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.
The volume of a cone of height 9cm is 1848cm\(^3\). Find its radius. [Take π = 22/7]
7cm
14cm
28cm
98cm
196cm
Correct answer is B
1/3 πr\(^2\) x 9 = 1848
3 x πr2 = 1848 r2 = 1848/3 x 7/22 r = 14
\(\text{Volume of a cone} = \frac{1}{3} \pi r^2 h\)
\(\frac{1}{3} \times \frac{22}{7} \times r^2 \times 9 = 1848\)
\(r^2 = \frac{1848 \times 7}{22 \times 3}\)
\(r^2 = 196 \therefore r = 14cm\)
3.5cm
7cm
14cm
28cm
32cm
Correct answer is C
Curved surface area of a cylindrical tin = \(2\pi rh\)
\(\therefore 2\pi rh = 704cm^2\)
\(2 \times \frac{22}{7} \times 8 \times h = 704\)
\(h = \frac{704 \times 7}{2 \times 22 \times 8}\)
\(h = 14cm\)
16/3cm
15/3cm
16/5cm
8/3cm
16/10cm
Correct answer is A
L = θ/360 x 2πR = 240/360 x 2 x 22/7 x 8/1 ... (1)
This must be equal to the circumference of the circle which is 2πr = 44R/7 .... (2)
equate (1) and (2)
r = 16/3 = 51/3
Solve the equation \(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
-3
-2
2
3
4
Correct answer is D
\(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
\(\frac{m}{3} - \frac{m}{4} = \frac{3}{4} - \frac{1}{2}\)
\(\frac{m}{12} = \frac{1}{4}\)
\(4m = 12 \implies m = 3\)
What must be added to the expression x\(^2\) - 18x to make it a perfect square?
3
9
36
72
81
Correct answer is E
x\(^2\) - 18x to be a perfect square.
\((\frac{b}{2})^2\) is added to ax\(^2\) + bx + c in order to make it a perfect square.
\(x^2 - 18x + (\frac{-18}{2})^2\)
= \(x^2 - 18x + 81\)