PLE 12.
A number is increased by 20% and then decreased by 20%. Find the net increase
or decrease per cent?
Answers
Answer:
Let us suppose the number to be x.
As per the first case, the number is increased by 20%
Hence: x+ ((x*20)/100)
Resultant number is ‘6x/5’
As per the second case, resultant number is then decreased by 20%
Hence : (6x/5) - (((6x/5) * 20)/100)
Which is (6x/5) - (6x/25) = 24x/25.
The original number being x, we witness a net decrease in the overall process as (24x/25) is less than x.
Therefore Net decrease_
x - (24x/25)= x/25
Net decrease in percentage= (Net Decrease/ Initial Value) * 100%
Thus, ((x/25)/x) * 100% = 4%
So, the answer is 4%.
Answer:
4%
Step-by-step explanation:
Let us suppose the number to be x.
As per the first case, the number is increased by 20%
Hence: x+ ((x*20)/100)
Resultant number is ‘6x/5’
As per the second case, resultant number is then decreased by 20%
Hence : (6x/5) - (((6x/5) * 20)/100)
Which is (6x/5) - (6x/25) = 24x/25.
The original number being x, we witness a net decrease in the overall process as (24x/25) is less than x.
Therefore Net decrease_
x - (24x/25)= x/25
Net decrease in percentage= (Net Decrease/ Initial Value) * 100%
Thus, ((x/25)/x) * 100% = 4%
So, the answer is 4%.