Mathematics (Core)
Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1989
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1989
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
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| 1. |
Which of the following is in descending order? A. \(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) B. \(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) C. \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) D. \(\frac{4}{5} \frac{9}{10} \frac{17}{20} \frac{3}{4}\)
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Detailed Solution\(\frac{9}{10} \frac{4}{5} \frac{3}{4} \frac{17}{20}\) = \(\frac{18, 16, 15, 17}{20}\)\(\frac{4}{5} \frac{9}{10} \frac{3}{4} \frac{17}{20}\) = \(\frac{16, 18, 15, 17}{20}\) \(\frac{9}{10} \frac{17}{20} \frac{4}{5} \frac{3}{4}\) = \(\frac{18, 17, 16, 15}{20}\) ∴ \(\frac{9}{10}; \frac{17}{20}; \frac{4}{5}; \frac{3}{4}\) is in descending order. |
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| 2. |
Evaluate 2700 000 x 0.03 ÷ 18 000 A. 4.5 x 100 B. 4.5 x 101 C. 4.5 x 102 D. 4.5 x 103
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Detailed Solution\(2700000 \times 0.03 \div 18000\)= \(\frac{2700000 \times 0.03}{18000}\) = \(\frac{81000}{18000} = 4.5 \times 10^{0}\) |
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| 3. |
The prime factors of 2520 are A. 2, 9, 5 B. 2, 9, 7 C. 2, 3, 5, 7 D. 2, 3, 7, 9
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Detailed Solution\(2520 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7\) in terms of its prime factors. Hence, the prime factors of 2520 are 2, 3, 5, 7. |
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| 4. |
If \(12_{e} = X_{7}\), where e = 12, find X. A. 20 B. 15 C. 14 D. 12
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Detailed Solution\(12_{12} = 1 \times 12^{1} + 2 \times 12^{0} = 12 + 2 = 14_{10}\)\(\therefore 12_{e} = 20_{7}\) |
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| 5. |
Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\) A. \(\frac{r}{2}\) B. 2r C. \(\frac{1}{2r}\) D. \(\frac{2}{r}\)
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Detailed Solution\(\sqrt[3]{(64r^6)^{\frac{1}{2}}}\)= \(\sqrt[3]{((8r^{3})^{2})^{\frac{1}{2}}}\) =\(2r\) |
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| 6. |
What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal? A. 10-5 B. 7 x 10-4 C. 8 x 10-5 D. 10-6
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Detailed Solution0.007685 = 0.00769 (three significant figures)0.007685 = 0.0077(4d.p) the difference = 0.0077 - 0.00769 = 0.00001 = 1.0 x 10-5 |
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| 7. |
If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\) A. 125 : 128 B. 3 : 5 C. 3 : 4 D. 2 : 5
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Detailed Solutiona : b = 5 : 8 = 2.5 : 40x : y = 25 : 16 \(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\) = 1 : \(\frac{40}{16}\) = 16 : 40 = 2 : 5 |
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| 8. |
Oke deposited N800.00 in the bank at the rate of 12\(\frac{1}{2}\)% simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank? A. 2\(\frac{2}{3}\) B. 4 C. 5\(\frac{1}{3}\) D. 8
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Detailed SolutionP = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125After sometimes, A = 1.5 x 800 = N1,200 A = P(1 + Tr) = 1200 = 800(1 + 0.125T) = 1200 = 800(1 + 0.125T) 1200 = 800(1 + 0.125T) 0.125T = 0.5 T = \(\frac{0.5}{0.125}\) = 4 |
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| 9. |
If the surface area of a sphere increased by 44%, find the percentage increase in diameter A. 44 B. 30 C. 22 D. 20
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Detailed SolutionSurface Area of Sphere A = 4\(\pi r^2\)∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\) = \(\frac{(D)^2}{2}\) = \(\pi\)D2 When increased by 44% A = \(\frac{144 \pi D^2}{100}\) \(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\) Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)% = 20% |
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