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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?

Re: The value of an investment increases by x% during January [#permalink]

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16 Nov 2013, 12:41

So I have a couple of principles that I think we can apply here to solve the problem quickly. For successive % changes, we must realize the following:

1) if x > y, the change is going to be positive, negative, or zero. For example, 100 is increased by 10% so we get 110. Then 110 is decreased by 2%, the result is ~108. So we have a net positive change. If 110 is decreased by ~9.09%, the result is 100. So we have a net change of 0. If 110 is decreased by ~9.99%, the result is 99, so the net change is negative.

2) If x<y, the net change is always going to be negative. For example, 100 is increased by 10%, which gives us 110 and then 110 is decreased by 11% which gives us ~98.

3) If x=y, the net change is always going to be negative. For example, 100 is increased by 10%, which gives us 110 and then 110 is decreased by 10% which gives us 99. This reminds me of my investment in Apple's stock. One day I'm up 3% and then the next day I'm down 3% and initially, I might think that I'm at a net change of 0, but I actually lost money . If my initial $100 goes up by 3%, I now have 103 (w00t w00t!), but if my $103 goes down by 3% the next day, I have less than my initial 100 because 3% of 103 is more than 3% of 100.

Now, for this problem, we are told that the net change is zero, so using the principles above, we know that x > y and more importantly, we get a net change of zero whenever we increase by 10% and decrease by a little more than 9%. So therefore, all we need to do is plug in 10 for x and the answer choice that gives us ~9 is correct.

I think if we remember these very easy principles, it will help us in the long run, because we are able to solve problems such as this one, in under 2 minutes.

Re: The value of an investment increases by x% during January [#permalink]

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03 Aug 2014, 18:34

Bunuel wrote:

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: \(I=I(1+\frac{x}{100})(1-\frac{y}{100})\) --> \(100^2=(100+x)(100-y)\) --> \(100^2=100^2-100y+100x-xy\) --> \(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

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).

Hi Bunuel,

How did you make the leap from ?

\(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

Re: The value of an investment increases by x% during January [#permalink]

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03 Aug 2014, 23:16

russ9 wrote:

Bunuel wrote:

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: \(I=I(1+\frac{x}{100})(1-\frac{y}{100})\) --> \(100^2=(100+x)(100-y)\) --> \(100^2=100^2-100y+100x-xy\) --> \(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

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).

Hi Bunuel,

How did you make the leap from ?

\(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

Re: The value of an investment increases by x% during January [#permalink]

Show Tags

03 Aug 2014, 23:37

russ9 wrote:

Bunuel wrote:

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: \(I=I(1+\frac{x}{100})(1-\frac{y}{100})\) --> \(100^2=(100+x)(100-y)\) --> \(100^2=100^2-100y+100x-xy\) --> \(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

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).

Hi Bunuel,

How did you make the leap from ?

\(y=\frac{100x}{100+x}\) or which is the same \(y=100-\frac{10,000}{100+x}\).

Re: The value of an investment increases by x% during January [#permalink]

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15 Aug 2015, 19:42

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