Mathematics (Core), 2010, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2010

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 49 Questions

#	Question	Ans
31.	If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k A. \(\frac{8}{3}\) B. \(\frac{7}{3}\) C. \(\frac{5}{3}\) D. \(\frac{2}{3}\) Show Content Detailed Solution x2 + kx + \(\frac{16}{9}\); Perfect square But, b2 - 4ac = 0, for a perfect square where a - 1; b = k; c = \(\frac{16}{9}\) k2 - 4(1) x \(\frac{16}{9}\) = 0 k2 - \(\frac{64}{9}\) = 0 k2 = \(\frac{64}{9}\) k = \(\sqrt{\frac{64}{9}}\) k = \(\frac{8}{3}\)	A
32.	If x km/h = y m/s, then y = A. \(\frac{7}{18}\)x B. \(\frac{11}{20}\)x C. \(\frac{4}{15}\)x D. \(\frac{5}{18}\)x Show Content Detailed Solution x kmh^-1 = y ms^-1 \(\frac{x km}{1 hr}\) = y ms^-1 \(x \times \frac{1km}{1hr}\) = y ms^-1 \(x \times \frac{1000m}{60 \times 60s}\) = y ms^-1 \(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms^-1 \(x \times \frac{5}{18} ms^{-1}\) \(x \times \frac{5}{18} ms^{-1}\) = y ms^-1 y = \(\frac{5}{18}\)x	D
33.	The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x A. x \(\leq\) 3 B. x \(\geq\) 3 C. x < 3 D. x > 3 Show Content Detailed Solution mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5 = \(\frac{14 + 2x}{4} \leq 5\) = 14 + 2x \(\leq\) 5 x 4 14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14 2x \(\leq\) 20 - 14 2x \(\leq\) 6 x \(\leq\) \(\frac{6}{2}\) x \(\leq\) 3	A
34.	In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events? A. \(\frac{3}{32}\) B. \(\frac{7}{3}\) C. \(\frac{5}{3}\) D. \(\frac{5}{16}\) Show Content Detailed Solution Pr. (winning 100m race) = \(\frac{1}{8}\) Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\) Pr. (winning high jump) = \(\frac{1}{4}\) Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump) = (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\)) = \(\frac{3}{32} + \frac{7}{32}\) = \(\frac{10}{32}\) = \(\frac{5}{16}\)	D
35.	In the diagram, < PSR = 220^o, < SPQ = 58^o and < PQR = 41^o. Calculate the obtuse angle QRS. A. 90^o B. 100^o C. 121^o D. 60^o Show Content Detailed Solution Joining SQ. In \(\bigtriangleup\) SPQ, (22^o + a) + 55^o + (41^o + b) = 180^o 121^o + a + b = 180^o a + b = 180 - 121 a + b = 59^o.....(1) In \(\bigtriangleup\) SRPQ; R + a + b = 180^o R + 59^o = 180^o (in (1), a + b = 59^o) R = 180 - 59 R = 121^o	C
36.	The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test? A. 100 B. 85 C. 80 D. 70 Show Content Detailed Solution Pass mark = 50% No. of students that passed = f₅₀ + f₆₅ + f₈₀ = 45 + 25 + 15 = 85	B
37.	The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50? A. 55\(\frac{1}{3}\)% B. 60% C. 65% D. 66\(\frac{2}{3}\)% Show Content Detailed Solution Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\) = \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100% = \(\frac{120}{180}\) x 100% = 66\(\frac{2}{3}\)%	D
38.	What is the value of m in the diagram? A. 20^o B. 30^o C. 40^o D. 50^o Show Content Detailed Solution 4m - 15^o = m + 75^o (Vertically opposite angles are equal) 4m - m = 75 + 15 3m = 90 m = \(\frac{90}{3}\) m = 30^o	B
39.	In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75^o. Find the value of y A. 105^o B. 110^o C. 130^o D. 150^o Show Content Detailed Solution In \(\bigtriangleup\) PQR, Q = S = 75^o (Corresponding angle) R = Q = 75^o (Base angles of an isosceles \(\bigtriangleup\)) But, y + 75^o = 180^o (Sum of angles in a straight line) y = 180 - 75 y = 105^o	A
40.	In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\) A. \(\frac{KL}{JI}\) B. \(\frac{HK}{JK}\) C. \(\frac{JI}{KL}\) D. \(\frac{HK}{LK}\) Show Content Detailed Solution \(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ < HKL = HJI = x^o Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\) \(\frac{LH}{JH} = \frac{KL}{JI}\)	A

