Real life problems on percentage will help us to solve different types of problems related to the real-life situations. To understand the procedures follow step-by-step explanation so that you can solve any other similar type of percentage problems.
Solved real life problems on percentage:
1. Mike needs 30% to pass. If he scored 212 marks and falls short by 13 marks, what was the maximum marks he could have got?
Solution:
If Mike had scored 13 marks more, he could have scored 30%
Therefore, Mike required 212 + 13 = 225 marks
Let the maximum marks be m.
Then 30 % of m = 225
(30/100) × m = 225
m = (225 × 100)/30
m = 22500/30
m = 750
2. A number is increased by 40 % and then decreased by 40 %. Find the net increase or decrease per cent.
Solution:
Let the number be 100.
Increase in the number = 40 % = 40 % of 100
= (40/100 × 100)
= 40
Therefore, increased number = 100 + 40 = 140
This number is decreased by 40 %
Therefore, decrease in number = 40 % of 140
= (40/100 × 140)
= 5600/100
= 56
Therefore, new number = 140 - 56 = 84
Thus, net decreases = 100 - 84 = 16
Hence, net percentage decrease = (16/100 × 100) %
= (1600/100) %
= 16 %
3. Max scored 6 marks more than what he did in the previous examination in which he scored 30. Maria scored 30 marks more than she did in the previous examination in which she scored 60. Who showed less improvement?
Solution:
Max percentage improvement in the first exam = (6/30 × 100) %
= (600/30) %
= 20 %
Maria percentage improvement in the first exam = (30/60 × 100) %
= (3000/60) %
= 50 %
Hence, 20 % < 50 %
Therefore, Max showed less improvement.
Percentage of the given Quantity
How much Percentage One Quantity is of Another?
Real Life Problems on Percentage
8th Grade Math Practice
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