 # 15 Most Important SSC CGL Quantitative Aptitude Percentage Question & Answer

As you know that the SSC CGL 2021 exams have started from 13 August to 25 August quantitative Aptitude math is a very important section of  SSC Exams. So in this post, we are with a new topic in which I will explain some of the most important SSC CGL Quantitative Aptitude Percentage questions and answers frequently asked in SSC and other government competitive exams.

## Some Important SSC CGL Quantitative Aptitude Percentage Questions and Answers

Here are 15 important SSC CGL quantitative aptitude percentage questions and answers. These questions will give you some basic ideas about which types of questions are asked in exams. Basically in the aptitude part, the questions are now too difficult to solve you have to manage time i mean you should try to solve them quickly.

*Q1. Two numbers are respectively 20% and 50% more than a third number. Then the ratio of two numbers is:

(A) 2:5
(B) 3:5
(C) 4:5
(D) 6:7

Solution
Let the third number is 100
Then the second number will be 120
The third number will be 150

Then the ratio of these two numbers will be 120:150
$$\Rightarrow \mathbf{4:5}$$

*Q2. X has twice as much money as that of Y and Y has 50% more money than that of Z. If the average money of all of them is 110, then the money, which X has is:

(A) 55 Rs
(B) 60 Rs
(C) 90 Rs
(D) 180 Rs

Solution
Let Z have money = 2 Unit
then Y have money = 3 Unit ( 50% of 2 will be 3)
and then X have money = 6 Unit ( X has twice money as that of Y)
Given that $$\left ( \frac{X+Y+X}{3} \right )= 110$$ Rs
$$\Rightarrow (2+3+6) U = \times 110$$
then 1 Unit = $$\frac{3\times 110}{11}= 30$$ Rs
So therefore, X have total money = $$6 \times 30 =\mathbf{180}$$

*Q3. In a village  30% of the population is literate, If the total population of the village is 6,600, then the number of literate is:

(A) 1980
(B) 4620
(C) 2200
(D) 3280

Solution
100% = 6600

then, 30 % = $$\frac{6600\times 30}{100}$$

$$\Rightarrow 66\times 30=\mathbf{1980}$$

*Q4. Two successive price increases of 10% and 10% on an article are equivalent to a single price increase of:

(A) 19%
(B) 20%
(C) 21%
(D) 22%

Solution
Trick-
$$\left ( a+b+\frac{ab}{100} \right )\%$$

Where a and b are the encreasement

So, $$\left ( 10+10+\frac{10\times 10}{100} \right )\%$$

=$$20+ 1= \mathbf{21\}$$

*Q5. In an examination, there were 640 boys and 360 girls, 60% of boys and 80% of girls were successful. The percentage of failure was:

(A) 20%
(B) 60%
(C) 30.5%
(D) 32.8%

Solution
It means that out of 640 boys, 40 % failed
than the total failed Boyes will be $$\frac{640\times 40}{100}= 256$$

Similarly, out of 360 girls , 20% failed

than the total failed Boyes will be $$\frac{360\times 20}{100}= 72$$

So out of 1000 (640+ 360) students the total failed student = 256 + 72 = 328

% Failure = $$\frac{328\times 100}{1000}= \mathbf{32.8}$$%

*Q6. In a test, a student got 30% marks and failed by 25 marks. In the same test, another student got 40%  marks and secured 25% marks more than the essential minimum pass marks. The maximum pass marks for the test were?

(A) 400
(B) 500
(C) 520
(D) 580

Solution
Since in both the case the passing marks will be same
and passing marks will be 30%+25

So according to the question $$(30\% + 25 ) = \left [ 40\% -(30\% + 25)\times 25\% \right ]$$

=$$(30\% + 25 ) = \left [ 40\% -(30\% + 25)\times \frac{1}{4} \right ]$$

= $$(30\% + 25 )\times 4 = \left [ 160\% -(30\% + 25 \right ]$$

=$$(120\% + 100 ) = \left [ 130\% + 25 \right ]$$

=$$10\% = 125$$

Total marks $$\Rightarrow (100\% = 1250$$

so maximum passing marks will be $$1250\times 30\% + 25$$

$$375+ 25= \mathbf{400}$$

*Q7. Bhuvnesh spends 30% of his salary on food and donates 3% in a Charitable Trust. He spends Rs 2,310 on these two items, then total salary for that month is:

(A) 6000
(B) 8000
(C) 9000
(D) 7000

Solution
It means that $$30\% + 3\% \rightarrow 2310$$
= $$33\% = 2310$$

$$100\% = \frac{2310 \times 100}{33} = \mathbf{7000}$$

*Q8. Out of his total income, Mr Kapoor spends 20% on house rent and 70% of the rest on household expenses. If he saves Rs 1,800 what is the total income ( in rupees )?

