Mathematics (Core), 2016, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2016

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 49 Questions

#	Question	Ans
11.	Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10? A. 1 B. 3 C. 7 D. 17 Show Content Detailed Solution 2x + y = 7...(1) 3x - 2y = 3...(2) From (1), y = 7 - 2x for y in (2) 3x - 2(7 - 2x) = 3 3x - 14 + 4x = 3 7x + 3 + 14 = 17 x = \(\frac{17}{7}\) Hence, 7 x \(\frac{17}{7}\) = 17 - 10 = 7	C
12.	Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\) A. \(\frac{x + 1}{x(1 - x)}\) B. \(\frac{3x - 1}{ x(1 - x)}\) C. \(\frac{3x + 1}{ x(1 - x)}\) D. \(\frac{x + 1}{ x(1 - x)}\) Show Content Detailed Solution \(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\) = \(\frac{2x - 1(1 + x)}{x(1 - x)}\) = \(\frac{3x - 1}{x(1 - x)}\)	B
13.	Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\) A. s = \(\frac{mrp}{nr + m^2}\) B. s = \(\frac{nr + m^2}{mrp}\) C. s = \(\frac{nrp}{mr + m^2}\) D. s = \(\frac{nrp}{nr + m^2}\) Show Content Detailed Solution P = S + \(\frac{sm^2}{nr}\) P = S(1 + \(\frac{m^2}{nr}\)) P = S(1 + \(\frac{nr + m^2}{nr}\)) nrp = S(nr + m²) S = \(\frac{nrp}{nr + m^2}\)	D
14.	Factorize; (2x + 3y)² - (x - 4y)² A. (3x - y)(x + 7y) B. (3x + y)(2x - 7y) C. (3x + y)(x - 7y) D. (3x - y)(2x + 7y) Show Content Detailed Solution (2x + 3y)² - (x - 4y)² = (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y) = 4x² + 12xy + 9y² - (x² - 8xy+ 16y²) = 4x² + 12xy + 19y² - x² + 8xy - 16y² = 3x² + 20xy - 7y² = 3x² + 21xy - xy - 7y² = 3x(x + 7y) - y(x + 7y) = (3x - y)(x + 7y)	A
15.	The curve surface area of a cylinder, 5cm high is 110cm ². Find the radius of its base. [Take \(\pi = \frac{22}{7}\)] A. 2.6cm B. 3.5cm C. 3.6cm D. 7.0cm Show Content Detailed Solution Curved surface area of cylinder = 2\(\pi\)rh 110 = 2 x \(\frac{22}{7}\) x r x 5 r = \(\frac{110 \times 7}{44 \times 5}\) = 3.5cm	B
16.	The volume of a pyramid with height 15cm is 90cm³. If its base is a rectangle with dimension xcm by 6cm, find the value of x A. 3 B. 5 C. 6 D. 8 Show Content Detailed Solution Volume of pyramid = \(\frac{1}{3}\)lbh 90 = \(\frac{1}{3} \times x \times 6 \times 15\) x = \(\frac{90 \times 33}{6 \times 15}\) = 3	A
17.	A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ A. \(\frac{3}{5}\) B. \(\frac{2}{3}\) C. \(\frac{3}{2}\) D. \(\frac{5}{3}\) Show Content Detailed Solution Let: (x1, y1) = (1, 2) (x2, y2) = (5, 8) The gradient m of \(\bar{PQ}\) is given by m = \(\frac{y_2 y_1}{x_2 - x_1}\) = \(\frac{8 - 2}{5 - 1}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)	C
18.	A straight line passes through the point P(1,2) and Q (5,8). Calculate the length PQ A. \(4\sqrt{11}\) B. \(4\sqrt{10}\) C. \(2\sqrt{17}\) D. \(2\sqrt{13}\) Show Content Detailed Solution \|PQ\| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\) = \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\) = \(\sqrt{4^2 + 6^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)	D
19.	If cos \(\theta\) = x and sin 60^o = x + 0.5 0^o < \(\theta\) < 90^o, find, correct to the nearest degree, the value of \(\theta\) A. 32^o B. 40^o C. 60^o D. 69^o Show Content Detailed Solution sin 60^o = x + 0.5 0^o(given) 0.8660 = x + 0.5 0.8660 - 0.5 = x x = 0.3660 cos\(\theta\) = x(given) cos\(\theta\) = 0.3660 Hence, \(\theta\) = cos^-1(0.3660) = 68.53^o = 69^o (nearest degree)	D
20.	\(\begin{array}{c\|c} Age(years) & 13 & 14 & 15 & 16 & 17 \\ \hline Frequency & 10 & 24 & 8 & 5 & 3 \end{array}\) The table shows the ages of students in a club. How many students are in the club? A. 50 B. 55 C. 60 D. 65 Show Content Detailed Solution Number of students in the club is 10 + 24 + 8 + 5 + 3 = 50	A

