Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2021
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2021
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Tips to make the best study space and quality time for studies are essential.
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There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
| # | Question | Ans |
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| 1. |
Correct 0.007985 to three significant figures. A. 0.0109 B. 0.0800 C. 0.00799 D. 0.008
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Detailed Solution
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| 2. |
Simplify: (11\(_{two}\))\(^2\) A. 1001\(_2\) B. 1101\(_2\) C. 101\(_2\) D. 10001\(_2\)
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Detailed Solution(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\) = \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\) = \({2 + 1} \times ({2 + 1})\) = 3 \(\times\) 3 = 9\(_{10}\) or 1001\(_2\) from |
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| 3. |
Solve: 2\(^{√2x + 1}\) = 32 A. 13 B. 24 C. 12 D. 11
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Detailed Solution2\(^{√2x + 1}\) = 322\(^{√2x + 1}\) = 2\(^5\) √2x + 1 = 5 square both sides 2x + 1 = 5\(^2\) 2x + 1 = 25 2x = 25 - 1 2x = 24 x = \(\frac{24}{2}\) x = 12 |
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| 4. |
If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n. A. 3m + n B. m + 3n C. 4mn D. 3mn
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Detailed Solutionlog\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m and log\(_{10}\) 3 = n : log\(_{10}\) 24 = 3m + n |
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| 5. |
Find the 5th term of the sequence 2,5,10,17....? A. 22 B. 24 C. 36 D. 26
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Detailed SolutionSimply add odd number starting from '3' to the next number2 2 + 3 = 5 5 + 5 = 10 10 + 7 = 17 17 + 9 = 26 The fifth term = 26 |
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| 6. |
If p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q. A. 0 B. -3, -2, -1, 0 and 1 C. -2, -1 and 0 D. -1, 0 and 1
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Detailed Solutionp = {-3<x<1} = {-2,-1 and 0}Q = {-1<x<3} = {0,1 and 2} P n Q = {0} or {-1<x<1} |
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| 7. |
Factorize 6pq-3rs-3ps+6qr A. 3(r -p)(2q + s) B. 3(p + r)( 2q - 2q - s) C. 3(2q - s)(p + r) D. 3(r - p)(s - 2q)
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Detailed Solution6pq-3rs-3ps+6qr = 3 (2pq - rs - ps + 2qr)= 3 ({2pq + 2qr} {-ps - rs}) = 3 (2q{ p + r} -s{p + r}) = 3 ({2q - s}{p + r}) |
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| 8. |
What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)? A. \(\frac{1}{3}\) B. 1\(\frac{1}{2}\) C. 1\(\frac{1}{6}\) D. \(\frac{1}{2}\)
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Detailed SolutionThe sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)= \(\frac{13}{6}\) + \(\frac{31}{12}\) = \(\frac{13 \times 2 + 31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12}\) What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\) \(\frac{57}{12}\) - y = 3\(\frac{1}{4}\) : y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\) y = \(\frac{57 - 3 \times 13}{12}\) = \(\frac{57 - 39}{12}\) y = \(\frac{18}{12}\) y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\) |
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| 9. |
Mensah is 5 years old and joyce is thrice as old as mensah. In how many years will joyce be twice as old as Mensah? A. 3 years B. 10 years C. 5 years D. 15 years
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Detailed SolutionMensah’s age is 5. Thus,Joyce’s age is 15 (5*3=15) The difference between their ages is 10 (15–5=10) As we ought to find how many years Joyce’s age will be twice of Mensah’s age, we should write down the following : 15+X=2*(5+X) 15+X=10+2X lets add (-10-X) to both sides of the equation and 15+X-10-X = 10+2X-10-X 5=X —-> X=5 After 5 years Joyce’s age will be 20 (15+5=20) After 5 years Mensah’s age will be 10 (5+5=10) After 5 years Joyce will be twice as old as Mensah (10*2=20) |
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| 10. |
If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x. A. -4 B. 4 C. 1 D. -1
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Detailed Solution16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\)= 2\(^4\) * 2\(^{(x + 1)}\) = 2\(^{2x}\) * 2\(^{3(1 - x)}\) --> 4 + x + 1 = 2x + 3 - 3x collect like terms --> x - 2x + 3x = 3 - 1 - 4 --> 2x = -2 --> x = -1 |
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