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.

831.

In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and <TOS = 76\(^o\). Using the diagram, calculate the value of angle PTR.

A.

73o

B.

67o

C.

57o

D.

37o

Correct answer is C

In the diagram given, < PRT = 3\(^o\) (Change in same segment)

< TPR = 90\(^o\) (angle in a semicircle)

Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)

= 180\(^o\) - 123\(^o\)

= 57\(^o\)

832.

In the diagram, PQ//RT, QR//Su,

A.

134o

B.

132o

C.

96o

D.

48o

Correct answer is B

In the diagram; a = b = 48o (alternate < S)

x = 180o - b (angles on a str. line)

x = 180o - 48o

= 132o

833.

In the diagram, PTR is a tangent to the centre O. If angles TON = 108°, Calculate the size of angle PTN

A.

132o

B.

126o

C.

108o

D.

102o

Correct answer is B

In the diagram; 108° + x + x = 180° (sum of angle in a triangle)

108° + 2x = 180°

x = 180° - 108°

= 72°

x = \(\frac{72^o}{2}\)

= 36°

(Angle between tangent and a chord through the point of contact)

Hence, angle PTN = 90 + 36

= 126°

834.

In the diagram, O is the centre. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.

A.

40o

B.

50o

C.

60o

D.

80o

Correct answer is A

In the diagram above,

∠MNO = 140° and angles on a straight line is 180°

: ∠NMO = (180 - 140)° = 40°

Hence; ∠POM = 40° ( alternate angle ∠S)

835.

The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take \(\pi = \frac{22}{7}\)]

A.

14cm

B.

7m

C.

3\(\frac{1}{2}\)m

D.

1\(\frac{3}{4}\)m

Correct answer is C

Volume of rectangular tank = L x B x H

= 2 x 7 x 11

= 154cm3

volume of cylindrical tank = \(\pi r^2h\)

154 = \(\frac{22}{7} \times r^2 \times 4\)

r2 = \(\frac{154 \times 7}{22 \times 4}\)

= \(\frac{49}{4}\)

r = \(\sqrt{\frac{49}{4}} = \frac{7}{2}\)

= 3\(\frac{1}{2}\)m