Year :
1987
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
1 - 10 of 47 Questions
| # | Question | Ans |
|---|---|---|
| 1. |
Convert 241 in base 5 to base 8, A. 718 B. 1078 C. 1768 D. 2418 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 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 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 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 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 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 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 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% 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 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 |
| 1. |
Convert 241 in base 5 to base 8, A. 718 B. 1078 C. 1768 D. 2418 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 |
|
| 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 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 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 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 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 |
| 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 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 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 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 |
|
| 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% 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 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 |