords AB and CD intersect at E. If arc AC = 120° and arc BD = 20°, find ∠AEC. Solution: In two intersecting chords, the arc AC and arc BD are the intercepts of the vertical angles ∠AEC and ∠BED, respectively, and ∠AEC = ∠BED = 1 2 (arc AC + arc BD). Therefore, ∠AEC = 1 2 (120 + 20) = 1 2 (140) = 70°. 37. Find the area (in square units) of the region between two concentric circles if a chord of the larger circle which is tangent to the smaller circle has length 7. Solution: Draw the radius of the smaller circle that is perpendicular to the chord and draw the radius of the larger circle that meets one of the endpoints of a chord. We formed a
  14. 14. MMC 2018 Elimination Round Grade 10 mbalvero Page14 right triangle with these two radii and a chord with half of its length (which, in the given, is 7 2 ). Let x be the radius of the smaller circle. Using Pythagorean Theorem, we can find the radius of the larger circle which is the hypotenuse. a2 + b2 = c2 ( 7 2 ) 2 + x2 = c2 49 4 + x2 = c2 c = √ 49 4 + x2 Recall the area of the circle A = πr2. The area of a smaller circle (with r = x) is πx2 while the area of the larger circle (with r =√ 49 4 + x2) is π(√ 49 4 + x2) 2 = π( 49 4 + x2) To get the area of the said region, subtract the two areas. Area = (area of larger circle) – (area of the smaller circle) = π( 49 4 + x2) – πx2 = π( 49 4 +x2 –x2 ) = π( 49 4 ) = 49 4 π square units 38. A triangle having sides 3√7, 9, and 12 units is inscribed in a circle. Find the circumference of the circle. Solution: To determine what kind of triangle given its side lengths, we use the relation a2 + b2 ??c2 such that ?? can be replaced with > which is an obtuse triangle, < an acute triangle and = a right triangle. The longest side should be the value of c which is 12. a2 + b2 ?? c2 (3√7)2 + 92 ?? 122 63 + 81 ?? 144 144 = 144 Since 144 = 144, the inscribed triangle is a right triangle. The hypotenuse is the diameter of the circle which has the length 12. The radius of the circle is 6 which gives the area of a circle of π(6)2 = 36 square units. 39. In the figure, quadrilateral ABCD is inscribed in a circle. Lines AB and CD intersect at P, while lines AD and BC intersect at Q. If ∠APD = 50° and ∠AQB = 34°, find ∠PAQ. Solution: Let x be the measure of ∠PAQ
  15. 15. MMC 2018 Elimination Round Grade 10 mbalvero Page15 Notice that ΔBAQ and ΔDAP are seen in the figure. The sum of the interior angles of a triangle is 180°. In ΔBAQ, ∠BQA = 34° and ∠BAQ = x. ∠QBA should be 180 – 34 – x = 146 – x. On the other hand, the ΔDAP has measures ∠DPA = 50° and ∠DAP = x. PDA should be 180 – 50 – x = 130 – x. Take note that ABCD is inscribed in a circle. ∠CBA and ∠CDA are supplementary angles. ∠CBA = ∠QBA and ∠CDA = ∠PDA. ∠QBA + ∠PDA = 180 (146 – x) + (130 – x) = 180 276 – 2x = 180 -2x = -96 x = ∠PAQ = 48° 40. How many books (all different) are there if there are 5,040 ways of arranging them in a shelf? Solution: If a set of n objects are arranged, the number of ways in arranging them is n!. By trial and error method, we have 7 different books because 7! = 5,040. 41. Lila and Dolf each counted the number of three-digit positive integers that can be formed using the digits 1, 2, 3, 4, and 5. Lila assumed a digit to be repeated, while Dolf assumed no digit can be repeated. If they counted correctly, what is the difference of their answers? Solution: Lila counts the 3-digit numbers whose any digit can be repeated. The hundreds digit can have 5 possible digits, the tens has also 5 and ones has 5 which give 5 × 5 × 5 = 125 such numbers. Dolf counts also 3-digit numbers but no digit id repeated. Here, there are 5 possible digits for hundreds digit. Since no digit will be repeated, there must be 4 possible digits in tens digit because one digit is used. Ones digit should have 3 possible digits because 2 digits are used. There should be 5 × 4 × 3 = 60 such numbers for Dolf. Hence, the difference of their answers is 125 – 60 = 65. 42. In how many ways can a photographer arrange 7 students for a club picture if Ana, Carol and Iris must always be together, but Rey and Ruel should not beside each other? Solution: Since Ana, Carol and Iris should be together in a picture, they will count as a single unit. There are now 5 units instead of 7. And since Rey and Ruel refused to be together, they should have at least 1 unit between them.
