Paper 1 | Objectives | 62 Questions
JAMB Exam
Year: 2015
Level: SHS
Time:
Type: Question Paper
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1. |
In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate A. 3% B. 1.0% C. 2.5% D. 5.40%
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Detailed SolutionPercentage birthrate = \(\frac{62}{6250} \times 100%\)= \(\frac{6200}{6250}\) = 0.992% \(\approx\) 1.0%. There is an explanation video available below. |
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2. |
Simplify 1¼ ÷ (2 ÷ ¼ of 28) A. 1 \( \frac{3}{8}\) B. 2 \( \frac{3}{4}\) C. 4 \( \frac{3}{8}\) D. 3 \( \frac{1}{5}\) |
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4. |
If three staff of Myschool Limited agreed to share their salary arrears in the ration of their ages, which are 18 years, 20 years, 22 years respectively. If the sum of the money collected is N120,000.00K, How much does the second staff received? A. N36,000 B. N44,000 C. N40,000 D. N15,000
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Detailed SolutionTotal of their ages: 18 + 20 + 22 = 60The second staff will get \(\frac{20}{60} \times 120,000\) = N40,000.00K There is an explanation video available below. |
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5. |
X and Y are two sets such that n(X) = 15, n(Y) = 12 and n{X ∩ Y} = 7. Find ∩{X ∪ Y} A. 21 B. 225 C. 15 D. 20
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Detailed Solutionn(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y)= 15 + 12 − 7 ∴ n(X ∪ Y) = 20 There is an explanation video available below. |
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6. |
Find the x and z intercepts of the graph of 3x - z \(\leq\) 9 A. (3, -9) B. (-3, 9) C. (-3, -9) D. (-3, 0)
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Detailed SolutionStarting from 3x − z \(\leq\) 9x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z \(\leq\) 9 3(0) − z \(\leq\) 9 z \(\leq\) − 9 If z = 0 Then, 3x − 0 \(\leq\) 9 3x \(\leq\) 9 x \(\leq\) 3 Intercept of x and z i.e (x,z) = (3, -9) There is an explanation video available below. |
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8. |
Find the total surface area of a cylinder of base radius 5cm and length 7cm ( π = 3.14) A. 17.8cm2 B. 15.8cm2 C. 75.4cm2 D. 54.7cm2 E. \(377.0cm^{2}\)
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Detailed SolutionThe total surface area of a cylinder = 2πrl + 2πr2= 2πr(l + r) = 2 × 3.14 x 5(7+5) 2 × 3.14 × 12 x 5 = 377.1cm (1DP) There is an explanation video available below. |
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9. |
If x + y = 90 simplify (sinx + siny)2−2sinxsiny A. 1 B. 0 C. 2 D. -1
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Detailed SolutionGiven: \(x + y = 90° ... (1)\)\((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\) = \(\sin^{2} x + \sin^{2} y ... (2)\) Recall: \(\sin x = \cos (90 - x) ... (a)\) From (1), \(y = 90 - x ... (b)\) Putting (a) and (b) in (2), we have \(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\) = 1 There is an explanation video available below. |
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10. |
A man with an annual salary of N2000, has allowances of N600. If Income Tax is 5%. How much income tax expenses does he pay each year? A. 15 B. 50 C. 70 D. 25
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Detailed SolutionHis annual salary = N2000His allowances = N600 So his taxable income = Annual salary − allowance = N2000 − N600 = N1400 He pay at 5% Then, his allowance income tax 5/100 × 1400 = N70 There is an explanation video available below. |
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11. |
Solve the equation \( 3x^2 − 4x − 5 = 0 \) A. x = 1.75 or − 0.15 B. x = 2.12 or − 0.79 C. x = 1.5 or − 0.34 D. x = 2.35 or −1.23
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Detailed SolutionUsing the quadratic formula, we have\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\) There is an explanation video available below. |
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12. |
The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms A. nth = 3 B. nth = 4 C. nth = 5 D. nth = 6
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Detailed Solutionnth term of a linear sequence (AP) = a+(n − 1)dfirst term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l) S\(_{n}\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5 There is an explanation video available below. |
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13. |
Find the equation of a line which is form origin and passes through the point (−3, −4) A. y = \( \frac{3x}{4} \) B. y = \( \frac{4x}{3} \) C. y = \( \frac{2x}{3} \) D. y = \( \frac{x}{2} \)
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Detailed SolutionThe slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)= \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\) There is an explanation video available below. |
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