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Mathematics (Core)
Paper 1 | Objectives | 50 Questions
JAMB Exam
Year: 2009
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 50 Questions
JAMB Exam
Year: 2009
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
If 55\(_x\) + 52\(_x\) = 77\(_{10}\) find X A. 5 B. 6 C. 7 D. 10
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Detailed Solution5 \(\times\) x\(^1\) + 5 \(\times\) x\(^0\) + 5 \(\times\) x\(^1\) + 2 \(\times\) x\(^0\) = 77(change all to base 10) 5x + 5 + 5x + 2 = 77 10x + 7 = 77 10x = 77-7 10x = 70 x = 70/10 x = 7 There is an explanation video available below. |
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| 2. |
Simplify \(7\frac{1}{12}-4\frac{3}{4}+2\frac{1}{2}\) A. 4 B. 41/6 C. 45/6 D. 51/6
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Detailed Solution\(7\frac{1}{12}-4\frac{3}{4}+2\frac{1}{2}=5\left(\frac{1-9+6}{12}\right)\\5\left(\frac{-2}{12}\right)\\ =4\left(\frac{12-2}{12}\right)\) (carry one from 5 and call it 12) \(4\frac{10}{12}\\ =4\frac{5}{6}\) There is an explanation video available below. |
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| 3. |
Evaluate \(\frac{81.81+99.44}{20.09+36.16}\) correct to 3 significant figures. A. 6.24 B. 3.22 C. 2.78 D. 2.13
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Detailed Solution\(\frac{81.81 + 99.44}{20.09 + 36.16}\\=\frac{181.25}{56.25}\\ =\frac{18125}{5625}\\ \frac{29}{9}\\ =3\frac{2}{9}\\ ≅ 3.22\) There is an explanation video available below. |
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| 4. |
A man bought a second-hand photocopying machine for N34,000. He serviced it at a cost of N2,000 and then sold it at a profit of 15%. What was the selling price? A. N37,550 B. N40,400 C. N41,400 D. N42,400
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Detailed SolutionC.P = N34000 + N2000 = N36000Gain = 100 + 15 = 115% S.P = \(\frac{115}{100}\times \frac{N3600}{1}\) = N41400 There is an explanation video available below. |
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| 5. |
A student spent 1/5 of his allowance on books, 1/2 of remainder on food and kept the rest for contingencies. What fraction was kept? A. 7/15 B. 8/15 C. 2/5 D. 4/5
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Detailed SolutionLet his allowance = y.\(Books = \frac{y}{5}\) \(Remainder: y - \frac{y}{5} = \frac{4y}{5}\) \(Food: \frac{1}{2} \times \frac{4y}{5} = \frac{2y}{5}\) \(Contingencies: \frac{4y}{5} - \frac{2y}{5} = \frac{2y}{5}\) Therefore, he kept \(\frac{2}{5}\) of his allowance for contingencies. There is an explanation video available below. |
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| 6. |
If log\(_{10}\)2 = 0.3010 and log\(_{10}\)7 = 0.8451, evaluate log\(_{10}\)280 A. 3.4471 B. 2.4471 C. 1.4471 D. 1.4071
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Detailed SolutionLog 2 = 0.3010, Log 7 = 0.8451∴Log 280 = Log 28 x 10 = Log 7x4x10 =Log7 + Log4 + Log10 = Log7 + Log2\(^2\) + Log10 = Log7 + 2Log2 + Log10 = 0.8451 + 2(0.3010) + 1 = 0.8451 + 0.6020 + 1 = 2.4471 There is an explanation video available below. |
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| 7. |
Simplify \(\frac{5+\sqrt{7}}{3+\sqrt{7}}\) A. 17-√7 B. 4-√7 C. 15+√7 D. 7-√7
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Detailed Solution\(\frac{5+\sqrt{7}}{3+\sqrt{7}}=\frac{5+\sqrt{7}}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}\\=\frac{(5+\sqrt{7})(3-\sqrt{7})}{3^2 - \sqrt{7}^2}\\ =\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}\\ =\frac{8-2\sqrt{7}}{2}\) Factorize then divide by 2 \(=\frac{2(4-\sqrt{7}}{2}\\ =4-\sqrt{7}\) There is an explanation video available below. |
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| 8. |
If x = {n\(^2\)+1:n is a positive integer and 1 \(\leq\) n \(\leq\) 5}, A. {5,10} B. {5, 10, 15} C. {2, 5, 10} D. {5, 10, 15, 20}
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Detailed SolutionX = {n(^2\)+1:n is a positive integer and 1 \(\leq\) n \(\leq\) 5}Implies X = {2, 5, 10, 17, 26} i.e. put n= 1, 2, 3, 4 and 5 Y = {5n:n is a positive integer and 1 \(\leq\) n \(\leq\) 5} Put X = 1, 2, 3, 4, and 5 Y = {5, 10, 15, 20, 25} X \(\cap\) Y = {2, 5, 10, 17, 26} \(\cap\) {5, 10, 15, 20, 25} = {5, 10} There is an explanation video available below. |
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| 9. |
I.S∩T∩W=S II. S ∪ T ∪ W = W A. I and II B. I and III C. II and III D. I, II and III
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Detailed SolutionIf S \(\subset\) T \(\subset\) W,S \(\cap\) T \(\cap\) W = S is true since S \(\cap\) T = S and S \(\cap\) W = S. S \(\cup\) T \(\cup\) W = W is also true. S \(\cup\) T = T and T \(\cup\) W = W. However, to say that T \(\cap\) W = S is not very true mathematically. Instead, it is safe to say S \(\subset\) (T \(\cap\) W). There is an explanation video available below. |
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| 10. |
If \(p=\sqrt{\frac{rs^3}{t}}\), express r in terms of p, s and t? A. \(\frac{p^2 t}{s^3}\) B. \(\frac{p^3 t}{s^3}\) C. \(\frac{p^3 t}{s^2}\) D. \(\frac{p^ t}{s^3}\)
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Detailed Solution\(p =\sqrt{\frac{rs^3}{t}}\\=p^2 =\frac{rs^3}{t}\\ tp^2 = rs^3\\ r = \frac{p^2 t}{s^3}\) There is an explanation video available below. |
Preview displays only 10 out of the 50 Questions