Mathematics (Core), 2012, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2012

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 49 Questions

#	Question	Ans
11.	If x + y = 2y - x + 1 = 5, find the value of x A. 3 B. 2 C. 1 D. -1 Show Content Detailed Solution x + y = 2y - x + 1 = 5 x + y = 2y - x + 1 x + x + y - 2y = 1 2x - y = 1....(i) 2y - x + 1 = 5 -x + 2y = 5 + 1 -x = 2y = 4 x - 2y = -4 .....(ii) solve simultaneously (i) x 2x - y = 1 (ii) x x - 2y = -4 2x - y = 1 =2x - 4y = -8 3y = 9 y = \(\frac{9}{3}\) y = 3 substitute y = 3 into equation (i) 2x - y = 1 2x - 3 = 1 2x = 1 + 3 2x = 4 &	B
12.	The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation A. 12n - 6 = 0 B. 3n - 12 = 0 C. 2n - 35 = 0 D. 5n - 33 = 0 Show Content Detailed Solution 12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3 36 + n - 6n = 3 -5n = 3 - 36 -5n = -33 -5n + 33 = 0 5n - 33 = 0	D
13.	Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) A. m \(\leq \frac{5}{4}\) B. m \(\geq \frac{5}{4}\) C. m \(\leq - \frac{1}{11}\) D. m \(\geq - \frac{1}{11}\) Show Content Detailed Solution \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) = \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\) 12(-2m - 5) \(\geq\) 4(5m - 14) -24m - 60 \(\geq\) 20m - 56 -24m - 20m \(\geq\) -56 + 60 44m \(\geq\) 4 m \(\leq \frac{4}{-44}\) m \(\leq \frac{-1}{11}\)	C
14.	The curved surface area of a cylindrical tin is 704cm². If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]` A. 14cm B. 9cm C. 8cm D. 7cm Show Content Detailed Solution Curved surface area = 2\(\pi h\) 704 = 2 x \(\frac{22}{7} \times 8 \times h\) 704 = \(\frac{352}{7}\)h 352h = 704 x 7 h = \(\frac{704 \times 7}{352}\) = \(\frac{4928}{352}\) h = 14cm	A
15.	The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector A. 360^o B. 252^o C. 246^o D. 234^o Show Content Detailed Solution Let 0 = angle of the minor sector angle of the major sector = 360 - \(\theta\)(angle at a point) 2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc) 2\(\pi r = 180^o\) r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\) Lenght of ninor arc = \(\frac{\theta}{360} \times 2 \pi r\) 54 = \(\frac{\theta}{360} \times 3 \pi r\) \(\theta = \frac{360 \times 54}{2 \pi r}\) but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r \(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\) \(\thet	B
16.	In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true? A. /XY/ + /YZ/ = /XZ/ B. /XY/ - /YZ/ = /XZ/ C. /XY/² = /Y/² - /XY/² D. /YZ/² = /XZ/² + /XY/²	D
17.	If cos(x + 40)^o = 0.0872, what is the value of x? A. 85^o B. 75^o C. 65^o D. 45^o Show Content Detailed Solution cos(x + 40)^o = 0.0872 x + 40 = cos^-10.0872 x + 40^o = 84.99^o x = 84.99^o - 40^o x = 44.99 x = 45^o	D
18.	A kite flies on a taut string of length 50m inclined at tan angle 54^o to the horizontal ground. The height of the kite above the ground is A. 50 tan 30^o B. 50 sin 54^o C. 50 tan 54^o D. 50 sin 36^o	B
19.	Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores A. \(\sqrt{10}\) B. 4 C. 5 D. \(\sqrt{30}\) Show Content Detailed Solution \(\begin{array}{c\|c} x & x - x & (x - \bar{x})^2\\ \hline 15 & -6 & 36\\21 & 0 & 0\\17 & -4 & 16\\ 26 & 5 & 25 \\ 18 & -3 &9 \\ 29 & 8 & 64 \end{array}\) \(E(x - \bar{x})^2\) = 150 N = 6 S.D = \(\sqrt{\frac{(x - x)^2}{N}}\) S.D = \(\sqrt{\frac{150}{6}}\) = 5	C
20.	A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours A. \(\frac{8}{15}\) B. \(\frac{13}{25}\) C. \(\frac{11}{15}\) D. \(\frac{13}{15}\) Show Content Detailed Solution Prob(RB + BR) = Total balls = 4 + 6 = 10 = prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\) = \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)	A

