Year :
1982
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
What is the least possible value of \(\frac{9}{1 + 2x^2}\) if 0 \(\geq\) x \(\geq\) 2? A. 9 B. 5 C. 1 D. 2 Detailed Solution0 \(\geq\) x \(\geq\) 2 \(\to\) 0, 1, 2If x = 0, \(\frac{9}{1 + 2x^2}\) \(\frac{9}{1 + 2(0)^2}\) = \(\frac{9}{1}\) = 3 If x = 2, \(\frac{9}{1 + 2(1)^2}\) = \(\frac{9}{3}\) = 3 If x = 2, \(\frac{9}{1 + 2(2)^2}\) = \(\frac{9}{9}\) = 1 The least value of \(\frac{9}{1 + 2x^2}\) is 1 when x = 2 |
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| 12. |
A man bought wrist watch for N150 but was only able to sell it for N120. Find the loss per cent on the transaction A. 25% B. 11% C. 20% D. 80% E. 30% Detailed SolutionLoss% = \(\frac{\text{actual loss}}{\text{cost price}}\) x 100%\(\frac{N150 - N120}{N150} \times 100%\) \(\frac{N30}{N150} \times 100%\) = 20% |
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| 13. |
Find the median of the set set of numbers 110, 116, 113, 119, 118, 127, 118, 117, 113 A. 117.5 B. 118 C. 117 D. 116 E. 113 Detailed SolutionRe-arrange in ascending order: 110, 113, 113, 116, |117|, 118, 119, 127a= 117 |
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| 14. |
If f(x - 2) = 3x2 + 4x + 1. Find the area of the sector A. 8 B. 40 C. 7 D. 24 E. 32 Detailed Solutionf(x - 2) = 3x2 + 4x + 1f(1) will be f(3 - 2) When x = 3, f(1) = 3(3)2 + 4 x 3 = 1 27 + 12 +1 = 40 |
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| 15. |
A sector of a circle is bounded by two radii 7cm long and an arc length 6cm. Find the area of the sector. A. 42cm2 B. 3cm2 C. 21cm2 D. 24cm2 E. 12cm2 Detailed SolutionLength of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6\(\theta\) x 2\(\pi\)r = 360 x 6 \(\theta\) = \(\frac{360 \times 6}{2\pi r}\) Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r2 \(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r2 = r = 3 x 7 = 21cm2 |
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| 16. |
Given that p:q = \(\frac{1}{3}\):\(\frac{1}{2}\) and q:r = \(\frac{2}{5}\), find p:r A. 4:105 B. 7:15 C. 20:21 D. 2:35 E. 3:20 Detailed Solution\(p : q = \frac{1}{3} : \frac{1}{2}\)\(\frac{p}{q} = \frac{2}{3}\) \(2q = 3p ... (1)\) \(q : r = \frac{2}{5} : \frac{4}{7}\) \(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\) \(10q = 7r ... (2)\) Eliminating q, we have (1) : \(2q = 3p \) \(10q = 15p\) \(\implies 15p = 7r\) \(\therefore \frac{p}{r} = \frac{7}{15}\) \(p : r = 7 : 15\) |
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| 17. |
If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\) A. \(\frac{(m^2 + n^2)}{m + n}\) B. \(\frac{(m^2 + n^2 + 2mn)}{4mn}\) C. \(\frac{2(m^2 + n^2 + mn)}{m + n}\) D. \(\frac{(m^2 + n^2 + mn)}{m + n}\) Detailed Solution\((m + n)^{2} = (m - n)^{2} + x^{2}\)\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\) \(x^{2} = 4mn\) \(x = \sqrt{4mn} = 2\sqrt{mn}\) 1 + tan2\(\theta\) = sec2\(\theta\) = \(\frac{1}{cos^2\theta}\) \(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\) \(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\) \(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\) = \(\frac{(m^2 + n^2 + 2mn)}{4mn}\) |
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| 18. |
The currency used in a country is called 'Matimalik'(M) and is of base seven. A lady in that country bought 4 bags of rice at M56 per bag and and 3 tins of milk at M4 per tin. What is the total cost of the item she bought? A. M2467 B. M2427 C. M2367 D. M3417 E. M3387 Detailed Solution4 bags of rice - M 56 each3 tins of milk - M 4 each \(M 56 \times 4 = M 323\) \(M 4 \times 3 = M 15\) \(M (323 + 15) = M 341\) |
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| 19. |
Solve for x, If \(\frac{\frac{2}{x}}{\frac{p^2 + p^2}{p^2 + p^2}}\) = m A. \(\frac{4pq}{m(p + q)}\) B. \(\frac{2p^2q^2}{m(q^2 + p^2)}\) C. \(\frac{2pq}{m(q^2 + p^2)}\) D. \(\frac{2p^2q^2}{m(p^2)}\) Detailed Solution\(\frac{1}{p^2}\) + \(\frac{1}{q^2}\) = \(\frac{q^2 + p^2}{p^2 + q^2}\)\(\frac{\frac{2}{x}}{\frac{p^2 + p^2}{p^2 + p^2}}\) m = \(\frac{2p^2q^2}{x(p^2 + q^2)}\) = m2p2q2 = m x (p2 + q2) x = \(\frac{2p^2q^2}{m(q^2 + p^2)}\) |
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| 20. |
Express 37.05 x 0.0042 in standard form A. 15.561 x 102 B. 1.5561 x 10-4 C. 1.556 x 10-1 D. 1.5561 x 101 E. 1.55 x 101 Detailed Solution37.05 x 0.0042 in standard form\(\begin{array}{c|c}No. & log \\\hline 37.05 & 1.5688\\ 0.0042 & 3.6232 \\ \hline & 1.1920\end{array}\) = 0.1556 = 1.556 x 10-1 |
