Quantitative Aptitude - SPLessons

Percentages Practice Set 9

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Percentages Practice Set 9

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Percentages: A Percentage is a dimensionless ratio or number expressed as a fraction of 100. It is often denoted by the character (%).


Example: 10% = \(\frac{10}{100}\) = \(\frac{1}{10}\)


Percentages is one of the important topic in the Quantitative Aptitude section. The article Percentages Practice Set 9 consists of different models of questions with answers useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and other examinations across the globe that include Quantitative Aptitude section. Prepare better for all exams with this Percentages Practice Set 9 for SSC CGL & Railways. This Percentages Practice Set 9 for SSC, Railways Exams will help you learn concepts of mensuration.


shape Quiz

1. What percent of a day in 6 hours?

    A. 25%
    B. 52%
    C. 50%
    D. 5%


Answer: Option A


Explanation:
(\(\frac{6}{24}\) × 100 = 25%


2. Two-fifth of one-third of three seventh of a number is 15.What is 40% of that number?

    A. 501
    B. 105
    C. 525
    D. 150


Answer: Option C


Explanation:
X * \(\frac{2}{5}\) x \(\frac{1}{3}\) x \(\frac{3}{7}\) × x = 15
X = 15 x (\(\frac{7}{3}\)) x (\(\frac{5}{2}\)) x 3 = \(\frac{525}{2}\)
40% of x = \(\frac{40}{100}\) \(\frac{525}{2}\)=>105


3. The population of of a town is 176000.If it increase at the rate of 5% per annum,what will be it’s population 2 years hence?

    A. 194000
    B. 194040
    C. 190440
    D. 194104


Answer: Option B


Explanation:

176000 x \(\frac{105}{100}\) = 88 × 105 × 21 = 194040


4. The length and breath of a rectangle are increased by 20% and 30%.The area of the resulting rectangle exceeds the area of the original rectangle?

    A. 50%
    B. 65%
    C. 56%
    D. 156%


Answer: Option C


Explanation:
\(\frac{120}{100}\) x \(\frac{130}{100}\) x 150 = 156.
156 – 100 = 56.


5. Water tax is is increased by 20% but its consumption is decreased by 20%.Then increase or decrease in the expenditure of the money is

    A. 40%
    B. 4%
    C. 96%
    D. None of the above


Answer: Option B


Explanation:
X – \(\frac{20 × 20 }{100}\) = 4

1. Sugar contain 5% water .What quantity of pure Sugar should be added to 10 liters of water to reduce this to 2%.

    A. 5 lit
    B. 6 lit
    C. 10 lit
    D. 15 lit


Answer: Option D


Explanation:
\(\frac{ 0.5 }{x + 10}\) = \(\frac{ 2 }{100}\)
2x = 30
X = 15


2. In an election between 2 candidates,Candidate who gets 40% of the total vote defeated by 15000.The no of votes polled by winner

    A. 10000
    B. 45000
    C. 30000
    D. 6000


Answer: Option B


Explanation:
40% – looser, 60% – winner, defeated 15000 = 20%
Winner = 60% = 15000 × 3 = 45000.


3. The price of a fan is decreased by 20%.as a result of wh ich the sale increased by 40%.What will be the effect on the total revenue of the shop?

    A. 12%
    B. 10%
    C. 20%
    D. 30%


Answer: Option A


Explanation:

Original revenue = 100 × 100 = 10000.
After 20% decrease = 80 × 140 = 11200.

X * \(\frac{(11200-10000)}{10000}\)) ×100 = \(\frac{1200}{10000}\)) × 100
= 12%.


4. Raj save 10%,after 2 years when his income is increased by 10%,he could save the same money then how much his expenditure increased?

    A. 10%
    B. 20%
    C. 50%
    D. 15%


Answer: Option A


Explanation:
Let income = 100, saving = 10 and expenditure =90.
New income = 110, saving = 11 and e = 99.

alcohol = \(\frac{99-90}{90}\) × 100 = \(\frac{9}{90}\) × 100 = 10%


5. 45% of ? = 25% of 355

    A. 195
    B. 176
    C. 127
    D. 197.22


Answer: Option D


Explanation:
\(\frac{45}{100}\) x X = \(\frac{25}{100}\) × 355

\(\frac{100}{45}\) x \(\frac{25}{100}\) = 197.22

1. Out of 500 students of a school 35% students plays football, 25% plays cricket and 20% neither play football nor cricket.How many students play football and cricket ?

    A. 200
    B. 100
    C. 50
    D. 94


Answer: Option A


Explanation:

Football = n(A) = (\(\frac{35}{100}\)) × 500 = 175
Cricket = n(B) = (\(\frac{25}{100}\)) × 500 = 125
neither play football nor cricket = ( \(\frac{20}{100}\)) × 500 = 100
both football and cricket = n(A U B) = 175 + 125 -100 = 200


2. The price of sugar is reduced by 3%.How many kg of sugar can now be bought for the money which was sufficient to buy 50kg of rice earlier ?

    A. 50 kg
    B. 55 kg
    C. 51.5 kg
    D. 56 kg


Answer: Option C


Explanation:
Let one kg of sugar earlier = Rs. 100
50 kg of sugar earlier = Rs. 5000
Now 1 kg of sugar = Rs.97
Quantity to buy now = \(\frac{5000}{97}\) = 51.5 kg


3. Fresh fruits contains 70% of water and dry fruits contain 20% of water.How much dry fruit can be obtained from 100 kg of fresh fruits ?

    A. 35
    B. 37
    C. 37.5
    D. 40


Answer: Option C


Explanation:

Quantity of pulb in 100 kg of fresh fruit = (100 – 70) × 100 = 30 kg
Quantity of dry fruit be x kg
(100 – 20 ) % of x = 30
(\(\frac{80}{100}\)) x = 30
X = \(\frac{(30 × 100)}{80}\) = 37.5


4. In an examination , full mark is 500, A gets 25% more than B, B gets 40% more than C and C gets 60% more than D.If A got 320, what percentage of full marks did D got (approx)?

    A. 21
    B. 22
    C. 23
    D. 24


Answer: Option C


Explanation:
A = \(\frac{125}{100}\) of B => B = \(\frac{4}{5}\) A
B = \(\frac{140}{100}\) of C => C = \(\frac{5}{7}\) B
C = \(\frac{160}{100}\) 0f D => D = \(\frac{5}{8}\) C
B = (\(\frac{4}{5}\) ) × 320 = 256
C = (\(\frac{5}{7}\) ) × 256 = 183
D = (\(\frac{5}{8}\) ) × 183 = 114
D percentage = (\(\frac{114}{500}\) ) × 100 = 23%


5. The population of a city is increased 5% ,7% and 11% in the last three years, What will be the present population if the population of a town is 2,40,000 three years ago ?

    A. 2,99,600
    B. 2,99,500
    C. 2,99,400
    D. 2,99,300


Answer: Option D


Explanation:
2,40,000 × (\(\frac{105}{100}\)) × (\(\frac{107}{100}\)) × (\(\frac{111}{100}\)) = 2,99,300

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