Mathematics (Core)
Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1979
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1979
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
| # | Question | Ans |
|---|---|---|
| 1. |
The mean of the numbers 1.2, 1.0, 0.9, 1.4, 0.8, 0.8, 1.2 and 1.1 is A. 1.5 B. 0.8 C. 1.0 D. 1.02 E. 1.05
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Detailed Solution\(\frac{1.2 + 1.0 + 0.9 + 1.4 + 0.8 + 0.8 + 1.2}{8}\) = \(\frac{8.4}{8}\)= 1.05 |
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| 2. |
\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\) equals A. 2 x 10-5 B. 2 x 10-1 C. 2 x 101 D. 2 x 10-4
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Detailed Solution\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\)= \(\frac{1.28 \times 10^{4}}{6.4 \times 10^{2}}\) \(\equiv \frac{12.8 \times 10^{3}}{6.4 \times 10^{2}}\) = \(2 \times 10^{1}\) |
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| 3. |
If the value of \(\pi\) is taken to be \(\frac{22}{7}\), the area of a semi-circle of diameter 42m is A. 5544m2 B. 1386m2 C. 132m2 D. 264m2 E. 693m2
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Detailed Solution\(\pi\) = \(\frac{22}{7}\), Diameter D = 42mArea of a circle = \(\frac{\pi D^2}{4}\) when a diameter is given area of a semicircle \(\frac{\pi D^2}{4}\) x \(\frac{1}{2}\) = \(\frac{22}{7}\) x \(\frac{4262}{4}\) x \(\frac{1}{2}\) = \(\frac{11}{7} \times 441\) = 693m\(^{2}\) |
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| 4. |
(3.2)2 - (1.8)2 equals A. 7.0 B. 2.56 C. 13.48 D. 2.0 E. 0.07
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Detailed Solution(3.2)2 - (1.8)2 = (3.2 + 1.8)(3.2 - 1.8)= 5 x 1.4 = 7 |
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| 5. |
In \(\bigtriangleup\)PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS A. 4cm B. 32cm C. \(\frac{30}{7}\) D. \(\frac{40}{7}\) E. 4.8cm
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Detailed SolutionCos Q = \(\frac{r^2 + p^2 - q^2}{2rp}\)= \(\frac{10^2 + 8^2 - 6^2}{2(10)(8)}\) = \(\frac{164 - 36}{160}\) = \(\frac{128}{160}\) = 0.8 Q = Cos-1 o.8 = 37o x rep. from rt< RSQ, Let RS = x \(\frac{x}{sin 37^o}\) = \(\frac{8}{sin 90^o}\) but sin 90o = 1 x = 8 sin 37o x = 4.8cm |
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| 6. |
After getting a rise of 15%, a man's new monthly salary is N345. How much per month did he earn before the increase? A. N330 B. N396.75 C. N300 D. N293.25 E. N360
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Detailed SolutionLet x represent his monthly salary before increase 15% ofx = 345 % profit = (100 + 15)% = 115% \(\frac{115}{100}\)x = 345 115x = 34500 x = N300.00 |
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| 7. |
In base ten, the number 101101 (base 2) equals A. 15 B. 4 C. 45 D. 32 E. 90
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Detailed Solution(101101)2 \(\begin{array}{c|c} 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^o\\ \hline 1 & 0 & 1 & 1 & 0 & 1\end{array}\)= 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 2o = 32 + 0 + 8 + 4 + 0 + 1 = (45)10 |
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| 8. |
The annul profits of a transport business were divided between the partners A and B in the ratio 3 : 5. If B received N3000 more than A, the total profit was A. N5000 B. N1800 C. N12000 D. N24000 E. N8000
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Detailed SolutionA : B = 3 : 5Total ratio = 3 + 5 = 8 Let rept, the total profit, A receives \(\frac{3x}{8}\) \(\frac{5x}{8}\) - \(\frac{3}{x}\) = 3000 \(\frac{5x}{8}\) - \(\frac{3x}{8}\) = 3000 = \(\frac{2x}{8}\) = 3000 2x = 2400 x = 1200 N12,000 |
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| 9. |
X is directly proportional to y and inversely proportional to z. If x = 9 when y = 24 and z = 8, what is the value of x when y = 5 and z = 6? A. \(\frac{5}{6}\) B. 11 C. 3\(\frac{3}{5}\) D. 2\(\frac{1}{2}\) E. 1\(\frac{1}{5}\)
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Detailed Solutionx \(\alpha\) y = x \(\alpha\) \(\frac{1}{z}\)x \(\alpha\) \(\frac{1}{z}\) x = k \(\frac{y}{z}\) k = \(\frac{xz}{y}\) = \(\frac{9 \times 8}{24}\) = 3 x = \(\frac{xz}{y}\) = \(\frac{3 \times 5}{6}\) = \(\frac{15}{6}\) = \(\frac{5}{2}\) = 2\(\frac{1}{2}\) |
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| 10. |
The solution of the equation x2 - 2x = 8 is A. x = 0 or 2 B. x = -2 or 4 C. x = 2 D. x = -4 E. x = 2 or 4
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Detailed Solutionx2 - 2x = 8 = 8x2 - 2x - 8 = 0 (x2 - 4x) + (2x - 8) = 0 When x + 2 = 0 x = -2, when x - 4 = 0 x = 4 -2 or 4 |
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| 11. |
A trader goes to Ghana for y days with y cedis. For the first x days, he spends X cedis per day. The amount he has to spend per day for the rest of his stay is A. \(\frac{y(y - x)}{y - x}\) cedis B. \(\frac{Yy - Xx)}{y - x}\) cedis C. \(\frac{Y - Xy)}{y - x}\) cedis D. \(\frac{Y - X}{y - x}\) cedis E. \(\frac{Y - Xx}{y - x}\) cedis
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Detailed SolutionThe amount he has to spend per day for the rest of his stay is \(\frac{Y - Xx}{y - x}\) cedis |
Preview displays only 11 out of the 51 Questions