How to Solve Successive Percentage Problems?

This time I thought of giving you the crux of the whole article at the very beginning. See the formulae above and you will know what I am talking about.

In almost all competitive examinations there are questions based on percentage. Many of such problems can be solved using one of the above formulae. Here we are going to talk about successive increases or decreases of a quantity. Problem which otherwise takes many steps and becomes difficult to answer can be solved instantly using these formulae. It gives a very short and fast calculating method. The only thing is to remember the formula well. The above four formulae are variation of each other and you need to understand where to use which one. I will share some examples to illustrate the use of each. You will yourself understand where to use which one.

Simple Example The population of a town is 6000. It increases by 10% during the first year and by 20% during the second year. What is the population after 2 years?

Answer: We use the third formula because we have to increase the given number, hence in numerator we will have larger number and in denominator you will have smaller number, i.e. 100

(A*(100+x)*(100+y))/(100*100)

(Since, there are only two increases given, we modify the formula to suit our purpose)

(6000*110*120)/(100*100)=7920

Note: To easily multiply numbers like 110 and 120, we can use base method of multiplication

If the same problem is reversed and we are given with the final number 7920, after successive increase of 10% and 20%, then we would use the first formula to find out the original number.

(A*100*100)/((100+x)*(100+y) )

= (7920*100*100)/(110*120)=6000

If instead of increase there are successive decreases, we use second or fourth formula as the case may be. If you have seen any question which can be solved using this method, share that with all by posting a comment below.

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