Year :
1991
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
1 - 10 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 1. |
Simplify 3\(\frac{1}{3}\) - 1\(\frac{1}{4}\) x \(\frac{2}{3}\) + 1\(\frac{2}{5}\) A. 2 B. 3 C. 4 D. 5 Detailed Solution3 - \(\frac{1}{3}\) - (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\)= \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\) = \(\frac{100 - 25 + 42}{30}\) = \(\frac{117}{30}\) = 3.9 \(\approx\) 4 |
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| 2. |
If 2257 is the result of subtracting 4577 from 7056 in base n, find n A. 8 B. 9 C. 10 D. 11 Detailed Solution\(\begin{array}{c|c} 7056 \\ \text{- 4577} \\\hline 2257 \end{array}\)By trial and error method Let the base to 8 i.e. Let n = 8 and it is easily verified that the subtraction holds. The subtraction does not hold when other values of n are tried n = 8 |
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| 3. |
Find correct to 3 decimal places (\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05) A. 99.998 B. 98.999 C. 89.899 D. 9.998 Detailed Solution(\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05)(\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05) \(\frac{5.005}{0.05}\) - 0.1025 100.1 - 0.1025 = 99.9975 99.998 to 3 decimal place. |
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| 4. |
Express 62 \(\div\) 3 as a decimal correct to 3 significant figures A. 20.6 B. 20.667 C. 20.67 D. 20.7 Detailed Solution\(\frac{62}{3}\) = 20.6666.....= 20.7 to 3 sig. fig. |
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| 5. |
Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories A. 250 B. 750 C. 1000 D. 1250 Detailed SolutionP produces 20,000 bagsP reduces production by 5% i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\) = 1000 Q produces 15000 bags Q increases production by 5% i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\) = 750 Effective loss by the two = 1000 - 750 = 250 bags |
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| 6. |
Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe? A. N10.80 B. N10.67 C. N2.80 D. N2.67 Detailed SolutionI = \(\frac{PRT}{100}\)= \(\frac{10 \times 2 \times 4}{100}\) = \(\frac{4}{5}\) = 0.8 Total amount = N10.80 He pays N8.00 Remainder = 10.80 - 8.00 = N2.80 |
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| 7. |
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water? A. 2\(\frac{4}{5}\)% B. 16\(\frac{7}{8}\)% C. 18\(\frac{1}{8}\)% D. 18\(\frac{7}{8}\)% Detailed Solution% of water in the mixture= \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\) \(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\) = \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\) = \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\) = \(\frac{1.35}{8}\) x \(\frac{100}{1}\) = \(\frac{33.75}{2}\) = 16.875 = 16\(\frac{7}{8}\) |
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| 8. |
What is the product of \(\frac{27}{5^1}\)(3)-3 and \(\frac{(1)^{-1}}{5}\)? A. 5 B. 3 C. 1 D. \(\frac{1}{25}\) Detailed Solution\(\frac{27}{5^1}\)(3)-3 x \(\frac{(1)^{-1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\)= \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\) = \(\frac{1}{25}\) |
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| 9. |
Simplify 2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 A. 1 - 4 log3 B. -1 + 2 log 3 C. -1 + 5 log2 D. 1 - 2log 2 Detailed Solution2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9[\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 10 - log10 22 = 1 - 2 log2 |
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| 10. |
Simplify \(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\) A. - \(\frac{1}{2}\sqrt{5}\) B. \(\frac{1}{2}\sqrt{5}\) C. -- \(\frac{1}{4}\sqrt{5}\) D. 5 Detailed Solution\(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\)= \(\frac{1 - \sqrt{5} - 1 - \sqrt{5}}{(1 + \sqrt{5}) (1 - \sqrt{5}}\) = \(\frac{-2\sqrt{5}}{1 - 5}\) = \(\frac{-2\sqrt{5}}{- 4}\) = \(\frac{1}{2}\sqrt{5}\) |
