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.

301.

What is the rate of change of the volume V of a hemisphere with respect to its radius r when r = 2?

A.

B.

16π

C.

D.

Correct answer is A

\(V = \frac{2}{3} \pi r^{3}\)

\(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\)

\(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\)

= \(8\pi\)

302.

In how many ways can 2 students be selected from a group of 5 students in a debating competition?

A.

25 ways

B.

10 ways

C.

15 ways

D.

20 ways

Correct answer is B

\(In\hspace{1mm} ^{5}C_{2}\hspace{1mm}ways\hspace{1mm}=\frac{5!}{(5-2)!2!}\\=\frac{5!}{3!2!}\\=\frac{5\times4\times3!}{3!\times2\times1}\\=10\hspace{1mm}ways\)

303.

Solve the following equation: \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) =  \(\frac{1}{(r + 2)}\)

A.

( -1,\(\frac{5}{2}\) )

B.

( 1, - \(\frac{5}{2}\) )

C.

( \(\frac{5}{2}\), 1 )

D.

(2,1)

Correct answer is B

\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) =  \(\frac{1}{(r + 2)}\)

\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\)  = \(\frac{5}{3}\)

The L.C.M.: (2r - 1) (r + 2) 

\(\frac{2(r + 2) - 1(2r - 1)}{(2r - 1) (r + 2)}\) = \(\frac{5}{3}\)

\(\frac{2r + 4 - 2r + 1}{ (2r - 1) (r + 2)}\) = \(\frac{5}{3}\)

cross multiply the solution

3 * 5 = (2r - 1) (r + 2) * 5

divide both sides 5

3 =  2r\(^2\) + 3r - 2 (when expanded)

collect like terms

2r\(^2\) + 3r - 2 - 3 = 0

2r\(^2\) + 3r - 5 = 0

Factors are -2r and +5r

2r\(^2\) -2r + 5r - 5 = 0

[2r\(^2\) -2r] [+ 5r - 5] = 0

2r(r-1) + 5(r-1) = 0

(2r+5) (r-1) = 0

 r = 1 or - \(\frac{5}{2}\)

304.

Find the least value of x which satisfies the equation 4x = 7(mod 9)

A.

7

B.

6

C.

5

D.

4

Correct answer is D

4x = 7 (mod 9)

4x = 7 + 9 (mod 9)

\(\frac{4x}{4} = \frac{16}{4}\) (mod 9)

x = 4

305.

In the diagram, O is the centre of the circle. If < NLM = 74\(^o\), < LMN = 39\(^o\) and < LOM = x, find the value of x. 

A.

134\(^o\)

B.

126\(^o\)

C.

113\(^o\)

D.

106\(^o\)

Correct answer is A

L\(N\)M = 180\(^o\) - (74\(^o\) + 39\(^o\)) 

180\(^o\) - 113\(^o\) 

= 67\(^o\)

L\(^O\)

M = x = 2 x 67\(^o\) 

= 134\(^o\)