Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 2008
Level: SHS
Time:
Type: Question Paper
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1. | ||
2. |
If 125x = 2010 find x A. 2 B. 3 C. 4 D. 5
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Detailed Solution125x = 201xX2 + 2xX1 + 5xX0 = 20 X2 + 2X + 5 = 20 X2 + 2X - 15 = 0 (X + 5)( X - 3) = 0 X + 5 implies X = -5 X - 3 implies X = 3 But X cannot be negative ∴X = 3 There is an explanation video available below. |
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3. |
Evaluate \(\frac{\left(\frac{3}{8}\div\frac{1}{2}+\frac{1}{2}\right)}{\left(\frac{1}{8}\times\frac{2}{3}+\frac{1}{3}\right)}\) A. 1/4 B. 1/3 C. 1/2 D. 3 |
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5. |
Calculate the simple interest on N7,500 for 8 years at 5% per annum. A. N3,000 B. N600 C. N300 D. N150
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Detailed Solution\(I = \frac{P \times T \times R}{100}\\=\frac{7500 \times 8 \times 5}{100}\\ =N3000\) There is an explanation video available below. |
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6. |
The cost of kerosene per liter increased from N60 to N85. What is the percentage rate of increase? A. 42% B. 41% C. 40% D. 25%
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Detailed SolutionN85 - N60 = N25 increase∴percentage increase = \(\frac{25}{60}\times \frac{100}{1}\\=\frac{125}{3}\\=41.67%\\ =42%\) There is an explanation video available below. |
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7. |
Simplify \(16^{\frac{-1}{2}}\times 4^{\frac{-1}{2}} \times 27^{\frac{1}{3}}\) A. 3/8 B. 2/3 C. 3/4 D. 3/2
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Detailed Solution\(16^{\frac{-1}{2}}\times 4^{\frac{-1}{2}} \times 27^{\frac{1}{3}}\\=\frac{1}{16^{\frac{1}{2}}}\times \frac{1}{4^{\frac{1}{2}}}\times 27^{\frac{1}{3}}\\ =\frac{1}{\sqrt{16}} \times \frac{1}{\sqrt{4}} \times sqrt[3]{27}\\ =\frac{1}{4} \times \frac{1}{2} \times \frac{3}{1}\\ = \frac{3}{8}\) There is an explanation video available below. |
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8. |
If log\(_{x^{1/2}}^{64}\) = 3, find the value of x A. 4 B. 16 C. 32 D. 64
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Detailed SolutionIf logx1/264 = 3(X 1/2)3 = 64 (X 1/2)3 = 4 3 X 1/2 = 4 X = 42 X = 16 There is an explanation video available below. |
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9. |
If \(\frac{1+\sqrt{2}}{1-\sqrt{2}}\) is expressed in the form of x+y√2 find the values of x and y A. (-3, -2) B. (-2, 3) C. (3,2) D. (2,-3)
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Detailed Solution\(\frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}\\=\frac{1+(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})}{1^2 - (\sqrt{2})^2}\\ =\frac{(1+\sqrt{2}+\sqrt{2}+2)}{1-2}\\ =\frac{3+2\sqrt{2}}{-1}\\ =-3-2\sqrt{2}\\ ∴X and Y = -3 and -2\) There is an explanation video available below. |
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10. |
If X = {n\(^2\) + 1:n = 0,2,3} and Y = {n+1:n=2,3,5}, find X∩Y. A. {1,3} B. {5,10} C. ∅ D. {4,6} |
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