How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
Evaluate \(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} \times \frac{1}{4}\)
- \(\frac{3}{40}\)
\(\frac{3}{40}\)
\(\frac{7}{40}\)
-\(\frac{263}{40}\)
Correct answer is C
\(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} x \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{3}{4} ÷ \frac{5}{12}) \times \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\)
⇒ \(\frac{5}{8} - \frac{9}{20}\)
∴ \(\frac{7}{40}\)
Find the area and perimeter of a square whose length of diagonals is 20\(\sqrt2\) cm
800 cm\(^2\), 80 cm
400 cm, 80 cm\(^2\)
80 cm, 800 cm\(^2\)
400 cm\(^2\), 80 cm
Correct answer is D
Using Pythagoras theorem
⇒ \((20\sqrt2)2 = x^2 + x^2\)
⇒ 800 = \(2x^2\)
⇒ 400 = \(x^2\)
⇒ x = \(\sqrt400\) = 20 cm
∴ Area of a square = \(x^2 = 20^2 = 400 cm^2\)
∴ Perimeter of a square = 4x = 4 x 20 = 80 cm
Find the volume of the composite solid above.
2048 cm\(^3\)
2568 cm\(^3\)
2672 cm\(^3\)
1320 cm\(^3\)
Correct answer is B
Volume of the composite solid = Volume of A + Volume of B
Volume of a cuboid = length x breadth x height
Volume of A = 6 x 26 x 8 = 1248 cm\(^3\)
Volume of B = 6 x 10 x 22 = 1320 cm\(^3\)
∴ Volume of the composite solid = 1248 + 1320 = 2568 cm\(^3\)
Two dice are tossed. What is the probability that the total score is a prime number.
\(\frac{5}{12}\)
\(\frac{5}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{3}\)
Correct answer is A
Total possible outcome = 6 x 6 = 36
Required outcome = 15
∴ Pr(E) = \(\frac{15}{36} = \frac{5}{12}\)
11
13
12
14
Correct answer is B
An exterior angle of a n-sided regular polygon = \(\frac{360}{n}\)
For (n - 1) sided regular polygon = \(\frac{360}{n - 1}\)
For (n + 2) sided regular polygon = \(\frac{360}{n + 1}\)
⇒ \(\frac{360}{n - 1} - \frac{360}{n + 2}\) = 6 9Given)
⇒ \(\frac{360(n + 2) - 360(n - 1)}{(n - 1)(n + 2)}\)
⇒ \(\frac{360n + 720 - 360n + 360}{(n - 1)(n + 2)}\)
⇒ \(\frac{1080}{(n - 1)(n + 2)} = \frac{6}{1}\)
⇒ 1080 = 6 (n - 1)(n + 2)
⇒ 180 = (n - 1)(n + 2)
⇒ 180 = n\(^2\)+ 2n - n - 2
⇒ 180 = n\(^2\) + n - 2
⇒ n\(^2\)b+ n - 2 - 180 = 0
⇒ n\(^2\) + n - 182 = 0
⇒ n\(^2\) + 14n - 13n - 182 = 0
⇒ n (n + 14) - 13 (n + 14) = 0
⇒ (n - 13) (n + 14) = 0
⇒ n - 13 = 0 or n + 14 = 0
⇒ n = 13 or n = -14
∴ n = 13 (We can't have a negative number of side)
In the last five seasons, what was the difference in the average performance ratings between Team B and Team A?
1.2
6.4
4.6
1.8
Correct answer is A
Average performance rating of Team B = \(\frac{7+9+1+9+6}{5} = \frac{32}{5}\) = 6.4
Average performance rating of Team A = \(\frac{5+3+6+10+2}{5} = \frac{26}{5}\) = 5.2
∴ The difference in the average performance ratings between Team B and Team A = 6.4 - 5.2 = 1.2
Express 16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\) in standard form
1.66 x 10\(^{-4}\)
1.66 x 10\(^{-5}\)
1.65 x 10\(^{-5}\)
1.65 x 10\(^{-4}\)
Correct answer is A
16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\)
⇒ 1.654 x 10\(^{-4}\) - 6.76 x 10\(^{-8}\) + 2.3 x 10\(^{-7}\)
⇒ 1.654 x 10\(^{-4}\) - 0.000676 x 10\(^{-4}\) + 0.0023 x 10\(^{-4}\)
⇒ (1.654 - 0.000676 + 0.0023) x 10\(^{-4}\)
∴ 1.655624 x 10\(^{-4}\) ≃ 1.66 x 10\(^{-4}\)
The third term of an A.P is 6 and the fifth term is 12. Find the sum of its first twelve terms
201
144
198
72
Correct answer is C
T\(_3\) = 6
T\(_5\) = 12
S\(_{12}\) = ?
T\(_n\) = a + (n - 1)d
⇒ T\(_3\) = a + 2d = 6 ----- (i)
⇒ T\(_5\) = a + 4d = 12 ----- (ii)
Subtract equation (ii) from (i)
⇒ -2d = -6
⇒ d\(\frac{-6}{-2}\) = 3
Substitute 3 for d in equation (i)
⇒ a + 2(3) = 6
⇒ a + 6 = 6
⇒ a = 6 - 6 = 0
S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\)
⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\)
⇒ S\(_{12}\) = 6(0 + 11 x 3)
⇒ S\(_{12}\) = 6(33)
∴ S\(_{12}\) = 198
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