Year :
1983
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
The value of (0.03)3 - (0.02)3 is A. 0.019 B. 0.0019 C. 0.00019 D. 0.000019 E. 0.000035 Detailed SolutionUsing the method of difference of two cubes,\(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\) \((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\) = \((0.01)(0.0009 + 0.0006 + 0.0004)\) = \(0.01 \times 0.0019\) = \(0.000019\) |
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| 12. |
Y varies partly as the square of x and partly as the inverse of the square root of x.Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4 A. y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\) B. y = x2 + \(\frac{1}{\sqrt{x}}\) C. y = x2 + \(\frac{1}{x}\) D. y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\) Detailed Solutiony = kx2 + \(\frac{c}{\sqrt{x}}\)y = 2when x = 1 2 = k + \(\frac{c}{1}\) k + c = 2 y = 6 when x = 4 6 = 16k + \(\frac{c}{2}\) 12 = 32k + c k + c = 2 32k + c = 12 = 31k + 10 k = \(\frac{10}{31}\) c = 2 - \(\frac{10}{31}\) = \(\frac{62 - 10}{31}\) = \(\frac{52}{31}\) y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\) |
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| 13. |
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\) A. \(\frac{x}{(x - 3)(x - 7)}\) B. \(\frac{x}{(x + 3)(x - 7)}\) C. \(\frac{x}{(x + 3)(x + 7)}\) D. \(\frac{x}{(x - 3)(x + 7)}\) Detailed Solution\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)= \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\) = \(\frac{x}{(x + 3)(x + 7)}\) |
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| 14. |
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x A. 2 B. 6 C. 18 D. 12 E. 5 Detailed Solution\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\) x2 - 12x = 0 x(x - 12) = 0 x = 0 or 12 The sides cannot be 0 hence x = 12. |
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| 15. |
The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores. A. 47 B. 50 C. 48\(\frac{1}{2}\) D. 48 E. 49 Detailed SolutionBy re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73The median = \(\frac{47 + 50}{2}\) \(\frac{97}{2}\) = 48\(\frac{1}{2}\) |
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| 16. |
Find x if (x\(_4\))\(^2\) = 100100\(_2\) A. 6 B. 12 C. 100 D. 210 E. 110 Detailed Solutionx\(_4\) = x \(\times\) 4\(^0\) = x in base 10.100100\(_2\) = 1 x 2\(^5\) + 1 x 2\(^2\) = 32 + 4 = 36 in base 10 \(\implies\) x\(^2\) = 36 x = 6 in base 10. Convert 6 to a number in base 4. = 12\(_4\) |
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| 17. |
Simplify log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 A. 1 B. \(\frac{7}{6}\)log10 a C. zero D. 10 Detailed Solutionlog10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\)= log10 a\(\frac{7}{12}\) - log10 a\(\frac{7}{12}\) = log10 1 = 0 |
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| 18. |
If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2 A. u = \(\frac{8}{v^3}\) B. v = \(\frac{8}{u^2v^3}\) C. u = 8v3 D. v = 8u2 Detailed SolutionW \(\alpha\) \(\frac{1}{v}\)u \(\alpha\) w3w = \(\frac{k1}{v}\) u = k2w3 u = k2(\(\frac{k1}{v}\))3 = \(\frac{k_2k_1^2}{v^3}\) k = k2k1k2 u = \(\frac{k}{v^3}\) k = uv3 = (1)(2)3 = 8 u = \(\frac{8}{v^3}\) |
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| 19. |
Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0 A. -28, 7 B. 6, -28 C. 6, -1 D. -1, 7 E. 3, 2 Detailed Solutionx2 + y - 8 = 0, y + 5x - 2 = 0Rearranging, x2 + y = 8.....(i) 5x + y = 2.......(ii) Subtract eqn(ii) from eqn(i) x2 - 5x - 6 = 0 (x - 6)(x + 1) = 0 x = 6, -1 |
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| 20. |
Find the missing value in the following table \(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline Y = x^3 - x + 3 & & 3 & 3 & 3 & 9 & 27\end{array}\) A. -3 B. 3 C. -9 D. 13 E. 9 Detailed SolutionWhen x = -2, y = x3 - x + 3= -23 - (-2) + 3 = -8 + 2 + 3 = -3 |
