Mathematics (Core), 1990, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

1990

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 47 Questions

#	Question	Ans
11.	Find the values of x for which y = 5 A. -2.5, 2.0 B. -2.6, 2.1 C. -2.0, 1.5 D. -2.0, 2.0 E. 0, 0	B
12.	An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7] A. 70^o B. 84^o C. 96^o D. 156^o E. 168^o Show Content Detailed Solution \(2\pi r = 2 \times 3.142 \times 15 = 94.28cm\) \(94.28cm = 360°\) \(1cm = \frac{360°}{94.28}\) \(\therefore 22cm = \frac{360°}{94.28} \times 22\) = 84°	B
13.	Find the sum of the first five terms of the G.P 2,6, 18 .... A. 484 B. 243 C. 242 D. 130 E. 121 Show Content Detailed Solution \(S_{n} = \frac{a(r^n - 1)}{r - 1}\) a = 2; r = 6/2 = 3. \(S_{5} = \frac{2(3^5 - 1)}{3 - 1}\) = \(243 - 1 = 242\)	C
14.	Let J be the set of positive integers, If H = {x: x∈J, x\(^2\) < 3 and x ≠ 0}, then A. H = {1} B. H is an infinite set C. H = {0, 1, 2} D. H = {} E. J ≤ H Show Content Detailed Solution H = {x: x is a positive integer, x\(^2\) < 3 and x \(\neq\) 0} H = {1}	A
15.	In a class of 80 students, every students had to study economics or geography, or both economics and geography, if 65 students studied economics and 50 studied geography, how many studied both subjects? A. 15 B. 30 C. 35 D. 45 E. 50 Show Content Detailed Solution ∩(E∩G) = ∩(E) + ∩(G) - ∩(E∩G) 80 = 65 + 50 - ∩(F∩G) ∴∩(E∩G) = 115 - 80 = 35	C
16.	The area shaded with horizontal lines is the solution set of the inequalities; A. y ≥ x, y + 3 ≥ 2x, x ≤ 3 B. y ≤ x, y + 2x ≥ -3, x ≤ 3 C. y ≤ -x, y + 2x ≤ 3, x ≥ -3 D. y ≥ -x, y + 3 ≤ 2x, x ≥ -3 E. y ≤ x, y ≤ 2x - 3, x ≥ 3	B
17.	The area shaded with vertical lines is the solution set of the inequalities; A. y ≤ x, y + 2x ≤ -3 B. y ≥ x, y + 2x ≥ - 3 C. y ≤ x, y < 2x - 3, x ≤ 3 D. y ≤ -x, y ≤ 2x - 3, x ≤ 3 E. y ≥ x, y + 2x ≥ 3, x ≤ 3	C
18.	In the diagram above, O is the center of the circle with radius 10cm, and ∠ABC = 30°. Calculate, correct to 1 decimal place, the length of arc AC [Take π = 22/7] A. 5.2cm B. 10.5cm C. 13.2cm D. 20.6cm E. 31.4cm Show Content Detailed Solution In the figure, < AOC = 2 x < ABC = 60° (angle subtended at the centre) \(\therefore\) Arc AC = \(\frac{60}{360} \times 2 \times 10 \times 3.14\) = \(\frac{31.4}{3}\) = 10.466 cm \(\approxeq\) 10.5 cm	B
19.	Factorize x\(^2\) + 4x - 192 A. (x-4) (x+48) B. (x-48) (x+4) C. (x-12) (x+16) D. (x-12) (x-16) E. (x+12) (x+16) Show Content Detailed Solution x\(^2\) + 4x - 192 x\(^2\) + 16x - 12x - 192 x(x + 16) - 12(x + 16) (x + 16)(x - 12)	C
20.	Factorize 2e\(^2\) - 3e + 1 A. (2e-1) (e-1) B. (e+1) (2e+1) C. (2e+3) (e+2) D. (2e-3) (e-1) E. (e²-3(2e-1) Show Content Detailed Solution 2e\(^2\) - 3e + 1 2e\(^2\) - 2e - e + 1 2e(e - 1) - 1(e - 1) (2e - 1)(e - 1)	A

