Mathematics (Core)
Paper 1 | Objectives | 46 Questions
JAMB Exam
Year: 1995
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 46 Questions
JAMB Exam
Year: 1995
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
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| 1. |
Calculate \(3310_5 - 1442_5\) A. 13135 B. 21315 C. 43025 D. 11035
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Detailed Solution\(3310_{5} - 1442_{5} = 1313_{5}\)You can either do a direct subtraction (which is easier), or convert both numbers in base 10, subtract and convert back to base 5 (if you're not sure). |
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| 2. |
Convert 3.1415926 to 5 decimal places A. 3.14160 B. 3.14159 C. 0.31415 D. 3.14200
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Detailed Solution3.1415926 \(\approxeq\) 3.14159 (to 5 d.p) |
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| 3. |
The length of a notebook 15cm, was measured as 16.8cm. Calculate the percentage error to 2 significant figures. A. 12.00% B. 11.00% C. 10.71% D. 0.12%
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Detailed SolutionError = (16.8cm - 15cm = 1.8cm\)\(\text{% error} = \frac{1.8}{15} \times 100% = \frac{180}{15}\) = \(12.00%\) |
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| 4. |
A worker's present salary is N24,000 per annum. His annual increment is 10% of his basic salary. What would be his annual salary at the beginning of the third year? A. N28,800 B. N29,040 C. N31,200 D. N31,944
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Detailed SolutionPresent salary = N24,000.Increment after first year = \(\frac{10}{100} \times N24,000 = N2,400\) Salary at the beginning of second year = N(24,000 + 2,400) = N26,400. Increment after second year = \(\frac{10}{100} \times N26,400 = N2,640\) Salary at the beginning of third year = N(26,400 + 2,640) = N29,040. \(\therefore\) His salary at the beginning of the third year = N29,040. |
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| 5. |
Express the product of 0.0014 and 0.011 in standard form A. 1.54 x 10-2 B. 1.54 x 10-3 C. 1.54 x 10-4 D. 1.54 x 10-5
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Detailed Solution\(0.0014 = 1.4 \times 10^{-3}\)\(0.011 = 1.1 \times 10^{-2}\) \(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\) = \(1.54 \times 10^{-5}\) |
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| 6. |
Evaluate \(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\) A. 27 B. 1 C. \(\frac{1}{3}\) D. \(\frac{1}{8}\)
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Detailed Solution\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)= \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\) = \(\frac{3^{3} - 3^{1}}{24}\) = \(\frac{27 - 3}{24} = 1\) |
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| 7. |
Simplify \(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\) A. \(\frac{1}{3}\) B. 9 C. 16cm D. 3
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Detailed Solution\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\) |
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| 8. |
Four members of a school first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams? A. 25 B. 21 C. 16 D. 3
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Detailed SolutionNumber of people playing both rugby and cricket = 4Number that play cricket only = 11 - 4 = 7 Number that play rugby only = 14 - 4 = 10. Number that play for at least one of the teams = 4 + 7 + 10 = 21. |
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| 9. |
If S = (x : x\(^2\) = 9, x > 4), then S is equal to A. 4 B. {0} C. \(\emptyset\) D. {\(\emptyset\)} |
C |
| 10. |
If x - 1 and x + 1 are both factors of the equation x\(^3\) + px\(^2\) + qx + 6 = 0, evaluate p and q A. -6, -1 B. 6, 1 C. 1, -1 D. 6, -6
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Detailed Solution\(x^{3} + px^{2} + qx + 6 = 0\)f(x - 1) = 0; f(1) = 0. \(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\) f(x + 1) = 0; f(-1) = 0. \((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\) Subtract (2) from (1). \(2q = -2 \implies q = -1\) \(p - (-1) = -5 \implies p = -5 - 1 = -6\) \((p, q) = (-6, -1)\) |
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