**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**.

**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?
**

**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
**

**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?
**

**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
**

**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%

**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.**

**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
**

**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:
**

**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?
**

**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

**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
**

**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:
**

**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?**

**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**

**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%