Mathematics (Core)
Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1984
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 1984
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
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. |
Simplify \(\frac{(\frac{2}{3} - \frac{1}{5}) - \frac{1}{3} \text{of} \frac{2}{5}}{3 - \frac{1}{1 \frac{1}{2}}}\) A. \(\frac{1}{7}\) B. 7 C. \(\frac{1}{3}\) D. 3
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Detailed Solution\(\frac{2}{3} - \frac{1}{5}\) = \(\frac{10 - 3}{15}\)= \(\frac{7}{15}\) \(\frac{1}{3}\) Of \(\frac{2}{5}\) = \(\frac{1}{3}\) x \(\frac{2}{5}\) = \(\frac{2}{15}\) (\(\frac{2}{3} - \frac{1}{5}\)) - \(\frac{1}{3}\) of \(\frac{2}{5}\) = \(\frac{7}{15} - \frac{2}{15}\) = \(\frac{1}{3}\) 3 - \(\frac{1}{1 \frac{1}{2}}\) = 3 - \(\frac{2}{3}\) = \(\frac{7}{3}\) \(\frac{\frac{2}{3} - \frac{1}{5} \text{of} \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}\) = \(\frac{\frac{1}{3}}{\frac{7}{3}}\) = \(\frac{1}{3}\) |
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| 2. |
\(\frac{0.0001432}{1940000}\) = k x 10n where 1 \(\leq\) k < 10 and n is a whole number. The values K and n are A. 7.381 qnd -11 B. 2.34 and 10 C. 3.871 and 2 D. 7.831 and -11
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Detailed Solution\(\frac{0.0001432}{1940000}\) = k x 10nwhere 1 \(\leq\) k \(\leq\) 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11 k = 7.381, n = -11 |
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| 3. |
P sold his bicycle to Q at a profit of 10%. Q sold it to R for N209 at a loss of 5%. How much did the bicycle cost P? A. N200 B. N196 C. N180 D. N205 E. N150
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Detailed SolutionLet the selling price(SP from P to Q be represented by xi.e. SP = x When SP = x at 10% profit CP = \(\frac{100}{100}\) + 10 of x = \(\frac{100}{110}\) of x when Q sells to R, SP = N209 at loss of 5% Q's cost price = Q's selling price CP = \(\frac{100}{95}\) x 209 = 220.00 x = 220 = \(\frac{2200}{11}\) = 200 = N200.00 |
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| 4. |
If the price of oranges was raised by \(\frac{1}{2}\)k per orange. The number of oranges a customer can buy for N2.40 will be less by 16. What is the present price of an orange? A. 2\(\frac{1}{2}\) B. \(\frac{2}{15}\) C. 5\(\frac{1}{3}\) D. 20
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Detailed SolutionLet x represent the price of an orange andy represent the number of oranges that can be bought xy = 240k, y = \(\frac{240}{x}\).....(i) If the price of an oranges is raised by \(\frac{1}{2}\)k per orange, number that can be bought for N240 is reduced by 16 Hence, y - 16 = \(\frac{240}{x + \frac{1}{2}\) = \(\frac{480}{2x + 1}\) = \(\frac{480}{2x + 1}\).....(ii) subt. for y in eqn (ii) \(\frac{240}{x}\) - 16 = \(\frac{480}{2x + 1}\) = \(\frac{240 - 16x}{x}\) = \(\frac{480}{2x + 1}\) = (240 - 16x)(2x + 1) |
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| 5. |
A man invested a total of N50000 in two companies. If these companies pay dividends of 6% and 8% respectively, how much did he invest at 8% if the total yield is N3700? A. N15 000 B. N29 600 C. N27 800 D. N21 400 E. N35 000
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Detailed SolutionTotal yield = N3,700Total amount invested = N 50,000 Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest then yield on x = \(\frac{6}{100}\) x and yield on y = \(\frac{8}{100}\)y Hence, \(\frac{6}{100}\)x + \(\frac{8}{100}\)y = 3,700.........(i) x + y = 50,000........(ii) 6x + 8y = 370,000 x 1 x + y = 50,000 x 6 6x + 8y = 370,000.........(iii) 6x + 6y = 300,000........(iv) eqn (iii) - eqn(2) 2y = 70,000< |
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| 6. |
Thirty boys and x girls sat for a rest. The mean of the boys scores and that of the girls were respectively 6 and 8. Find x if the total scores was 468 A. 38 B. 24 C. 36 D. 22 E. 41
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Detailed SolutionNumber of boy's = 30Number of girls = x Mean of boys' score = 6, for girls = 8 Total score for boys = 30 x 6 Total score for both boys and girls = 468 180 + 8x = 468 x = \(\frac{288}{8}\) = 36 |
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| 7. |
The cost of production of an article is made up as follows: Labour N70, Power N15, Materials N30, Miscellaneous N5. Find the angle of the sector representing Labour in a pie chart A. 210o B. 105o C. 105o D. 175o E. 90o
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Detailed SolutionTotal cost of production = N120.00Labour Cost = \(\frac{70}{120}\) x \(\frac{360^o}{1}\) = 210o |
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| 8. |
Bola choose at random a number between 1 and 300. What is the probability that the number is divisible by 4? A. \(\frac{1}{4}\) B. \(\frac{4}{4}\) C. \(\frac{6}{4}\) D. \(\frac{7}{4}\) E. \(\frac{37}{149}\)
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Detailed SolutionNumbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.PLet x represent numbers divisible by 4, nth term = a + (n - 1)d a = 4, d = 4 Last term = 4 + (n - 1)4 296 = 4 + 4n - 4 = \(\frac{296}{4}\) = 74 rn(Note: 296 is the last Number divisible by 4 between 1 and 300) Prob. of x = \(\frac{74}{298}\) = \(\frac{37}{149}\) |
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| 9. |
Find, without using logarithm tables, the value of \(\frac{log_3 27 - log_{\frac{1}{4}} 64}{log_3 \frac{1}{81}}\) A. \(\frac{-3}{9}\) B. \(\frac{-3}{2}\) C. \(\frac{6}{11}\) D. \(\frac{43}{78}\)
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Detailed Solution\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\)\(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\) \(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\) \(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\) \(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\) = \(\frac{6}{-4} = \frac{-3}{2}\) |
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