11.	Find the values of x for which y = 5 A. -2.5, 2.0 B. -2.6, 2.1 C. -2.0, 1.5 D. -2.0, 2.0 E. 0, 0	B
12.	An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7] A. 70^o B. 84^o C. 96^o D. 156^o E. 168^o Show Content Detailed Solution \(2\pi r = 2 \times 3.142 \times 15 = 94.28cm\) \(94.28cm = 360°\) \(1cm = \frac{360°}{94.28}\) \(\therefore 22cm = \frac{360°}{94.28} \times 22\) = 84°	B
13.	Find the sum of the first five terms of the G.P 2,6, 18 .... A. 484 B. 243 C. 242 D. 130 E. 121 Show Content Detailed Solution \(S_{n} = \frac{a(r^n - 1)}{r - 1}\) a = 2; r = 6/2 = 3. \(S_{5} = \frac{2(3^5 - 1)}{3 - 1}\) = \(243 - 1 = 242\)	C
14.	Let J be the set of positive integers, If H = {x: x∈J, x\(^2\) < 3 and x ≠ 0}, then A. H = {1} B. H is an infinite set C. H = {0, 1, 2} D. H = {} E. J ≤ H Show Content Detailed Solution H = {x: x is a positive integer, x\(^2\) < 3 and x \(\neq\) 0} H = {1}	A
15.	In a class of 80 students, every students had to study economics or geography, or both economics and geography, if 65 students studied economics and 50 studied geography, how many studied both subjects? A. 15 B. 30 C. 35 D. 45 E. 50 Show Content Detailed Solution ∩(E∩G) = ∩(E) + ∩(G) - ∩(E∩G) 80 = 65 + 50 - ∩(F∩G) ∴∩(E∩G) = 115 - 80 = 35	C

16.	The area shaded with horizontal lines is the solution set of the inequalities; A. y ≥ x, y + 3 ≥ 2x, x ≤ 3 B. y ≤ x, y + 2x ≥ -3, x ≤ 3 C. y ≤ -x, y + 2x ≤ 3, x ≥ -3 D. y ≥ -x, y + 3 ≤ 2x, x ≥ -3 E. y ≤ x, y ≤ 2x - 3, x ≥ 3	B
17.	The area shaded with vertical lines is the solution set of the inequalities; A. y ≤ x, y + 2x ≤ -3 B. y ≥ x, y + 2x ≥ - 3 C. y ≤ x, y < 2x - 3, x ≤ 3 D. y ≤ -x, y ≤ 2x - 3, x ≤ 3 E. y ≥ x, y + 2x ≥ 3, x ≤ 3	C
18.	In the diagram above, O is the center of the circle with radius 10cm, and ∠ABC = 30°. Calculate, correct to 1 decimal place, the length of arc AC [Take π = 22/7] A. 5.2cm B. 10.5cm C. 13.2cm D. 20.6cm E. 31.4cm Show Content Detailed Solution In the figure, < AOC = 2 x < ABC = 60° (angle subtended at the centre) \(\therefore\) Arc AC = \(\frac{60}{360} \times 2 \times 10 \times 3.14\) = \(\frac{31.4}{3}\) = 10.466 cm \(\approxeq\) 10.5 cm	B
19.	Factorize x\(^2\) + 4x - 192 A. (x-4) (x+48) B. (x-48) (x+4) C. (x-12) (x+16) D. (x-12) (x-16) E. (x+12) (x+16) Show Content Detailed Solution x\(^2\) + 4x - 192 x\(^2\) + 16x - 12x - 192 x(x + 16) - 12(x + 16) (x + 16)(x - 12)	C
20.	Factorize 2e\(^2\) - 3e + 1 A. (2e-1) (e-1) B. (e+1) (2e+1) C. (2e+3) (e+2) D. (2e-3) (e-1) E. (e²-3(2e-1) Show Content Detailed Solution 2e\(^2\) - 3e + 1 2e\(^2\) - 2e - e + 1 2e(e - 1) - 1(e - 1) (2e - 1)(e - 1)	A