Year :
1981
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 47 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Marks scored by some children in an arithmetic test are:5, 3, 6, 9, 4, 7, 8, 6, 2, 7, 8, 4, 5, 2, 1, 0, 6, 9, 0, 8. A. 6 B. 5 C. 8 D. 7 E. 10 Detailed SolutionAdd all the No.s together to give 100\(\sum\)xf = 100 N = 20 x = \(\frac{\sum xf}{N}\) = \(\frac{100}{20}\) = 5 |
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| 12. |
The weights of 30 new-born babies are given as follow: 6, 9, 5, 7, 6, 7, 5, 8, 9, 5, 7, 5, 8, 7, 8, 7, 5, 6, 5, 7, 6, 9, 9, 7, 8, 8, 7, 8, 9, 8. The mode is A. 6 B. 5 C. 8 D. 7 E. 10 Detailed SolutionMode is the No having the highest frequency = 7 |
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| 13. |
If \(\sin x° = \frac{a}{b}\), what is \(\sin (90 - x)°\)? A. \(\frac{\sqrt{b^2 - a^2}}{b}\) B. 1\(\frac{-a}{b}\) C. \(\frac{b^2 - a^2}{b}\) D. \(\frac{a^2 - b^2}{b}\) E. \(\sqrt{b^2 - a^3}\) Detailed Solution\(\sin x = \frac{a}{b}\)\(\sin^{2} x + \cos^{2} x = 1\) \(\sin^{2} x = \frac{a^{2}}{b^{2}}\) \(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\) \(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\) \(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\) = \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\) = \(\frac{\sqrt{b^{2} - a^{2}}}{b}\) |
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| 14. |
7 pupils of average age 12 years leave a class of 25 pupils of average age 14 years. If 6 new pupils of average age 11years join the class, what is the average age of the pupils now in the class? A. 13years B. 12 years 7\(\frac{1}{2}\) months C. 13 years 5 months D. 13 years 10 months E. 11 years Detailed SolutionTotal age of the 7 pupils = 7 x 12 = 84Total age of the 25 pupils = 25 x 14 = 350 Total age of the 6 pupils = 6 x 11 = 66 7 pupils leaving a class of 25 pupils given 18 pupils Total age of the 18 pupils = 350 - 84 = 266 6 pupils joining 18 pupils = 24 Total age of 24 pupils = 266 + 66 = 332 Average age of 24 pupils now in class \(\frac{332}{24}\) = 13yrs 10 months |
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| 15. |
A sum of money invested at 5% per annum simple interest amounts to $285.20 after 3 years. How long will it take the same sum to amount to $434.00 at 7\(\frac{1}{2}\)% per annum simple interest? A. 7\(\frac{1}{2}\) years B. 10 years C. 5 years D. 12 years E. 14 years Detailed Solution\(A = P(1 + \frac{R}{100})^{T}\)\(285.20 = P(1 + \frac{5}{100})^{3}\) \(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\) \(P = \frac{285.20}{1.16} = $245.86\) Given Amount = $434.00 Interest = Amount - Principal Interest = $434.00 - $245.86 = $188.14 \(T = \frac{100I}{PR}\) \(T = \frac{100 \times 188.14}{245.86 \times 7.5}\) \(T = \frac{18814}{1843.95} = 10.2\) Approximately 10 years. |
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| 16. |
By selling an article for N45.00 a man makes a profit of 8%. For how much should he have sold it in order to make a profit of 32%? A. N180.00 B. N59.00 C. N63.00 D. N58.00 E. N55.00 Detailed SolutionWhen C.P = N45.00 at 8% profitC.P = \(\frac{100}{108}\) x 45 at profit of 32%, S.P = \(\frac{132}{100}\) x c.p = \(\frac{132}{100}\) x \(\frac{100 \times 45}{108}\) = \(\frac{5940}{108}\) = N55.00 |
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| 17. |
