# 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:

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 %

Answer & Explanation Answer: A) 21.6 %

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 %.

63 27736
Q:

How many litres of pure acid are there in 8 litres of a 20% solution ?

 A) 1.4 B) 1.5 C) 1.6 D) 1.7

Explanation:

Quantity of pure acid = 20% of 8 litres = ((20/100)*8) litres = 1.6 litres.

44 27083
Q:

Two numbers are 20% and 30% less than the third number . How much percent is the second number less than first?

Let the third number be 100.

Then, the first number = 100 - 20 = 80 and second number = 100 - 30 = 70.

Difference between the first and second number = 80-70 = 10

$\inline&space;\therefore$ Required percentage = % = 100/8 = 12.5%

26540
Q:

A student multiplied a number by 2/5 instead of 5/2. What is the percentage error in evaluation ?

 A) 52% B) 64% C) 84% D) 77%

Explanation:

Let the number be 'x'

Then, according to the given data,

$52x-25x52xx100$

$2125x100$

= 84%

18 25966
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

47 25786
Q:

In some quantity of ghee, 60% is pure ghee and 40% is vanaspati. If 10 kg of pure ghee is added, then the strength of vanaspati ghee becomes 20%. The original quantity was :

 A) 10 kg B) 15 kg C) 20 kg D) 25 kg

Answer & Explanation Answer: A) 10 kg

Explanation:

Let the original quantity be x kg. Vanaspati ghee in x kg  =  (40x / 100 )kg = (2x / 5) kg.

Now, (2x/5)/(x + 10) = 20/100

=> 2x / (5x + 50) = 1/5

=> 5x = 50

=> x = 10.

56 25783
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.

59 25531
Q:

The difference between a number and its two-fifth is 510. What is 10% of that number ?

 A) 75 B) 85 C) 95 D) 105

Explanation:

Let the number be x. Then, x-(2/5)x = 510

=> 3x/5 =510

=>x =[510 * ( 5/3)] =850

10 % 0f 850 = 85.