Percentage Questions and Answers

Q1 :

In an election, one of the two candidates got 40% of the total votes cast and lost by 100 votes. Find total number of votes in the election is -

A
  

400

B
  

500

C
  

600

D
  

700

Check Answer
Right Answer: B

500

Description:

100 votes = 20% of the total votes cast difference % between two candidates.
∴thereforeThe total number of votes cast = 500

Q2 :

If 20% of a number is 7.2, then the number is

A
  

34

B
  

35

C
  

36

D
  

37

Check Answer
Right Answer: C

36

Description:

20% is 7.2

20%×5=100% , so the number is 7.2×5=3620% times 5 = 100% text{ , so the number is } 7.2 times 5 = 36

Q3 :

The population of a town increased from 70,000 to 71,050. Find the increase per cent.

A
  

3%

B
  

2%

C
  

2.5%

D
  

1.5%

Check Answer
Right Answer: D

1.5%

Description:

 Increase %=105070,000×100=1.5%text{ Increase } % = frac{1050}{70,000} times 100 = 1.5%

Q4 :

The population of a city is 64,000 and its yearly expand is 10 percent. The expand in the population at the end of 3 years is

A
  

20,184

B
  

22,184

C
  

21,184

D
  

31,184

Check Answer
Right Answer: C

21,184

Description:

Population after 3 years

=64000(1+10100)3=64000×13311000= 64000 Big(1 + frac{10}{100}Big)^3 = 64000 times frac{1331}{1000}

Increase = 64000 ×13311000−64000times frac{1331}{1000} - 64000 = 21184

Q5 :

A shopkeeper falsely claims to sell items at their cost price but uses a weighing scale that measures only 800 grams instead of the standard 1 kilogram. What is his percentage profit?

A
  

25%

B
  

20%

C
  

16%

D
  

8%

Check Answer
Right Answer: A

25%

Description:

Let cost price = Rs. 100 per kg
= Rs. 100 per 1000 gm
and he use 1 kg as 800 gm.
 ∴therefore  His gain = cost of 200 gm = Rs. 20
 ∴therefore His gain  %=2080×100=25%% = frac{20}{80} times 100 = 25%

Q6 :

A person's salary has increased from Rs. 7200 to Rs. 8100. By what percentage has his salary increased?

A
  

25%25%

B
  

18%18%

C
  

1623%16frac{2}3{}%

D
  

1212%12frac{1}{2}%

Check Answer
Right Answer: D

1212%12frac{1}{2}%

Description:

Percentage increase in salary =8100−72007200×100%= frac{8100 - 7200}{7200} times 100%

=9007200×100%=12.5%=1212%begin{aligned} &= frac{900}{7200} times 100% &= 12.5% &= 12 frac{1}{2}% end{aligned}

Q7 :

When the price of cooking gas goes up by 15%, what percentage reduction in consumption is required for a family to maintain the same gas expenditure?

A
  

12123%12frac{1}{23}%

B
  

13123%13frac{1}{23}%

C
  

14123%14frac{1}{23}%

D
  

None of these

Check Answer
Right Answer: B

13123%13frac{1}{23}%

Description:

Let initial price of cooking gas be Rs. 100.
Price after increase = Rs. 115
On Rs. 115 he should reduce Rs. 15 on Rs. 100, he should reduce

=15100×100=13123%=frac{15}{100} times 100 = 13frac{1}{23}%

Q8 :

If 90% of A = 30% of B and B = xx % of A, the value of xx is

A
  

700

B
  

600

C
  

300

D
  

1100

Check Answer
Right Answer: C

300

Description:

Given, 90% of A=30% of B90A100=30B100⇒AB=30⇒B=3A Now, B=x% of A,3A=xA100∴x=300begin{aligned} & text{Given, } 90% text{ of } A = 30% text{ of } B & frac{90A}{100} = frac{30B}{100} & Rightarrow frac{A}{B} = frac{3}{0} Rightarrow B = 3A & text{ Now, } B = x% text{ of } A, 3A = frac{xA}{100} & therefore x = 300 end{aligned}

Q9 :

38 liters of milk was poured into a tub, but the tub was still 5% empty. How much more milk is needed to completely fill the tub?

A
  

1 liter

B
  

2 liters

C
  

3 liters

D
  

4 liters

Check Answer
Right Answer: B

2 liters

Description:

Amount of milk in tub = 38 liters
 ∵because Tub found to be 5% empty
 ∴therefore Total quantity of milk in per cent = 95%
Now, completely fill the tub, total amount of additional milk

=38×595=2 liters= frac{38 times 5}{95} = 2 text{ liters}

Q10 :

10% of a city's population leaves. Then, another 10% of the remaining population also leaves. What percentage of the original population still remains in the city?

