Mathematics (Core), 1995, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

1995

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

41 - 48 of 48 Questions

#	Question	Ans
41.	Without using tables, find the value of \(\frac{\sin 20°}{\cos 70°} + \frac{\cos 25°}{\sin 65°}\) A. 2 B. 1 C. o D. -1 E. -2 Show Content Detailed Solution Note: sin 20° = cos 70° hence \(\frac{\sin 20}{\cos 70} = 1\) Also, cos 65° = sin 25° \(\frac{\cos 65}{\sin 25} = 1\) \(\therefore \frac{\sin 20}{\cos 70} + \frac{\cos 65}{\sin 25}\) = 1 + 1 = 2	A
42.	In the diagram above, ∠PRQ = 90°, ∠QPR = 30° and /PQ/ = 10 cm. Find y. A. 6cm B. 5cm C. 4cm D. 3cm E. 2cm Show Content Detailed Solution \(\sin 30° = \frac{y}{10}\) \(y = 10 \sin 30° = 10 \times 0.5 = 5 cm\)	B
43.	The bearing of two points Q and R from a point P are 030° and 120° respectively, lf /PQ/ = 12 m and /PR/ = 5 m, find the distance QR. A. 13m B. 11m C. 9m D. 7m E. 5m Show Content Detailed Solution x\(^2\) = 12\(^2\) + 5\(^2\) x\(^2\) = 144 + 25 x\(^2\) = 169 x = \(\sqrt{169}\) = 13m.	A
44.	What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14? A. 8 B. 9 C. 10 D. 11 E. 14 Show Content Detailed Solution The mode = 8 with the highest frequency of 4 times.	A
45.	The mean of 20 observations in an experiment is 4, lf the observed largest value is 23, find the mean of the remaining observations. A. 4 B. 3 C. 2.85 D. 2.60 E. 2.56 Show Content Detailed Solution Mean of 20 observations = 4 All the observations = 20 x 4 = 80 Largest observed value = 23 Remaining observation = 80 - 23 = 57 Mean of observation = 57/19 = 3	B
46.	Find the median of the following numbers 2.64, 2.50, 2.72, 2.91 and 2.35. A. 2.91 B. 2.72 C. 2.64 D. 2.50 E. 2.35 Show Content Detailed Solution Arranging the numbers in ascending order, we have 2.35, 2.50, 2.64, 2.72, 2.91. The median = 2.64	C
47.	Two fair dice are tossed together once. Find the probability that the sum of the outcome is at least 10. A. 1/12 B. 5/ 36 C. 1/ 6 D. 1/4 E. 5/18 Show Content Detailed Solution at least 10 => ≥ 10 6/36 = 1/6	C
48.	From a box containing 2 red, 6 white and 5 black balls, a ball is randomly selected. What is the probability that the selected ball is black? A. 2/13 B. 5/13 C. 5/11 D. 5/6 E. 11/13 Show Content Detailed Solution 2r, 6w, 5b balls Total balls = 2 + 6 + 5 = 13 balls p(black) = 5/13	B

41.	Without using tables, find the value of \(\frac{\sin 20°}{\cos 70°} + \frac{\cos 25°}{\sin 65°}\) A. 2 B. 1 C. o D. -1 E. -2 Show Content Detailed Solution Note: sin 20° = cos 70° hence \(\frac{\sin 20}{\cos 70} = 1\) Also, cos 65° = sin 25° \(\frac{\cos 65}{\sin 25} = 1\) \(\therefore \frac{\sin 20}{\cos 70} + \frac{\cos 65}{\sin 25}\) = 1 + 1 = 2	A
42.	In the diagram above, ∠PRQ = 90°, ∠QPR = 30° and /PQ/ = 10 cm. Find y. A. 6cm B. 5cm C. 4cm D. 3cm E. 2cm Show Content Detailed Solution \(\sin 30° = \frac{y}{10}\) \(y = 10 \sin 30° = 10 \times 0.5 = 5 cm\)	B
43.	The bearing of two points Q and R from a point P are 030° and 120° respectively, lf /PQ/ = 12 m and /PR/ = 5 m, find the distance QR. A. 13m B. 11m C. 9m D. 7m E. 5m Show Content Detailed Solution x\(^2\) = 12\(^2\) + 5\(^2\) x\(^2\) = 144 + 25 x\(^2\) = 169 x = \(\sqrt{169}\) = 13m.	A
44.	What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14? A. 8 B. 9 C. 10 D. 11 E. 14 Show Content Detailed Solution The mode = 8 with the highest frequency of 4 times.	A

45.	The mean of 20 observations in an experiment is 4, lf the observed largest value is 23, find the mean of the remaining observations. A. 4 B. 3 C. 2.85 D. 2.60 E. 2.56 Show Content Detailed Solution Mean of 20 observations = 4 All the observations = 20 x 4 = 80 Largest observed value = 23 Remaining observation = 80 - 23 = 57 Mean of observation = 57/19 = 3	B
46.	Find the median of the following numbers 2.64, 2.50, 2.72, 2.91 and 2.35. A. 2.91 B. 2.72 C. 2.64 D. 2.50 E. 2.35 Show Content Detailed Solution Arranging the numbers in ascending order, we have 2.35, 2.50, 2.64, 2.72, 2.91. The median = 2.64	C
47.	Two fair dice are tossed together once. Find the probability that the sum of the outcome is at least 10. A. 1/12 B. 5/ 36 C. 1/ 6 D. 1/4 E. 5/18 Show Content Detailed Solution at least 10 => ≥ 10 6/36 = 1/6	C
48.	From a box containing 2 red, 6 white and 5 black balls, a ball is randomly selected. What is the probability that the selected ball is black? A. 2/13 B. 5/13 C. 5/11 D. 5/6 E. 11/13 Show Content Detailed Solution 2r, 6w, 5b balls Total balls = 2 + 6 + 5 = 13 balls p(black) = 5/13	B