11.	Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10? A. 1 B. 3 C. 7 D. 17 Show Content Detailed Solution 2x + y = 7...(1) 3x - 2y = 3...(2) From (1), y = 7 - 2x for y in (2) 3x - 2(7 - 2x) = 3 3x - 14 + 4x = 3 7x + 3 + 14 = 17 x = \(\frac{17}{7}\) Hence, 7 x \(\frac{17}{7}\) = 17 - 10 = 7	C
12.	Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\) A. \(\frac{x + 1}{x(1 - x)}\) B. \(\frac{3x - 1}{ x(1 - x)}\) C. \(\frac{3x + 1}{ x(1 - x)}\) D. \(\frac{x + 1}{ x(1 - x)}\) Show Content Detailed Solution \(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\) = \(\frac{2x - 1(1 + x)}{x(1 - x)}\) = \(\frac{3x - 1}{x(1 - x)}\)	B
13.	Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\) A. s = \(\frac{mrp}{nr + m^2}\) B. s = \(\frac{nr + m^2}{mrp}\) C. s = \(\frac{nrp}{mr + m^2}\) D. s = \(\frac{nrp}{nr + m^2}\) Show Content Detailed Solution P = S + \(\frac{sm^2}{nr}\) P = S(1 + \(\frac{m^2}{nr}\)) P = S(1 + \(\frac{nr + m^2}{nr}\)) nrp = S(nr + m²) S = \(\frac{nrp}{nr + m^2}\)	D
14.	Factorize; (2x + 3y)² - (x - 4y)² A. (3x - y)(x + 7y) B. (3x + y)(2x - 7y) C. (3x + y)(x - 7y) D. (3x - y)(2x + 7y) Show Content Detailed Solution (2x + 3y)² - (x - 4y)² = (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y) = 4x² + 12xy + 9y² - (x² - 8xy+ 16y²) = 4x² + 12xy + 19y² - x² + 8xy - 16y² = 3x² + 20xy - 7y² = 3x² + 21xy - xy - 7y² = 3x(x + 7y) - y(x + 7y) = (3x - y)(x + 7y)	A
15.	The curve surface area of a cylinder, 5cm high is 110cm ². Find the radius of its base. [Take \(\pi = \frac{22}{7}\)] A. 2.6cm B. 3.5cm C. 3.6cm D. 7.0cm Show Content Detailed Solution Curved surface area of cylinder = 2\(\pi\)rh 110 = 2 x \(\frac{22}{7}\) x r x 5 r = \(\frac{110 \times 7}{44 \times 5}\) = 3.5cm	B

16.	The volume of a pyramid with height 15cm is 90cm³. If its base is a rectangle with dimension xcm by 6cm, find the value of x A. 3 B. 5 C. 6 D. 8 Show Content Detailed Solution Volume of pyramid = \(\frac{1}{3}\)lbh 90 = \(\frac{1}{3} \times x \times 6 \times 15\) x = \(\frac{90 \times 33}{6 \times 15}\) = 3	A
17.	A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ A. \(\frac{3}{5}\) B. \(\frac{2}{3}\) C. \(\frac{3}{2}\) D. \(\frac{5}{3}\) Show Content Detailed Solution Let: (x1, y1) = (1, 2) (x2, y2) = (5, 8) The gradient m of \(\bar{PQ}\) is given by m = \(\frac{y_2 y_1}{x_2 - x_1}\) = \(\frac{8 - 2}{5 - 1}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)	C
18.	A straight line passes through the point P(1,2) and Q (5,8). Calculate the length PQ A. \(4\sqrt{11}\) B. \(4\sqrt{10}\) C. \(2\sqrt{17}\) D. \(2\sqrt{13}\) Show Content Detailed Solution \|PQ\| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\) = \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\) = \(\sqrt{4^2 + 6^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)	D
19.	If cos \(\theta\) = x and sin 60^o = x + 0.5 0^o < \(\theta\) < 90^o, find, correct to the nearest degree, the value of \(\theta\) A. 32^o B. 40^o C. 60^o D. 69^o Show Content Detailed Solution sin 60^o = x + 0.5 0^o(given) 0.8660 = x + 0.5 0.8660 - 0.5 = x x = 0.3660 cos\(\theta\) = x(given) cos\(\theta\) = 0.3660 Hence, \(\theta\) = cos^-1(0.3660) = 68.53^o = 69^o (nearest degree)	D
20.	\(\begin{array}{c\|c} Age(years) & 13 & 14 & 15 & 16 & 17 \\ \hline Frequency & 10 & 24 & 8 & 5 & 3 \end{array}\) The table shows the ages of students in a club. How many students are in the club? A. 50 B. 55 C. 60 D. 65 Show Content Detailed Solution Number of students in the club is 10 + 24 + 8 + 5 + 3 = 50	A