Year :
1988
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error? A. 20 B. 2 C. 1.1 D. 0.2 Detailed Solution% error = \(\frac{\text{actual error}}{\text{true value}}\) x 100Where actual error = 5.1 - 5.0 = 0.1 true value = 5.0g % error = \(\frac{0.1}{5.0}\) x 100 = \(\frac{10}{5}\) = 2% |
|
| 22. |
Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720.00. Find the value of the store. A. N1,080.00 B. N2,400.00 C. N3,000.00 D. N3,600.00 Detailed SolutionLet value of store = XRatio of Taiwo's share to kehine's is 11:9 Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720 \(\frac{2}{3}\) of 9 = 6 ∴ Sum of the ratio = 11 + 9 = 20 \{\frac{6}{20}\) of x = N720 \(\frac{6x}{20}\) = 720 ∴ x = \(\frac{720 \times 20}{6}\) x = N24,000 |
|
| 23. |
A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket A. 1 : 1 : 1 B. 4 : 2 : 1 C. 5 : 1 : 1 D. 4 : 1 : 1 Detailed SolutionLet x represent total number of balls in the basket.If there are 10 blue balls, \(\frac{1}{8}\) of x = 10 x = 10 x 8 = 80 balls Green balls will be \(\frac{5}{8}\) x 80 = 50 and black balls = \(\frac{2}{8}\) x 80 = 20 Ratio = Green : black : blue 50 : 20 : 10 -10 : -10 : - ------------------ New Ratio 40 : 10 : 10 4 : 1 : 1 |
|
| 24. |
A tax payer is allowed \(\frac{1}{8}\)th of his income tax-free, and pays 20% on the remainder. If he pays N490.00 tax, what is his income? A. N560.00 B. N2,450.00 C. N2,800.00 D. N3,920.00 Detailed SolutionHe pays tax on 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)th of his income20% is 490, 100% is \(\frac{100}{20}\) x 490, N2,450.00 = \(\frac{7}{8}\) of his income = N2,450.00 \(\frac{1}{\frac{7}{8}}\) x 2450 = \(\frac{8 \times 2450}{7}\) = \(\frac{19600}{7}\) = N2800.00 |
|
| 25. |
Evaluate \(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) = \(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) A. \(\frac{2}{5}\) B. \(\frac{5}{3}\) C. 3\(\sqrt{5}\) D. 3\(\sqrt{2}\) Detailed Solution\(\frac{8^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{\frac{2}{10^3}}\) = \(\frac{(2^3)^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{(2 \times 5)^{\frac{2}{3}}}\)= \(\frac{2 \times 5}{2^{\frac{2}{3}} \times 5^{\frac{3}{2}}}\) = 21 - \(\frac{2}{3}\) = 2\(\frac{1}{3}\) = 3\(\sqrt{2}\) |
|
| 26. |
If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\), evaluate; without using logarithm tables, \(log_{10} 4.5\) A. 0.3010 B. 0.4771 C. 0.6532 D. 0.9542 Detailed SolutionIf \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\),\(\log_{10} 4.5 = \log_{10} (\frac{3 \times 3}{2})\) \(log_{10} 3 + log_{10} 3 - log_{10} 2 = 0.4771 + 0.4771 - 0.3010\) = 0.6532 |
|
| 27. |
Find m such that (m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\) A. 1 B. 2 C. 3 D. 4 Detailed Solution(m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\)(m + \(\sqrt{3}\))(4 - 2\(\sqrt{3}\)) = 6 - 2\(\sqrt{2}\) = 6 - 2\(\sqrt{3}\) 4m - 6 + 4 - 2m\(\sqrt{3}\) = 6 - 2\(\sqrt{3}\) comparing co-efficients, 4m - 6 = 6.......(i) 4 - 2m = -2.......(ii) in both equations, m = 3 |
