Mathematics (Core), 2014, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2014

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 49 Questions

#	Question	Ans
31.	If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime A. \(\frac{5}{9}\) B. 1\(\frac{4}{9}\) C. \(\frac{1}{3}\) D. \(\frac{2}{9}\) Show Content Detailed Solution p = { 1, 2, 3} Q = {2, 3, 5} prob(prime number) = prob(1 and 2) or = prob(2 and 3) or = prob(3 and 2) There are three possibilities of the sum being prime Total possibility = 9 probability(sum being prime) = \(\frac{3}{9}\) = \(\frac{1}{3}\)	C
32.	If log 5.957 = 0.7750, find log \(3 \sqrt{0.0005957}\) A. 4.1986 B. 2.9250 C. 1.5917 D. 1.2853 Show Content Detailed Solution \(3 \sqrt{0.0005957}\) \(\begin{array}{c\|c} \log No(0.0005957) & \frac{1}{3} \log \frac{1}{3} \\ \hline (0.0005957)^{\frac{1}{3}} & 4.7750 \times \frac{1}{3} \\ & 6 + 2.7750 \\\hline & 3 \\\hline & 2.9250\end{array}\)	B
33.	The probability of an event P happening is \(\frac{1}{5}\) and that of event Q is \(\frac{1}{4}\). If the events are independent, what is the probability that neither of them happens? A. \(\frac{4}{5}\) B. \(\frac{3}{4}\) C. \(\frac{3}{5}\) D. \(\frac{1}{20}\) Show Content Detailed Solution prob(p) = \(\frac{1}{5}\) prob(Q) = \(\frac{1}{4}\) Prob(neither p) = 1 - \(\frac{1}{5}\) \(\frac{5 - 1}{5} = \frac{4}{5}\) prob(neither Q) = 1 - \(\frac{1}{4}\) \(\frac{4 - 1}{4} = \frac{3}{4}\) prob(neither of them) = \(\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}\) = \(\frac{3}{5}\)	C
34.	Each exterior angle of a polygon is 30^o. Calculate the sum of the interior angles A. 540^o B. 720^o C. 1080^o D. 1800^o Show Content Detailed Solution number of sides = \(\frac{360^o}{\theta} = \frac{360^o}{306o}\) n = 12^o Sum of interior angle = (n - 2) 180^o (12 - 2) 180v = 10 x 180^o = 1800^o	D
35.	Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141. A. 11 B. 12 C. 13 D. 14 Show Content Detailed Solution T₁, T₂, T₃ 2, -9, -20 .... -141 l = a + (n - 1)d first term, a = 2 common difference d = T₃ - T₂ = T₂ - T₁ = -20 - (-9) = -9 -2 = -20 + 9 = -9 -2 = -20 + 9 = -11 -11 = -11 d = -1 last term l = -141 -141 = 2 + (\(\cap\) - 1) (-11) -141 = 2 + (-11 \(\cap\) + 11)	D
36.	In what modulus is it true that 9 + 8 = 5? A. mod 10 B. mod 11 C. mod 12 D. mod 13 Show Content Detailed Solution 9 + 8 = 17 17 + mod 12 = 1 rem 5 it is mod 12	C
37.	The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q? A. 2.5cm B. 5.0cm C. 7.5cm D. 20.0cm Show Content Detailed Solution volume of cylinder = \(\pi r^2h\) volume of cylinder p = \(\pi r^2 \times 10\) = 10\(\pi r^2\) volume of cylinder Q = \(\pi (2r)^2 h\) = 4\(\pi r^2\)h 4\(\pi r^2 = 10 \pi r^2 h\) h = \(\frac{10}{4} = 4.5cm\)	A
38.	The bar chart shows the scores of some students in a test. How many students took the test? A. 18 B. 19 C. 20 D. 22 Show Content Detailed Solution score 0 = 4 students score 1 = 2 students score 2 = 5 students score 3 = 6 students score 4 = 3 students = 20 students	C
39.	The bar chart shows the scores of some students in a test. If one students is selected at random, find the probability that he/she scored at most 2 marks A. \(\frac{11}{18}\) B. \(\frac{11}{20}\) C. \(\frac{7}{22}\) D. \(\frac{5}{19}\) Show Content Detailed Solution at most 2 marks = 5 + 2 + 4 students = 11 students probability(at most 2 marks) = \(\frac{11}{20}\)	B
40.	In the diagram, the value of x + y = 220^o. Find the value of n A. 20^o B. 40^o C. 60^o D. 80^o Show Content Detailed Solution y + z = 180^o z = 180^o - y x + k = 180^o k = 180^o - x the sum of angles in the \(\bigtriangleup\) are: n + z + k = 180 but n + 180^o - y + 180^o - x = 180^o n + 180^o - y - x = 0 but we want to find n n + 180^o = x + y but x + y = 220^o n + 180^o = 220^o &l	B

