Mathematics Questions & Answers - Free Online Practice

1,671.

In the diagram, HK is parallel to QR, PH = 4cm and HQ = 3cm. What is the ratio of KR:PR?

7:3

3:7

3:4

4:3

View answer & explanation

Correct answer is B

In \(\bigtriangleup\)PHK and \(\bigtriangleup\)PAR = \(\frac{PH}{PQ} = \frac{PK}{PR}\)

\(\frac{4}{7} \times \frac{PR-KR}{PR}\)

4PR = 7(PR - KR) = 7PR - 7KR

\(\frac{KR}{PR} = \frac{3}{7}\)

KR:PR = 3:7

1,672.

MN is tangent to the given circle at M, MR and MQ are two chords. IF QNM is 60^o and MNQ is 40^o. Find RMQ

120^o

110^o

60^o

20^o

View answer & explanation

Correct answer is D

QMN = 60^o

MRQ = 60^o(angle in the alternate segment are equal)

MQN = 80^o(angle sum of a \(\bigtriangleup\) = 180^o)

60 = x = 80^o(exterior angle = sum of opposite interior angles)

x = 80^o - 60^o = 20^o

RMQ = 20^o

1,673.

On the curve, the points at which the gradient of the curve is equal to zero are

c, d, f. i, l

b, e, g, j, m

a, b, c, d, f, i, j, l

c, d, f, h, i, l

View answer & explanation

Correct answer is B

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

1,674.

In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid

36\(\pi\)cm ³

54\(\pi\)cm ²

18\(\pi\)cm ²

108\(\pi\)cm ²

View answer & explanation

Correct answer is A

The volume of the solid = vol. of cone + vol. of hemisphere

volume of cone = \(\frac{1}{3} \pi r^2 h\)

= \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\)

vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\)

= \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\)

vol. of solid = 18\(\pi\) + 18\(\pi\)

= 36\(\pi\)cm3