Mathematics (Core)
Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1979
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
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
Managing Stress and anxiety in exams, 10 tips and strategies for Coping exam stress and test anxiety.
Middle East and North Africa MENA Scholarship Program (MSP) initiative provides scholarships
Adequate preparation and effective revision strategies tips to get a high score 320 in JAMB in 2022
| # | 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
Show Content
Detailed Solution\(\frac{1.2 + 1.0 + 0.9 + 1.4 + 0.8 + 0.8 + 1.2}{8}\) = \(\frac{8.4}{8}\)= 1.05 |
|
| 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
Show Content
Detailed Solution\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\)= \(\frac{1.28 \times 10^{4}}{6.4 \times 10^{2}}\) \(\equiv \frac{12.8 \times 10^{3}}{6.4 \times 10^{2}}\) = \(2 \times 10^{1}\) |
|
| 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
Show Content
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\(^{2}\) |
|
| 4. |
(3.2)2 - (1.8)2 equals A. 7.0 B. 2.56 C. 13.48 D. 2.0 E. 0.07
Show Content
Detailed Solution(3.2)2 - (1.8)2 = (3.2 + 1.8)(3.2 - 1.8)= 5 x 1.4 = 7 |
|
| 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
Show Content
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 |
|
| 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
Show Content
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 |
|
| 7. |
In base ten, the number 101101 (base 2) equals A. 15 B. 4 C. 45 D. 32 E. 90
Show Content
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 |
|
| 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
Show Content
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 |
|
| 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\(\frac{1}{5}\)
Show Content
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\(\frac{1}{2}\) |
|
| 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
Show Content
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 |
|
| 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
Show Content
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