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.
If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two
1110
10111
11101
111100
Correct answer is B
\(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13\\
101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\\
2 & 11 &R1\\
2 & 5 & R1\\
2 & 2 & R1\\
2 & 1 & R0\\
& 0& R1 \uparrow\\
\end{matrix} \\
=y=10111_2\)
Evaluate \( 202^2_{three} - 112^2_{three}\)
21120
21121
21112
21011
Correct answer is A
\(202^2_{three}\)when converted to base ten \(=(202_3)^2\\
202_3 = 2 \times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\\
=20_{ten}; (202_3)^2 = (20)^2_{ten} = 400\\
112^2_{three}\)when converted to base ten \(= (112_3)^2\\
112_3 = 1 \times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\\
(112_3)^2 = (14)^2_{ten} = 196_{ten}\\
Evaluate \Longrightarrow 400-196 = 204\)
Reconvert to base three
\(\begin{matrix}
3 & 204\\
3 & 69 &R0\\
3 & 22 & R2\\
3 & 7 & R1\\
2 & 2 & R1\\
& 0& R2 \uparrow\\
\end{matrix} \\
=21120_3\)
The bearing of P from Q is x, where 270o < x < 360o. Find the bearing of Q from P
(x - 90)o
(x-270)o
(x - 135)o
(x - 180)o
Correct answer is D
Bearing of Q and P
\(180^{\circ} – (360^{\circ}-x)\\
180^{\circ}-360^{\circ}+x\\
-180^{\circ}+x\\
x-180^{\circ}\)
A man made a loss of 15% by selling an article for N595. Find the cost price of the article
N600.00
N684.25
N700.00
N892.50
Correct answer is C
15% loss = N595
\(\implies\) (100 - 15)% of CP = N595
85% of CP = N595
CP = \(\frac{595 \times 100}{85}\)
= N700.00
A right circular cone is such that its radius r is twice its height h. Find its volume in terms of h
\(\frac{2}{3}\pi h^2\)
\(\frac{1}{12}\pi h^3\)
\(\frac{4}{3}\pi h^2\)
\(\frac{4}{3}\pi h^3\)
Correct answer is D
Volume of a cone = \(\frac{\pi r^2 h}{3}\)
r = 2h.
V = \(\frac{\pi \times (2h)^2 \times h}{3}\)
= \(\frac{4}{3} \pi h^3\)