Mathematics (Core), 2013, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2013

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 48 Questions

#	Question	Ans
21.	If p = (y : 2y \(\geq\) 6) and Q = (y : y -3 \(\geq\) 4), where y is an integer, find p\(\cap\)Q A. {3, 4} B. {3, 7} C. {3, 4, 5, 6, 7} D. {4, 5, 6} Show Content Detailed Solution p = (y : 2y \(\geq\) 6) 2y \(\leq\) 6 y \(\leq \frac{6}{2}\) y = \(\leq\) 3 and Q = (y : y -3 \(\geq\) 4) y - 3 \(\geq\) 4 y \(\geq\) 4 + 3 y \(\geq\) 7 therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....} P\(\cap\)Q = {3, 4, 5, 6, 7}	C
22.	Find the values of k in the equation 6k² = 5k + 6 A. {\(\frac{-2}{3}, \frac{-3}{2}\)} B. {\(\frac{-2}{3}, \frac{3}{2}\)} C. {\(\frac{2}{3}, \frac{-3}{2}\)} D. {\(\frac{2}{3}, \frac{3}{2}\)} Show Content Detailed Solution 6k² = 5k + 6 6k² - 5k - 6 = 0 6k² - 0k + 4k - 6 = 0 3k(2k - 3) + 2(2k - 3) = 0 (3k + 2)(2k - 3) = 0 3k + 2 = 0 or 2k - 3 = 0 3k = -2 or 2k = 3 k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\) k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))	B
23.	If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9 A. 8 B. 7 C. 6 D. 5 Show Content Detailed Solution y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\) 6 = k\(\sqrt{3 + 1}\) 6 = k\(\sqrt{4}\) 6 = 2k k = \(\frac{6}{2}\) = 3 y = \(\sqrt{(x + 1)}\) 9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3) \(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\) 3 = \(\sqrt{x + 1}\)(square both sides) 9 = x + 1 x = 9 - 1 x = 8	A
24.	The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k A. zero B. -2 C. -4 D. -8 Show Content Detailed Solution y = x² + 2x + k at point(2,0) x = 2, y = 0 0 = (2)² + 2(20 + k) 0 = 4 + 4 + k 0 = 8 + k k = -8	D
25.	What is the locus of the point X which moves relative to two fixed points P and M on a plane such that < PXM = 30^o A. thebisector of the straight line joining P and M B. an arc of a circle with PM as a chord C. the bisector of angle PXM D. a circle centre X and radius PM	B
26.	When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number A. 11 B. \(\frac{15}{2}\) C. 5 D. \(\frac{5}{2}\) Show Content Detailed Solution Let the number be y, subtract y from 2 i.e 2 - y 2 - y = 4 < \(\frac{1}{5}\) y, 2 - y = \(\frac{y}{5}\) - 4 2 - y + 4 = \(\frac{y}{5}\) 6 = \(\frac{y}{5}\) + y 6 = \(\frac{y + 5y}{5}\) 6 = \(\frac{6y}{5}\) multiplying through by 5 6 * 5 = 6y \(\frac{30}{6}\) = y = 5	C
27.	Express \(\frac{2}{x + 3} - \frac{1}{x - 2}\) as a simple fraction A. \(\frac{x - 7}{x^2 + x - 6}\) B. \(\frac{x - 1}{x^2 + x - 6}\) C. \(\frac{x - 2}{x^2 + x - 6}\) D. \(\frac{x - 27}{x^2 + x - 6}\) Show Content Detailed Solution \(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\) = \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\) = \(\frac{x -7}{x^2 + x - 6}\) = \(\frac{x - 7}{x^2 + x - 6}\)	A
28.	An interior angle of a regular polygon is 5 times each exterior angle. How many sides has the polygon? A. 15 B. 12 C. 9 D. 6 Show Content Detailed Solution Let the interior angle = x^o interior angle = 5x^o (sum of int. angle ann exterior) (angles = angle or straight line) 6x = 180 x = \(\frac{180}{6}\) x = 30^o no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\) = \(\frac{360}{30}\) = 12	B
29.	Given that P = x² + 4x - 2, Q = 2x - 1 and Q - p = 2, find x A. -2 B. -1 C. 1 D. 2 Show Content Detailed Solution P = x² + 4x - 2, Q = 2x - 1 Q - p = 2, (2x - 1) - (x² + 4x - 2) = 2 2x - 1 - x² - 4x + 2 = 2 -2x - x² + 1 -x² - 2x - 1 = 0 x² + 2x + 1 = 0 x² + x + x + 1 = 0 x(x + 1) + 1(x + 1) = 0 (x + 1)(x + 1) = 0 x + 1 = 0 or x + 1 = 0 x = -1 or x = -1 x = -1	B
30.	A pyramid has a rectangular base with dimensions 12m by 8m. If its height is 14m, calculate the volume A. 322m³ B. 448m³ C. 632m² D. 840m² Show Content Detailed Solution Volume of pyramid = \(\frac{1}{3}\) x base area x height = \(\frac{1}{3} \times 12^4 \times 8 \times 14\) = 4 x 8 x 14 = 448m³	B

