Year :
2003
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
31 - 40 of 46 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
In a ∆ XYZ, /YZ/ = 6cm YXZ = 60o and XYZ is a right angle. Calculate /XZ/in cm, leaving your answer in surd form A. 2√3 B. 4√3 C. 6√3 D. 12√3 Detailed Solution
 sin 60^o = \frac{|YZ|}{|XZ|}=\frac{6}{P}\\ P sin 60^o = 6\\ P = \frac{6}{sin60^o}\\ =\frac{6}{\sqrt{\frac{3}{2}}}=4\sqrt{3}\) |
|
| 32. |
If \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula. A. \(R = \frac{3Qt}{P^2 - Q}\) B. \(R = \frac{P^2 – 3t}{Q+1}\) C. \(R = \frac{P^2 + 3t}{Q - 1}\) D. \(R = \frac{P^2-3Qt}{Q}\) |
D |
| 33. |
 From the Venn diagram below, how many elements are in P∩Q? A. 1 B. 2 C. 4 D. 6 Detailed SolutionP \(\cap\) Q = {f, e} = 2 |
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| 34. |
From the Venn Diagram below, find Q' ∩ R. A. (e) B. (c, h) C. (c, g, h) D. (c, e, g, h) Detailed SolutionQ' ∩ RQ' = U - Q Q' = {a, b, c, d, g, h, i} R = {c, e, h, g} Q' ∩ R = {c, h, g} |
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| 35. |
The square root of a number is 2k. What is half of the number A. \(\sqrt{\frac{k}{2}}\) B. \(\sqrt{k}\) C. \(\frac{1}{2}k^2\) D. 2k2 Detailed SolutionLet the number be x.\(\sqrt{x} = 2k \implies x = (2k)^2\) = \(4k^2\) \(\frac{1}{2} \times 4k^2 = 2k^2\) |
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| 36. |
Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r? A. p varies as the square of r B. p varies as the square root of r C. p varies inversely as the square of r D. p varies inversely as r Detailed Solution\(p \propto q^2\)\(q \propto \frac{1}{\sqrt{r}\) \(p = kq^2\) \(q = \frac{c}{\sqrt{r}}\) where c and k are constants. \(q^2 = \frac{c^2}{r}\) \(p = \frac{kc^2}{r}\) If k and c are constants, then kc\(^2\) is also a constant, say z. \(p = \frac{z}{r}\) p varies inversely as r. |
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| 37. |
The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests A. 1-(x-y)+xy B. 1-(x+y)-xy C. 1-(x+y)+xy D. 1 - (x - y) + x Detailed SolutionProb (passing English) = xProb (passing Maths) = Y Prob (failing English) = 1 - x Prob (failing Maths) = 1 - y Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y) =1 - y - x + xy =1 - (y + x) + xy |
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| 38. |
The locus of points equidistant from two intersecting straight lines PQ and PR is A. a circle centre P radius Q. B. a circle centre P radius PR C. the point of intersection of the perpendicular bisectors of PQ and PR D. the bisector of angle QPR |
C |
| 39. |
Find the equation whose roots are \(-\frac{2}{3}\) and 3 A. 3x2+11x-6=0 B. 3x2+7x+6=0 C. 3x2-11x-6=0 D. 3x2-7x-6=0 Detailed Solution\(x = -\frac{2}{3} \implies x + \frac{2}{3} = 0\)\(x = 3 \implies x - 3 = 0\) \(\implies (x - 3)(x + \frac{2}{3}) = 0\) \(x^2 - 3x + \frac{2}{3}x - 2 = 0\) \(x^2 - \frac{7}{3}x - 2 = 0\) \(3x^2 - 7x - 6 = 0\) |
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| 40. |
Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form A. \(\frac{\sqrt{2}-1}{2}\) B. \(\frac{\sqrt{3}-\sqrt{2}}{4}\) C. \(\frac{\sqrt{6}-\sqrt{2}}{2}\) D. \(\frac{\sqrt{6}-\sqrt{2}}{4}\) Detailed Solution\(cos45^o \times cos30^o - sin45^o \times sin30^o\\\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\\ \frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\) |
