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Example 18: A student has to score 40% of the maximum marks to pass. Priya scored 165 marks
and failed by 35 marks. Calculate the maximum marks.

Solution: Marks scored by Priya = 165
Shortage of marks out of passing marks = 35

\ Minimum passing marks = 165 + 35 = 200
According to the question,

40% of maximum marks = 200
200
\ 100% of maximum marks = × 100 = 500
40
Thus, the maximum marks is 500.

Example 19: The population of a small town last year was 2500. This year it is 3000. Find the
percentage increase in the population.
Solution: Original population = 2500. New population = 3000.

Increase in population = 3000 − 2500 = 500.

 Increase in value 
\ Percentage increase in population =   Original value × 100 %


 500  500
=  × 100 % = % = 20%

 2500  25
Thus, the population of the town is increased by 20% in this year.

Practice Time 10B

1. Evaluate the following:
1
(a) 42% of 225 (b) 3.8% of 170 (c) 15 3 % of 345 m (d) 80% of 15 kg
1
(e) 2 % of 190 L
2
2. Find the number/quantity whose:
(a) 51% is `170 (b) 24% is 192 kg (c) 26% is 1300 km (d) 33% is 600 L
(e) 8% is 10 hr

3. What percent of
(a) 800 is 180? (b) 12 kg is 480 g? (c) 105 km is 900 m? (d) 1 hr is 45 min?
(e) 150 L is 135 L?

4. Find the value of x.
(a) x% of 72 is 12 (b) x% of 1250 is 1824 (c) 12% of x is 600 (d) x% of 372 L is 272 L
(e) 64% of x is 480
5. What per cent of 5 day is 10 min?

6. If 13% of a number is 65, find the number.

227 Percentage and Its Applications