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.

1,426.

In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.

A.

25o

B.

65o

C.

115o

D.

130o

Correct answer is B

x = 65o (An interior angle of a cyclic quadrilateral = opposite exterior angle).

1,427.

In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of \(\bigtriangleup\)YZR?

A.

1:2

B.

2:1

C.

1:2

D.

3:1

Correct answer is B

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A2

Area of \(\bigtriangleup\)PQR = A = A1 + 2A2

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A2 But, A = A1 + 2A

A1 = 4A2 - 2A2

A1 = 2A2

\(\frac{A_1}{A_2}\) = 2

A1:A2 = 2:1


Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

1,428.

In the diagram, PQ//RS, QU//PT and < PSR = 42o. Find angle x.

A.

84o

B.

48o

C.

42o

D.

32o

Correct answer is C

From the diagram, < QPS = xo (Corresponding angles)

Also, < QPS = < PSR(Alternate angles)

x = 42o

1,429.

In the figure shown, PQs is a straight line. What is the value of < PRQ?

A.

128o

B.

108o

C.

98o

D.

78o

Correct answer is D

< QPR + < PRQ = < RQS

(Sum of two interior angles of a triangle = Opposite exterior angles)

70o + < PRQ = 148

< PRQ = 148o - 70o

= 78o

1,430.

In the diagram above, PQRS is a rhombus. /PR/ = 10cm and /QS/ = 24cm. Calculate the perimeter of the rhombus.

A.

34cm

B.

52cm

C.

56cm

D.

96cm

Correct answer is B

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 52 + 122

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm