Mathematics (Core), 2005, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2005

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

1 - 10 of 47 Questions

#	Question	Ans
1.	Correct 0.04945 to two significant figures A. 0.040 B. 0.049 C. 0.050 D. 0.49 Show Content Detailed Solution 0.04945 \(\approxeq\) 0.049. (to 2 sig. figs)	B
2.	Simplify \(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}\) A. \(\frac{1}{6}(5\sqrt{3}-3\sqrt{2}\) B. \(\frac{1}{6}(15\sqrt{3}-6\sqrt{2}\) C. \(\frac{1}{6}(3\sqrt{2}-\sqrt{3}\) D. \(\frac{1}{6}(10\sqrt{3}-9\sqrt{2}\) Show Content Detailed Solution \(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\\ =\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\\ \frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\\ \frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\\ \frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)	D
3.	Evaluate, correct to the nearest whole number \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\) A. 33 B. 8 C. 7 D. o Show Content Detailed Solution \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\\ =\frac{15}{2}-\left(\frac{5}{2}+\frac{3}{1}\right)\times\frac{2}{33}\\ =\frac{15}{2}-\left(\frac{5+6}{2}\right)\times \frac{2}{33}=\frac{15}{2}-\frac{11}{2}\times \frac{2}{33}=\frac{15}{2}-\frac{1}{3}\\ =\frac{45-2}{6}=\frac{43}{6}\)	C
4.	Simplify the expression \(log_{10}18 - log_{10}2.88+log_{10}16\) A. 31.12 B. 3.112 C. 2 D. 1 Show Content Detailed Solution \(log_{10}18 - log_{10}2.88+log_{10}16\\ =log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\\ =log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)	C
5.	Find the equation whose roots are 2 and \(-3\frac{1}{2}\) A. 2x² + 3x + 14 = 0 B. 2x² + 5x + 7 = 0 C. 2x² + 5x - 7 = 0 D. 2x² + 3x - 14 = 0 Show Content Detailed Solution x² (sum of roots)x + (product of roots) = 0 Sum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2}=-\frac{1}{2}\) Product of roots \(=2 \times -3\frac{1}{2}=-7\\ x^2-\left(\frac{-3}{2}\right)x+(-7)=0\Rightarrow 2x^2 + 3x - 14 = 0\)	D
6.	A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x A. 2 B. 5 C. 6 D. 10 Show Content Detailed Solution The cost price of the whole mangoes = N5x The sold amount of the mangoes = 3x * 220 = N6.60x The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)	B
7.	From a point P, R is 5km due west and 12km due south. Find the distance between P and R A. 5km B. 12km C. 13km D. 17km Show Content Detailed Solution Using Pythagoras theorem \(\|PR\|^2=\|PO\|^2+\|OR\|^2\\ \|PR\|^2=5^2+12^2\\ \|PR\|=\sqrt{25+144}=\sqrt{169}=13km\)	C
8.	A fair die is tossed once, what is the probability of obtaining neither 5 or 2 A. \(\frac{5}{6}\) B. \(\frac{2}{3}\) C. \(\frac{1}{2}\) D. \(\frac{1}{6}\) Show Content Detailed Solution Probability of obtaining a 5 is P(5)=\(\frac{1}{6}\) Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\) Probability of obatining either 2 or 5 = P(2∪5) = \(\frac{2}{6}\) Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)	B
9.	In the diagram, KL//MN, ∠LKP = 30^o and ∠NMP = 45^o. Find the size of the reflex ∠KPM. A. 285^o B. 255^o C. 225^o D. 210^o Show Content Detailed Solution ∠KPM = 360^o - (30 + 45)^o = 360^o - 75^o = 285^o	A
10.	In the diagram, PR is a diameter, ∠PRQ = (3x-8)^o and ∠RPQ = (2y-7)^o. s x in terms of y A. \(x=\frac{75-2y}{3}\) B. \(x=\frac{105-3y}{2}\) C. \(x=\frac{105-2y}{3}\) D. \(x=\frac{75-3y}{2}\) Show Content Detailed Solution 180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle) 180 = ∠RPQ + ∠PRQ + 90 => 180^o = (3x-8)^o+(2y-7)^o+90^o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)	C

