# Percentage Questions

FACTS  AND  FORMULAE  FOR  PERCENTAGE  QUESTIONS

I.Concept of Percentage : By a certain percent , we mean that many hundredths. Thus x percent means x hundredths, written as x%.

To express x% as a fraction : We have , x% = x/100.

Thus, 20% = 20/100 = 1/5;

48% = 48/100 = 12/25, etc.

To express a/b as a percent : We have, $\frac{a}{b}=\left(\frac{a}{b}×100\right)%$ .

Thus, $\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%$

II. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is $\left[\frac{R}{\left(100+R\right)}×100\right]%$

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}×100\right]%$

III. Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :

1. Population after n years = $P{\left(1+\frac{R}{100}\right)}^{n}$

2. Population n years ago =  $\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}$

IV. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,

1. Value of the machine after n years = $P{\left(1-\frac{R}{100}\right)}^{n}$

2. Value of the machine n years ago = $\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}$

V. If A is R% more than B, then B is less than A by

$\left[\frac{R}{\left(100+R\right)}×100\right]%$

If A is R% less than B , then B is more than A by

$\left[\frac{R}{\left(100-R\right)}×100\right]%$

Q:

Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :

 A) 20 % B) 40 % C) 60 % D) 80 %

Explanation:

Let the total original sale be Rs. 100. Then, original number of visitors = 100.

New number of visitors  =  120/0.75 = 160.

Increase % = 60 %.

94 27550
Q:

Of the 1000 inhabitants of a town, 60 % are males of whom 120 % are literate. If, of all the inhabitants,  25% are literate, then what percent of the females of the town are literate ?

 A) 32.5 % B) 43 % C) 46.6 % D) 53.2 %

Explanation:

Number of males = 60% of 1000 = 600. Number of females = (1000 - 600) = 400.

Number of literates = 25% of 1000 = 250.

Number of literate males = 20% of 600 = 120.

Number of literate females = (250 - 120) = 130.

Required pecentage = (130/400 * 100 ) % = 32.5 %.

68 27533
Q:

What percent decrease in salaries would exactly cancel out the 20 percent increase ?

 A) (16 + 2/3) % B) (15 + 2/3)% C) (14 +1/3)% D) (13 + 4/5)%

Explanation:

Let the orginal salary = Rs. 100. New's salary = Rs.120.

Decrease on 120 = 20. Decrease on 100 = [(20/120)*100]% = (16 + 2/3) %.

54 25704
Q:

In a City, 35% of the population is composed of migrants, 20% of whom are from rural areas. Of the local population, 48% is female while this figure for rural and urban migrants is 30% and 40% respectively. If the total population of the city is 728400, what is its female population ?

 A) 324138 B) 349680 C) 509940 D) None of these

Explanation:

Total Population = 728400

Migrants = 35 % of 728400  =  254940

$\inline \therefore$local population = (728400 - 254940) = 473460.

Rural migrants = 20% of 254940 = 50988

Urban migrants = (254940 - 50988) = 203952

Female population = 48% of 473460 + 30% of 50988 + 40% of 203952 = 324138

44 25292
Q:

A bag contains 600 coins of 25 p denomination and 1200 coins of 50 p denomination. If 12% of 25 p coins and 24% of 50 p coins are removed, the percentage of money removed from the bag is nearly :

 A) 21.6 % B) 15.3 % C) 14.6 % D) 12.5 %

Explanation:

Total money = Rs.[600*(25/100)+1200*(50/100)]= Rs. 750.

25 paise coins removed = Rs. (600*12/100) = 72.

50 paise coins removed = Rs. (1200*24/100)= 288.

Money removed =Rs.(72*25/100+288*50/100)  = Rs.162.

Required percentage = (162/750*100)% = 21.6 %.

55 23792
Q:

What will come in place of the question mark(?) in the following question ?

56% of 870 + 82% of 180 = 32% of 90 + ?

 A) 777 B) 666 C) 606 D) 789

Explanation:

56% of 870 = 56x870/100 = 487.20

82% of 180 = 82x180/100 = 147.60

32% of 90 = 32x90/100

487.20 + 147.6 - 28.8  =  ?

? = 634.8  - 28.8

? = 606.

8 22809
Q:

The sum of the number of boys and girls in a school is 150. if the number of boys is x, then the number of girls becomes x% of the total number of students. The number of boys is :

 A) 60 B) 70 C) 80 D) 90

Explanation:

We have : x + (x% of 150 )= 150

=> x+(x/100)*150] = 150

=>

=> x = (150*2)/5 = 60

43 22786
Q:

The diference of two numbers is 20% of the larger number, if the smaller number is 20, then the larger number is :

 A) 15 B) 25 C) 35 D) 45

Explanation:

Let the large number be x.

Then x - 20 = 20x/100

=> x - x/5 = 20  => x = 25.