Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1998
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1998
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
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Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
| # | Question | Ans |
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| 1. |
If 10112 + x7 = 2510, solve for X. A. 207 B. 14 C. 20 D. 24
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Detailed Solution10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o= 8 + 0 + 2 + 1 = 1110 x7 = 2510 - 1110 = 1410 \(\begin{array}{c|c} 7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2 \end{array}\) X = 207 |
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| 2. |
Evaluate [\(\frac{1}{0.03}\) \(\div\) \(\frac{1}{0.024}\)]-1 correct to 2 decimal places A. 3.76 B. 1.25 C. 0.94 D. 0.75
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Detailed Solution[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)]= [\(\frac{1}{0.03 \times 0.024}\)]-1 = [\(\frac{0.024}{0.003}\)]-1 = \(\frac{0.03}{0.024}\) = \(\frac{30}{24}\) = 1.25 |
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| 3. |
If b3 = a-2 and c\(\frac{1}{3}\) = a\(\frac{1}{2}\)b, express c in terms of a A. a-\(\frac{1}{2}\) B. a\(\frac{1}{3}\) C. a\(\frac{3}{2}\) D. a\(\frac{2}{3}\)
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Detailed Solutionc\(\frac{1}{3}\) = a\(\frac{1}{2}\)b= a\(\frac{1}{2}\)b x a-2 = a-\(\frac{3}{2}\) = (c\(\frac{1}{3}\))3 = (a-\(\frac{3}{2}\))\(\frac{1}{3}\) c = a-\(\frac{1}{2}\) |
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| 4. |
Given that log4(Y - 1) + log4(\(\frac{1}{2}\)x) = 1 and log2(y + 1) + log2x = 2, solve for x and y respectively A. 2, 3 B. 3, 2 C. -2, -3 D. -3, -2
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Detailed Solutionlog4(y - 1) + log4(\(\frac{1}{2}\)x) = 1log4(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1) log2(y + 1) + log2x = 2 log2(y + 1)x = 2 \(\to\) (y + 1)x = 22 = 4.....(ii) From equation (ii) x = \(\frac{4}{y + 1}\)........(iii) put equation (iii) in (i) = y (y - 1)[\(\frac{1}{2}(\frac{4}{y - 1}\))] = 4 = 2y - 2 = 4y + 4 2y = -6 y = -3 x = \(\frac{ |
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| 5. |
Find the value of k if \(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\) A. 3 B. 2 C. \(\sqrt{3}\) D. \(\sqrt 2\)
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Detailed Solution\(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\)\(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = k\(\sqrt{3 - 2}\) = k(\(\sqrt{3} - \sqrt{2}\)) = k\(\sqrt{3 - 2}\) = k\(\sqrt{3}\) - k\(\sqrt{2}\) = k\(\sqrt{3 - 2}\) k2 = \(\sqrt{2}\) k = \(\frac{2}{\sqrt{2}}\) = \(\sqrt{2}\) |
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| 6. |
A market woman sells oil in cylindrical tins 10cm deep and 6cm in diameter at N15.00 each. If she bought a full cylindrical jug 18cm deep and 10cm in diameter for N50.00, how much did she make by selling all the oil? A. N62.50 B. N35.00 C. N31.00 D. N25.00
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Detailed SolutionV\(\pi\)r2h = \(\pi\)(3)2(10) = 90\(\pi\)cm3V = \(\pi\)(5)2 x 18 = 450\(\pi\)cm3 No of volume = \(\frac{450\pi}{90\pi}\) = 5 selling price = 5 x N15 = N75 profit = N75 - N50 = N25.00 |
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| 7. |
A man is paid r naira per hour for normal work and double rate for overtime. if he does a 35-hour week which includes q hours of overtime, what is his weekly earning in naira? A. r(35 + q) B. q(35r - q) C. q(35 + r) D. r(35 + 2q)
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Detailed SolutionThe cost of normal work = 35rThe cost of overtime = q x 2r = 2qr The man's total weekly earning = 35r + 2qr = r(35 + 2q) |
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| 8. |
When the expression pm2 + qm + 1 is divided by (m - 1), it has a remainder is 4, Find p and q respectively A. 2, -1 B. -1, 2 C. 3, -2 D. -2, 3
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Detailed Solutionpm2 + qm + 1 = (m - 1) Q(x) + 2p(1)2 + q(1) + 1 = 2 p + q + 1 = 2 p + q = 1.....(i) pm2 + qm + 1 = (m - 1)Q(x) + 4 p(-1)2 + q(-1) + 1 = 4 p - q + 1 = 4 p - q = 3....(ii) p + q = 1, p - q = -3 2p = -2, p = -1 -1 + q = 1 q = 2 |
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| 9. |
Factorize r2 - r(2p + q) + 2pq A. (r - 2q)(2r - p) B. (r - p)(r + p0 C. (r - q)(r - 2p) D. (2r - q)(r + p)
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Detailed Solutionr2 - r(2p + q) + 2pq = r2 - 2pr -qr + 2pq= r(r - 2p) - q(r - 2p) = (r - q)(r - 2p) |
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| 10. |
Solve for the equation \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0 A. \(\frac{3}{2}\) B. \(\frac{2}{3}\) C. \(\frac{4}{9}\) D. \(\frac{9}{4}\)
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Detailed Solution\(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0= \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) = 1 = (\(\sqrt{x}\) - \(\sqrt{(x - 2)}\))2 = 1 = x - 2 \(\sqrt{x(x - 2)}\) + x -2 = 1 = (2x - 3)2 = [2 \(\sqrt{x(x - 4)}\)]2 = 4x2 - 12x + 9 = 4(x2 - 2x) = 4x2 - 12x + 9 = 4x2 - 8x 4x = 9 x = \(\frac{9}{4}\) |
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