SSC Practice test(1) Questions and answers

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Exams

This SSC Exam Sample question paper consists of the below sections.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension

101). The percentage increase in the surface area of a cube when each side is doubled is :

A). 1.5

B). 0.5

C). 2

D). 3

View Answer

Correct Answer: 3

Surface area of a cube = \( 6 X \left(side\right)^{2} \)

Let the side of cube = 'a'

then its surface area = \( 6a^{2} \)

new side of a cube = 2a

then its surface area = \( 6 \left(2a\right)^{2} = 24a^{2} \)

Therefore, Percentage increase = \( \Large \frac{24a^{2}-6a^{2}}{6a^{2}} \times 100 \)

= \( \Large \frac{18a^{2}}{6a^{2}} \times 100 = 300\% \)

102). Find the square root of:
\( \Large \frac{ \left(0.064-0.008\right) \left(0.16-0.04\right) }{ \left(0.16+0.08+0.04\right) \left(0.4+0.2\right)^{3} } \)

A). 3

B). \( \Large \frac{3}{2} \)

C). \( \Large \frac{2}{3} \)

D). \( \Large \frac{1}{3} \)

View Answer

Correct Answer: \( \Large \frac{1}{3} \)

To find :

\( \Large \sqrt{\frac{ \left(0.064-0.008\right) \left(0.16-0.04\right) }{ \left(0.16+0.08+0.04\right) \left(0.4+0.2\right)^{3} }} \)

Let x = 0.4 ; y = 0.2

\( \Large \sqrt\frac{ \left(x^{3}-y^{3}\right) \left(x^{2}-y^{2}\right) }{ \left(x^{2}+xy+y^{2}\right) \left(x+y\right)^{3} } \)

since \( \Large \left(x^{3}-y^{3}\right) = \left(x-y\right) \left(x^{2}+xy+y^{2}\right) \)

Therefore, \( \Large \sqrt \frac{ \left(x-y\right) \left(x+y\right) \left(x-y\right) }{ \left(x+y\right)^{3} } = \frac{ \left(x-y\right)^{2} }{ \left(x+y\right)^{2} } \)

= \( \Large \frac{x-y}{x+y} = \frac{0.4-0.2}{0.4+0.02} = \frac{1}{3} \)

103). The angle elevation of a tower from a distance 100 mtr. from its foot is 30o . Then the height of the tower is :

A). \( \Large \frac{50}{\sqrt{3}}mtr \)

B). \( \Large 50\sqrt{3} mtr \)

C). \( \Large \frac{100}{\sqrt{3}} mtr \)

D). \( \Large 100\sqrt{3} mtr \)

View Answer

104). A sum of money lent out at simple interest amounts to Rs.720 after 2 years and Rs.1020 after a further period of 5 years. Find the principal.

A). Rs.1740

B). Rs.120

C). Rs.6000

D). Rs.600

View Answer

105). A can do a work in 10 days and B in 20 days. If they together work on it for 5 days, then the fraction of the work that is left is :

A). \( \Large \frac{1}{4} \)

B). \( \Large \frac{3}{4} \)

C). \( \Large \frac{3}{20} \)

D). \( \Large \frac{4}{3} \)

View Answer

Correct Answer: \( \Large \frac{1}{4} \)

Work done by A and B for 5 days =

= \( \Large 5 \left(\frac{1}{10}+\frac{1}{20}\right) \)

= \( \Large 5 \left(\frac{3}{20}\right) = \frac{3}{4} \)

Therefore, left work = \( \Large 1 - \frac{3}{4} = \frac{1}{4}\)

106). A man spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. His savings are increased by:

A). \( \Large 37\frac{1}{2}\% \)

B). 0.5

C). 0.1

D). 0.25

View Answer

Correct Answer: 0.5

Let the income be Rs.100

Amout spent = 75% = Rs.75

Therefore, savings = Rs.25

new income = \( 100 + \left(100 \times \frac{20}{100}\right) \)

new expenditure = 10% increase

(i.e) \( \Large 75 + \left(75 \times \frac{10}{100}\right) \) = 82.5%

Therefore, new savings = Rs.120 - Rs.82.5 = Rs.37.5

Therefore, increase percent in savings = \( \Large \left(\frac{37.5-25}{25} \times 100\right) \)=50%

107). Pipe A can fill a tani in 4 hours and Pipe B can fill it in 6 hours. If they are opened on alternate opurs ad if pipe A is opene first then in how many hours, the tank shall be full?

A). \( \Large 3\frac{1}{2} \)

B). \( \Large 3\frac{1}{4} \)

C). \( \Large 4\frac{1}{2} \)

D). \( \Large 4\frac{21}{3} \)

View Answer

Correct Answer: \( \Large 4\frac{21}{3} \)

Part of work done in two hours = \( \Large \frac{1}{4}+\frac{1}{6}=\frac{5}{12} \)

Part of work done in 4 hours = \( \Large 2 \left(\frac{5}{12}\right) = \frac{5}{6} \)

Therefore, required time = \( \Large 4 + \frac{\frac{1}{6}}{\frac{1}{4}} = 4\frac{2}{3} hours \)

108). A shopkeeper gains 17% after allowing a discount of 10% on the marked price of an article. Find his profit percent if the article is sold at marked price allowing no discount.

A). 0.3

B). 0.37

C). 0.23

D). 0.27

View Answer

Correct Answer: 0.3

Let the cost price = Rs.100

Therefore, Selling price = \( 100 \times \frac{117}{100} = Rs.117 \)

Therefore, Marked price = \( 117 \times \frac{100}{90} = 130 \)

Therefore, Profit Percentage = \( \frac{ \left(130-100\right) }{100} \times 100 = 30\% \)

109). ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If \( \angle CBE = \angle ACF = 130 ^{\circ} \), then the value of \( \angle GAB \) is

A). \( 130 ^{\circ} \)

B). \( 90 ^{\circ} \)

C). \( 100 ^{\circ} \)

D). \( 80 ^{\circ} \)

View Answer

110). A person of height 6 ft. wants to pluck a fruit which is on a 26/3 ft. high tree. lf the person is standing 8/ 3 ft. away from the base of the tree, then at what angle should he throw a stone so that, it hits the fruit?

A). \( 60 ^{\circ} \)

B). \( 75 ^{\circ} \)

C). \( 30 ^{\circ} \)

D). \( 45 ^{\circ} \)

View Answer

Correct Answer: \( 30 ^{\circ} \)

\( \Large tan \theta = \frac{AB}{EB} = \frac{ \left(AC-BC\right) }{DC} \)

= \( \Large \frac{ \left(\frac{26}{3}-6\right) }{\frac{8}{\sqrt{3}}} \)

= \( \Large \frac{\frac{ \left(26-18\right) }{3}}{\frac{8}{\sqrt{3}}} = \frac{1}{\sqrt{3}} \)

= \( \Large \theta = tan^{-1} \left(\frac{1}{\sqrt{3}}\right) = 30 ^{\circ} \)

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