Year :
2005
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 47 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Convert 101101two to a number in base ten A. 61 B. 46 C. 45 D. 44 Detailed Solution1011012 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 32 + 0 + 8 + 4 + 0 + 1 = 4510 |
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| 12. |
Solve the equation \(2^7 = 8^{5-x}\) A. \(\frac{5}{8}\) B. \(\frac{8}{3}\) C. \(\frac{3}{2}\) D. \(\frac{15}{4}\) Detailed Solution\(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x\\\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\) |
|
| 13. |
Expand (2x-3y)(x-5y) A. \(2x^2 + 13xy - 15y^2\) B. \(2x^2 - 13xy - 15y^2\) C. \(2x^2 + 13xy + 15y^2\) D. \(2x^2 - 13xy + 15y^2\) Detailed Solution\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\\=2x^2 - 13xy + 15y^2\) |
|
| 14. |
Make f the subject of the relation \(v = u + ft\) A. \(\frac{v-u}{t}\) B. \(\frac{u-v}{t}\) C. \(t(v+u\) D. \(\frac{v}{u}-t\) Detailed Solution\(v=u+ft \Rightarrow v-u=ft\Rightarrow f=\frac{v-u}{t}\) |
|
| 15. |
The lengths of the adjacent sides of a right - angled triangle are xcm, (x-1)cm. If the length of the hypotenuse is \(\sqrt{13}cm\), find the value of x A. 2 B. 3 C. 4 D. 5 Detailed Solution\(x^2+(x-1)^2 = (\sqrt{13})^2 \Rightarrow x^2 + x^2 - 2x + 1 = 13\\2x^2 - 2x - 12 = 0\) dividing through by 2 \(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 \) |
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| 16. |
The probability that John and James pass an examination are 3/4 and 3/5 respectively, find the probability of both boys failing the examination A. \(\frac{1}{10}\) B. \(\frac{3}{10}\) C. \(\frac{9}{20}\) D. \(\frac{11}{20}\) Detailed SolutionProb.(John pass)\(=frac{3}{4}\) prob.(John fail) \(=1-frac{3}{4}=\frac{1}{4}\)Prob.(James pass) \(=frac{3}{5}\) Prob.(James fail) \(=1-frac{3}{5}=\frac{2}{5}\) Prob(both boys fail)\(=frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\) |
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| 17. |
 In the diagram PQ and MN are straight lines. Find the value of x A. 13o B. 17o C. 28o D. 30o Detailed SolutionThe straight line MN = 180 = 2(x+30o) + 86o =>180 = 2x + 60o + 86o => 180o = 2x + 146o => 2x = 180 - 46 = 34o => x = 34/2 = 17o |
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| 18. |
 In the diagram, PQRS is a parallelogram and ∠QRT = 30o. Find x A. 95o B. 100o C. 120o D. 150o Detailed SolutionSince ∠QRS = 180o - 30o = 150o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150o |
|
| 19. |
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee A. 026o B. 045o C. 210o D. 240o Detailed Solution
 \theta = cos^{-1}(0.5000)=60^{\circ}\) The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\) |
|
| 20. |
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)] A. \(\frac{\sqrt{2}}{7}cm\) B. \(3\frac{1}{2}cm\) C. 7cm D. \(7\sqrt{2}cm\) Detailed Solution\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\) |
| 11. |
Convert 101101two to a number in base ten A. 61 B. 46 C. 45 D. 44 Detailed Solution1011012 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 32 + 0 + 8 + 4 + 0 + 1 = 4510 |
|
| 12. |
Solve the equation \(2^7 = 8^{5-x}\) A. \(\frac{5}{8}\) B. \(\frac{8}{3}\) C. \(\frac{3}{2}\) D. \(\frac{15}{4}\) Detailed Solution\(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x\\\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\) |
|
| 13. |
Expand (2x-3y)(x-5y) A. \(2x^2 + 13xy - 15y^2\) B. \(2x^2 - 13xy - 15y^2\) C. \(2x^2 + 13xy + 15y^2\) D. \(2x^2 - 13xy + 15y^2\) Detailed Solution\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\\=2x^2 - 13xy + 15y^2\) |
|
| 14. |
Make f the subject of the relation \(v = u + ft\) A. \(\frac{v-u}{t}\) B. \(\frac{u-v}{t}\) C. \(t(v+u\) D. \(\frac{v}{u}-t\) Detailed Solution\(v=u+ft \Rightarrow v-u=ft\Rightarrow f=\frac{v-u}{t}\) |
|
| 15. |
The lengths of the adjacent sides of a right - angled triangle are xcm, (x-1)cm. If the length of the hypotenuse is \(\sqrt{13}cm\), find the value of x A. 2 B. 3 C. 4 D. 5 Detailed Solution\(x^2+(x-1)^2 = (\sqrt{13})^2 \Rightarrow x^2 + x^2 - 2x + 1 = 13\\2x^2 - 2x - 12 = 0\) dividing through by 2 \(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 \) |
| 16. |
The probability that John and James pass an examination are 3/4 and 3/5 respectively, find the probability of both boys failing the examination A. \(\frac{1}{10}\) B. \(\frac{3}{10}\) C. \(\frac{9}{20}\) D. \(\frac{11}{20}\) Detailed SolutionProb.(John pass)\(=frac{3}{4}\) prob.(John fail) \(=1-frac{3}{4}=\frac{1}{4}\)Prob.(James pass) \(=frac{3}{5}\) Prob.(James fail) \(=1-frac{3}{5}=\frac{2}{5}\) Prob(both boys fail)\(=frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\) |
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| 17. |
In the diagram PQ and MN are straight lines. Find the value of x A. 13o B. 17o C. 28o D. 30o Detailed SolutionThe straight line MN = 180 = 2(x+30o) + 86o =>180 = 2x + 60o + 86o => 180o = 2x + 146o => 2x = 180 - 46 = 34o => x = 34/2 = 17o |
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| 18. |
In the diagram, PQRS is a parallelogram and ∠QRT = 30o. Find x A. 95o B. 100o C. 120o D. 150o Detailed SolutionSince ∠QRS = 180o - 30o = 150o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150o |
|
| 19. |
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee A. 026o B. 045o C. 210o D. 240o Detailed Solution
\theta = cos^{-1}(0.5000)=60^{\circ}\) The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\) |
|
| 20. |
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)] A. \(\frac{\sqrt{2}}{7}cm\) B. \(3\frac{1}{2}cm\) C. 7cm D. \(7\sqrt{2}cm\) Detailed Solution\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\) |