Quantitative Aptitude - SPLessons

SSC CPO Percentages Quiz 2

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

SSC CPO Percentages Quiz 2

shape Introduction

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 SSC CPO Percentages Quiz 2 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 SSC CPO Percentages Quiz 2 for SSC CGL & Railways. This SSC CPO Percentages Quiz 2 for SSC, Railways Exams will help you learn concepts of mensuration. Candidates can check the daily updates on SSC Official Website.


shape Samples

1. Candidates in a competitive examination consisted of 60% men and 40% women 70% men and 75% women cleared the qualifying test and entered the final test where 80% men and 70% women were successful? Which of the following statements is correct?

    A. Success rate is higher for women
    B. Overall success rate is below 50%
    C. More men cleared the examination than women
    D. Both ‘A’ and ‘B’


Answer: Option C

Explanation:
Let there are 100 persons in which 60 men and 40 women number of men who cleared the qualifying test

= 70 x 60/100 = 42

Number of women who cleared the qualifying test

= 40 x ¾ = 30

Number of men who get success in final test

= 42 x \(\frac{4 }{5}\) = 33.6

Number of women who get success in final test

= 30 x \(\frac{70 }{100}\) = 21

Hence more men cleared the examination than a woman.


2. In a test a candidate attempted only 8 Questions and secured 50% marks in each of the questions if the obtained a total of 40% in the test and all questions in the test carried equal marks, how many questions were there in the test?

    A. 8
    B. 10
    C. 15
    D. 16


Answer: Option B
Let the marks of each question be 10

Total marks got by the candidate = 8 x 5 = 40 marks

40% = 40 : 100 = 100

Therefore, Total number of questions = \(\frac{100 }{10}\) =10


3. A city has a population of 300000 out of which 180000 are males 50% of the population is illiterate if 70 % of the males are literate, then the number of literate females is

    A. 24000
    B. 30000
    C. 54000

    D. 60000


Answer: Option A

Explanation:
Total population = 300000

Total number of males = 180000

Total literates = 5% of total population = 150000

Number of literate males = 70% of males = 126000


4. In a company, 60% of the employees are men, of this 40 % are drawing more than Rs. 50000 per year. 36% of the total employees of the company draw more than Rs.50000 per year then what is the percentage of women who are drawing less than Rs. 5000 per year?

    A. 70%

    B. 60%
    C. 40%

    D. 30%


Answer: Option A

Explanation:
Total number of employees be 100

Then number of men = \(\frac{6000 }{100}\) = 60

Number of women = \(\frac{4000 }{100}\) = 40

Therefore, number of men drawing more than Rs. 50000 = \(\frac{2400 }{100}\) = 24 men

Since the number of total employees drawing more than Rs. 50000 = 3600/100 = 36

Number of women who draw more than Rs. 50000 = 36- 24 = 12

Number of women who draw less than Rs. 50000 = 40 -12 = 28

Therefore, Percentage of the women who draw less than Rs. 50000 per year = 28/40 x 100% = 70%


5. 10 % of the electorate did not cast their votes in an election between two candidates. 10 % of the votes polled were found invalid. The successful candidate got 54 % of the valid votes and won by a majority of 1620 votes. The number of voters enrolled on the voter’s list was

    A. 25000

    B. 33000

    C. 15%

    D. Can’t be determined


Answer: Option C

Explanation:
As per the given information, two equations can be written

2000 \(\frac{x }{100}\) + 2000 \(\frac{y }{100}\) = 700
2000 \(\frac{x }{100}\) + 2000 \(\frac{y }{100}\) = 900
The equations can be simplified to

x + y = 35

and 2x + 3y = 90.

After Solving this equation, we get

x = 15%

6. The sum of two numbers is 2490. If 6.5 % of one number is equal to 8.5 % of the other then the numbers are:

    A. 989, 1501

    B. 1011, 1479
    C. 1401, 1089

    D. 1411, 1079


Answer: Option D

Explanation:
Let the number be x and (2490 – x).

Then 6.5 % of x = 8.5 % of (2490 – x)

=> 6.5/100 × x = \(\frac{8.5}{100}\) × (2490 – x)

=> \(\frac{65x}{1000}\) = 85(2490 – x)/1000

=> 65x = (85 × 2490) – 85x

=> 150x =(85 × 2490)

=> x = \(\frac{211650}{10}\) = 1411

Hence the numbers are 1411 and (2490 – 1411) = 1079


7. 8 % of the voters in an election did not cast their votes in the election, there were only two candidates. The winner by obtaining 48 % of the total votes defeated his contestant by 1100 votes. The total number of voters in the election was.

