# 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:

40% of 75 + 80% of 25 = K% of 250

Find the value of K?

 A) 10 B) 20 C) 30 D) 35

Explanation:

From the given data,

9 34212
Q:

1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of the girls pass. The percentage of the total who failed is :

 A) 58 B) 62 *2/3 C) 64 D) 67

Explanation:

Total number of students = 1100 + 700 = 1800.

Number of students passed = (42% of 1100 + 30% of 700) - (462 + 210) = 672.

Number of failues = 1800-672 = 1128.

Percentage failure = (1128/1800 * 100 )% = 62 * 2/3 %.

102 32662
Q:

If 15% of x = 20% of y, then x:y is  ___?

 A) 4:3 B) 5:4 C) 4:5 D) 3:4

Explanation:

Given 15% of x = 20% of y

=> 15x = 20y

=> x/y = 20/15

=> x : y = 4 : 3

81 32361
Q:

56% of 225 + 20% of 150 = ? - 109

Find the value of '?' in the above equation

 A) 264 B) 220 C) 241 D) None

Explanation:

Given 56% of 225 + 20% of 150 = ? - 109

we know that a% of b = b% of a

Now,

225% of 56 + 20% of 150 = ? - 109

200% of 56 + 25% of 56 + 20% of 150 = ? - 109

112 + 14 + 30 = ? - 109

? = 112 + 14 + 30 + 109

? = 264

29 31895
Q:

In a History examination, the average for the entire class was 80 marks. If 10% of the students scored 35 marks and 20% scored 90 marks, what was the average marks of the remaining students of the class ?

 A) 25 B) 50 C) 75 D) 100

Explanation:

Let the number of students in the class be 100 and let this required average be x.

Then, (10 * 95) + (20 * 90) + (70 * x) = (100 * 80)

=> 70x = 8000 - (950 + 1800) = 5250

=> x = 75.

115 30990
Q:

The population of a town was 1,60,000 three years ago, If it increased by 3%, 2.5% and 5% respectively in the last three years, then the present population in

 A) 155679 B) 167890 C) 179890 D) 177366

Explanation:

Present population =  160000 *  (1 + 3/100)(1 + 5/200)(1 + 5/100)= 177366.

75 29252
Q:

If x is 80% of y, then what percent of 2x is y?

 A) 65.5 % B) 64.5 % C) 63.5 % D) 62.5 %

Explanation:

x = 80 % of y

=> x =(80/100 )y

=> y/x = 5/4

Required percentage =  [(y/2x)*100] % = (5/8*100) % =62.5%

93 28571
Q:

If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 350%, the resultant fraction is 25/51. What is the original fraction ?

 A) 31/25 B) 15/17 C) 14/25 D) 11/16

Explanation:

The original fraction is    = 15/17.