31.	If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k A. \(\frac{8}{3}\) B. \(\frac{7}{3}\) C. \(\frac{5}{3}\) D. \(\frac{2}{3}\) Show Content Detailed Solution x2 + kx + \(\frac{16}{9}\); Perfect square But, b2 - 4ac = 0, for a perfect square where a - 1; b = k; c = \(\frac{16}{9}\) k2 - 4(1) x \(\frac{16}{9}\) = 0 k2 - \(\frac{64}{9}\) = 0 k2 = \(\frac{64}{9}\) k = \(\sqrt{\frac{64}{9}}\) k = \(\frac{8}{3}\)	A
32.	If x km/h = y m/s, then y = A. \(\frac{7}{18}\)x B. \(\frac{11}{20}\)x C. \(\frac{4}{15}\)x D. \(\frac{5}{18}\)x Show Content Detailed Solution x kmh^-1 = y ms^-1 \(\frac{x km}{1 hr}\) = y ms^-1 \(x \times \frac{1km}{1hr}\) = y ms^-1 \(x \times \frac{1000m}{60 \times 60s}\) = y ms^-1 \(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms^-1 \(x \times \frac{5}{18} ms^{-1}\) \(x \times \frac{5}{18} ms^{-1}\) = y ms^-1 y = \(\frac{5}{18}\)x	D
33.	The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x A. x \(\leq\) 3 B. x \(\geq\) 3 C. x < 3 D. x > 3 Show Content Detailed Solution mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5 = \(\frac{14 + 2x}{4} \leq 5\) = 14 + 2x \(\leq\) 5 x 4 14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14 2x \(\leq\) 20 - 14 2x \(\leq\) 6 x \(\leq\) \(\frac{6}{2}\) x \(\leq\) 3	A
34.	In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events? A. \(\frac{3}{32}\) B. \(\frac{7}{3}\) C. \(\frac{5}{3}\) D. \(\frac{5}{16}\) Show Content Detailed Solution Pr. (winning 100m race) = \(\frac{1}{8}\) Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\) Pr. (winning high jump) = \(\frac{1}{4}\) Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump) = (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\)) = \(\frac{3}{32} + \frac{7}{32}\) = \(\frac{10}{32}\) = \(\frac{5}{16}\)	D
35.	In the diagram, < PSR = 220^o, < SPQ = 58^o and < PQR = 41^o. Calculate the obtuse angle QRS. A. 90^o B. 100^o C. 121^o D. 60^o Show Content Detailed Solution Joining SQ. In \(\bigtriangleup\) SPQ, (22^o + a) + 55^o + (41^o + b) = 180^o 121^o + a + b = 180^o a + b = 180 - 121 a + b = 59^o.....(1) In \(\bigtriangleup\) SRPQ; R + a + b = 180^o R + 59^o = 180^o (in (1), a + b = 59^o) R = 180 - 59 R = 121^o	C

36.	The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test? A. 100 B. 85 C. 80 D. 70 Show Content Detailed Solution Pass mark = 50% No. of students that passed = f₅₀ + f₆₅ + f₈₀ = 45 + 25 + 15 = 85	B
37.	The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50? A. 55\(\frac{1}{3}\)% B. 60% C. 65% D. 66\(\frac{2}{3}\)% Show Content Detailed Solution Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\) = \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100% = \(\frac{120}{180}\) x 100% = 66\(\frac{2}{3}\)%	D
38.	What is the value of m in the diagram? A. 20^o B. 30^o C. 40^o D. 50^o Show Content Detailed Solution 4m - 15^o = m + 75^o (Vertically opposite angles are equal) 4m - m = 75 + 15 3m = 90 m = \(\frac{90}{3}\) m = 30^o	B
39.	In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75^o. Find the value of y A. 105^o B. 110^o C. 130^o D. 150^o Show Content Detailed Solution In \(\bigtriangleup\) PQR, Q = S = 75^o (Corresponding angle) R = Q = 75^o (Base angles of an isosceles \(\bigtriangleup\)) But, y + 75^o = 180^o (Sum of angles in a straight line) y = 180 - 75 y = 105^o	A
40.	In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\) A. \(\frac{KL}{JI}\) B. \(\frac{HK}{JK}\) C. \(\frac{JI}{KL}\) D. \(\frac{HK}{LK}\) Show Content Detailed Solution \(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ < HKL = HJI = x^o Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\) \(\frac{LH}{JH} = \frac{KL}{JI}\)	A