Mathematics Questions & Answers - Free Online Practice

1,771.

A solid sphere of radius 3cm, a solid right cone of radius 3cm and height 12cm and a solid right circular cycular of radius 3cm and height 4cm.Which of the following statements is true?

The volume of the sphere is greater than the volume of the cone

The volume of the cone is less than the volume of the cylinder

The total surface area of the cone is greater than that of the sphere

The total surface area of the cylinder is less than that of the sphere

The total sufeac area of the cone is equal to that of the cylinder

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

1,772.

The set of value of x and y which satisfies the equations x² - y - 1 = 0 and y - 2x + 2 = 0 is

1, 0

1, 1

2, 2

0, 2

1, 2

View answer & explanation

Correct answer is A

x² - y - 1 = 0.......(i)

y - 2x + 2 = 0......(ii)

By re-arranging eqn. (ii)

y = 2x - 2........(iii)

Subst. eqn. (iii) in eqn (i)

x² - (2x - 2) - 1 = 0

x² - 2x + 1 = 0

= (x - 1) = 0

When x - 1 = 0

x = 1

Sub. for x = 1 in eqn. (iii)

y = 2 - 2 = 0

x = 1, y = 0

1,773.

A hollow right prism of equilateral triangular base of side 4cm is filled with water up to a certain height. If a sphere of radius \(\frac{1}{2}\)cm is immersed in the water, then the rise of water is

1cm

\(\sqrt{\frac{3\pi}{24}}\)

\(\frac{\pi}{24\sqrt{3}}\)

24\(\sqrt{3}\)

View answer & explanation

Correct answer is C

The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\)

Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))³

= \(\frac{1}{8}\)

= \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\)

= \(\frac{\pi}{24\sqrt{3}}\)

1,774.

If sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3, then the angle \(\theta\) is equal to

30^o

45^o

60^o

90^o

105^o

View answer & explanation

Correct answer is B

sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\)

1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

2 tan\(^2\) \(\theta\) = 2

tan\(^2\) \(\theta\) = 1

tan\(\theta\) = √1

where √1 = 1

tan\(\theta\) = 1

And tan 45° = 1

∴ \(\theta\) = 45°

1,775.

In a soccer competition in one season, a club had scored the following goals: 2, 0, 3, 3, 2, 1, 4, 0, 0, 5, 1, 0, 2, 2, 1, 3, 1, 4, 1 and 1. The mean, median and mode are respectively

1, 1.8 and 1.5

1.8, 1.5 and 1

1.8, 1 and 1.5

1.51, 1 and 1.8

1.5, 1.8 and 1

View answer & explanation

Correct answer is B

By re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5.

Mean = \(\frac{36}{20}\) = 1.8

Median = \(\frac{1 + 2}{2}\)

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

= 1.5

Mode = 1

= 1.8, 1.5 and 1