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.

2,561.

The sum of the ages of Musa and Lawal is 28 years. After sharing a certain sum of money in the ratio of their ages, Musa gets N600 and Lawal N800. How old is Lawal?

A.

14 years

B.

20 years

C.

12 years

D.

16 years

Correct answer is D

M + L = 28,

M : L = 600 : 800

= 3 : 4

\(\frac{M}{L}\) = \(\frac{3}{4}\) \(\to\) M = \(\frac{3}{4}\)L

\(\frac{3}{4}\)L + L = 28

\(\frac{7L}{4}\) = 28

L = \(\frac{4 \times 28}{7}\)

= 16

2,562.

If y = (1 + x)2, find \(\frac{dy}{dx}\)

A.

x - 1

B.

2 + 2x

C.

1 + 2x

D.

2x - 1

Correct answer is B

If y = (1 + x)2, find \(\frac{dy}{dx}\)

y = (1 + x)2

\(\frac{dy}{dx}\) = 2(1 + x)

= 2 + 2x

2,563.

A student sitting on a tower 68 metres high observes his principal's car at the angle of depression of 20°. How far is the car from the bottom of the tower to the nearest metre?

A.

184m

B.

185m

C.

186m

D.

187m

Correct answer is D

Tan 20° = \(\frac{68m}{x}\)

x tan 20° = 68

x = \(\frac{68}{tan 20}\) = \(\frac{68}{0.364}\)

x = 186.8

= 187m

2,564.

Calculate the distance between points L(-1, -6) and M(-3, -5)

A.

√5

B.

2√3

C.

√20

D.

√50

Correct answer is A

L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\)

D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\)

D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\)

D = \(\sqrt{(-2)^2 + 1^2}\)

D = \(\sqrt{4 + 1}\)

D = \(\sqrt{5}\)

2,565.

\(\begin{pmatrix} -2 & 1 \\ 2 & 3 \end{pmatrix}\) \(\begin{pmatrix}p & q \\ r & s\end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\). What is the value of r?

A.

-\(\frac{1}{8}\)

B.

\(\frac{3}{8}\)

C.

\(\frac{5}{8}\)

D.

\(\frac{1}{4}\)

Correct answer is D

-2p + r = 1.......(i)

2p + 3r = 0.......(ii)

r - 1 + 2p ........(iii)

2p + 3(1 + 2p) = 0

2p + 3(1 + 2p) = 0

2p + 3 + 6p = 0

3 - 8p = 0 \(\to\) 8p = 3

p = \(\frac{3}{8}\)

6 = 1 - 2 \(\frac{3}{8}\)

= 1 - \(\frac{6}{8}\)

\(\frac{8 - 6}{8}\) = \(\frac{2}{8}\)

= \(\frac{1}{4}\)