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.

1,616.

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

A.

18o

B.

54o

C.

90o

D.

108o

Correct answer is B

The total angle in a regular pentagon = \((2(5) - 4) \times 90\)

= \(6 \times 90 = 540°\)

Each interior angle = \(\frac{540}{5} = 108°\)

While the interior of a quadrilateral = \(\frac{360}{4} = 90°\)

\(PTU + TPU + TUP = 180°\)

\(108° + (180° - 90°) + TUP = 180°\)

\(TUP = 180° - (108° + 18°) = 54°\)

1,617.

The figure represents the graphs of y = x(2 - x) and y = (x - 1)(x - 3). What are the x-coordinates of P, Q and F respectively?

A.

1, 3, 2

B.

0, 0, 0

C.

0, 2, 3

D.

1, 2, 3

Correct answer is D

The two equations are y = x(2 - x) and y = (x - 1)(x - 3)

The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0

put y = 0; 0 = x(2 - x)

0 = (x - 1)(x - 3)

x = 0 or x = 2; x = 1 or x = 3

The values of x at P, Q, R are increasing towards the positive direction of x-axis

at P, x = 1 at Q, x = 2 at R, x = 3

P, Q, R are respectively (1, 2, 3)

1,618.

The shaded portion in the venn diagram is

A.

x \(\cap\) z

B.

xo \(\cap\) y \(\cap\) z

C.

x \(\cap\) yo \(\cap\) z

D.

x \(\cap\) y \(\cap\) zo

Correct answer is C

The shaded part exists on x \(\cap\) z but not in y

1,619.

In the figure, if XZ is 10cm, calculate RY in cm

A.

10

B.

10(1 - \(\frac{1}{\sqrt{3}}\))

C.

10(1 - \(\frac{1}{\sqrt{2}}\))

D.

10[1 - \(\sqrt{3}\)]

Correct answer is B

RY = YZ - RZ

= 10 tan 45 - 10 tan 30

= 10(1 - \(\frac{1}{\sqrt{3}}\))

1,620.

The diagram is a circle with centre O. Find the area of the shaded portion

A.

9(\(\pi - 2) cm^2\)

B.

18\(\pi cm^2\)

C.

9\(\pi cm^2\)

D.

36\(\pi cm^2\)

Correct answer is A

Area of the quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (6)^2\)

= \(\frac{36 \pi}{4} = 9 \pi \)

Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\)

Area of shaded portion =(9 \(\pi \) - 18)cm^2\) =

9(\(\pi - 2)cm^2\).