Year :
1984
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
1 - 10 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 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 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 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 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 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 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 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 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. \(\frac{37}{149}\) 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. \(\frac{43}{78}\) 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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| 10. |
A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is A. -3 = \(\frac{y + 4}{x + 1}\) B. 4y = -3 + x C. \(\frac{y}{x}\) = \(\frac{-3}{4}\) D. 4x = y + 3 Detailed SolutionP(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4\(\frac{x}{1}\) = \(\frac{y}{4}\) y = 4x Hence the sign of the graph is y + 3 = 4x |
| 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 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 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 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 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 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< |
| 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 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 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. \(\frac{37}{149}\) 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. \(\frac{43}{78}\) 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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| 10. |
A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is A. -3 = \(\frac{y + 4}{x + 1}\) B. 4y = -3 + x C. \(\frac{y}{x}\) = \(\frac{-3}{4}\) D. 4x = y + 3 Detailed SolutionP(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4\(\frac{x}{1}\) = \(\frac{y}{4}\) y = 4x Hence the sign of the graph is y + 3 = 4x |