SOLVING WORD PROBLEMS ON PERCENTAGES

Problem 1 :

What is the value of 3 1/3 % of 1.5 % of smallest six digit number?

Solution :

Smallest six digit number is 100000.

1.5 % of the smallest six digit number

= (1.5/100) × 100000

= 0.015 × 100000

= 1500

3 1/3 % of 1500 = 10/3%

= 3.3% × 1500

= (3.3/100) × 1500

= 0.033 × 1500

= 49.5

Problem 2 :

What percent of 2/7 is 1/35?

Solution :

Let ‘x’ be required percentage.

x% of 2/7 = 1/35

(x/100) of (2/7) = 1/35

2x/700 = 1/35

x/350 = 1/35

Using cross multiplication.

35x = 350

x = 350/35

x = 10

So, 10 per cent of 2/7 is 1/35.

Problem 3 :

If the C.P. of 6 articles is equal to the S.P. of 4 articles, find the gain percent?

Solution :

Let cost price of one article be ‘x’.

The cost price of 6 articles = 6x

Cost price of 6 articles = Selling price of 4 articles

The selling price of 4 articles = 6x (because cost price of 6 articles)

Gain = 6x - 4x ==> 2x

So, gain percent is 50%

Problem 4 :

If 40% of a number is 256, then what is the 25% of that number?

Solution :

Let x be the original number.

If 40% of a number x = 256

(40/100) × x = 256

(2/5) × x = 256

x = 256 × (5/2)

x = 1280/2

x = 640

So, 25% of a number is 640.

Problem 5 :

If x% of 25/2 is 150, then what is the value of x?

Solution :

If x% of 25/2 =150

(x/100) × (25/2) = 150

x/100 = 150 × (2/25)

x/100 = 300/25

x/100 = 12

x = 12 × 100

x = 1200

So, the value of x is 1200.

Problem 6 :

What percentage of n umbers from 1 to 30 has the digit 9 in the unit place?

Solution :

Total number of numbers from 1 to 30 = 30

Numbers which has 9 at the unit place.

9,19 and 29.

Total numbers = 3

Required Percentage = (3/30) ×100

= 10%

Problem 7 :

If the cost price is 25% of selling price, then find the profit percent?

Solution :

Let x be the selling price.

Profit % = (profit / cost price) x 100%

Profit = Selling price - cost price

So, profit per cent = 300%

Problem 8 :

In how much time would the simple interest on a certain sum be 0.125 times the Principal at 10% per annum?

Solution :

Simple interest = (Principle value × rate × time)/100

0.125P = (P × 10 × T)/100

0.125P = (P × T)/10

T = (10 × 0.125P)/P

T = 10 × 0.125

T = 1.25

T = 125/100

T = 5/4

1 year 3 months is the investment period.

Problem 9 :

A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water shall i put in the tank, so that it is 80% full?

Solution :

Capacity of tank = 50 litres of water.

Tank full of the water = 30%

= (30/100) × 50

= 15 litres

80% full of the water is

= (80/100) × 50

= 40 litres

Number of liters to be put = 40 – 15

= 25 litres

25 litres of water i need to put.

Problem 10 :

A 60 lit tank was full of oil. Ram used 40% of it and poured the rest into a 40 lit tank.

(i) What percent of the 40 lit tank was filled with oil?

(ii) If Ram used 3 lit of oil daily, what percent of oil in the 40 lit tank would be used in 10 days?

Solution :

Quantity of oil in tank = 60 litres

40% of it is used.

Quantity of oil used = 40% of 60 lit

Quantity of oil remaining = 40 lit

= (40/100) × 60

= 24 litres

(i) Quantity of oil left = 60 - 24 ==> 36

Percent of the 40 lit tank = (36/40) × 100

= 90%  

(ii) Every day he uses 3 lit of oil

Quantity of oil used in 10 days.

= 10 × 3

= 30 lit

Quantity of oil in the 40 lit tank = 36 liter.

Percent of oil in the 40 lit in the tank = (30/36) × 100%

= 83.3%