| 1. |
Simplify 3\(\frac{1}{3}\) - 1\(\frac{1}{4}\) x \(\frac{2}{3}\) + 1\(\frac{2}{5}\) A. 2 B. 3 C. 4 D. 5 Detailed Solution3 - \(\frac{1}{3}\) - (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\)= \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\) = \(\frac{100 - 25 + 42}{30}\) = \(\frac{117}{30}\) = 3.9 \(\approx\) 4 |
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| 2. |
If 2257 is the result of subtracting 4577 from 7056 in base n, find n A. 8 B. 9 C. 10 D. 11 Detailed Solution\(\begin{array}{c|c} 7056 \\ \text{- 4577} \\\hline 2257 \end{array}\)By trial and error method Let the base to 8 i.e. Let n = 8 and it is easily verified that the subtraction holds. The subtraction does not hold when other values of n are tried n = 8 |
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| 3. |
Find correct to 3 decimal places (\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05) A. 99.998 B. 98.999 C. 89.899 D. 9.998 Detailed Solution(\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05)(\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05) \(\frac{5.005}{0.05}\) - 0.1025 100.1 - 0.1025 = 99.9975 99.998 to 3 decimal place. |
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| 4. |
Express 62 \(\div\) 3 as a decimal correct to 3 significant figures A. 20.6 B. 20.667 C. 20.67 D. 20.7 Detailed Solution\(\frac{62}{3}\) = 20.6666.....= 20.7 to 3 sig. fig. |
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| 5. |
Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories A. 250 B. 750 C. 1000 D. 1250 Detailed SolutionP produces 20,000 bagsP reduces production by 5% i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\) = 1000 Q produces 15000 bags Q increases production by 5% i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\) = 750 Effective loss by the two = 1000 - 750 = 250 bags |
| 6. |
Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe? A. N10.80 B. N10.67 C. N2.80 D. N2.67 Detailed SolutionI = \(\frac{PRT}{100}\)= \(\frac{10 \times 2 \times 4}{100}\) = \(\frac{4}{5}\) = 0.8 Total amount = N10.80 He pays N8.00 Remainder = 10.80 - 8.00 = N2.80 |
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| 7. |
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water? A. 2\(\frac{4}{5}\)% B. 16\(\frac{7}{8}\)% C. 18\(\frac{1}{8}\)% D. 18\(\frac{7}{8}\)% Detailed Solution% of water in the mixture= \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\) \(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\) = \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\) = \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\) = \(\frac{1.35}{8}\) x \(\frac{100}{1}\) = \(\frac{33.75}{2}\) = 16.875 = 16\(\frac{7}{8}\) |
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| 8. |
What is the product of \(\frac{27}{5^1}\)(3)-3 and \(\frac{(1)^{-1}}{5}\)? A. 5 B. 3 C. 1 D. \(\frac{1}{25}\) Detailed Solution\(\frac{27}{5^1}\)(3)-3 x \(\frac{(1)^{-1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\)= \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\) = \(\frac{1}{25}\) |
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| 9. |
Simplify 2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 A. 1 - 4 log3 B. -1 + 2 log 3 C. -1 + 5 log2 D. 1 - 2log 2 Detailed Solution2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9[\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 10 - log10 22 = 1 - 2 log2 |
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| 10. |
Simplify \(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\) A. - \(\frac{1}{2}\sqrt{5}\) B. \(\frac{1}{2}\sqrt{5}\) C. -- \(\frac{1}{4}\sqrt{5}\) D. 5 Detailed Solution\(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\)= \(\frac{1 - \sqrt{5} - 1 - \sqrt{5}}{(1 + \sqrt{5}) (1 - \sqrt{5}}\) = \(\frac{-2\sqrt{5}}{1 - 5}\) = \(\frac{-2\sqrt{5}}{- 4}\) = \(\frac{1}{2}\sqrt{5}\) |