| 11. |
The value of (0.03)3 - (0.02)3 is A. 0.019 B. 0.0019 C. 0.00019 D. 0.000019 E. 0.000035 Detailed SolutionUsing the method of difference of two cubes,\(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\) \((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\) = \((0.01)(0.0009 + 0.0006 + 0.0004)\) = \(0.01 \times 0.0019\) = \(0.000019\) |
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| 12. |
Y varies partly as the square of x and partly as the inverse of the square root of x.Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4 A. y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\) B. y = x2 + \(\frac{1}{\sqrt{x}}\) C. y = x2 + \(\frac{1}{x}\) D. y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\) Detailed Solutiony = kx2 + \(\frac{c}{\sqrt{x}}\)y = 2when x = 1 2 = k + \(\frac{c}{1}\) k + c = 2 y = 6 when x = 4 6 = 16k + \(\frac{c}{2}\) 12 = 32k + c k + c = 2 32k + c = 12 = 31k + 10 k = \(\frac{10}{31}\) c = 2 - \(\frac{10}{31}\) = \(\frac{62 - 10}{31}\) = \(\frac{52}{31}\) y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\) |
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| 13. |
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\) A. \(\frac{x}{(x - 3)(x - 7)}\) B. \(\frac{x}{(x + 3)(x - 7)}\) C. \(\frac{x}{(x + 3)(x + 7)}\) D. \(\frac{x}{(x - 3)(x + 7)}\) Detailed Solution\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)= \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\) = \(\frac{x}{(x + 3)(x + 7)}\) |
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| 14. |
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x A. 2 B. 6 C. 18 D. 12 E. 5 Detailed Solution\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\) x2 - 12x = 0 x(x - 12) = 0 x = 0 or 12 The sides cannot be 0 hence x = 12. |
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| 15. |
The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores. A. 47 B. 50 C. 48\(\frac{1}{2}\) D. 48 E. 49 Detailed SolutionBy re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73The median = \(\frac{47 + 50}{2}\) \(\frac{97}{2}\) = 48\(\frac{1}{2}\) |
| 16. |
Find x if (x\(_4\))\(^2\) = 100100\(_2\) A. 6 B. 12 C. 100 D. 210 E. 110 Detailed Solutionx\(_4\) = x \(\times\) 4\(^0\) = x in base 10.100100\(_2\) = 1 x 2\(^5\) + 1 x 2\(^2\) = 32 + 4 = 36 in base 10 \(\implies\) x\(^2\) = 36 x = 6 in base 10. Convert 6 to a number in base 4. = 12\(_4\) |
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| 17. |
Simplify log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 A. 1 B. \(\frac{7}{6}\)log10 a C. zero D. 10 Detailed Solutionlog10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\)= log10 a\(\frac{7}{12}\) - log10 a\(\frac{7}{12}\) = log10 1 = 0 |
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| 18. |
If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2 A. u = \(\frac{8}{v^3}\) B. v = \(\frac{8}{u^2v^3}\) C. u = 8v3 D. v = 8u2 Detailed SolutionW \(\alpha\) \(\frac{1}{v}\)u \(\alpha\) w3w = \(\frac{k1}{v}\) u = k2w3 u = k2(\(\frac{k1}{v}\))3 = \(\frac{k_2k_1^2}{v^3}\) k = k2k1k2 u = \(\frac{k}{v^3}\) k = uv3 = (1)(2)3 = 8 u = \(\frac{8}{v^3}\) |
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| 19. |
Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0 A. -28, 7 B. 6, -28 C. 6, -1 D. -1, 7 E. 3, 2 Detailed Solutionx2 + y - 8 = 0, y + 5x - 2 = 0Rearranging, x2 + y = 8.....(i) 5x + y = 2.......(ii) Subtract eqn(ii) from eqn(i) x2 - 5x - 6 = 0 (x - 6)(x + 1) = 0 x = 6, -1 |
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| 20. |
Find the missing value in the following table \(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline Y = x^3 - x + 3 & & 3 & 3 & 3 & 9 & 27\end{array}\) A. -3 B. 3 C. -9 D. 13 E. 9 Detailed SolutionWhen x = -2, y = x3 - x + 3= -23 - (-2) + 3 = -8 + 2 + 3 = -3 |