|
| 28. |
The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form A. 2.25 x 10-4m B. 50 x 10-4m C. 2.25 x 10-5m D. 4.50 x 10-5m Detailed SolutionThickness of an 800 pages book = 18mm to meter18 x 103m = 1.8 x 10-2m One leaf = \(\frac{1.8 \times 10^{-2}}{800}\) = \(\frac{1.8 \times 10^{-2}}{8 \times 10^{2}}\) = \(\frac{-1.8}{8}\) x 10-4 = 0.225 x 10-4 = 2.25 x 10-5m one leaf contains 2 pages : 2 * 2.25 x 10\(^{-5}\)m = 4.5 * 10\(^{-5}\)m |
|
| 29. |
Simplify \(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) A. \(\frac{3}{x + 1}\) B. \(\frac{3x + 2}{(x + 1)(x + 2)}\) C. \(\frac{5x + 6}{(x + 1)(x + 2)}\) D. \(\frac{2x^2 + 5x + 2}{(x + 1)(x + 2)}\) Detailed Solution\(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) = \(\frac{(x + 2)(x + 2) - (x -2) - (x - 2)(x + 1)}{(x + 1)(x + 2)}\)= \(\frac{(x^2 + 4x + 4) - (x^2 - x - 2)}{(x + 1)(x + 2)}\) = \(\frac{x^2 + 4x + 4 - x^2 + x + 2}{(x + 1)(x + 2)}\) = \(\frac{5x + 6}{(x + 1)(x + 2)}\) |
|
| 30. |
Solve the following equation equation for \(x^2 + \frac{2x}{r^2} + \frac{1}{r^4}\) = 0 A. r2 B. \(\frac{1}{r^4}\) C. -\(\frac{1}{r^2}\) D. 1 - r Detailed Solution\(x^2 + \frac{2x}{r^2}\) + \(\frac{1}{r^4}\) = 0(x + \(\frac{1}{r^2}\)) = 0 x + \(\frac{1}{r^2}\) = 0 x = \(\frac{-1}{r^2}\) |
| 21. |
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error? A. 20 B. 2 C. 1.1 D. 0.2 Detailed Solution% error = \(\frac{\text{actual error}}{\text{true value}}\) x 100Where actual error = 5.1 - 5.0 = 0.1 true value = 5.0g % error = \(\frac{0.1}{5.0}\) x 100 = \(\frac{10}{5}\) = 2% |
|
| 22. |
Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720.00. Find the value of the store. A. N1,080.00 B. N2,400.00 C. N3,000.00 D. N3,600.00 Detailed SolutionLet value of store = XRatio of Taiwo's share to kehine's is 11:9 Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720 \(\frac{2}{3}\) of 9 = 6 ∴ Sum of the ratio = 11 + 9 = 20 \{\frac{6}{20}\) of x = N720 \(\frac{6x}{20}\) = 720 ∴ x = \(\frac{720 \times 20}{6}\) x = N24,000 |
|
| 23. |
A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket A. 1 : 1 : 1 B. 4 : 2 : 1 C. 5 : 1 : 1 D. 4 : 1 : 1 Detailed SolutionLet x represent total number of balls in the basket.If there are 10 blue balls, \(\frac{1}{8}\) of x = 10 x = 10 x 8 = 80 balls Green balls will be \(\frac{5}{8}\) x 80 = 50 and black balls = \(\frac{2}{8}\) x 80 = 20 Ratio = Green : black : blue 50 : 20 : 10 -10 : -10 : - ------------------ New Ratio 40 : 10 : 10 4 : 1 : 1 |
|
| 24. |
