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Re: Nice question on percentage
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29 Oct 2013, 10:28
AccipiterQ 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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\)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)*(100y)=100, since when multiplying by 100 you get rid of the 100 beneath the Y as well? \(1=(1+\frac{x}{100})(1\frac{y}{100})\); \(1=(\frac{100+x}{100})(\frac{100y}{100})\); \(100*100=(100+x)(100y)\). Hope it's clear.
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Re: The value of an investment increases by x% during January
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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.



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Re: The value of an investment increases by x% during January
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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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\) > \(100^2=100^2100y+100xxy\) > \(y=\frac{100x}{100+x}\) 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 \(y=100\frac{10,000}{100+x}\).



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Re: The value of an investment increases by x% during January
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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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\) > \(100^2=100^2100y+100xxy\) > \(y=\frac{100x}{100+x}\) 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 \(y=100\frac{10,000}{100+x}\). \(y = 100  \frac{10000}{100+x}\) \(y = \frac{100 * 100 + 100x  100000}{100+x}\) \(y = \frac{100x}{100+x}\)
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Re: The value of an investment increases by x% during January
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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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\) > \(100^2=100^2100y+100xxy\) > \(y=\frac{100x}{100+x}\) 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 \(y=100\frac{10,000}{100+x}\). \(y=\frac{100x}{100+x}\) Add / subtract 100 to RHS\(y = 100 + \frac{100x}{100+x}  100\) \(y = 100 + \frac{100x  10000 100x}{100+x}\) \(y = 100  \frac{10000}{100+x}\)
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Re: The value of an investment increases by x% during January
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21 Mar 2017, 15:43
I tried to pick smart numbers and then to test the answers
so: 100 * x>(100y)*100*x we need y
let's x=25 125>125*0,8 hence y=20% if we test the cases, E works well



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Re: The value of an investment increases by x% during January
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13 Jun 2017, 02:02
Bunuel y= 100 100(10000 + x)  I don't understand why I keep getting this: here are my steps 100[y/100]= 1/ (100 + [x/100]) 100y= 100 /(10000 +x) y= 100 /(10000 +x) 100 y= 100/(10000 +x) + 100 y= 100  100/(10000 +x) Please help



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Re: The value of an investment increases by x% during January
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12 Jul 2017, 15:43
AccipiterQ 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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\)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)*(100y)=100, since when multiplying by 100 you get rid of the 100 beneath the Y as well? Can someone answer this for me? thanks
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Re: The value of an investment increases by x% during January
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12 Jul 2017, 21:29
Smokeybear00 wrote: AccipiterQ wrote: Bunuel wrote: 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: \(I=I(1+\frac{x}{100})(1\frac{y}{100})\) > \(100^2=(100+x)(100y)\)
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)*(100y)=100, since when multiplying by 100 you get rid of the 100 beneath the Y as well? Can someone answer this for me? thanks Please read the whole thread before posting a question: https://gmatclub.com/forum/thevalueof ... l#p1285304
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Re: The value of an investment increases by x% during January
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13 Jul 2017, 02:47
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 ?
A. (200x)\(100 + 2x) B. x(2 + x)\(1 + x)^2 C. 2x\1 + 2x D. x(200 + x)\10,000 E. 100 –( 10,000 \ 100 + x) Let initial investment at the beginning of January be I . So after January total value of investment = (1+x/100)I After February total value of investment = (1+x/100)(1y/100) I Now initial investment at the beginning of January = value of investment at the end of February I = (1+x/100)(1y/100) I 1= (100+x)/100 * (100y )/100 10000= (100+x)(100y ) 100  y = 10000/(100+x) y = 100  10000/(100+x) Answer E.
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The value of an investment increases by x% during January
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14 Jul 2017, 07:11
Quote:
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)*(100y)=100, since when multiplying by 100 you get rid of the 100 beneath the Y as well?
Quote: Can someone answer this for me? thanks Quote: Please read the whole thread before posting a question: https://gmatclub.com/forum/thevalueof ... l#p1285304I did and no one has answered his question.
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Re: The value of an investment increases by x% during January
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Re: The value of an investment increases by x% during January
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