Year :
2001
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
1 - 10 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 1. |
Which of the following correctly expresses 48 as a product of prime factors? A. 3 x 4 x 4 B. 2 x 3 x 8 C. 2 x 2 x 2 x 3 x 4 D. 2 x 2 x 2 x 2 x 3 Detailed Solution48 = 2 x 2 x 2 x 2 x 3 |
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| 2. |
Evaluate \((20_{three})^2 - (11_{three})^2\) in base three A. 101 B. 121 C. 202 D. 2020 Detailed Solution\((20_{three})^2 - (11_{three})^2\)= \((20_{3} - 11_{3})(20_{3} + 11_{3})\) = \((2_{3})(101_{3})\) = \(202_{3}\) |
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| 3. |
Evaluate \(log_{10}5 + log_{10}20\) A. 2 B. 3 C. 4 D. 5 Detailed Solution\(log_{10}5 + log_{10}20\\log_{10}[5\times 20]\\ log_{10}100\\ log_{10}10^2\\ 2log_{10}10=2\) |
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| 4. |
In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects? A. 15 B. 30 C. 35 D. 45 Detailed SolutionLet c = no of students that offered both subjects\(\therefore\) No of students offering Economics = 65 - c No of students offering Geography = 50 - c 65 - c + c + 50 - c = 80 115 - c = 80 c = 35 35 students offer both Economics and Geography. |
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| 5. |
If N varies directly as M and N = 8 when M = 20 find M when N = 7 A. 13 B. 15 C. \(17\frac{1}{2}\) D. \(18\frac{1}{2}\) Detailed SolutionN ∝ M∴N = kM where k is a constant N = 8 when M = 20 ∴ 8 = k x 20 \(k = \frac{8}{20} = \frac{2}{5}\\ ∴ N = \frac{2}{5}M\\ If N = 7\\ ∴ 7 = \frac{2}{5}M\\ M = \frac{35}{2} = 17\frac{1}{2}\) |
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| 6. |
Express 0.0462 in standard form A. 0.460 x 10-1 B. 0.462 x 10-2 C. 4.62 x 10-1 D. 4.62 x 10-2 |
D |
| 7. |
In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether? A. 81 B. 72 C. 54 D. 45 Detailed SolutionRatio of good ones to bad ones is 5:4; If 36 is bad;∴ the good ones = \(\frac{5\times 36}{4}=45\) oranges. The total number of oranges is 36 + 45 = 81. |
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| 8. |
A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate A. 95% B. 47.5% C. 5.26% D. 5% Detailed SolutionThe percentage error is \(\frac{error}{actual}\times \frac{100}{1}\%\\=\frac{200-190}{200}\times \frac{100}{1}\\ =\frac{10}{200}\times \frac{100}{1}\% = 5\%\) |
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| 9. |
Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\) A. \(\frac{8}{25}\) B. \(\frac{12}{25}\) C. \(3\frac{3}{5}\) D. \(4\frac{1}{8}\) Detailed Solution\(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 ÷ \left(1\frac{1}{2}\right)^{-1}\\\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5} \) |
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| 10. |
The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\) A. 3rd B. 4th C. 5th D. 6th Detailed Solution\(T_{n} = 2^{2n - 1}\)\(2^{2n - 1} = 2^9\) \(2n - 1 = 9 \implies 2n = 9 + 1\) \(2n = 10 \implies n = 5\) The 5th term = 2\(^9\) |
| 1. |
Which of the following correctly expresses 48 as a product of prime factors? A. 3 x 4 x 4 B. 2 x 3 x 8 C. 2 x 2 x 2 x 3 x 4 D. 2 x 2 x 2 x 2 x 3 Detailed Solution48 = 2 x 2 x 2 x 2 x 3 |
|
| 2. |
Evaluate \((20_{three})^2 - (11_{three})^2\) in base three A. 101 B. 121 C. 202 D. 2020 Detailed Solution\((20_{three})^2 - (11_{three})^2\)= \((20_{3} - 11_{3})(20_{3} + 11_{3})\) = \((2_{3})(101_{3})\) = \(202_{3}\) |
|
| 3. |
Evaluate \(log_{10}5 + log_{10}20\) A. 2 B. 3 C. 4 D. 5 Detailed Solution\(log_{10}5 + log_{10}20\\log_{10}[5\times 20]\\ log_{10}100\\ log_{10}10^2\\ 2log_{10}10=2\) |
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| 4. |
In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects? A. 15 B. 30 C. 35 D. 45 Detailed SolutionLet c = no of students that offered both subjects\(\therefore\) No of students offering Economics = 65 - c No of students offering Geography = 50 - c 65 - c + c + 50 - c = 80 115 - c = 80 c = 35 35 students offer both Economics and Geography. |
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| 5. |
If N varies directly as M and N = 8 when M = 20 find M when N = 7 A. 13 B. 15 C. \(17\frac{1}{2}\) D. \(18\frac{1}{2}\) Detailed SolutionN ∝ M∴N = kM where k is a constant N = 8 when M = 20 ∴ 8 = k x 20 \(k = \frac{8}{20} = \frac{2}{5}\\ ∴ N = \frac{2}{5}M\\ If N = 7\\ ∴ 7 = \frac{2}{5}M\\ M = \frac{35}{2} = 17\frac{1}{2}\) |
| 6. |
Express 0.0462 in standard form A. 0.460 x 10-1 B. 0.462 x 10-2 C. 4.62 x 10-1 D. 4.62 x 10-2 |
D |
| 7. |
In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether? A. 81 B. 72 C. 54 D. 45 Detailed SolutionRatio of good ones to bad ones is 5:4; If 36 is bad;∴ the good ones = \(\frac{5\times 36}{4}=45\) oranges. The total number of oranges is 36 + 45 = 81. |
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| 8. |
A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate A. 95% B. 47.5% C. 5.26% D. 5% Detailed SolutionThe percentage error is \(\frac{error}{actual}\times \frac{100}{1}\%\\=\frac{200-190}{200}\times \frac{100}{1}\\ =\frac{10}{200}\times \frac{100}{1}\% = 5\%\) |
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| 9. |
Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\) A. \(\frac{8}{25}\) B. \(\frac{12}{25}\) C. \(3\frac{3}{5}\) D. \(4\frac{1}{8}\) Detailed Solution\(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 ÷ \left(1\frac{1}{2}\right)^{-1}\\\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5} \) |
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| 10. |
The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\) A. 3rd B. 4th C. 5th D. 6th Detailed Solution\(T_{n} = 2^{2n - 1}\)\(2^{2n - 1} = 2^9\) \(2n - 1 = 9 \implies 2n = 9 + 1\) \(2n = 10 \implies n = 5\) The 5th term = 2\(^9\) |