## Percentages Practice Questions – Set 1

Percentages Practice Questions – Set 1
1. In an election between two candidates, 75% of the voters cast their votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.
a) 9000
b) 15680
c) 16800
d) 17200
e) None of these
Explanation:
Let the number of votes enrolled be x. Then ,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100) * (98/100) * (75/100) * x] = 9261.
X = [(9261*100*100*100)/(75*98*75)] =16800.
2. If the numerator of a fraction be increased by 15% and its denominator be diminished by 8%, the value of the fraction is 15/16. Find the original fraction.
a) 5/4
b) 7/2
c) 8/3
d) 9/5
e) None of these
Explanation:
Let the original fraction be x/y.
Then (115% of x)/(92% of y)=15/16 => (115x/92y)=15/16
((15/16)*(92/115))=3/4
3. In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure  on it does not increase ?
a) 20%
b) 27%
c) 30%
d) 40%
e) 42%
Explanation:
Reduction in consumption = [((R/(100+R))*100]%
[(25/125)*100]%=20%.
4. During one year, the population of town increased by 5% . If the total population is 9975 at the end of the second year , then what was the population size in the beginning of the first year ?
a) 8000
b) 9700
c) 10000
d) 11500
e) None of these
Explanation:
Population in the beginning of the first year
= 9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.
5. Shobha’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems. Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the geometry problems correctly.  she did not pass the test because she got less than  60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?
a) 5%
b) 13%
c) 18%
d) 20%
e) 25%
Explanation:
Number  of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)
=7 + 12+21= 45
questions to be answered correctly for 60% grade=60% of 75 = 45
therefore required number of questions= (45-40) = 5.
6. How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to 10% solution?
a) 9/4
b) 7/2
c) 8/3
d) 7/5
e) None of these
Explanation:
Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then , (0.6+x)/(30+x)=10/100
==> 60+100x=300+10x
==> 90x=240 ==> x=8/3.
7. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
a) 57%
b) 60%
c) 65%
d) 90%
e) None of these
Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Therefore, Required percentage = (11628/20400 x 100) % = 57%
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
e) None of these
Explanation:
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.
9. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?
a) 14
b) 21
c) 33
d) 35
e) None of these
Explanation:
Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.
10. A marketing manager is able to manage an average of Rs240 per week during 40h week . But in the final week of the month, his rates are increased by 50%. Then find out the commission the marketing manager makes in that week, if he puts in 60h during that final week?
a) 500
b) 540
c) 520
d) 600
e) None of these