Year :
1990
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 44 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
A car painter charges N40.00 per day for himself and N10.00 per day for his assistant. if a fleet of cars were painted for N2000.00 and the painter worked 10days more than his assistant, how much did the assistant receive? A. N32.00 B. N320.00 C. N420.00 D. N1680.00 Detailed SolutionLet his assistant work for x days∴ his master worked (x + 10) day. Amount received by master = 40(x + 10), amount got by his assistance = 10x Total amount collected = N2000.00 ∴ 40(x + 10) + 10x = 2000 = 40x + 400 + 10x = 2000 50x + 400 = 2000 50x = 2000 - 400 50x = 1600 x = \(\frac{1600}{50}\) x = 32 days |
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| 22. |
Simplify \(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\) A. \(\frac{x}{x^2 - y^2}\) B. \(\frac{y^2}{x^2 - y^2}\) C. \(\frac{x^2}{x^2 - y^2}\) D. \(\frac{y}{x^2 - y^2}\) Detailed Solution\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\)\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\) = \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\) = \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\) = \(\frac{y^2}{(x + y)(x - y)}\) = \(\frac{y^2}{(x^2 - y^2)}\) |
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| 23. |
Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3 A. \(\sqrt{10}\) B. 8 C. 12.2 D. 13.4 Detailed SolutionGiven that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3x = 7 ∴ x2 = 49 \(\sqrt{y}\) = 3 ∴ y2 = 81 = x2 + y2 + z2 = 194 49 + 81 + z2 = 194 130 + z2 = 194 z2 = 194 - 130 = 64 z = \(\sqrt{64}\) = 8 |
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| 24. |
Find the sum of the first twenty terms of the progression log a, log a2, log a3..... A. log a20 B. log a21 C. log a200 D. log a210 |
D |
| 25. |
Find the sum of the first 18 terms of the progression 3, 6, 12...... A. 3(217 - 1) B. 3(218 - 1) C. 3(218 + 1) D. 3(217 - 1) Detailed Solution3 + 6 + 12 + .....18thy term1st term = 3, common ratio \(\frac{6}{3}\) = 2 n = 18, sum of GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\) s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\) = 3(2^18 - 1) |
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| 26. |
At what value of x is the function x\(^2\) + x + 1 minimum? A. -1 B. \(-\frac{1}{2}\) C. \(\frac{1}{2}\) D. 1 Detailed Solutionx\(^2\) + x + 1\(\frac{dy}{dx}\) = 2x + 1 At the turning point, \(\frac{dy}{dx}\) = 0 2x + 1 = 0 x = -\(\frac{1}{2}\) |
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| 27. |
The angle of a sector of s circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector A. 8.8cm B. 25.4cm C. 25.6cm D. 29.8cm Detailed SolutionLength of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm |
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| 28. |
Find the length of a side of a rhombus whose diagonals are 6cm and 8cm A. 8cm B. 5cm C. 4cm D. 3cm Detailed SolutionThe diagonal of a rhombus is a line segment that joins any two non-adjacent vertices.A rhombus has two diagonals that bisect each other at right angles. i.e this splits 6cm into 3cm each AND 8cm to 4cm Using Hyp\(^2\) = adj\(^2\) + opp\(^2\) Hyp\(^2\) = 3\(^2\) + 4\(^2\) Hyp\(^2\) = 25 Hyp = 5 ∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths |
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| 29. |
Each of the interior angles of a regular polygon is 140°. How many sides has the polygon? A. 9 B. 8 C. 7 D. 5 Detailed SolutionFor a regular polygon of n sidesn = \(\frac{360}{\text{Exterior angle}}\) Exterior < = 180° - 140° = 40° n = \(\frac{360}{40}\) = 9 sides |
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| 30. |
If Cos \(\theta\) = \(\frac{12}{13}\). Find \(\theta\) + cos2\(\theta\) A. \(\frac{169}{25}\) B. \(\frac{25}{169}\) C. \(\frac{169}{144}\) D. \(\frac{144}{169}\) Detailed SolutionCos \(\theta\) = \(\frac{12}{13}\)x2 + 122 = 132 x2 = 169- 144 = 25 x = 25 = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cos2\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\) |
