Mathematics (Core)
Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1999
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
Answers provided
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Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1999
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group? A. 25 B. 19 C. 18 D. 17
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Detailed SolutionLet the three items be M, Y and P.n{M ∩ Y} only = 4-3 = 1 n{M ∩ P) only = 5-3 = 2 n{ Y ∩ P} only = 2 n{M} only = 12-(1+3+2) = 6 n{Y} only = 10-(1+2+3) = 4 n{P} only = 14-(2+3+2) = 7 n{M∩P∩Y} = 3 Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women. |
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| 2. |
If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X. A. \(\frac{1}{2}\)X B. X-\(\frac{1}{4}\) C. X-\(\frac{1}{3}\) D. X-\(\frac{1}{2}\)
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Detailed Solution\(log_810\) = X = \(log_8{2 x 5}\)\(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks |
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| 3. |
Find the value of X if \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) A. 3√2+4 B. 3√2-4 C. 3-2√2 D. 4+2√2
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Detailed SolutionStart your solution by cross-multiplying,\(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - √2] √2 = x + √2 then collect like terms x√2 - x = √2 + 2 and factorize accordingly to get the unknown. x(√2 - 1) = √2 + 2 x = {√2 + 2} / {√2 - 1} rationalize x = {√2 + 2} / {√2 - 1} * { |
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| 4. |
If \(\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}\) What is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10
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Detailed SolutionHint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive atp = \(\frac{5}{2}\), q = -\(\frac{25}{4}\) and r = -\(\frac{17}{4}\) NUMERATOR: a\(^{\frac{2}{1}}*{\frac{3}{4}}\) b\(^{\frac{-3}{1}}*{\frac{3}{4}}\) c\(^{\frac{1}{1}}*{\frac{3}{4}}\) a\(\frac{3}{2}\) b\(\frac{-9}{4}\) c\(\frac{3}{4}\) Using index method for numerator & denominator : a\(^{\frac{3}{2}}-{\frac{-1}{1}}\) b\(^{\frac{-9}{4}}-{\frac{4}{1}}\) c\(^{\frac{3}{4}}-{\frac{5}{1}}\) : a\(\frac{5}{2}\) b\(\frac{-25}{4}\) c\(\frac{-17}{4}\) Then p+2q will giv |
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| 5. |
If {(a2b-3c)3/4/a-1b4c5} = apbqcr; what is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10
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Detailed SolutionHint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = -25/4 and r = -9/2.Then substitute the values of p and q into p+2q to get -10. |
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| 6. |
A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss. A. 30% gain B. 25% gain C. 30% loss D. 25% loss
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Detailed SolutionCost price (cp) = (100/5) x N1.20 =N24.00Selling price (sp) = 100-20 = 80 (80/4) x N1.50 = N30.00 Gain = SP-CP = N30.00-N24.00 = N6.00 Gain% = (gain/CP) x 100 = 25% |
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| 7. |
What is the answer when 2434\(_6\) is divided by 42\(_6\)? A. 236 B. 356 C. 526 D. 556
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Detailed SolutionHint: Two methods can be used;1. Direct division (if you know division in number bases) 2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356 2434\(_6\) to base 10 2 \(\times\) 6\(^3\) + 4 \(\times\) 6\(^2\) + 3 \(\times\) 6\(^1\) + 4 \(\times\) 6\(^0\) 2 \(\times\) 216 + 4 \(\times\) 36 + 3 \(\times\) 6 + 4 \(\times\) 1 432 + 144 + 18 + 4 = 598 And: 42\(_6\) to base 10 4 \(\times\) 6\(^1\) + 2 \(\times\) 6\(^0\) 4 \(\times\) 6 + 2 \(\times\) 1 24 + 2 = 26 Then divide 598 by 26 = 23 Then convert 23\(_{10}\) to base 6 23\(_{10}\) = 35\(_6\) |
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| 8. |
If 2\(_9\) x (Y3)\(_9\) = 3\(_5\) x (Y3)\(_5\), find the value of Y. A. 4 B. 3 C. 2 D. 1
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Detailed SolutionThe correct answer is 1.PROVE==> (2x9\(^0\)) (3x9\(^0\) + Yx9\(^1\)) = (3x5\(^0\))(3x5\(^0\) + Yx5\(^1\)). You multiply to get 2(3+9Y)=3(3+5Y). You further multiply to get 6+18Y=9+15Y. ==> Collect like terms 18Y-15Y=9-6. 3Y/3=3/3. Y=1 Explanation Credit: endowednuel |
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| 9. |
Simplify \(\sqrt{\frac{(0.0023 \times 750)}{(0.00345 \times 1.25)}}\) A. 15 B. 20 C. 40 D. 75
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Detailed SolutionMultiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\)Dividing, => \(\sqrt{400} = 20\). |
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