11.	If x + y = 2y - x + 1 = 5, find the value of x A. 3 B. 2 C. 1 D. -1 Show Content Detailed Solution x + y = 2y - x + 1 = 5 x + y = 2y - x + 1 x + x + y - 2y = 1 2x - y = 1....(i) 2y - x + 1 = 5 -x + 2y = 5 + 1 -x = 2y = 4 x - 2y = -4 .....(ii) solve simultaneously (i) x 2x - y = 1 (ii) x x - 2y = -4 2x - y = 1 =2x - 4y = -8 3y = 9 y = \(\frac{9}{3}\) y = 3 substitute y = 3 into equation (i) 2x - y = 1 2x - 3 = 1 2x = 1 + 3 2x = 4 &	B
12.	The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation A. 12n - 6 = 0 B. 3n - 12 = 0 C. 2n - 35 = 0 D. 5n - 33 = 0 Show Content Detailed Solution 12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3 36 + n - 6n = 3 -5n = 3 - 36 -5n = -33 -5n + 33 = 0 5n - 33 = 0	D
13.	Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) A. m \(\leq \frac{5}{4}\) B. m \(\geq \frac{5}{4}\) C. m \(\leq - \frac{1}{11}\) D. m \(\geq - \frac{1}{11}\) Show Content Detailed Solution \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\) = \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\) 12(-2m - 5) \(\geq\) 4(5m - 14) -24m - 60 \(\geq\) 20m - 56 -24m - 20m \(\geq\) -56 + 60 44m \(\geq\) 4 m \(\leq \frac{4}{-44}\) m \(\leq \frac{-1}{11}\)	C
14.	The curved surface area of a cylindrical tin is 704cm². If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]` A. 14cm B. 9cm C. 8cm D. 7cm Show Content Detailed Solution Curved surface area = 2\(\pi h\) 704 = 2 x \(\frac{22}{7} \times 8 \times h\) 704 = \(\frac{352}{7}\)h 352h = 704 x 7 h = \(\frac{704 \times 7}{352}\) = \(\frac{4928}{352}\) h = 14cm	A
15.	The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector A. 360^o B. 252^o C. 246^o D. 234^o Show Content Detailed Solution Let 0 = angle of the minor sector angle of the major sector = 360 - \(\theta\)(angle at a point) 2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc) 2\(\pi r = 180^o\) r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\) Lenght of ninor arc = \(\frac{\theta}{360} \times 2 \pi r\) 54 = \(\frac{\theta}{360} \times 3 \pi r\) \(\theta = \frac{360 \times 54}{2 \pi r}\) but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r \(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\) \(\thet	B

16.	In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true? A. /XY/ + /YZ/ = /XZ/ B. /XY/ - /YZ/ = /XZ/ C. /XY/² = /Y/² - /XY/² D. /YZ/² = /XZ/² + /XY/²	D
17.	If cos(x + 40)^o = 0.0872, what is the value of x? A. 85^o B. 75^o C. 65^o D. 45^o Show Content Detailed Solution cos(x + 40)^o = 0.0872 x + 40 = cos^-10.0872 x + 40^o = 84.99^o x = 84.99^o - 40^o x = 44.99 x = 45^o	D
18.	A kite flies on a taut string of length 50m inclined at tan angle 54^o to the horizontal ground. The height of the kite above the ground is A. 50 tan 30^o B. 50 sin 54^o C. 50 tan 54^o D. 50 sin 36^o	B
19.	Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores A. \(\sqrt{10}\) B. 4 C. 5 D. \(\sqrt{30}\) Show Content Detailed Solution \(\begin{array}{c\|c} x & x - x & (x - \bar{x})^2\\ \hline 15 & -6 & 36\\21 & 0 & 0\\17 & -4 & 16\\ 26 & 5 & 25 \\ 18 & -3 &9 \\ 29 & 8 & 64 \end{array}\) \(E(x - \bar{x})^2\) = 150 N = 6 S.D = \(\sqrt{\frac{(x - x)^2}{N}}\) S.D = \(\sqrt{\frac{150}{6}}\) = 5	C
20.	A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours A. \(\frac{8}{15}\) B. \(\frac{13}{25}\) C. \(\frac{11}{15}\) D. \(\frac{13}{15}\) Show Content Detailed Solution Prob(RB + BR) = Total balls = 4 + 6 = 10 = prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\) = \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)	A