Mathematics (Core)
Paper 1 | Objectives | 47 Questions
JAMB Exam
Year: 1985
Level: SHS
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Type: Question Paper
Answers provided
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Paper 1 | Objectives | 47 Questions
JAMB Exam
Year: 1985
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 three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of \(\frac{3p - q}{4q + r}\) A. \(\frac{3}{2}\) B. \(\frac{2}{3}\) C. 2 D. 3 E. 18
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Detailed SolutionP : Q : r = 6 : 4 : 55 = 6 + 4 + 5 = 15 P = \(\frac{6}{15}\), q = \(\frac{4}{15}\), r = \(\frac{5}{15}\) = \(\frac{1}{3}\) To find \(\frac{3p - q}{4q + r}\) 3p - q = 3 x \(\frac{6}{15}\) - \(\frac{4}{15}\) \(\frac{18}{15}\) - \(\frac{4}{15}\) = \(\frac{14}{15}\) ∴ 4q + r = 4 x \(\frac{4}{15}\) + \(\frac{5}{15}\) \(\frac{16}{15}\) = \(\frac{16}{15}\) + \(\frac{5}{15}\) = \(\frac{21}{15}\) \(\frac{14}{15}\) x \(\frac{15}{21}\) = \(\frac{14}{21}\) = \(\frac{2}{3}\) |
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| 2. |
Arrange the following numbers in ascending order of magnitude \(\frac{6}{7}\), \(\frac{13}{15}\), 0.8650 A. \(\frac{6}{7}\) < 0.865 < \(\frac{13}{15}\) B. \(\frac{13}{15}\) < \(\frac{6}{7}\) < 0.865 C. \(\frac{6}{7}\) < \(\frac{13}{15}\) < 0.865 D. 0.865 < \(\frac{6}{7}\) < \(\frac{13}{15}\)
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Detailed Solution\(\frac{6}{7}\), \(\frac{13}{15}\), 0.8650In ascending order, we have 0.8571, 0.8650, 0.8666 i.e. \(\frac{6}{7}\) < 0.8650 < \(\frac{13}{15}\) |
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| 3. |
A sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to N330.00. Find the amount originally invested A. N180.00 B. n165.00 C. N150.00 D. N250.00 E. N200.00
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Detailed SolutionS.I = \(\frac{PTR}{100}\)T = 4yrs, R = 8%, a = N330.00 330 - P = \(\frac{PTR}{100}\), A = P + I i.e. A = P + \(\frac{PTR}{100}\) 330 = P + \(\frac{P(4) (8)}{100}\) 33000 = 32P + 100p 132P = 33000 P = N250.00 |
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| 4. |
In the equation below, Solve for x if all the numbers are in base 2: \(\frac{11}{x}\) = \(\frac{1000}{x + 101}\) A. 101 B. 11 C. 110 D. 111 E. 10
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Detailed Solution\(\frac{11}{x}\) = \(\frac{1000}{x + 101}\) = 11(x + 101)1000x = 11x + 1111 1000x - 11x = 1111 101x = 1111 x = \(\frac{1111}{101}\) x = 11 |
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| 5. |
List all integers satisfying the inequality -2 \(\leq\) 2 x -6 < 4 A. 2, 3, 4, 5 B. 2, 3, 4 C. 2, 5 D. 3, 4, 5 E. 4, 5
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Detailed Solution-2 \(\leq\) 2x - 6 < 4 = 2x - 6 < 4= 2x < 10 = x < 5 2x \(\geq\) -2 + 6 \(\geq\) = x \(\geq\) 2 ∴ 2 \(\leq\) x < 5 [2, 3, 4] |
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| 6. |
Find correct to two decimals places 100 + \(\frac{1}{100}\) + \(\frac{3}{1000}\) + \(\frac{27}{10000}\) A. 100.02 B. 1000.02 C. 100.22 D. 100.01 E. 100.51
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Detailed Solution100 + \(\frac{1}{100}\) + \(\frac{3}{1000}\) + \(\frac{27}{10000}\)\(\frac{1000,000 + 100 + 30 + 27}{10000}\) = \(\frac{1,000.157}{10000}\) = 100.02 |
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| 7. |
John gives one-third of his money to Janet who has N105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first A. N45.00 B. N48.00 C. N52.00 D. N60.00 E. N52.00
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Detailed SolutionLet x be John's money, Janet already had N105, \(\frac{1}{3}\) of x was given to JanetJanet now has \(\frac{1}{3^2}\)x + 105 = \(\frac{x + 315}{3}\) John's money left = \(\frac{2}{3}\)x = \(\frac{\frac{1}{4}(x + 315)}{3}\) = \(\frac{2}{3}\) 24x = 3x + 945 ∴ x = 45 |
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| 8. |
Find x if log\(_9\)x = 1.5 A. 72.0 B. 27.0 C. 36.0 D. 3.5 E. 24.5
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Detailed SolutionIf log\(_9\)x = 1.5,9\(^1.5\) = x 9^\(\frac{3}{2}\) = x (√9)\(^3\) = 3 ∴ x = 27 |
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| 9. |
Write h in terms of a, b, c, d if a = \(\frac{b(1 - ch)}{a - dh}\) A. h = \(\frac{a - b}{ad}\) B. h = \(\frac{1 - b}{ad - bc}\) C. h = \(\frac{(a - b)^2}{ad - bc}\) D. h = \(\frac{a - b}{ad - bc}\) E. h = \(\frac{b - a}{ab - dc}\)
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Detailed Solutiona = \(\frac{b(1 - ch)}{a - dh}\)a = \(\frac{b - bch}{1 - dh}\) = a - adh = b - bch a - b = bch + adn a - b = adh a - b = h(ad - bc) h = \(\frac{a - b}{ad - bc}\) |
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| 10. |
22\(\frac{1}{2}\)% of the Nigerian Naira is equal to 17\(\frac{1}{10}\)% of a foreign currency M. What is the conversion rate of the M to the Naira? A. 1M = 1\(\frac{15}{57}\)N B. 1M = 38\(\frac{1}{4}\)N C. 1M = 1\(\frac{18}{57}\)N D. 1M = 384\(\frac{3}{4}\)N
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Detailed SolutionN = 22\(\frac{1}{2}\)%, M = 17\(\frac{1}{10}\)%M = \(\frac{171}{10}\)%, N = \(\frac{45}{2}\) \(\frac{45}{2}\) x \(\frac{10}{171}\) = \(\frac{225}{171}\) = 1 \(\frac{54}{171}\) = 1 \(\frac{18}{57}\) |
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