Mathematics (Core)
Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2000
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2000
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
A guide to passing your exams without studying hard, how to pass any test and top your class
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
| # | Question | Ans |
|---|---|---|
| 1. |
Express \(\frac{7}{19}\) as a percentage, correct to one decimal place A. 2.7% B. 3.7% C. 27.1% D. 36.8%
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Detailed Solution\(\frac{7}{19}\) as a percentage will be\(\frac{7}{19}\times \frac{100}{1}\% = 36.8\%\) |
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| 2. |
Express 398753 correct to three significant figures A. 398000 B. 398700 C. 398800 D. 399000
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Detailed Solution398753 \(\approxeq\) 399000 (to 3 s.f) |
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| 3. |
Simplify \(\frac{10}{\sqrt{32}}\) A. \(\frac{5}{4}\sqrt{2}\) B. \(\frac{4}{5}\sqrt{2}\) C. \(\frac{5}{16}\sqrt{2}\) D. \(\frac{16}{5}\sqrt{2}\)
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Detailed Solution\(\frac{10}{\sqrt{32}}=\frac{10}{\sqrt{16\times 2}} = \frac{10}{4\sqrt{2}}\\\frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}=\frac{10\sqrt{2}}{4\times 2}= \frac{5}{4}\sqrt{2}\) |
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| 4. |
Find the missing number in the addition of the following numbers, in base seven A. 3453 B. 5556 C. 6016 D. 13453
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Detailed Solution4321\(_7\) + 1234\(_7\) = 5555\(_7\)Missing number = 12341\(_7\) - 5555\(_7\) = 3453\(_7\) |
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| 5. |
What fraction must 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. \(\frac{1}{2}\) C. \(1\frac{1}{6}\) D. \(1\frac{1}{2}\)
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Detailed Solution\(2\frac{1}{6} + 2\frac{7}{12}\)= \(\frac{13}{6} + \frac{31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12} = \frac{19}{4}\) \(\frac{19}{4} - 3\frac{1}{4}\) = \(\frac{19}{4} - \frac{13}{4}\) = \(\frac{6}{4}\) = \(1\frac{1}{2}\) |
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| 6. |
Simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\) A. \(\frac{80}{243}\) B. \(\frac{1}{64}\) C. \(\frac{25}{6}\) D. \(\frac{15}{4}\)
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Detailed Solution\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\\\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}=\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\\ =\frac{27}{8}\times \frac{10}{9}=\frac{15}{4}\) |
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| 7. |
Which of the following numbers is perfect cube? A. 350 B. 504 C. 950 D. 1728 |
D |
| 8. |
If \(104_x = 68\), find the value of x A. 5 B. 7 C. 8 D. 9
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Detailed Solution\(104_x = 68\\1 \times x^2 + 0 \times x + 4 \times x^0 = 68\\ x^2 = 68 - 4; x^2 = 64\\ x = \sqrt{64}=8\) |
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| 9. |
The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men A. 45 years B. 72 years C. 108 years D. 216 years
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Detailed SolutionGiven that the ages are in the ratio 3: 4: 5.Let the sum of their ages be t. \(\therefore\) The youngest age = \(\frac{3}{12} t\) The eldest age = \(\frac{5}{12} t\) \(\implies \frac{5}{12} t - \frac{3}{12} t = 18\) \(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\) t = 108 years. The sum of their ages = 108 years. |
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| 10. |
Given that \(log_4 x = -3\), find x. A. \(\frac{1}{81}\) B. \(\frac{1}{64}\) C. 64 D. 81
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Detailed Solution\(\log_4 x = -3\)\(x = 4^{-3}\) = \(\frac{1}{64}\) |
Preview displays only 10 out of the 48 Questions