  16. 16. MMC 2018 Elimination Round Grade 10 mbalvero Page16 For Problems 43 and 44: A committee is to be formed from 7 City A residents and 9 City B residents. Here are possible positions for Rey (X) and Ruel (Y): X __ Y __ __ X __ __ Y __ X __ __ __ Y __ X __ Y __ __ X __ __Y __ __ X __ Y There are 6 possible positions for Rey and Ruel. The single unit of three girls (Ana, Carol and Iris) can be placed in any of the blanks in each position. The remaining two students can be arranged freely. We have - 6 positions for Rey and Ruel, - 2! = 2 ways to arrange Rey and Ruel their positions, - 3! = 6 arrangements for the 3 units (a single unit of three girls and other two students), and - 3! = 6 ways to arrange Ana, Carol and Iris their positions in a single unit. Therefore, the number of ways (or possible club pictures) is (6)(2)(6)(6) = 432. 43. How many 6-member committees with 3 members from each city can be formed? Solution: This means that 3 members belong to City A while another 3 belong to City B. Solving the number ways for each city using Combination formula C(n, r) = n! r!(n – r)! , we have City A: n = 7, r = 3 C(7, 3) = 7! 3!(7 – 3)! = (7)(6)(5)(4!) 6(4!) = (7)(5) = 35 City B: n = 9, r = 3 C(9, 3) = 9! 3!(9 – 3)! = (9)(8)(7)(6!) 6(6!) = (3)(4)(7) = 84 Since two cities are independent, multiply the number of ways from both cities. Hence, (35)(84) = 2,940. 44. How many 5-member committees with more City A than City B members can be formed? Solution: Since there are more members from City A from City B, we have multiple cases to deal with. Here are those cases: - Case 1: 3 from City A, 2 from City B - Case 2: 4 from City A, 1 from City B - Case 3: 5 from City A
  17. 17. MMC 2018 Elimination Round Grade 10 mbalvero Page17 For Problems 45 and 46: The numbers 1, 2, …, 25 are written on slips of paper which are placed in a box. Solving for the 3 cases in the same manner as the previous one, we have the following solutions below. Case 1: 3 from City A, 2 from City B C(7, 3) C(9, 2) = 7! 3!(7 – 3)! × 9! 2!(9 – 2)! = (7)(6)(5)(4!) 6(4!) × (9)(8)(7!) 2(7!) = (7)(5) × (9)(4) = 1,260 Case 2: 4 from City A, 1 from City B C(7, 4) C(9, 1) = 7! 4!(7 – 4)! × 9! 1!(9 – 1)! = (7)(6)(5)(4)(3!) 24(3!) × (9)(8!) 1(8!) = (7)(5) × 9 = 315 Case 1: 5 from City A C(7, 5) = 7! 5!(7 – 5)! = (7)(6)(5)(4)(3)(2!) 120(2!) = (7)(3) = 21 Adding all 3 cases, we have 1,260 + 315 + 21 = 1,596 such committees. 45. If two slips are picked at the same time from the box, find the probability that one number is a multiple of 7 and the other is a multiple by 8. Solution: First, we’re going to find the number of ways to pick 2 slips of paper at the same time. Using Combination formula with n = 25 and r = 2, we have C(25, 2) = 25! 2!(25 – 2)! = (25)(24)(23!) 2(23!) = (25)(12) = 300 ways From 1 to 25, there are 3 multiples of 7 (7, 14, 21) and also 3 multiples of 8 (8, 16, 24). Using counting principle, there are 3 × 3 = 9 ways to pick such pair of slips. Hence, the probability is 9 300 = 3 100 or 3%. 46. If three slips are picked at the same time from the box, find the probability that the sum of the numbers is odd. Solution: Using Combination formula again, we’re going to find the number of ways to pick 3 slips at the same time. With n = 25 and r = 3, we have C(25, 3) = 25! 3!(25 – 3)! = (25)(24)(23)(22!) 6(22!) = (25)(4)(23) = 2,300 ways One way to express an odd number is the sum of 3 odd numbers (ex. 3 + 5 + 7 = 15). An odd number can also be the sum of one odd and two even numbers(ex. 2 + 4 + 5 = 11). We have two cases to express an odd number as a sum.
