Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2011
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2011
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira A. 0.13 B. 7.49 C. 8.00 D. 13.00
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Detailed SolutionD14.95 = N112.00D1.00 = \(\frac{N112}{D14.95} \times\) D1.00 = 7.49 |
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| 2. |
Solve for x in the equation; \(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3) A. zero B. 1 C. 2 D. 3
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Detailed Solution\(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3)\(\frac{6x}{5} - \frac{3}{5} = \frac{5x}{4} - \frac{3}{4}\) \(\frac{6x}{5} - \frac{5x}{4} = \frac{3}{5} - \frac{3}{4}\) \(\frac{24x - 25x}{20} = \frac{12 - 15}{20}\) \(\frac{-x}{20} = \frac{-3}{20}\) -20x = -60 x = \(\frac{-60}{-20}\) x = 3 |
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| 3. |
Given that cos xo = \(\frac{1}{r}\), express tan x in terms of r A. \(\frac{1}{\sqrt{r}}\) B. \(\sqrt{r}\) C. \(\sqrt{r^2 + 1}\) D. \(\sqrt{r^2 - 1}\)
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Detailed Solutioncos xo = \(\frac{1}{r}\); \(\sqrt{r^2 - 1}\)By Pythagoras r2 = 12 + x2 - 1 x = \(\sqrt{r^2 - 1}\) tan xo = \(\sqrt{r^2 - 1}\) = \(\sqrt{r^2 - 1}\) |
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| 4. |
Solve the equation; 3x - 2y = 7, x + 2y = -3 A. x = 1, y = -2 B. x = 1, y = 3 C. x = -2, y = -1 D. x = 4, y = -3
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Detailed Solution3x - 2y = 7; Solve by elimination methodx + 2y = -3 3x - 2y = 7.....(i) -3x + 6y = -9.....(ii) -8y = 16 y = \(\frac{16}{-8} = -2\) Put y = -2 into equation (i); 3x - 2y = 7 3x - 2(-2) = 7 3x + 4 = 7 3x = 7 - 4 3x = 3 x = \(\frac{3}{3}\) = 1 |
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| 5. |
If a number is chosen at random from the set {x: 4 \(\leq x \leq 15\)}. Find the probability that it is a multiple of 3 or a multiple of 4 A. \(\frac{1}{12}\) B. \(\frac{5}{12}\) C. \(\frac{7}{12}\) D. \(\frac{11}{12}\)
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Detailed Solutionset = 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15multiple of 3 = 6, 9, 12, 15 multiple of 4 = 4, 8, 12 Prob(multiple of 3 or 4) = Prob(multiple of 3) + pro(multiple of 4) = \(\frac{4}{12} + \frac{3}{12} = \frac{7}{12}\) |
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| 6. |
One of the factors of (mn - nq - n2 + mq) is (m - n). The other factor is? A. (n - q) B. (q - n) C. (n + q) D. (q - m)
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Detailed Solutionmn - nq - n\(^2\) + mqmn - n\(^2\) - nq + mq (mn - n\(^2\)) (mq - nq) n(m - n) + q(m - n) = (n +q) (m - n) |
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| 7. |
A cylindrical container has a base radius of 14cm and height 18cm. How many litres of liquid can it hold? correct to the nearest litre [Take \(\pi = \frac{22}{7}\)] A. 11 B. 14 C. 16 D. 18
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Detailed SolutionVolume of cylinder = \(\pi r^2 h\)= \(\frac{22}{7} \times 14^2\) x 14 x 18 = 22 x 28 x 18 = 11088cm3 1000cm3 = \(\frac{1}{1000}\) x 11088cm3 = 11.088 = 11 |
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| 8. |
A regular polygon of n sides has each exterior angle to 45o. Find the value of n A. 6 B. 8 C. 12 D. 15
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Detailed SolutionEach interior angle = \(\frac{360^o}{n}\)45o = \(\frac{360^o}{n}\) 45on = 360o n = \(\frac{360^o}{45}\) = 8 |
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| 9. |
Esther was facing S 20° W. She turned 90° in the clock wise direction. What direction is she facing? A. N 70o W B. S 70o W C. N 20o W D. S 20o E
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Detailed SolutionThe bearing S 20° W is equivalent to a bearing of 200°.Moving clockwise 90° ⇒ 200° + 90° = 290° This is equivalent to N 70° W. |
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| 10. |
The cross section section of a uniform prism is a right-angled triangle with sides 3cm. 4cm and 5cm. If its length is 10cm. Calculate the total surface area A. 142cm2 B. 132cm2 C. 122cm2 D. 112cm2
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Detailed SolutionA prism has 3 rectangular faces and 2 triangular faces and 2 rectangular faces = 10(3 + 4 + 5) = 120Area of triangular faces = \(\sqrt{s(s - a) (s - b) (s - c)}\) where s = \(\frac{a + b + c}{2}\) = \(\frac{3 + 4 + 5}{2}\) = \(\frac{12}{2}\) = 6 Area of \(\bigtriangleup\) = \(\sqrt{6(6 - 30(6 - 4)(6 - 5)}\) = \(\sqrt{6 \times 3 \times 2 \times 1}\) = 6 Area of triangle faces = 2 x 6 = 12cm2 Total surface area = Area of rectangular face + Area of \(\bigtriangleup\) = 120 + 12 = 132cm2 & |
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