Question

) A number is first increased by 20% and then decreased by 20%. Find the net increase and
decrease percent​

Answer 1

Answer:

Let us suppose the number to be x.

Let us suppose the number to be x.As per the first case, the number is increased by 20%

Let us suppose the number to be x.As per the first case, the number is increased by 20%Hence: x+ ((x*20)/100)

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’

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%

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)

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.

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.

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_

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

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/25Net decrease in percentage= (Net Decrease/ Initial Value) * 100%

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/25Net decrease in percentage= (Net Decrease/ Initial Value) * 100%Thus, ((x/25)/x) * 100% = 4%

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/25Net decrease in percentage= (Net Decrease/ Initial Value) * 100%Thus, ((x/25)/x) * 100% = 4%So, the answer is 4%.

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