| 11. |
What is the least possible value of \(\frac{9}{1 + 2x^2}\) if 0 \(\geq\) x \(\geq\) 2? A. 9 B. 5 C. 1 D. 2 Detailed Solution0 \(\geq\) x \(\geq\) 2 \(\to\) 0, 1, 2If x = 0, \(\frac{9}{1 + 2x^2}\) \(\frac{9}{1 + 2(0)^2}\) = \(\frac{9}{1}\) = 3 If x = 2, \(\frac{9}{1 + 2(1)^2}\) = \(\frac{9}{3}\) = 3 If x = 2, \(\frac{9}{1 + 2(2)^2}\) = \(\frac{9}{9}\) = 1 The least value of \(\frac{9}{1 + 2x^2}\) is 1 when x = 2 |
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| 12. |
A man bought wrist watch for N150 but was only able to sell it for N120. Find the loss per cent on the transaction A. 25% B. 11% C. 20% D. 80% E. 30% Detailed SolutionLoss% = \(\frac{\text{actual loss}}{\text{cost price}}\) x 100%\(\frac{N150 - N120}{N150} \times 100%\) \(\frac{N30}{N150} \times 100%\) = 20% |
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| 13. |
Find the median of the set set of numbers 110, 116, 113, 119, 118, 127, 118, 117, 113 A. 117.5 B. 118 C. 117 D. 116 E. 113 Detailed SolutionRe-arrange in ascending order: 110, 113, 113, 116, |117|, 118, 119, 127a= 117 |
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| 14. |
If f(x - 2) = 3x2 + 4x + 1. Find the area of the sector A. 8 B. 40 C. 7 D. 24 E. 32 Detailed Solutionf(x - 2) = 3x2 + 4x + 1f(1) will be f(3 - 2) When x = 3, f(1) = 3(3)2 + 4 x 3 = 1 27 + 12 +1 = 40 |
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| 15. |
A sector of a circle is bounded by two radii 7cm long and an arc length 6cm. Find the area of the sector. A. 42cm2 B. 3cm2 C. 21cm2 D. 24cm2 E. 12cm2 Detailed SolutionLength of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6\(\theta\) x 2\(\pi\)r = 360 x 6 \(\theta\) = \(\frac{360 \times 6}{2\pi r}\) Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r2 \(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r2 = r = 3 x 7 = 21cm2 |
| 16. |
Given that p:q = \(\frac{1}{3}\):\(\frac{1}{2}\) and q:r = \(\frac{2}{5}\), find p:r A. 4:105 B. 7:15 C. 20:21 D. 2:35 E. 3:20 Detailed Solution\(p : q = \frac{1}{3} : \frac{1}{2}\)\(\frac{p}{q} = \frac{2}{3}\) \(2q = 3p ... (1)\) \(q : r = \frac{2}{5} : \frac{4}{7}\) \(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\) \(10q = 7r ... (2)\) Eliminating q, we have (1) : \(2q = 3p \) \(10q = 15p\) \(\implies 15p = 7r\) \(\therefore \frac{p}{r} = \frac{7}{15}\) \(p : r = 7 : 15\) |
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| 17. |
If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\) A. \(\frac{(m^2 + n^2)}{m + n}\) B. \(\frac{(m^2 + n^2 + 2mn)}{4mn}\) C. \(\frac{2(m^2 + n^2 + mn)}{m + n}\) D. \(\frac{(m^2 + n^2 + mn)}{m + n}\) Detailed Solution\((m + n)^{2} = (m - n)^{2} + x^{2}\)\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\) \(x^{2} = 4mn\) \(x = \sqrt{4mn} = 2\sqrt{mn}\) 1 + tan2\(\theta\) = sec2\(\theta\) = \(\frac{1}{cos^2\theta}\) \(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\) \(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\) \(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\) = \(\frac{(m^2 + n^2 + 2mn)}{4mn}\) |
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| 18. |
The currency used in a country is called 'Matimalik'(M) and is of base seven. A lady in that country bought 4 bags of rice at M56 per bag and and 3 tins of milk at M4 per tin. What is the total cost of the item she bought? A. M2467 B. M2427 C. M2367 D. M3417 E. M3387 Detailed Solution4 bags of rice - M 56 each3 tins of milk - M 4 each \(M 56 \times 4 = M 323\) \(M 4 \times 3 = M 15\) \(M (323 + 15) = M 341\) |
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| 19. |
Solve for x, If \(\frac{\frac{2}{x}}{\frac{p^2 + p^2}{p^2 + p^2}}\) = m A. \(\frac{4pq}{m(p + q)}\) B. \(\frac{2p^2q^2}{m(q^2 + p^2)}\) C. \(\frac{2pq}{m(q^2 + p^2)}\) D. \(\frac{2p^2q^2}{m(p^2)}\) Detailed Solution\(\frac{1}{p^2}\) + \(\frac{1}{q^2}\) = \(\frac{q^2 + p^2}{p^2 + q^2}\)\(\frac{\frac{2}{x}}{\frac{p^2 + p^2}{p^2 + p^2}}\) m = \(\frac{2p^2q^2}{x(p^2 + q^2)}\) = m2p2q2 = m x (p2 + q2) x = \(\frac{2p^2q^2}{m(q^2 + p^2)}\) |
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| 20. |
Express 37.05 x 0.0042 in standard form A. 15.561 x 102 B. 1.5561 x 10-4 C. 1.556 x 10-1 D. 1.5561 x 101 E. 1.55 x 101 Detailed Solution37.05 x 0.0042 in standard form\(\begin{array}{c|c}No. & log \\\hline 37.05 & 1.5688\\ 0.0042 & 3.6232 \\ \hline & 1.1920\end{array}\) = 0.1556 = 1.556 x 10-1 |