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.
17.1cm2
27.2cm2
47.1cm2
37.3cm2
Correct answer is C
Find the slant height
\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)
\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)
\( l^2 = √ 25 \)
Squaring both sides
l = 5cm
The area of curved surface (s) =π(3)(5)
15π = 15 × 3.14
= 47.1cm2
Integral ∫\( (5x^3 + 7x^2 − 2x + 5)\)dx
\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)
\( \frac{5x^4}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)
\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)
\( \frac{2x^2}{3} + \frac{x}{5} - C \)
Correct answer is B
\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\)
= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\)
= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)
Evaluate log5(\( y^2x^5 ÷ 125b½) \)
2 log5y + 5log5 y2 − 3
log5 y2 + 5log5 x + 3
25logy 5 + 3
2log5y + 5log5x − ½ log5b −3
Correct answer is D
\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\)
= \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\)
= \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\)
= \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)
100112 + *****2 + 111002 + 1012 = 10011112
11112
110112
101112
110012
Correct answer is B
Convert the binary to base 10 and they convert back to base two
100112 + xxxxx2 + 111002 + 1012 = 10011112
(1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20)
= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx2 + 33 = 79
xxxxx2 + 52 = 79
xxxxx2 = 79 − 52
xxxxx2 = 2710
\( \begin{array}{c|c}
2 & 27 \\
\hline
2 & 13 \text{ rem 1}\\
2 & 6 \text{ rem 1}\\
2 & 3 \text{ rem 0}\\
2 & 1 \text{ rem 1}\\
& 0 \text{ rem 1}\\
\end{array}\uparrow \)
2710 = 110112
Therefore xxxxx2 = 2710 = 110112
1.35
1.353
1.455
0.455
Correct answer is C
log717
= [log 17 ÷ log7]
= [1.2304 ÷ 0.8451]
[100.0899 ÷ 101.9270]
= 1.455(antilog)