  18. 18. MMC 2018 Elimination Round Grade 10 mbalvero Page18 Case 1: Three numbers are odd. Notice that from 1 to 25, 13 numbers are odd (1, 3, 5, …, 25). The number of ways to pick 3 numbers from 13 odd numbers is C(13, 3) = 13! 3!(13 – 3)! = (13)(12)(11)(10!) 6(10!) = (13)(2)(11) = 286 Case 2: An odd number and two even numbers Again, there are 13 odd numbers. 12 numbers are even (2, 4, 6, …, 24). We can find the number of ways to pick such numbers. Since they are independent, we multiply the two numbers. C(13, 1) C(12, 2) = 13! 1!(13 – 1)! × 12! 2!(12 – 2)! = (13)(12!) 1(12!) × (12)(11)(10!) 2(10!) = 13 × (6)(11) = 858 Adding both cases we have 286 + 858 = 1,144 and therefore, the probability is 1,144 2,300 = 286 575 . 47. Two fair dice are rolled. Find the probability that one die is 2 while the other is a prime number. Solution: There are only 5 possible outcomes for the event namely (2, 2), (2, 3), (3, 2), (2, 5) and (5, 2). Note there are 6 × 6 = 36 total outcomes for rolling a pair of dice. Hence, the probability is 5 36 . 48. In a class of Grade 10 students, the probability that a random chosen student likes dogs is 0.72, that a student likes cats is 0.54, and a student likes neither is 0.11. Find the probability that a student likes both cats and dogs. Solution: Let D be the students who likes dogs, P(D) = 0.54, while C are those who like cats, P(C) = 0.72. If D ∪ C are those who like dogs or cats, (D ∪ C)C don’t (maybe they like other animals). If P((D ∪ C)C) = 0.11, then P(D ∪ C) = 1 – 0.11 = 0.89. Using the Addition Rule with two non- mutually exclusive events (events that have intersection), we can solve for P(D ∩ C) where D ∩ C likes both cats and dogs. P(D ∪ C) = P(D) + P(C) – P(D ∩ C) 0.89 = 0.72 + 0.54 – P(D ∩ C) 0.89 = 1.26 – P(D ∩ C) -P(D ∩ C) = -0.37 P(D ∩ C) = 0.37 49. In a company, the men who do not wear glasses are thrice as many as the men who do, while the women who do not wear glasses are ten more than the women who do. The probability that an employee wears glasses is 3 10 . If a selected employee wears glasses, the probability that the employee is a woman is 2 3 . How
  19. 19. MMC 2018 Elimination Round Grade 10 mbalvero Page19 many are the employees of the company? Solution: Let M and F be the number of male and female employees, both wearing glasses, respectively. The table below shows the number of employees in a company. With glasses Without glasses Total Male M 3M 4M Female F F + 10 2F + 10 Total M + F 3M + F + 10 4M + 2F + 10 The total number of employees in a company seems to be 4M + 2F + 10. Out of them, M + F wears glasses. Thus, the probability that an employee wears glasses is M + F 4M + 2F + 10 which is equivalent to 3 10 . Therefore, we have M + F 4M + 2F + 10 = 3 10 as the first equation. Simplifying this, we have M + F 4M + 2F + 10 = 3 10 10(M + F) = 3(4M + 2F + 10) 10M + 10F = 12M + 6F + 30 -2M + 4F = 30 2M – 4F = -30 eq. 1 There are M + F employees who wear glasses, and F of them are women. Thus, the probability is F M + F which is also 2 3 . Therefore, we have F M + F = 2 3 as the second equation and this can be simplified as F M + F = 2 3 3F = 2(M + F) 3F = 2M + 2F F = 2M eq. 2 We have formed a system of equations in 2 variables. To solve for M and F, substitute the second equation into first one. 2M – 4F = -30  eq. 1 F = 2M  eq. 2 2M – 4(2M) = -30 2M – 8M = -30 -6M = -30 M = 5 F = 2(5) = 10 As said earlier, the total number of employees is 4M + 2F + 10. Therefore, the number of employees is 4(5) + 2(10) + 10 = 20 + 20 + 10 = 50. 50. The number of ways of selecting 6 objects from n distinct objects is the same as the number of ways of selecting and arranging 3 objects from n distinct objects. Find n. Solution: The keyword “selecting”means we use combination while “selecting and arranging” means we use permutation. We have the following working equation and its solution below. C(n, 6) = P(n, 3) n! 6!(n – 6)! = n! (n – 3)!
  20. 20. MMC 2018 Elimination Round Grade 10 mbalvero Page20 n!(n – 3)! = 720n!(n – 6)! n!(n – 3)(n – 4)(n – 5)(n – 6)! = 720n!(n – 6)! (n – 3)(n – 4)(n – 5) = 720 n3 – 12n2 + 47n – 60 = 720 n3 – 12n2 + 47n – 780 = 0 (n – 13)(n2 + n + 60) = 0 n – 13 = 0 n = 13

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