Mathematics (Core)
Paper 1 | Objectives | 47 Questions
JAMB Exam
Year: 1987
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 47 Questions
JAMB Exam
Year: 1987
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
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| 1. |
Convert 241 in base 5 to base 8, A. 718 B. 1078 C. 1768 D. 2418
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Detailed Solution2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o50 + 20 + 1 = 7110 Convert 7110 to base 8 \(\begin{array}{c|c} 8 & 71 \\ 8 & 8 R 7\\8 & 1 R 0\\8 & 0 R 1\end{array}\) = 1078 |
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| 2. |
Find the least length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length A. 120cm B. 240cm C. 360cm D. 480cm
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Detailed SolutionThe least length is \(\frac{40}{48}\) = \(\frac{5}{8}\)for the rod to be cut in exactly equal trips Ratio \(\frac{5}{6}\) : \(\frac{48}{40}\) \(\frac{\frac{5}{6}}{\frac{40}{48}}\) = 1 \(\frac{5}{6}\) x \(\frac{48}{40}\) = \(\frac{240}{240}\) = 1 The least length = 240cm |
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| 3. |
A rectangular lawn has an area of 1815 square yards. if its length is 50 meters, find its width in meters given that 1 metre equals 1.1 yards A. 39.93 B. 35.00 C. 33.00 D. 30.00
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Detailed Solution1m = 1.1 yard, length(L)= 50m= (50 x 1.1)yards = 55 yards Area(A) = length(L) y width (W) 1815 = 55y width (W) width (w) = \(\frac{1815}{55}\) = 33 yards But 33 yards = 30 meters |
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| 4. |
Reduce each number to two significant figures and then evaluate \(\frac{0.021741 \times 1.2047}{0.023789}\) A. 0.8 B. 0.9 C. 1.1 D. 1.2
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Detailed Solution\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.022 \times 1.2}{0.024}\) (to 216)= \(\frac{0.0264}{0.024}\) = 1.1 |
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| 5. |
A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey A. 55.00km/h B. 60.00km/h C. 67.50km\h D. 75.00km\h
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Detailed SolutionAverage speed from P to Q = 90km\hAverage speed from O to P = 45km/h Average for the entire journey = 90 + 45 \(\frac{135}{2}\) = 67.50 km/h |
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| 6. |
If the length of a square is increased by 20% while while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square? A. 6 : 5 B. 25 : 24 C. 5 : 6 D. 24 : 25
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Detailed SolutionLength and width of a square is 100%Length increased by 20% and Width decreased by 20% to form a rectangle Length of rectangle = 120% to form a rectangle Length of rectangle = 120% and Width of rectangle = 80% Area of rectangle = L x W Area of square = W Ratio of the area of the rectangle to the area of the square A = \(\frac{\text{Area of rectangle}}{\text{Area of square}}\) \(\frac{120 \times 30}{100 \times 100}\) = \(\frac{96}{100}\) = 24 : 25 |
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| 7. |
Two brothers invested a total of N5,000.00 on a farm project, the farm yield was sold for N 15,000.00 at the end of the season. If the profit was shared in the ratio 2 : 3, what is the difference in the amount to profit received by the brothers? A. N2,000.00 B. N4,000.00 C. N6,000.00 D. N10,000.00
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Detailed SolutionTotal amount invested by A and B = N5,000farm yield was sold for N15,000.00 profit = 15,000.00 - 5,000.00 = N10,000.00 Profit was shared in ratio 2 : 3 2 + 3 = 5 A received \(\frac{2}{5}\) of profit = \(\frac{2}{5}\) x 10,000 = N4,000.00 A receive \(\frac{3}{5}\) of profit = \(\frac{3}{5}\) x 10,000 = N6,000.00 Difference in profit received = N6,0000 - N4,000.00 = N2,000.00 |
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| 8. |
A man invests a sum of money at 4% per annum simple invest. After 3 years, the principal amounts to N7,000.00. Find the sum invested A. N7,840 B. N6,250.00 C. N616.00 D. N5,833.33
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Detailed SolutionI = \(\frac{PRT}{100}\)100(A - P) = prt P = \(\frac{100A}{100 + RT}\) = \(\frac{100 \times 7000}{100 + (4 \times 3)}\) p = \(\frac{700,000}{112}\) = N6,250.00 |
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| 9. |
By selling 20 oranges for N1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for N1.10? A. 8% B. 10% C. 12% D. 15%
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Detailed Solutionprofit 8% of N1.35 = \(\frac{8}{100}\) x N1.35 = N0.08Cost price = N1.35 - N0.10 = N1.25 If he sells the 20 oranges for N1.10 now %loss = \(\frac{\text{actual loss}}{\text{Cost price}}\) x 100 \(\frac{125 - 1.10}{1.25}\) x 100 = \(\frac{0.15 \times 100}{1.25}\) = \(\frac{15}{1.25}\) = 12% |
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| 10. |
Four boys and ten girls can cut a field in 5 hours if the boys work at \(\frac{5}{4}\) the rate at which the girls work. How many boys will be needed to cut the field in 3 hours? A. 180 B. 60 C. 25 D. 20
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Detailed SolutionLet x represents number of boys that can work at \(\frac{5}{4}\) the rate at which the 10 girls workFor 1hr. x boys will work for \(\frac{\frac{1}{5}}{4}\) x 10 x = \(\frac{4}{5}\) x 10 = 8 boys 8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs For 3 hrs, \(\frac{12 \times 5}{3}\) boys will be needed = 20 boys |
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