| 31. |
In a ∆ XYZ, /YZ/ = 6cm YXZ = 60o and XYZ is a right angle. Calculate /XZ/in cm, leaving your answer in surd form A. 2√3 B. 4√3 C. 6√3 D. 12√3 Detailed Solution
sin 60^o = \frac{|YZ|}{|XZ|}=\frac{6}{P}\\ P sin 60^o = 6\\ P = \frac{6}{sin60^o}\\ =\frac{6}{\sqrt{\frac{3}{2}}}=4\sqrt{3}\) |
|
| 32. |
If \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula. A. \(R = \frac{3Qt}{P^2 - Q}\) B. \(R = \frac{P^2 – 3t}{Q+1}\) C. \(R = \frac{P^2 + 3t}{Q - 1}\) D. \(R = \frac{P^2-3Qt}{Q}\) |
D |
| 33. |
From the Venn diagram below, how many elements are in P∩Q? A. 1 B. 2 C. 4 D. 6 Detailed SolutionP \(\cap\) Q = {f, e} = 2 |
|
| 34. |
From the Venn Diagram below, find Q' ∩ R. A. (e) B. (c, h) C. (c, g, h) D. (c, e, g, h) Detailed SolutionQ' ∩ RQ' = U - Q Q' = {a, b, c, d, g, h, i} R = {c, e, h, g} Q' ∩ R = {c, h, g} |
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| 35. |
The square root of a number is 2k. What is half of the number A. \(\sqrt{\frac{k}{2}}\) B. \(\sqrt{k}\) C. \(\frac{1}{2}k^2\) D. 2k2 Detailed SolutionLet the number be x.\(\sqrt{x} = 2k \implies x = (2k)^2\) = \(4k^2\) \(\frac{1}{2} \times 4k^2 = 2k^2\) |
| 36. |
Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r? A. p varies as the square of r B. p varies as the square root of r C. p varies inversely as the square of r D. p varies inversely as r Detailed Solution\(p \propto q^2\)\(q \propto \frac{1}{\sqrt{r}\) \(p = kq^2\) \(q = \frac{c}{\sqrt{r}}\) where c and k are constants. \(q^2 = \frac{c^2}{r}\) \(p = \frac{kc^2}{r}\) If k and c are constants, then kc\(^2\) is also a constant, say z. \(p = \frac{z}{r}\) p varies inversely as r. |
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| 37. |
The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests A. 1-(x-y)+xy B. 1-(x+y)-xy C. 1-(x+y)+xy D. 1 - (x - y) + x Detailed SolutionProb (passing English) = xProb (passing Maths) = Y Prob (failing English) = 1 - x Prob (failing Maths) = 1 - y Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y) =1 - y - x + xy =1 - (y + x) + xy |
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| 38. |
The locus of points equidistant from two intersecting straight lines PQ and PR is A. a circle centre P radius Q. B. a circle centre P radius PR C. the point of intersection of the perpendicular bisectors of PQ and PR D. the bisector of angle QPR |
C |
| 39. |
Find the equation whose roots are \(-\frac{2}{3}\) and 3 A. 3x2+11x-6=0 B. 3x2+7x+6=0 C. 3x2-11x-6=0 D. 3x2-7x-6=0 Detailed Solution\(x = -\frac{2}{3} \implies x + \frac{2}{3} = 0\)\(x = 3 \implies x - 3 = 0\) \(\implies (x - 3)(x + \frac{2}{3}) = 0\) \(x^2 - 3x + \frac{2}{3}x - 2 = 0\) \(x^2 - \frac{7}{3}x - 2 = 0\) \(3x^2 - 7x - 6 = 0\) |
|
| 40. |
Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form A. \(\frac{\sqrt{2}-1}{2}\) B. \(\frac{\sqrt{3}-\sqrt{2}}{4}\) C. \(\frac{\sqrt{6}-\sqrt{2}}{2}\) D. \(\frac{\sqrt{6}-\sqrt{2}}{4}\) Detailed Solution\(cos45^o \times cos30^o - sin45^o \times sin30^o\\\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\\ \frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\) |