Year :
1979
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 51 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If < PQS = 15o find < QRS A. 75o B. 37\(\frac{1}{2}\)o C. 127\(\frac{1}{2}\)o D. 105o Detailed SolutionPSO = 90o(angle in a semicircle)SPO = 180o - (90o + 15o) = 75o < QRS = 108o - 75o = 105o(opposite angles in cyclic quadrilateral are supplementary) |
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| 22. |
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0 A. c = \(\frac{2d - 5 - b}{ad}\) B. c = \(\frac{2d - 5 - abd}{ad}\) C. c = \(\frac{2d - 5 - ab}{ad}\) D. c = \(\frac{5 - 2d - b}{ad}\) Detailed Solutiona(b + c) + \(\frac{5}{d}\) - 2 = 0ab + ac + \(\frac{5}{d}\) - 2 = 0 abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d acd = 2d - 5 - abd c = \(\frac{2d - 5 - abd}{ad}\) |
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| 23. |
Which of the following values of the variable x, (a)x = 0, (b)x = -3, (c)x = 9, satisfy the inequalities 0 < \(\frac{x + 3}{x - 1}\) < 2? A. (a), (b), (c) B. (c) C. None of the choices D. All of the above Detailed Solution0 < \(\frac{x + 3}{x - 1}\) < 2Put x = 0, -3 and 9 0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2 i.e. 0 < 1.5 \(\leq\) 2 (true) but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2 i.e. 0 < -3 \(\leq\) 2 (not true) -3 \(\leq\) 2 -3 is not greater than 0 |
|
| 24. |
On each market day Mrs. Bassey walks to the market from her home at a steady speed. This journey normally takes her 2 hours to complete. She finds, however, that by increasing her usual speed by 1 km/hr she can save 20 minutes. Find her usual speed in km/hr A. 1\(\frac{2}{3}\) B. 2 C. 5 D. 6 E. 10 Detailed SolutionLet x rept. the usual speed = \(\frac{Distance}{Time}\)= \(\frac{20}{1}\) x \(\frac{1}{2}\) = 5 |
|
| 25. |
Solve the simultaneous linear equations: 2x + 5y = 11, 7x + 4y = 2 A. x = -8, y = 1 B. x = -2, y = 4 C. x = \(\frac{-34}{27}\), y = \(\frac{73}{27}\) D. x = 7, y = -9 Detailed Solution2x + 5y = 11.......(i)7x + 4y = 2.......(ii) (i) x 7 \(\to\) 14x + 35y = 77.........(iii) (ii) x 2 \(\to\) 14x + 8y = 4........(iv) (iii) - (iv) 27y = 73 y = \(\frac{73}{27}\) |
|
| 26. |
If x3 - 12x - 16 = 0 has x = -2 as a solution then the equaion has A. x = -4 as a solution also B. 3 roots all different C. 3 roots with two equal and the third different D. 3 roots all equal E. only one root |
B |
| 27. |
Find the value of (4\(\frac{1}{2}\))6 A. 1 B. 2 C. 4 D. 6 E. 64 |
E |
| 28. |
Find the value of log\(_{10}\)\(\frac{1}{40}\), given that log10\(_4\) = 0.6021 A. 1.3979 B. 2.3979 C. -1.6021 D. 2.6021 Detailed Solutionlog\(_{10}\)\(\frac{1}{40}\) = log\(_{10}\)1 - (log\(_{10}\)4 + log\(_{10}\)10)= 0 - 1.6021 = -1.6021 |
|
| 29. |
12 men complete a job in 9 days. How many men working t the same rate would be required to complete the job in 6 days? A. 24 B. 18 C. 12 D. 9 E. 8 Detailed Solution12 men in 9 days,1 day in \(12 \times 9\) men In 6 days we have \(\frac{12 \times 9}{6}\) men = \(\frac{108}{6}\) = 18 men |
|
| 30. |
For the set of numbers 2, 3, 5, 6, 7, 7, 8 A. The median is greater than the mode B. the mean is greater than the mode C. the mean is greater than the median D. the median is equal to the mean E. the mean is less than the median Detailed SolutionMean = 5, 42Median = 6 Mode = 7 The mean is less than the median |
