prashantbacchewar wrote:

The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?

(200x)\(100 + 2x)

x(2 + x)\(1 + x)^2

2x\1 + 2x

x(200 + x)\10,000

100 –( 10,000 \ 100 + x)

I got this question on MGMAT exam and got it wrong during the test. Later during the review I was able to solve this question correctly. I felt I should post this.

Let the value of investment at the beginning of January be

I.

Then the value of the investment at the end of February would be

I(1+\frac{x}{100})(1-\frac{y}{100}).

We are told that these values are equal: I=I(1+\frac{x}{100})(1-\frac{y}{100}) --> 100^2=(100+x)(100-y)Answer: E.

This question can also be done by picking some smart numbers for I, x and y (say I=10, x=100 and y=50).

how did you get rid of the 100s on the bottom of each fraction and end up with 10000 on the other side? If you took the (1+x/100) portion and multiplied teh whole equation by 100 to get rid of it wouldn't you end up with (100+X)*(100-y)=100, since when multiplying by 100 you get rid of the 100 beneath the Y as well?