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Mathematics (Core)
Paper 1 | Objectives | 40 Questions
JAMB Exam
Year: 2020
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper has been added to Cart!
Paper 1 | Objectives | 40 Questions
JAMB Exam
Year: 2020
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
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| # | Question | Ans |
|---|---|---|
| 1. |
Evaluate (212)\(_3\) - (121)\(_3\) + (222)\(_3\) A. (313)\(_3\) B. (1000)\(_3\) C. (1020)\(_3\) D. (1222)\(_3\)
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Detailed SolutionEvaluate (212)\(_3\) - (121)\(_3\) + (222)\(_3\)(212)\(_3\) + (222)\(_3\) = 1211\(_3\) → 1211\(_3\) - 121\(_3\) = 1020\(_3\) There is an explanation video available below. |
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| 2. |
Factorise (4a + 3) \(^2\) - (3a - 2)\(^2\) A. (a + 1)(a + 5) B. (a - 5)(7a - 1) C. (a + 5)(7a + 1) D. a(7a + 1)
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Detailed Solution[(4a + 3) \(^2\) - (3a - 2)\(^2\) = a\(^2\) - b\(^2\) = (a + b) (a - b)= [(4a + 3) + (3a - 2)] [(4a + 3) - (3a - 2)] = [4a + 3 + 3a - 2] [4a + 3 - 3a + 2] = (7a + 1)(a + 5) (a + 5) (7a + 1) There is an explanation video available below. |
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| 3. |
Find all median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119 A. 131 B. 125 C. 123 D. 120
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Detailed SolutionArrange in ascending order89, 100, 108, 119, 120,130, 131, 131, 141, 161 Median = \(\frac{120 + 130}{2}\) = 125 There is an explanation video available below. |
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| 4. |
Find all real number x which satisfy the inequality \(\frac{1}{3}\) (x + 1) - 1 > \(\frac{1}{5}\)(x + 4) A. x < 11 B. x < -1 C. x > 6 D. x > 11
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Detailed Solution\(\frac{1}{3}\) (x + 1) - 1 > \(\frac{1}{5}\)(x + 4) = \(\frac{x + 1}{3} - 1\) > \(\frac{x + 4}{5}\)\(\frac{x + 1}{3} - \frac{x + 4}{5} -1\) > 0 = \(\frac{5x + 5 - 3x - 12}{15}\) 2x - 7 > 15 2x > 22 = x > 11 There is an explanation video available below. |
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| 5. |
Reach each number to two significant figures and then evaluate \(\frac{0.02174 \times 1.2047}{0.023789}\) A. 0 B. 0.9 C. 1.1 D. 1.2
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Detailed Solution\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.0255 \times 1.2}{0.024}\) (to 216)= \(\frac{0.0264}{0.024}\) = 1.1 There is an explanation video available below. |
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| 6. |
Four boys and ten girls can cut a field first 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 rep. numbers of boys that can work at \(\frac{5}{4}\) the rate atwhich the 10 girls work For 1 hrs, x boys will work for \(\frac{\frac{1}{5} \times 10}{4}\) x = \(\frac{5}{4}\) x 10 = 8 boys 8 boys will do the work of ten girls at the same rate 4 + 8 = 12 bous cut the field in 5 hrs for 3 hrs, \(\frac{12 \times 5}{3}\) boys will be needed = 20 boys. There is an explanation video available below. |
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| 7. |
What is the circumference of latitude 0\(^o\)S if R is the radius of the earth? A. cos \(\theta\) B. 2\(\pi\)R cos \(\theta\) C. R sin \(\theta\) D. 2 \(\pi\) r sin \(\theta\)
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Detailed SolutionCircumstances of latitude 0\(^o\)s where R is the radius of the earth 2\(\pi\)r cos \(\theta\)There is an explanation video available below. |
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| 8. |
A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room A. \(\frac{15}{17}\) B. \(\frac{8}{17}\) C. \(\frac{8}{15}\) D. \(\frac{12}{17}\)
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Detailed SolutionABCD is the floor. By pathagorasAC\(^2\) = 144 + 81 = \(\sqrt{225}\) AC = 15cm Height of room 8m, diagonal of floor is 15m Therefore, the cosine of the angle which a diagonal of the room makes with the floor is EC\(^2\) = 15\(^2\) + 8\(^2\) cosine \(\frac{adj}{Hyp} = \frac{15}{17}\) EC\(^2\) = \(\sqrt{225 + 64}\) EC = \(\sqrt{289}\) = 17 There is an explanation video available below. |
Preview displays only 8 out of the 40 Questions