    A. 21000

    B. 2200

    C. 23500

    D. 27500


Answer: Option D

Explanation:
Let the total number of voters be x

Votes cast = 92 % of x = (92/100 × x) = \(\frac{23x}{25}\)

Votes in favour of winning candidate = \(\frac{48}{100}\) × x

= \(\frac{12x}{25}\)

Votes polled by defeated candidate = (\(\frac{23x}{25}\) – \(\frac{12x}{25}\))

= 11x/25 12x/25 – \(\frac{11x}{25}\) = 1100

=> 12x – 11x = 27500

=> x = 27500


8. In an election between two candidates. One got 55 % of the total valid votes. 20 % of the votes were invalid. If the total number of votes was 7500. The number of valid votes that the other candidate got was

    A. 2700

    B. 2900

    C. 3000

    D. 3100


Answer: Option A

Explanation:
Valid votes = (\(\frac{80}{100}\) × 7500) = 6000

Valid votes polled by one candidates

=(\(\frac{55 }{100}\) × 6000) = 3300

Valid votes polled by another candidate

= (6000 – 3300) = 2700


9. In an election between two candidates 75 % of the voters cast their votes. Out of which 2 % of the voters were declared invalid. A candidate got 9261 votes which were 75 % of the total valid votes. The total number of voters enrolled in that election was:

    A. 10000

    B. 16400

    C. 16800

    D. 18000


Answer: Option C

Explanation:
Let the enrolled votes be x,

Votes cast = (\(\frac{75}{100}\) × x) = \(\frac{3x}{4}\)

Valid votes = (\(\frac{98}{100}\)
× \(\frac{3x}{4}\)
)

= \(\frac{147x }{200}\) × \(\frac{75 }{100}\) × \(\frac{147x }{200}\) = 9261

=> 441x = (9261 × 800)

=> x = 16800


10. Fresh fruit contains 68 % water and dry fruit contains 20 % water. How much dry fruit can be obtained from 100 kg of fresh fruits?

    A. 32 kg

    B. 40 kg
    C. 52 kg

    D. 80 kg


Answer: Option B

Explanation:
Quantity of water in 100kg of fresh fruits =(68/100 × 100)kg

Quantity of pulp in it = (100 – 68)kg = 32 kg

Let the dry fruit be x kg

Water in it = (\(\frac{20}{100}\) × x)kg = \(\frac{x}{5}\) kg

Quantity of pulp in it = (x – \(\frac{x}{5}\))kg = \(\frac{4x}{5}\) kg

Therefore, \(\frac{4x}{5}\) = 32 => x = \(\frac{160}{4}\) = 40 kg

11. Fresh grapes contain 80 % water dry grapes contain 10 % water. If the weight of dry grapes is 250 kg. What was its total weight when it was fresh?

    A. 1000 kg
    B. 1100 kg
    C. 1125 kg

    D. 1225 kg


Answer: Option C

Explanation:
Let the weight of fresh grapes be x kg

Quantity of water in it = (\(\frac{80}{100}\) × x)kg = \(\frac{4x}{5}\) kg

Quantity of pulp in it = (x – \(\frac{4x}{5}\))kg = \(\frac{x}{5}\) kg

Quantity of water in 250 kg dry grapes

= (\(\frac{10 }{100}\) × 250)kg = 25kg

Quantity of pulp in it = (250 – 25)kg = 225 kg

Therefore, \(\frac{x}{5}\) = 225

=> x = 1125


12. The quality of water that should be added to reduce 9 ml. Lotion containing 50 % alcohol to a lotion containing 30 % alcohol is

    A. 3 ml

    B. 4 ml
    C. 5 ml
    D. 6 ml


Answer: Option D

Explanation:

Alcohol in 9 ml lotion = (\(\frac{50 }{100}\) × 9)ml = 4.5ml

Water in it = (9 – 4.5)ml = 4.5ml

Let x ml of water be added to it, then \(\frac{4.5}{9+x}\) × 100 = 30

=> \(\frac{4.5 }{9+x}\) = \(\frac{30 }{100}\) = \(\frac{3}{10}\)

=> 3(9+x) = 45 => 27 + 3x = 45

=> 3x = 18

=> x = 6

Water to be added = 6 ml


13. In two successive years, 100 and 75 students of a school appeared at the final examination. Respectively 75 % and 60 % of them passed. The average rate of pass is:

    A. 68 \(\frac{4}{7}\) %

    B. 78 %

    C. 80 %
    D. 80 \(\frac{4}{7}\) %


Answer: Option A

Explanation:
Total candidates = (100 + 75) = 175

Total passed = (\(\frac{75}{100}\) × 100) + (\(\frac{60}{100}\) × 75)

= (75 + 45) = 120

Therefore Pass % = (\(\frac{120}{175}\) × 100)%

= 480/7 % = 68 4/7 %


14. Of the 1000 inhabitants of a town 60 % are males of whom 20 % are literate. If of all the inhabitants 25 % are literate. Then what percent of the females of the town are literate?

    A. 22. 5 %
    B. 27.5 %

    C. 32. 5 %

    D. 37.5 %


Answer: Option C

Explanation:

Males = (60/100 × 1000) = 600,

females = (1000 – 600) = 400

Literate males = (\(\frac{20 }{100}\) × 600) = 120

Total literates = (\(\frac{25 }{100}\) × 1000) = 250

Female literates = (250 – 120) = 130

Required % = ( \(\frac{130}{140}\) × 100)% = 32.5 %

15. In a school, 40 % of the students play football and 50 % play cricket. If 18 % of the students play neither football nor cricket, the percentage of students playing both is

    A. 40 %

    B. 32 %
    C. 22 %

    D. 8 %


Answer: Option A

Explanation:
Let A = set of students who play football and

B = set of students play cricket.

Then n(A) = 40, n (B) = 50 and

n(A U B) = (100 – 18) = 82

n(A U B) = n(A) + n(B) – n(A ∩ B)

n(A∩B) = n(A) + n(B) – n(AUB) = (40 + 50 -82) = 8

Percentage of the students who play both = 8%




Geography – Related Information
Geography Quiz Practice Sets

Economy Quiz Practice Sets

Indian Dance Forms Quiz