Mathematics questions and answers

Mathematics Questions and Answers

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.

1,096.

Find the area of the curved surface of a cone whose base radius is 3cm and whose height is 4cm (π = 3.14)

A.

17.1cm2

B.

27.2cm2

C.

47.1cm2

D.

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

1,097.

Integral ∫\( (5x^3 + 7x^2 − 2x + 5)\)dx

A.

\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)

B.

\( \frac{5x^4}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)

C.

\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)

D.

\( \frac{2x^2}{3} + \frac{x}{5} - C \)

Correct answer is B

\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\)

= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\)

= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)

1,098.

Evaluate log5(\( y^2x^5 ÷ 125b½) \)

A.

2 log5y + 5log5 y2 − 3

B.

log5 y2 + 5log5 x + 3

C.

25logy 5 + 3

D.

2log5y + 5log5x − ½ log5b −3

Correct answer is D

\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\)

= \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\)

= \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\)

= \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)

1,099.

100112 + *****2 + 111002 + 1012 = 10011112

A.

11112

B.

110112

C.

101112

D.

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

1,100.

Evaluate log717

A.

1.35

B.

1.353

C.

1.455

D.

0.455

Correct answer is C

log717

= [log 17 ÷ log7]

= [1.2304 ÷ 0.8451]

[100.0899 ÷ 101.9270]

= 1.455(antilog)