| 21. |
A car painter charges N40.00 per day for himself and N10.00 per day for his assistant. if a fleet of cars were painted for N2000.00 and the painter worked 10days more than his assistant, how much did the assistant receive? A. N32.00 B. N320.00 C. N420.00 D. N1680.00 Detailed SolutionLet his assistant work for x days∴ his master worked (x + 10) day. Amount received by master = 40(x + 10), amount got by his assistance = 10x Total amount collected = N2000.00 ∴ 40(x + 10) + 10x = 2000 = 40x + 400 + 10x = 2000 50x + 400 = 2000 50x = 2000 - 400 50x = 1600 x = \(\frac{1600}{50}\) x = 32 days |
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| 22. |
Simplify \(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\) A. \(\frac{x}{x^2 - y^2}\) B. \(\frac{y^2}{x^2 - y^2}\) C. \(\frac{x^2}{x^2 - y^2}\) D. \(\frac{y}{x^2 - y^2}\) Detailed Solution\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{x^2 - y^2}\)\(\frac{x}{x + y}\) + \(\frac{y}{x - y}\) - \(\frac{x^2}{(x + y)(x - y}\) = \(\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y}\) = \(\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y}\) = \(\frac{y^2}{(x + y)(x - y)}\) = \(\frac{y^2}{(x^2 - y^2)}\) |
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| 23. |
Given that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3 A. \(\sqrt{10}\) B. 8 C. 12.2 D. 13.4 Detailed SolutionGiven that x2 + y2 + z2 = 194, calculate z if x = 7 and \(\sqrt{y}\) = 3x = 7 ∴ x2 = 49 \(\sqrt{y}\) = 3 ∴ y2 = 81 = x2 + y2 + z2 = 194 49 + 81 + z2 = 194 130 + z2 = 194 z2 = 194 - 130 = 64 z = \(\sqrt{64}\) = 8 |
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| 24. |
Find the sum of the first twenty terms of the progression log a, log a2, log a3..... A. log a20 B. log a21 C. log a200 D. log a210 |
D |
| 25. |
Find the sum of the first 18 terms of the progression 3, 6, 12...... A. 3(217 - 1) B. 3(218 - 1) C. 3(218 + 1) D. 3(217 - 1) Detailed Solution3 + 6 + 12 + .....18thy term1st term = 3, common ratio \(\frac{6}{3}\) = 2 n = 18, sum of GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\) s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\) = 3(2^18 - 1) |
| 26. |
At what value of x is the function x\(^2\) + x + 1 minimum? A. -1 B. \(-\frac{1}{2}\) C. \(\frac{1}{2}\) D. 1 Detailed Solutionx\(^2\) + x + 1\(\frac{dy}{dx}\) = 2x + 1 At the turning point, \(\frac{dy}{dx}\) = 0 2x + 1 = 0 x = -\(\frac{1}{2}\) |
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| 27. |
The angle of a sector of s circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector A. 8.8cm B. 25.4cm C. 25.6cm D. 29.8cm Detailed SolutionLength of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm |
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| 28. |
Find the length of a side of a rhombus whose diagonals are 6cm and 8cm A. 8cm B. 5cm C. 4cm D. 3cm Detailed SolutionThe diagonal of a rhombus is a line segment that joins any two non-adjacent vertices.A rhombus has two diagonals that bisect each other at right angles. i.e this splits 6cm into 3cm each AND 8cm to 4cm Using Hyp\(^2\) = adj\(^2\) + opp\(^2\) Hyp\(^2\) = 3\(^2\) + 4\(^2\) Hyp\(^2\) = 25 Hyp = 5 ∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths |
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| 29. |
Each of the interior angles of a regular polygon is 140°. How many sides has the polygon? A. 9 B. 8 C. 7 D. 5 Detailed SolutionFor a regular polygon of n sidesn = \(\frac{360}{\text{Exterior angle}}\) Exterior < = 180° - 140° = 40° n = \(\frac{360}{40}\) = 9 sides |
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| 30. |
If Cos \(\theta\) = \(\frac{12}{13}\). Find \(\theta\) + cos2\(\theta\) A. \(\frac{169}{25}\) B. \(\frac{25}{169}\) C. \(\frac{169}{144}\) D. \(\frac{144}{169}\) Detailed SolutionCos \(\theta\) = \(\frac{12}{13}\)x2 + 122 = 132 x2 = 169- 144 = 25 x = 25 = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cos2\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\) |