Mathematics (Core)
Paper 1 | Objectives | 45 Questions
WASSCE/WAEC MAY/JUNE
Year: 2015
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 45 Questions
WASSCE/WAEC MAY/JUNE
Year: 2015
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 {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x A. 2 B. 5 C. 15 D. 18 |
C |
| 2. |
The sum 110112, 11112 and 10m10n02. Find the value of m and n. A. m = 0, n = 0 B. m = 1, n = 0 C. m = 0, n = 1 D. m = 1, n = 1 |
C |
| 3. |
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in Naira? A. N250,000.00 B. N200,000.00 C. N150,000.00 D. N100,000.00
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Detailed SolutionN1.00 = $0.075N X = $15,000 X = \(\frac{1.00 \times 15000}{0.075}\) = N200,000.00 |
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| 4. |
If \(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\), find the value of x. A. 1 B. \(\frac{1}{2}\) C. -\(\frac{1}{2}\) D. -1
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Detailed Solution\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\)\(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\) \(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\) \(3^{(3x + 1 - x - 4x)} = 3^0\) \(3^{(1 - 2x)} = 3^0\) since the bases are equal, 1 - 2x = 0 - 2x = -1 x = \(\frac{1}{2}\) |
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| 5. |
Find the 7th term of the sequence: 2, 5, 10, 17, 6,... A. 37 B. 48 C. 50 D. 63 |
C |
| 6. |
Given that logx 64 = 3, evaluate x log8 A. 6 B. 9 C. 12 D. 24
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Detailed SolutionIf logx 64 = 3, then 64 = x343 = x3 Since the indices are equal, x = 4 Hence, x log28 = log28x = log284 = log2(23)4 = log2212 = 1 log2 = 1(1) = 12 |
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| 7. |
If 2n = y, Find 2\(^{(2 + \frac{n}{3})}\) A. 4y\(^\frac{1}{3}\) B. 4y\(^-3\) C. 2y\(^\frac{1}{3}\) D. 2y\(^-3\)
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Detailed SolutionIf 2n = y,then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\) = 4 x (2n)\(^{\frac{1}{3}}\) But y = 2n, hence 2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\) = 4y\(^\frac{1}{3}\) |
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| 8. |
Factorize completely: 6ax - 12by - 9ay + 8bx A. (2a - 3b)(4x + 3y) B. (3a + 4b)(2x - 3y) C. (3a - 4b)(2x + 3y) D. (2a + 3b)(4x -3y)
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Detailed Solution6ax - 12by - 9ay + 8bx= 6ax - 9ay + 8bx - 12by = 3a(2x - 3y) + 4b(2x - 3y) = (3a + 4b)(2x - 3y) |
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| 9. |
Find the equation whose roots are \(\frac{3}{4}\) and -4 A. 4x2 - 13x + 12 = 0 B. 4x2 - 13x - 12 = 0 C. 4x2 + 13x - 12 = 0 D. 4x2 + 13x + 12 = 0
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Detailed SolutionLet x = \(\frac{3}{4}\) or x = -4i.e. 4x = 3 or x = -4 (4x - 3)(x + 4) = 0 therefore, 4x2 + 13x - 12 = 0 |
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