21.	If p = (y : 2y \(\geq\) 6) and Q = (y : y -3 \(\geq\) 4), where y is an integer, find p\(\cap\)Q A. {3, 4} B. {3, 7} C. {3, 4, 5, 6, 7} D. {4, 5, 6} Show Content Detailed Solution p = (y : 2y \(\geq\) 6) 2y \(\leq\) 6 y \(\leq \frac{6}{2}\) y = \(\leq\) 3 and Q = (y : y -3 \(\geq\) 4) y - 3 \(\geq\) 4 y \(\geq\) 4 + 3 y \(\geq\) 7 therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....} P\(\cap\)Q = {3, 4, 5, 6, 7}	C
22.	Find the values of k in the equation 6k² = 5k + 6 A. {\(\frac{-2}{3}, \frac{-3}{2}\)} B. {\(\frac{-2}{3}, \frac{3}{2}\)} C. {\(\frac{2}{3}, \frac{-3}{2}\)} D. {\(\frac{2}{3}, \frac{3}{2}\)} Show Content Detailed Solution 6k² = 5k + 6 6k² - 5k - 6 = 0 6k² - 0k + 4k - 6 = 0 3k(2k - 3) + 2(2k - 3) = 0 (3k + 2)(2k - 3) = 0 3k + 2 = 0 or 2k - 3 = 0 3k = -2 or 2k = 3 k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\) k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))	B
23.	If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9 A. 8 B. 7 C. 6 D. 5 Show Content Detailed Solution y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\) 6 = k\(\sqrt{3 + 1}\) 6 = k\(\sqrt{4}\) 6 = 2k k = \(\frac{6}{2}\) = 3 y = \(\sqrt{(x + 1)}\) 9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3) \(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\) 3 = \(\sqrt{x + 1}\)(square both sides) 9 = x + 1 x = 9 - 1 x = 8	A
24.	The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k A. zero B. -2 C. -4 D. -8 Show Content Detailed Solution y = x² + 2x + k at point(2,0) x = 2, y = 0 0 = (2)² + 2(20 + k) 0 = 4 + 4 + k 0 = 8 + k k = -8	D
25.	What is the locus of the point X which moves relative to two fixed points P and M on a plane such that < PXM = 30^o A. thebisector of the straight line joining P and M B. an arc of a circle with PM as a chord C. the bisector of angle PXM D. a circle centre X and radius PM	B

26.	When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number A. 11 B. \(\frac{15}{2}\) C. 5 D. \(\frac{5}{2}\) Show Content Detailed Solution Let the number be y, subtract y from 2 i.e 2 - y 2 - y = 4 < \(\frac{1}{5}\) y, 2 - y = \(\frac{y}{5}\) - 4 2 - y + 4 = \(\frac{y}{5}\) 6 = \(\frac{y}{5}\) + y 6 = \(\frac{y + 5y}{5}\) 6 = \(\frac{6y}{5}\) multiplying through by 5 6 * 5 = 6y \(\frac{30}{6}\) = y = 5	C
27.	Express \(\frac{2}{x + 3} - \frac{1}{x - 2}\) as a simple fraction A. \(\frac{x - 7}{x^2 + x - 6}\) B. \(\frac{x - 1}{x^2 + x - 6}\) C. \(\frac{x - 2}{x^2 + x - 6}\) D. \(\frac{x - 27}{x^2 + x - 6}\) Show Content Detailed Solution \(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\) = \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\) = \(\frac{x -7}{x^2 + x - 6}\) = \(\frac{x - 7}{x^2 + x - 6}\)	A
28.	An interior angle of a regular polygon is 5 times each exterior angle. How many sides has the polygon? A. 15 B. 12 C. 9 D. 6 Show Content Detailed Solution Let the interior angle = x^o interior angle = 5x^o (sum of int. angle ann exterior) (angles = angle or straight line) 6x = 180 x = \(\frac{180}{6}\) x = 30^o no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\) = \(\frac{360}{30}\) = 12	B
29.	Given that P = x² + 4x - 2, Q = 2x - 1 and Q - p = 2, find x A. -2 B. -1 C. 1 D. 2 Show Content Detailed Solution P = x² + 4x - 2, Q = 2x - 1 Q - p = 2, (2x - 1) - (x² + 4x - 2) = 2 2x - 1 - x² - 4x + 2 = 2 -2x - x² + 1 -x² - 2x - 1 = 0 x² + 2x + 1 = 0 x² + x + x + 1 = 0 x(x + 1) + 1(x + 1) = 0 (x + 1)(x + 1) = 0 x + 1 = 0 or x + 1 = 0 x = -1 or x = -1 x = -1	B
30.	A pyramid has a rectangular base with dimensions 12m by 8m. If its height is 14m, calculate the volume A. 322m³ B. 448m³ C. 632m² D. 840m² Show Content Detailed Solution Volume of pyramid = \(\frac{1}{3}\) x base area x height = \(\frac{1}{3} \times 12^4 \times 8 \times 14\) = 4 x 8 x 14 = 448m³	B