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.

631.

The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.

A.

1

B.

2

C.

3

D.

9

Correct answer is A

Using the two - point from

\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)

\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)

\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)

\(\frac{-12(y -2)}{-4}\) = x - 4

3(y -2) = x -4

3y - 6 = x - 4

3y = x - 4 + 6

3y = x + 2...

By comparing the equations;

3y = px + , p = 1

632.

If M and N are the points (-3, 8) and (5, -7) respectively, find |MN|

A.

8 units

B.

11 units

C.

15 units

D.

17 units

Correct answer is D

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

= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\)

= \(\sqrt{(-8)^2 + (8 + 7)^2}\)

= \(\sqrt{64 + (15)^2}\)

= \(\sqrt{64 + 225}\)

= \(\sqrt{289}\)

= 17 units

633.

The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x

A.

\(45^o\)

B.

\(95^o\)

C.

\(84^o\)

D.

\(85^o\)

Correct answer is C

x +  2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)

8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\)

8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\)

8x = 720\(^o\) - 40\(^o\) = 680\(^o\)

x = \(\frac{680^o}{8}\)

= 85\(^o\)

634.

The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)]

A.

113\(cm^3\)

B.

131\(cm^3\)

C.

311\(cm^3\)

D.

414\(cm^3\)

Correct answer is A

Surface area of a sphere = \(4 \pi r^2\) \(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number)

635.

The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)]

A.

14m

B.

10m

C.

7m

D.

5m

Correct answer is C

Volume of a cylinder = \( \pi r^2\)h

385 = \(\frac{22}{7}\) x \(r^2\) x 10

385 x 7 = 22 x \(r^2\) x 10

\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)

= 12.25

r = \(\sqrt{12.25}\)

= 3.5m

Hence, diameter of tank = 2r

= 2 x 3.5 = 7m