Paper 1  Objectives  47 Questions
JAMB Exam
Year: 1981
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = \(\frac{1}{3}\). A. 81 B. 27 C. \(\frac{1}{9}\) D. \(\frac{1}{27}\) E. \(\frac{1}{81}\)
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Detailed Solution\(x \propto \frac{1}{y}\)\(x = \frac{k}{y}\) \(y \propto t^{2}\) \(y = ct^{2}\) k and c are constants. \(x = \frac{k}{ct^{2}}\) Let \(\frac{k}{c} = d\) (a constant) \(x = \frac{d}{t^{2}}\) \(1 = \frac{d}{3^{2}} \implies d = 9\) \(\therefore x = \frac{9}{t^{2}}\) \(x = 9 \div (\frac{1}{3})^{2} \) = \( 9 \div \frac{1}{9} = 9 \times 9 = 81\) 

2. 
If sine x equals cosine x, what is x in radians? A. \(\frac{\pi}{2}\) B. \(\frac{\pi}{3}\) C. \(\frac{\pi}{4}\) D. \(\frac{\pi}{6}\) E. \(\frac{\pi}{12}\)
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Detailed Solution\(\sin x = \cos x\)\(\implies x = 45°\) In radians, \(x = \frac{\pi}{4}\). 

3. 
The ratio of the price of a loaf of bread to the price of a packet of sugar in 1975 was r : t. In 1980, the price of a loaf went up by 25% and that of a packet of sugar went up by 10%. Their new ratio is now A. 40r:50t B. 44r : 50t C. 50r : 44t D. 44r:55t E. 55r:44t
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Detailed SolutionRatio of bread to sugar = r:t25% increase in bread = \(\frac{125r}{100}\) 10% increase in sugar = \(\frac{100t}{100}\) New ratio = \(\frac{125r}{100}\):\(\frac{110t}{100}\) = 25r:22t = 50r:44t 

4. 
Find a twodigit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits A. 24 B. 42 C. 74 D. 47 E. 72
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Detailed SolutionLet the tens digits of the number be x and the unit digit be y3x = 2y  2 3x  2y = 2.......(i) If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20 (20x + 2y)  (10y + x) = 20 19x  8y = 20.....(ii) Multiply eqn.(i) by 8 and eqn.(ii) by 2 24x  16y = 16......(iii) 38x  16y = 40........(iv) eqn(iv)  eqn(iii) 14x = 56 x = 4 Sub. for x = 4 in eqn(i) 3(4)  

5. 
Find the value of x satisfying \(\frac{x}{2}\)  \(\frac{1}{3}\) < \(\frac{2x}{5}\) + \(\frac{1}{6}\) A. x < 5 B. x < 7\(\frac{1}{2}\) C. x > 5 D. x > 7\(\frac{1}{2}\)
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Detailed Solution\(\frac{x}{2}  \frac{1}{3} < \frac{2x}{5} + \frac{1}{6}\)\(\frac{x}{2}  \frac{2x}{5} < \frac{1}{6} + \frac{1}{3}\) \(\frac{x}{10} < \frac{1}{2}\) \(2x < 10 \implies x < 5\) 

6. 
A group of 14 children children received the following scores in a reading test: 35, 35, 26, 26, 26, 29, 29, 29, 12, 25, 25, 25, 25, 17. What was the median score? A. 29 B. 26 C. 24 D. 25 E. 23
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Detailed SolutionArranging the scores in ascending order:12, 17, 25, 25, 25, 25, 26, 26, 26, 29, 29, 29, 35, 35. The median is the average of the 7th and 8th marks. = \(\frac{26 + 26}{2} = 26\) 

7. 
Which of the following fractions is less than onethird? A. \(\frac{22}{63}\) B. \(\frac{4}{11}\) C. \(\frac{15}{46}\) D. \(\frac{33}{98}\) E. \(\frac{122}{303}\)
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Detailed SolutionAll others are greater than 0.333 when converted to their fractions except \(\frac{15}{46}\) 

8. 
A cuboid has a diagonal of length 9cm and a square base of side 4cm. What is its height? A. 9cm B. \(\sqrt{65}\)cm C. \(4\sqrt{2}\)cm D. 7cm E. 6.5cm
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Detailed SolutionGiven a cuboid, the diagonal cuts a face of the cuboid into 2 rightangled triangles.Hence, using the Pythagoras theorem, we have \(9^{2} = 4^{2} + x^{2}\) \(81 = 16 + x^{2}\) \(x^{2} = 81  16 = 65\) \(\therefore x = \sqrt{65} cm\) 

9. 
Evaluate correct to 4 decimal places 827.51 x 0.015 A. 8.8415 B. 12.4127 C. 124.1265 D. 12.4120 E. 114.1265
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Detailed Solution827.51 x 0.015By normal multiplication or use of four figure table, 827.51 x 0.015 = 12.4127 (to 4 decimal places). 

10. 
What is the area between two concentric circles of diameters 26cm and 20cm? A. 100\(\pi\) B. 169\(\pi\) C. 69\(\pi\) D. 9\(\pi\) E. 269\(\pi\)
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Detailed SolutionArea of circle 1 with diameter 26cm:\(\pi r^{2} = \pi \times (\frac{26}{2})^{2} \) = \(169 \pi cm^{2}\) Area of circle 2 with diameter 20 cm: \(\pi R^{2} = \pi \times (\frac{20}{2})^{2}\) = \(100 \pi cm^{2}\) Area between the two circles = \((169  100) \pi cm^{2}\) = \(69 \pi cm^{2}\) 
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