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:

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

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.

55 22211
Q:

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

52.5% of 800 + 30.5% of 2800 = ? + 87.30

 A) 1096.30 B) 1226.70 C) 1124.20 D) 1186.70

Explanation:

52.5% of 800 + 30.5% of 2800 = ? + 87.30

420 + 854 - 87.30 = ?

1274 - 87.30 = ?

? = 1186.70

10 22153
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.

37 21691
Q:

In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?

 A) 4000 B) 8000 C) 12000 D) 16000

Explanation:

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

55 21227
Q:

A district has 64000 inhabitants. If the population increases at the rate of 2 * 1/2 % per annum, then the number of inhabitants at the end of 3 years will be :

 A) 65380 B) 68921 C) 70987 D) 72345

Explanation:

Population afer 3 years = 64000 * (1 +  5/(2 *100) )^3 = (64000 * 41/40 * 41/40 * 41/40 ) = 68921..

59 20903
Q:

A number is increased by 20% and then decreased by 20%, the final value of the number is ?

 A) increase by 2% B) decrease by 3% C) decrease by 4% D) increase by 5%

Explanation:

Here, x = 20 and y = - 20
Therefore, the net % change in value
= %
%  or  - 4%

Since the sign is negative, there is a decrease in value by 4%.

35 20693
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.

37 20484
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%