An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle? A. 7cm B. 12cm C. 8cm D. 7cm E. 69cm Detailed Solution In \(\Delta DAC, \stackrel\frown{DAC} = \theta\) \(\sin \theta = \frac{5}{13}\) \(\theta = 22.6°\) \(< DOC = 22.6° \times 2 = 45.2°\) \(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\) \(r = 7.046cm\) = \(7\frac{1}{24} cm\) |
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| 18. |
A micrometer is defined as one millionth of a millimeter. A length of 12,000 micrometres may be represented as A. 0.00012m B. 0.0000012m C. 0.000012m D. 0.00000012m E. 0.000000012m Detailed Solution1 UM = 10-6mm= 10-9m 1.2 x 104 x 10-9m = 1.2 x 10-5m (0.000012) |
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| 19. |
In one and a half hours, the minute hand of a clock rotates through an angle of A. 90o B. 180o C. 640o D. 450o E. 540o Detailed Solution1 hr, = 60 mins, 60 mins = 360o30 mins = \(\frac{360^o}{1}\) c \(\frac{30}{60}\) = 180o 90 mins = 360 + 180o = 540o |
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| 20. |
Simplify \(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{3 - 2}{\sqrt{3} + \sqrt{2}}\) A. 2\(\sqrt{2} - \sqrt{3}\) B. 3(\(\sqrt{6}\) - 1) C. \(\sqrt{6}\) - 3 D. -\(\frac{1}{2}\) E. \(\frac{-\sqrt{3}}{\sqrt{2} - \sqrt{2}}\) Detailed Solution\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{3 - 2}{\sqrt{3} + \sqrt{2}}\)\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = \(\frac{\sqrt{2}}{\sqrt{3}}\) - \(\frac{x}{\sqrt{2}}\) \(\frac{\sqrt{3} + \sqrt{2}}{3 + \sqrt{2}}\) = \(\sqrt{6}\) + 2 \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) = \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = 5 - 2\(\sqrt{6}\) \(\sqrt{6}\) + 2 - (5 - 2 \(\sqrt{6}\)) = \(\sqrt{6}\) + 2 - 5 + 2\(\sqrt{6}\) = 3\(\sqrt{6}\) - 3 = 3(\(\sqrt{6}\) - 1) |
| 11. |
Marks scored by some children in an arithmetic test are:5, 3, 6, 9, 4, 7, 8, 6, 2, 7, 8, 4, 5, 2, 1, 0, 6, 9, 0, 8. A. 6 B. 5 C. 8 D. 7 E. 10 Detailed SolutionAdd all the No.s together to give 100\(\sum\)xf = 100 N = 20 x = \(\frac{\sum xf}{N}\) = \(\frac{100}{20}\) = 5 |
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| 12. |
The weights of 30 new-born babies are given as follow: 6, 9, 5, 7, 6, 7, 5, 8, 9, 5, 7, 5, 8, 7, 8, 7, 5, 6, 5, 7, 6, 9, 9, 7, 8, 8, 7, 8, 9, 8. The mode is A. 6 B. 5 C. 8 D. 7 E. 10 Detailed SolutionMode is the No having the highest frequency = 7 |
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| 13. |
If \(\sin x° = \frac{a}{b}\), what is \(\sin (90 - x)°\)? A. \(\frac{\sqrt{b^2 - a^2}}{b}\) B. 1\(\frac{-a}{b}\) C. \(\frac{b^2 - a^2}{b}\) D. \(\frac{a^2 - b^2}{b}\) E. \(\sqrt{b^2 - a^3}\) Detailed Solution\(\sin x = \frac{a}{b}\)\(\sin^{2} x + \cos^{2} x = 1\) \(\sin^{2} x = \frac{a^{2}}{b^{2}}\) \(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\) \(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\) \(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\) = \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\) = \(\frac{\sqrt{b^{2} - a^{2}}}{b}\) |
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| 14. |
