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Percentage Objective questions and answer

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1.)  Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?

A. 90                         B. 140

C. 220                       D. None of these

View Answer

Answer: Option B


Sol.  Let the number be x.

Then, 4 * x/5 –(65% of x) = 21

4x/5 –65x/100 = 21

5 x = 2100

x = 140.

2.) 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. None of these

View Answer

Answer: Option C


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.

3.) 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. None of these

View Answer

Answer: Option D


Let the original fraction be x/y.

Then (115% of x)/(92% of y)=15/16 => (115x/92y)=15/16


4.) 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%

View Answer

Answer: Option A


Reduction  in consumption = [((R/(100+R))*100]%


5.) 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. None of these

View Answer

Answer: Option C


Population in the beginning of the first year

= 9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.

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