A
  

80%

B
  

80.4%

C
  

80.6%

D
  

81%

Check Answer
Right Answer: D

81%

Description:

=(1−10100)(1−10100)=90100×90100=81%begin{aligned} & =bigg(1 - frac{10}{100} bigg) bigg(1 - frac{10}{100} bigg) \ & = frac{90}{100} times frac{90}{100} = 81% end{aligned}

Q11 :

After 24% wastage, the net output of a coal mine is 68,400 quintals. What was the total original output of the coal mine in quintals?

A
  

70000

B
  

90000

C
  

80000

D
  

89000

Check Answer
Right Answer: B

90000

Description:

Let total output be xx quintal.
 ∴therefore Useful output = 100 − 24 = 76%

⇒76100x=68400⇒x=68400×10076∴x=90000 quintal begin{aligned} & Rightarrow frac{76}{100}x = 68400 & Rightarrow x = frac{68400 times 100}{76} & therefore x = 90000 text{ quintal } end{aligned}

Q12 :

If the price of an item is first increased by 20% and then decreased by 20%, what will be the final price as a percentage of the original price?

A
  

4% less

B
  

4% more

C
  

20% less

D
  

20% more

Check Answer
Right Answer: A

4% less

Description:

The price of item first increased by 20% and then decreased by 20%.

∴ Net effect (20−20+20×(−20)100)=−400100=−4%begin{aligned} & therefore text{ Net effect } bigg(20 - 20 + frac{20 times (-20)}{100} bigg) \ &= frac{-400}{100} = -4% end{aligned}

Q13 :

A man donated 4% of his salary to a charity and deposited 10% of the rest in the bank. If now he has Rs. 10800, then his income was

A
  

13500

B
  

14500

C
  

40000

D
  

12500

Check Answer
Right Answer: D

12500

Description:

Let total income be xx
Income deposited =10% of[x−4100x]= 10% text{ of} bigg[x - frac{4}{100} x bigg]

=10100(x−4100x)=961000x= frac{10}{100} bigg(x - frac{4}{100} x bigg) = frac{96}{1000}x

Remaining income = Rs. 10800∴4100x+96x1000+10800=x⇒10800=x−136x1000 ⇒10800=864x1000 ⇒x=10800×1000864 ∴x= Rs. 12500begin{aligned} & text{Remaining income } = text{ Rs. } 10800 & therefore frac{4}{100}x + frac{96x}{1000}+10800 = x & Rightarrow 10800 = x - frac{136x}{1000} ~ & Rightarrow 10800 = frac{864x}{1000} ~ & Rightarrow x = frac{10800 times 1000}{864} ~ & therefore x = text{ Rs. } 12500 end{aligned}

Q14 :

Rohit saves 30% of his salary. If his expenses increase by 30%, he is left with a savings of Rs. 1215 per month. What is his monthly salary?

A
  

Rs. 13500

B
  

Rs. 14500

C
  

Rs. 30000

D
  

Rs. 12500

Check Answer
Right Answer: A

Rs. 13500

Description:

Let the salary of Rohit be Rs. 100, then saving = Rs. 30
Expenses = Rs. 70
New expenses = (100 + 30)% of Rs. 70 = Rs. 91
New saving = Rs. (100 - 91) = Rs. 9
He saves Rs. 9, his salary = Rs. 100
If he saves Rs. 1215.
Then, his salary = Rs. (1009×1215)bigg(frac{100}{9} times 1215 bigg)
= Rs. 13500

Q15 :

If 1 liter of water is evaporated from a 6-liter solution containing 4% sugar, what will be the percentage of sugar in the remaining solution?

A
  

4%

B
  

5%

C
  

445%4frac{4}{5}%

D
  

45%frac{4}{5}%

Check Answer
Right Answer: C

445%4frac{4}{5}%

Description:

Amount of sugar in 6 liter of solution
=4100×6=0.24 Liter=frac{4}{100} times 6 = 0.24 text{ Liter}

After evaporation, sugar in 5 liter = 0.24 liter
 ∴therefore Percentage of Sugar

=(0.245×100)=445%= bigg(frac{0.24}{5} times 100 bigg) = 4frac{4}{5}%

Q16 :

Two numbers are 30% and 40% less than a third number, respectively. By what percentage is the second number smaller than the first?

A
  

35%

B
  

36%

C
  

1327%13frac{2}{7}%

D
  

1427%14frac{2}{7}%

Check Answer
Right Answer: D

1427%14frac{2}{7}%

Description:

Let the third number be zz.
 ∴therefore First number  x=(100−30)100×z=7z10x = frac{(100-30)}{100} times z = frac{7z}{10}

Second Number y=(100−40)100×z=6z10y = frac{(100-40)}{100} times z = frac{6z}{10}

Difference between first and second number
=(x−y)=7z10−6z10=z10= (x - y ) = frac{7z}{10} - frac{6z}{10} = frac{z}{10}

Hence, the required percent
=z107z10×100%=1427%= dfrac{frac{z}{10}}{frac{7z}{10}} times 100% = 14frac{2}{7}%