Real Life Problems on Percentage provided here covers different types of questions that you might come across regarding the concept Percentage. To be clear with the concept go through the step by step solutions provided for all the Problems on Percentage. Assess your preparation standards and identify the areas you are lagging and concentrate on them. Practice the Solved Examples on Percentage and enhance your conceptual knowledge.

1. Jim needs 20% to pass. If he scored 225 marks and falls short by 25 marks, what were the maximum marks he could have got?

Solution:

If Jim Scored 25 Marks he would have got 20% pass

Thus Jim required = 225+25

= 250 Marks

Let the Maximum Marks be m

20% of m = 250

20/100*m = 250

m = (250*100)/20

= 1250

2. A fruit seller had some oranges. He sells 30% oranges and still has 350 oranges. How many Oranges Originally, he had?

Solution:

Let suppose he had x oranges

x – 30% of x = 350

x(1-30/100) = 350

x = 350/(70/100)

= (350*100)/70

= 500

Therefore, fruit seller had 500 oranges originally.

3. If 10% of a = b, then b% of 10 is the same as

Solution:

10/100*a = b

a/10 = b

a = 10b ……(1)

b% of 10 = b/100*10

= 10b/100

= b/10

Substitute the value of b obtained in equation (1)

= (a/10)/10

= a/10*1/10

= a/100

= 1/100*a

= 1% of a

Therefore, b% of 10 is same as 1% of a.

4. In an election between two candidates, one got 50% of the total valid votes. If the total number of votes was 8000, the number of valid votes that the candidate got, was?

Solution:

Total Number of Votes Polled = 8000

Candidate got 50% of Total Votes

= 50% of 8000

= 50/100*8000

= 50*80

= 4000

5. Three candidates X, Y, Z contested in an election and received 1100, 7536, and 11500 votes respectively. What percentage of the total votes did the winning candidate get?

Solution:

The candidate Z with the highest votes will win

Total Votes = 1100+7536+11500

= 20136

Winning Percentage = (Votes Candidate Z got/ Total Votes)*100

= (11500/20136)*100

= 57.11%

Therefore, the Winning Percentage is 57.11%

6. The population of a town increased from 1,75,000 to 2,50,500 in a decade. What is the average percent increase of population per year is?

Solution:

Increase in Population in 10 years = 2, 50, 500- 1, 75, 000

= 75500

Average Percent Increase in Population per Decade = (75500/175000)*100

= 43.14%

Required Average per year = (43.14/10)%

= 4.314%

7. Robin scored 5 marks more than what he did in the previous examination in which he scored 25. John scored 40 marks more than she did in the previous examination in which she scored 30. Who showed less improvement?

Solution:

Robin Improvement = (5/25)*100

= 20%

John Improvement = 40/30*100

= 4000/30

= 133.33%

Therefore, Robin showed less percentage.