Could any one explain this math question from Princeton ?

[Devi]

The o/p of a factory was increased by 10% to keep up with the rising demand.To meet the holiday rush this new o/p was again increased by 20%. By approxmately what percentage the o/p now have to be reduced to restore to the original o/p


I always get this qn wrong ! Please help.

My method.

Initial O/p =x
After first increment o/p = 110/100 * x = 1.1x
After second increment o/p = 1.1 X * 120/100 = 1.32x

Increment is = 0.32x

Now the o/p must be reduced from 1.32 x to x right?

The answer is 24 % I cant seem to reach till there..

[Nermal]

Your method is correct, you just have to go on.

I usually pick numbers:
first: x=100
x+10%=110
second: (x+10%)+20%=132

Now we want to know by what percent we have to drop 132 in order to get 100.
In total numbers we have to drop by
132-100=32
Now we want to know what fraction 32 of 132 is:
32/132=0,24=24%

[Devi]

Thank you Nermal.

[Nermal]

You are welcome Devi!