7 pupils of average age 12 years leave a class of 25 pupils of average age 14 years. If 6 new pupils of average age 11years join the class, what is the average age of the pupils now in the class? A. 13years B. 12 years 7\(\frac{1}{2}\) months C. 13 years 5 months D. 13 years 10 months E. 11 years Detailed SolutionTotal age of the 7 pupils = 7 x 12 = 84Total age of the 25 pupils = 25 x 14 = 350 Total age of the 6 pupils = 6 x 11 = 66 7 pupils leaving a class of 25 pupils given 18 pupils Total age of the 18 pupils = 350 - 84 = 266 6 pupils joining 18 pupils = 24 Total age of 24 pupils = 266 + 66 = 332 Average age of 24 pupils now in class \(\frac{332}{24}\) = 13yrs 10 months |
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| 15. |
A sum of money invested at 5% per annum simple interest amounts to $285.20 after 3 years. How long will it take the same sum to amount to $434.00 at 7\(\frac{1}{2}\)% per annum simple interest? A. 7\(\frac{1}{2}\) years B. 10 years C. 5 years D. 12 years E. 14 years Detailed Solution\(A = P(1 + \frac{R}{100})^{T}\)\(285.20 = P(1 + \frac{5}{100})^{3}\) \(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\) \(P = \frac{285.20}{1.16} = $245.86\) Given Amount = $434.00 Interest = Amount - Principal Interest = $434.00 - $245.86 = $188.14 \(T = \frac{100I}{PR}\) \(T = \frac{100 \times 188.14}{245.86 \times 7.5}\) \(T = \frac{18814}{1843.95} = 10.2\) Approximately 10 years. |
| 16. |
By selling an article for N45.00 a man makes a profit of 8%. For how much should he have sold it in order to make a profit of 32%? A. N180.00 B. N59.00 C. N63.00 D. N58.00 E. N55.00 Detailed SolutionWhen C.P = N45.00 at 8% profitC.P = \(\frac{100}{108}\) x 45 at profit of 32%, S.P = \(\frac{132}{100}\) x c.p = \(\frac{132}{100}\) x \(\frac{100 \times 45}{108}\) = \(\frac{5940}{108}\) = N55.00 |
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| 17. |
An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle? A. 7cm B. 12cm C. 8cm D. 7cm E. 69cm Detailed SolutionIn \(\Delta DAC, \stackrel\frown{DAC} = \theta\) \(\sin \theta = \frac{5}{13}\) \(\theta = 22.6°\) \(< DOC = 22.6° \times 2 = 45.2°\) \(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\) \(r = 7.046cm\) = \(7\frac{1}{24} cm\) |
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| 18. |
A micrometer is defined as one millionth of a millimeter. A length of 12,000 micrometres may be represented as A. 0.00012m B. 0.0000012m C. 0.000012m D. 0.00000012m E. 0.000000012m Detailed Solution1 UM = 10-6mm= 10-9m 1.2 x 104 x 10-9m = 1.2 x 10-5m (0.000012) |
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| 19. |
In one and a half hours, the minute hand of a clock rotates through an angle of A. 90o B. 180o C. 640o D. 450o E. 540o Detailed Solution1 hr, = 60 mins, 60 mins = 360o30 mins = \(\frac{360^o}{1}\) c \(\frac{30}{60}\) = 180o 90 mins = 360 + 180o = 540o |
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| 20. |
Simplify \(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{3 - 2}{\sqrt{3} + \sqrt{2}}\) A. 2\(\sqrt{2} - \sqrt{3}\) B. 3(\(\sqrt{6}\) - 1) C. \(\sqrt{6}\) - 3 D. -\(\frac{1}{2}\) E. \(\frac{-\sqrt{3}}{\sqrt{2} - \sqrt{2}}\) Detailed Solution\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{3 - 2}{\sqrt{3} + \sqrt{2}}\)\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = \(\frac{\sqrt{2}}{\sqrt{3}}\) - \(\frac{x}{\sqrt{2}}\) \(\frac{\sqrt{3} + \sqrt{2}}{3 + \sqrt{2}}\) = \(\sqrt{6}\) + 2 \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) = \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = 5 - 2\(\sqrt{6}\) \(\sqrt{6}\) + 2 - (5 - 2 \(\sqrt{6}\)) = \(\sqrt{6}\) + 2 - 5 + 2\(\sqrt{6}\) = 3\(\sqrt{6}\) - 3 = 3(\(\sqrt{6}\) - 1) |