Mathematics (Core)
Paper 1 | Objectives | 47 Questions
JAMB Exam
Year: 1985
Level: SHS
Time:
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
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
The goal of mocks tests is to create a bench-marking tool to help students assess their performances.
Good jobs available to people without a college degree, how to get good job without a college degree?
| # | Question | Ans |
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| 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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