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.

2,376.

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

A.

5

B.

1

C.

6

D.

3

Correct answer is D

2x + 3 \(\neq\) 2x - 3 for any value of x

∴ for the \(\bigtriangleup\) to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6

2,377.

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of 22\(\frac{1}{2}\)o at the centre of the circle

A.

2\(\pi\)

B.

\(\pi\)

C.

\(\frac{2}{3}\)

D.

\(\frac{\pi}{2}\)

Correct answer is D

Diameter = 8cm

∴ Radius = 4cm

Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)

∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4

= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)

= \(\frac{180}{360}\)

= \(\frac{\pi}{2}\)

2,378.

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

A.

12

B.

10

C.

9

D.

8

Correct answer is A

A rectangular polygon has each interior angle to be 150o

let the polygon has n-sides

therefore, Total interior angle 150 x n = 150n

hence 150n = (2n - 4)90

150n = 180n - 360

360 = (180 - 150)n

30n = 360

n = 12

2,379.

A man's initial salary is N540.00 a month and increases after each period of six months by N36.00 a month. Find his salary in the eighth month of the third year

A.

N828.00

B.

N756.00

C.

N720.00

D.

N684.00

Correct answer is C

Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 yrs and 6 months amount(increment) = N36 x 5 = N180 The man's new salary = N540 = N180 = N720.00

2,380.

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

A.

0, 8

B.

-1, \(\frac{5}{3}\)

C.

2, 3

D.

1, -1

Correct answer is B

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

\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)

\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)

\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)

\(k^{2} - 8k = 0\)

\(k(k - 8) = 0\)

\(\therefore \text{k = 0 or 8}\)

The terms of the sequence given k = 0: (1, -1, 1)

\(\implies \text{The common ratio r = -1}\)

The terms of the sequence given k = 8: (9, 15, 25)

\(\implies \text{The common ratio r = } \frac{5}{3}\)

The possible values of the common ratio are -1 and \(\frac{5}{3}\).