(A) Rs 7800
(B) Rs 7000
(C) Rs 8000
(D) Rs 7500

Solution
Let’s consider that he has 100 Units out of 100 Units he spends 20 Units on house rent
Then the rest money is 80 Units
Now he spends 70% of 80 units = 56 Units
so the rest of money =24 Units

Now According to the question $$24~ Units \rightarrow 1800$$

= $$1~ Units \rightarrow \frac{1800}{6\times 4}=75$$

=$$100~ Units \rightarrow \mathbf{7500}$$

*Q9. Two persons contested an election of Parliament, The winning candidate secured 57% of the total votes polled and won by a majority of 42,000 votes. The number of total votes polled is:

(A) 5,00,000
(B) 6,00,000
(C) 3,00,000
(D) 4,00,000

Solution
The winning candidate gets 57% of the total votes and won by 42,000 votes
so the looser candidate will get 43 % of the total votes

Now According to the question $$57\% – 43\% \rightarrow 42,000$$

= $$14\% \rightarrow 42,000$$

=$$1\% \rightarrow 3,000$$

=$$100\% \rightarrow \mathbf{3,00,000}$$

*Q10. The population of a village has increased annually at the rate of 25%. If at the end of 3 years it is 10,000, the population in the beginning of the first year was:

(A) Rs 5120
(B) Rs 5000
(C) Rs 4900
(D) Rs 4500

Solution
The population is increasing by 25%
It means that it becomes 4 to 5 in a year

$$I^{st}~~~~~year ~~4\rightarrow ~5$$
$$II^{st} ~~~~year ~~4\rightarrow ~5$$
$$III^{st} ~~~year ~~4\rightarrow ~5$$

Given that $$5\times 5\times 5\rightarrow 10,000$$
Then we have to find $$5\times 5\times 5\rightarrow ?$$

$$4\times 4\times 4\rightarrow \frac{10,000 \times 64}{5\times 5\times 5}= \mathbf{5120}$$

*Q11. If the population of a town is 64000 and its annual increase is 10%, then its population at the end of 3 years will be :

(A) 80000
(B) 85184
(C) 85000
(D)85100

Solution
The population is increasing by 10% per year
It means that it becomes 10 to 11 in a year

$$I^{st}~~~~~year ~~10\rightarrow ~11$$
$$II^{st} ~~~~year ~~10\rightarrow ~11$$
$$III^{st} ~~~year ~~10\rightarrow ~11$$

Given that $$10\times 10\times 10\overset{64}{\rightarrow} 64000$$
So that $$11\times 11\times 11\overset{64}{\rightarrow} \mathbf{85184}$$

*Q12. Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100kgs of fresh fruits?

(A) 32 Kgs
(B) 40 Kgs
(C) 52 Kgs
(D) 80 Kgs

Solution
In the fresh fruit, 68% contains water and the rest 32% is pulp.
Then in 100 kg of fresh fruit, there will be 32kg of pulp.
The dry fruit contains 20% of water and the rest 80% is pulp
Since in the dry fruit, the quantity of pulp will be same
i.e. 32 kg pulp will be 80 % in the dry fruit.

$$80\%\rightarrow 32$$ $$100\%\rightarrow \mathbf{40}$$

*Q13. In 2 Kg mixture of copper and aluminium, 30% is copper. How much aluminium powder should be added to the mixture so that the quantity of copper becomes 20%

(A) 900 gms
(B) 800 gms
(C) 1000 gms
(D) 1200 gms

Solution
In 2 Kg mixture, 30% is copper and the rest 70% is aluminium.
It means that 600 gm is copper and 1400 gm is aluminium
Since after adding aluminium powder, the quantity of copper remains the same 600 gm.
And after adding aluminium powder, the copper becomes 20 %
It means that $$20\% \rightarrow 600~gm$$
So that $$100\% \rightarrow 3000~gm$$

So that extra added Aluminium powder is $$3000 – 2000 = \mathbf{1000 ~gm}$$

*Q14. The sum of two numbers is 520. If the bigger number is decreased by 4% and the smaller number increases by 12% then the numbers obtained are equal. The smaller number is

(A) 280
(B) 240
(C) 210
(D) 300

Solution
Let bigger number is A and the smaller number is B
Then According to the question, 96% of A =112% of B

$$\Rightarrow \frac{A}{B}= \frac{112}{96}=\frac{7}{6}$$

Since, $$(6+7) U \rightarrow 520$$
= $$1U \rightarrow 40$$
Smaller number $$6U \rightarrow \mathbf{240}$$

*Q15. Kamal has some apples. If he sold 40% apples more than he ate if he sold 70 apples then How many he ate.

(A) 18
(B) 42
(C) 50
(D) 99

Solution
Let Kamal ate 100% apples
Then he sold= $$100\%+40\%\rightarrow 70$$

It means that $$140\% \overset{\times \frac{1}{2}}{\rightarrow} 70$$

So that $$100\% \overset{\times \frac{1}{2}}{\rightarrow} \mathbf{50}$$