A tax payer is allowed \(\frac{1}{8}\)th of his income tax-free, and pays 20% on the remainder. If he pays N490.00 tax, what is his income? A. N560.00 B. N2,450.00 C. N2,800.00 D. N3,920.00 Detailed SolutionHe pays tax on 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)th of his income20% is 490, 100% is \(\frac{100}{20}\) x 490, N2,450.00 = \(\frac{7}{8}\) of his income = N2,450.00 \(\frac{1}{\frac{7}{8}}\) x 2450 = \(\frac{8 \times 2450}{7}\) = \(\frac{19600}{7}\) = N2800.00 |
|
| 25. |
Evaluate \(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) = \(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) A. \(\frac{2}{5}\) B. \(\frac{5}{3}\) C. 3\(\sqrt{5}\) D. 3\(\sqrt{2}\) Detailed Solution\(\frac{8^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{\frac{2}{10^3}}\) = \(\frac{(2^3)^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{(2 \times 5)^{\frac{2}{3}}}\)= \(\frac{2 \times 5}{2^{\frac{2}{3}} \times 5^{\frac{3}{2}}}\) = 21 - \(\frac{2}{3}\) = 2\(\frac{1}{3}\) = 3\(\sqrt{2}\) |
| 26. |
If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\), evaluate; without using logarithm tables, \(log_{10} 4.5\) A. 0.3010 B. 0.4771 C. 0.6532 D. 0.9542 Detailed SolutionIf \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\),\(\log_{10} 4.5 = \log_{10} (\frac{3 \times 3}{2})\) \(log_{10} 3 + log_{10} 3 - log_{10} 2 = 0.4771 + 0.4771 - 0.3010\) = 0.6532 |
|
| 27. |
Find m such that (m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\) A. 1 B. 2 C. 3 D. 4 Detailed Solution(m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\)(m + \(\sqrt{3}\))(4 - 2\(\sqrt{3}\)) = 6 - 2\(\sqrt{2}\) = 6 - 2\(\sqrt{3}\) 4m - 6 + 4 - 2m\(\sqrt{3}\) = 6 - 2\(\sqrt{3}\) comparing co-efficients, 4m - 6 = 6.......(i) 4 - 2m = -2.......(ii) in both equations, m = 3 |
|
| 28. |
The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form A. 2.25 x 10-4m B. 50 x 10-4m C. 2.25 x 10-5m D. 4.50 x 10-5m Detailed SolutionThickness of an 800 pages book = 18mm to meter18 x 103m = 1.8 x 10-2m One leaf = \(\frac{1.8 \times 10^{-2}}{800}\) = \(\frac{1.8 \times 10^{-2}}{8 \times 10^{2}}\) = \(\frac{-1.8}{8}\) x 10-4 = 0.225 x 10-4 = 2.25 x 10-5m one leaf contains 2 pages : 2 * 2.25 x 10\(^{-5}\)m = 4.5 * 10\(^{-5}\)m |
|
| 29. |
Simplify \(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) A. \(\frac{3}{x + 1}\) B. \(\frac{3x + 2}{(x + 1)(x + 2)}\) C. \(\frac{5x + 6}{(x + 1)(x + 2)}\) D. \(\frac{2x^2 + 5x + 2}{(x + 1)(x + 2)}\) Detailed Solution\(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) = \(\frac{(x + 2)(x + 2) - (x -2) - (x - 2)(x + 1)}{(x + 1)(x + 2)}\)= \(\frac{(x^2 + 4x + 4) - (x^2 - x - 2)}{(x + 1)(x + 2)}\) = \(\frac{x^2 + 4x + 4 - x^2 + x + 2}{(x + 1)(x + 2)}\) = \(\frac{5x + 6}{(x + 1)(x + 2)}\) |
|
| 30. |
Solve the following equation equation for \(x^2 + \frac{2x}{r^2} + \frac{1}{r^4}\) = 0 A. r2 B. \(\frac{1}{r^4}\) C. -\(\frac{1}{r^2}\) D. 1 - r Detailed Solution\(x^2 + \frac{2x}{r^2}\) + \(\frac{1}{r^4}\) = 0(x + \(\frac{1}{r^2}\)) = 0 x + \(\frac{1}{r^2}\) = 0 x = \(\frac{-1}{r^2}\) |