31.	If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime A. \(\frac{5}{9}\) B. 1\(\frac{4}{9}\) C. \(\frac{1}{3}\) D. \(\frac{2}{9}\) Show Content Detailed Solution p = { 1, 2, 3} Q = {2, 3, 5} prob(prime number) = prob(1 and 2) or = prob(2 and 3) or = prob(3 and 2) There are three possibilities of the sum being prime Total possibility = 9 probability(sum being prime) = \(\frac{3}{9}\) = \(\frac{1}{3}\)	C
32.	If log 5.957 = 0.7750, find log \(3 \sqrt{0.0005957}\) A. 4.1986 B. 2.9250 C. 1.5917 D. 1.2853 Show Content Detailed Solution \(3 \sqrt{0.0005957}\) \(\begin{array}{c\|c} \log No(0.0005957) & \frac{1}{3} \log \frac{1}{3} \\ \hline (0.0005957)^{\frac{1}{3}} & 4.7750 \times \frac{1}{3} \\ & 6 + 2.7750 \\\hline & 3 \\\hline & 2.9250\end{array}\)	B
33.	The probability of an event P happening is \(\frac{1}{5}\) and that of event Q is \(\frac{1}{4}\). If the events are independent, what is the probability that neither of them happens? A. \(\frac{4}{5}\) B. \(\frac{3}{4}\) C. \(\frac{3}{5}\) D. \(\frac{1}{20}\) Show Content Detailed Solution prob(p) = \(\frac{1}{5}\) prob(Q) = \(\frac{1}{4}\) Prob(neither p) = 1 - \(\frac{1}{5}\) \(\frac{5 - 1}{5} = \frac{4}{5}\) prob(neither Q) = 1 - \(\frac{1}{4}\) \(\frac{4 - 1}{4} = \frac{3}{4}\) prob(neither of them) = \(\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}\) = \(\frac{3}{5}\)	C
34.	Each exterior angle of a polygon is 30^o. Calculate the sum of the interior angles A. 540^o B. 720^o C. 1080^o D. 1800^o Show Content Detailed Solution number of sides = \(\frac{360^o}{\theta} = \frac{360^o}{306o}\) n = 12^o Sum of interior angle = (n - 2) 180^o (12 - 2) 180v = 10 x 180^o = 1800^o	D
35.	Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141. A. 11 B. 12 C. 13 D. 14 Show Content Detailed Solution T₁, T₂, T₃ 2, -9, -20 .... -141 l = a + (n - 1)d first term, a = 2 common difference d = T₃ - T₂ = T₂ - T₁ = -20 - (-9) = -9 -2 = -20 + 9 = -9 -2 = -20 + 9 = -11 -11 = -11 d = -1 last term l = -141 -141 = 2 + (\(\cap\) - 1) (-11) -141 = 2 + (-11 \(\cap\) + 11)	D

36.	In what modulus is it true that 9 + 8 = 5? A. mod 10 B. mod 11 C. mod 12 D. mod 13 Show Content Detailed Solution 9 + 8 = 17 17 + mod 12 = 1 rem 5 it is mod 12	C
37.	The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q? A. 2.5cm B. 5.0cm C. 7.5cm D. 20.0cm Show Content Detailed Solution volume of cylinder = \(\pi r^2h\) volume of cylinder p = \(\pi r^2 \times 10\) = 10\(\pi r^2\) volume of cylinder Q = \(\pi (2r)^2 h\) = 4\(\pi r^2\)h 4\(\pi r^2 = 10 \pi r^2 h\) h = \(\frac{10}{4} = 4.5cm\)	A
38.	The bar chart shows the scores of some students in a test. How many students took the test? A. 18 B. 19 C. 20 D. 22 Show Content Detailed Solution score 0 = 4 students score 1 = 2 students score 2 = 5 students score 3 = 6 students score 4 = 3 students = 20 students	C
39.	The bar chart shows the scores of some students in a test. If one students is selected at random, find the probability that he/she scored at most 2 marks A. \(\frac{11}{18}\) B. \(\frac{11}{20}\) C. \(\frac{7}{22}\) D. \(\frac{5}{19}\) Show Content Detailed Solution at most 2 marks = 5 + 2 + 4 students = 11 students probability(at most 2 marks) = \(\frac{11}{20}\)	B
40.	In the diagram, the value of x + y = 220^o. Find the value of n A. 20^o B. 40^o C. 60^o D. 80^o Show Content Detailed Solution y + z = 180^o z = 180^o - y x + k = 180^o k = 180^o - x the sum of angles in the \(\bigtriangleup\) are: n + z + k = 180 but n + 180^o - y + 180^o - x = 180^o n + 180^o - y - x = 0 but we want to find n n + 180^o = x + y but x + y = 220^o n + 180^o = 220^o &l	B