| 21. |
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If < PQS = 15o find < QRS A. 75o B. 37\(\frac{1}{2}\)o C. 127\(\frac{1}{2}\)o D. 105o Detailed SolutionPSO = 90o(angle in a semicircle)SPO = 180o - (90o + 15o) = 75o < QRS = 108o - 75o = 105o(opposite angles in cyclic quadrilateral are supplementary) |
|
| 22. |
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0 A. c = \(\frac{2d - 5 - b}{ad}\) B. c = \(\frac{2d - 5 - abd}{ad}\) C. c = \(\frac{2d - 5 - ab}{ad}\) D. c = \(\frac{5 - 2d - b}{ad}\) Detailed Solutiona(b + c) + \(\frac{5}{d}\) - 2 = 0ab + ac + \(\frac{5}{d}\) - 2 = 0 abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d acd = 2d - 5 - abd c = \(\frac{2d - 5 - abd}{ad}\) |
|
| 23. |
Which of the following values of the variable x, (a)x = 0, (b)x = -3, (c)x = 9, satisfy the inequalities 0 < \(\frac{x + 3}{x - 1}\) < 2? A. (a), (b), (c) B. (c) C. None of the choices D. All of the above Detailed Solution0 < \(\frac{x + 3}{x - 1}\) < 2Put x = 0, -3 and 9 0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2 i.e. 0 < 1.5 \(\leq\) 2 (true) but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2 i.e. 0 < -3 \(\leq\) 2 (not true) -3 \(\leq\) 2 -3 is not greater than 0 |
|
| 24. |
On each market day Mrs. Bassey walks to the market from her home at a steady speed. This journey normally takes her 2 hours to complete. She finds, however, that by increasing her usual speed by 1 km/hr she can save 20 minutes. Find her usual speed in km/hr A. 1\(\frac{2}{3}\) B. 2 C. 5 D. 6 E. 10 Detailed SolutionLet x rept. the usual speed = \(\frac{Distance}{Time}\)= \(\frac{20}{1}\) x \(\frac{1}{2}\) = 5 |
|
| 25. |
Solve the simultaneous linear equations: 2x + 5y = 11, 7x + 4y = 2 A. x = -8, y = 1 B. x = -2, y = 4 C. x = \(\frac{-34}{27}\), y = \(\frac{73}{27}\) D. x = 7, y = -9 Detailed Solution2x + 5y = 11.......(i)7x + 4y = 2.......(ii) (i) x 7 \(\to\) 14x + 35y = 77.........(iii) (ii) x 2 \(\to\) 14x + 8y = 4........(iv) (iii) - (iv) 27y = 73 y = \(\frac{73}{27}\) |
| 26. |
If x3 - 12x - 16 = 0 has x = -2 as a solution then the equaion has A. x = -4 as a solution also B. 3 roots all different C. 3 roots with two equal and the third different D. 3 roots all equal E. only one root |
B |
| 27. |
Find the value of (4\(\frac{1}{2}\))6 A. 1 B. 2 C. 4 D. 6 E. 64 |
E |
| 28. |
Find the value of log\(_{10}\)\(\frac{1}{40}\), given that log10\(_4\) = 0.6021 A. 1.3979 B. 2.3979 C. -1.6021 D. 2.6021 Detailed Solutionlog\(_{10}\)\(\frac{1}{40}\) = log\(_{10}\)1 - (log\(_{10}\)4 + log\(_{10}\)10)= 0 - 1.6021 = -1.6021 |
|
| 29. |
12 men complete a job in 9 days. How many men working t the same rate would be required to complete the job in 6 days? A. 24 B. 18 C. 12 D. 9 E. 8 Detailed Solution12 men in 9 days,1 day in \(12 \times 9\) men In 6 days we have \(\frac{12 \times 9}{6}\) men = \(\frac{108}{6}\) = 18 men |
|
| 30. |
For the set of numbers 2, 3, 5, 6, 7, 7, 8 A. The median is greater than the mode B. the mean is greater than the mode C. the mean is greater than the median D. the median is equal to the mean E. the mean is less than the median Detailed SolutionMean = 5, 42Median = 6 Mode = 7 The mean is less than the median |