1.	Correct 0.04945 to two significant figures A. 0.040 B. 0.049 C. 0.050 D. 0.49 Show Content Detailed Solution 0.04945 \(\approxeq\) 0.049. (to 2 sig. figs)	B
2.	Simplify \(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}\) A. \(\frac{1}{6}(5\sqrt{3}-3\sqrt{2}\) B. \(\frac{1}{6}(15\sqrt{3}-6\sqrt{2}\) C. \(\frac{1}{6}(3\sqrt{2}-\sqrt{3}\) D. \(\frac{1}{6}(10\sqrt{3}-9\sqrt{2}\) Show Content Detailed Solution \(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\\ =\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\\ \frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\\ \frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\\ \frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)	D
3.	Evaluate, correct to the nearest whole number \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\) A. 33 B. 8 C. 7 D. o Show Content Detailed Solution \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\\ =\frac{15}{2}-\left(\frac{5}{2}+\frac{3}{1}\right)\times\frac{2}{33}\\ =\frac{15}{2}-\left(\frac{5+6}{2}\right)\times \frac{2}{33}=\frac{15}{2}-\frac{11}{2}\times \frac{2}{33}=\frac{15}{2}-\frac{1}{3}\\ =\frac{45-2}{6}=\frac{43}{6}\)	C
4.	Simplify the expression \(log_{10}18 - log_{10}2.88+log_{10}16\) A. 31.12 B. 3.112 C. 2 D. 1 Show Content Detailed Solution \(log_{10}18 - log_{10}2.88+log_{10}16\\ =log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\\ =log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)	C
5.	Find the equation whose roots are 2 and \(-3\frac{1}{2}\) A. 2x² + 3x + 14 = 0 B. 2x² + 5x + 7 = 0 C. 2x² + 5x - 7 = 0 D. 2x² + 3x - 14 = 0 Show Content Detailed Solution x² (sum of roots)x + (product of roots) = 0 Sum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2}=-\frac{1}{2}\) Product of roots \(=2 \times -3\frac{1}{2}=-7\\ x^2-\left(\frac{-3}{2}\right)x+(-7)=0\Rightarrow 2x^2 + 3x - 14 = 0\)	D

6.	A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x A. 2 B. 5 C. 6 D. 10 Show Content Detailed Solution The cost price of the whole mangoes = N5x The sold amount of the mangoes = 3x * 220 = N6.60x The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)	B
7.	From a point P, R is 5km due west and 12km due south. Find the distance between P and R A. 5km B. 12km C. 13km D. 17km Show Content Detailed Solution Using Pythagoras theorem \(\|PR\|^2=\|PO\|^2+\|OR\|^2\\ \|PR\|^2=5^2+12^2\\ \|PR\|=\sqrt{25+144}=\sqrt{169}=13km\)	C
8.	A fair die is tossed once, what is the probability of obtaining neither 5 or 2 A. \(\frac{5}{6}\) B. \(\frac{2}{3}\) C. \(\frac{1}{2}\) D. \(\frac{1}{6}\) Show Content Detailed Solution Probability of obtaining a 5 is P(5)=\(\frac{1}{6}\) Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\) Probability of obatining either 2 or 5 = P(2∪5) = \(\frac{2}{6}\) Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)	B
9.	In the diagram, KL//MN, ∠LKP = 30^o and ∠NMP = 45^o. Find the size of the reflex ∠KPM. A. 285^o B. 255^o C. 225^o D. 210^o Show Content Detailed Solution ∠KPM = 360^o - (30 + 45)^o = 360^o - 75^o = 285^o	A
10.	In the diagram, PR is a diameter, ∠PRQ = (3x-8)^o and ∠RPQ = (2y-7)^o. s x in terms of y A. \(x=\frac{75-2y}{3}\) B. \(x=\frac{105-3y}{2}\) C. \(x=\frac{105-2y}{3}\) D. \(x=\frac{75-3y}{2}\) Show Content Detailed Solution 180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle) 180 = ∠RPQ + ∠PRQ + 90 => 180^o = (3x-8)^o+(2y-7)^o+90^o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)	C