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The value of an investment increases by x% during January [#permalink]
06 Oct 2010, 07:02
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Difficulty:
55% (hard)
Question Stats:
71% (02:58) correct
29% (02:32) wrong based on 233 sessions
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)
Re: Nice question on percentage [#permalink]
06 Oct 2010, 07:28
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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). _________________
Re: Nice question on percentage [#permalink]
06 Oct 2010, 07:39
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).
Ah good tips here..
I was almost there but didn't restate to \(y=100-\frac{10,000}{100+x}\) so I thought my math was wrong..
Re: Nice question on percentage [#permalink]
06 Oct 2010, 07:42
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krazo wrote:
I got to the OA but it took much longer than 2 minutes!!
At first I tired algebra but failed, so I switched to picking numbers...
I set the original amount to $100, x = 10% and solved for y using this formula:
\(100 = 100(1.1)(1-\frac{y}{100})\)
Then I looked at which of the answer choices gave me the same value.
10% for x is not a good choice. Chosen numbers should be easy to manipulate with.
Try initial investment to be $10 (I=$10) increase by 100% (x=100) --> you'll get $20. Now we need $20 to reduce back to $10, so it's 50% decrease (y=50).
Substitute \(x=100\) in answer choices to see which gives \(y=50\) --> E: \(100-\frac{10,000}{100+x}=100-\frac{10,000}{100+100}=50=y\) --> correct.
It might happen that for some choices of I, x and y, other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.
Re: Nice question on percentage [#permalink]
21 Feb 2011, 07:38
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 Brunel
For your solution, once you arrive at the answer y={100x}/{100+x}; how do you know intuitively to multiply to get y = 100 - 10,000/(100+x)? Or do you work backwards from y = 100 - 10,000/(100+x) to get y={100x}/{100+x}?
I am trying to solve this question using algebra only but was thinking if its better to use values instead...
Re: Nice question on percentage [#permalink]
21 Feb 2011, 18:26
I got this one right. Once you have y = ... , you can look through the choices and see that only E is remotely similar to the desired form. Just check by changing your answer to the form in E).
The formula gives you the overall percentage change. Since the value of the investment is the same after the two successive percentage changes, it means the overall percentage change is 0.
x - y - xy/100 = 0 (Since y is a decrease) x - y(1 + x/100) = 0 y = 100x/(100 + x) (which is the same as 100 - 10,000/(100 + x)) Answer (E) _________________
Re: Nice question on percentage [#permalink]
05 Mar 2011, 19:08
Good question.
Prashant , i appreciate you posting the question. Please post the multiple choices , if available. That way others can solve the problem too without looking at the solution directly.
The formula gives you the overall percentage change. Since the value of the investment is the same after the two successive percentage changes, it means the overall percentage change is 0.
x - y - xy/100 = 0 (Since y is a decrease) x - y(1 + x/100) = 0 y = 100x/(100 + x) (which is the same as 100 - 10,000/(100 + x)) Answer (E)
can someone please explain me how 100x/(100+x) == 100 - 10,000/(100+x)
This MGMAT question is not cool. I arrived at y=100x/(100+x) and stared at the 5 answer choices for like 4 mins before deciding to guess (which I was wrong) and move on. In the real test, how am I supposed to know which answer choice will reduce into my answer. This is not to mention there are chances that I made mistakes to arrive at my own answer too. So I wouldn't risk spending the time to work out on each of the answer (5 of them to reach E) choice and hope that it will arrive at my result. The only thing I agree here is when making a quick glance, only E has the same denominator, which makes it a good contender.
This MGMAT question is not cool. I arrived at y=100x/(100+x) and stared at the 5 answer choices for like 4 mins before deciding to guess (which I was wrong) and move on. In the real test, how am I supposed to know which answer choice will reduce into my answer. This is not to mention there are chances that I made mistakes to arrive at my own answer too. So I wouldn't risk spending the time to work out on each of the answer (5 of them to reach E) choice and hope that it will arrive at my result. The only thing I agree here is when making a quick glance, only E has the same denominator, which makes it a good contender.
There are many ways to solve a question. One important thing is exposure to different methods. When I saw this question, I knew I have seen a formula which could be easily used. Another method could be just plugging in values.
Say initial investment was $100 and final was also $100. Say it increased by 10% (x = 10) in Jan so it became 110. By what % should it decrease to become 100 again? By (10/110) *100= 100/11 %
Now plug in x = 10 and the option that gives you 100/11 is the answer. _________________
Re: The value of an investment increases by x% during January [#permalink]
03 Jan 2013, 14:54
What if I just plug in 0 for X and hence 0 for Y? Hence I also end up with E as this is the only answer choice that (1) gives me a value and (2) gives me the value 0. This takes like 10sec to figure out the solution...
Is this just coincidence that it worked in this case (i.e. 0 is normally a bad number to pick?) or is it okay to pick the number 0 and then to check the answer choices?
Re: The value of an investment increases by x% during January [#permalink]
03 Jan 2013, 19:36
1
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Expert's post
ln2282 wrote:
What if I just plug in 0 for X and hence 0 for Y? Hence I also end up with E as this is the only answer choice that (1) gives me a value and (2) gives me the value 0. This takes like 10sec to figure out the solution...
Is this just coincidence that it worked in this case (i.e. 0 is normally a bad number to pick?) or is it okay to pick the number 0 and then to check the answer choices?
There is nothing wrong with taking x = 0, y = 0 since it is a possible situation. IF there is no change in Jan, there will be no change in Feb to keep the value same at the end of Feb. Generally, x = 0 or x = 1 are great values to take.
The only problem is that it doesn't give you the answer here. Put x = 0 in all other options, you get y = 0 (numerator becomes 0 in every case and denominator is non-zero). So you cannot say what the answer is. _________________
Re: The value of an investment increases by x% during January [#permalink]
04 Sep 2013, 20:59
Expert's post
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)
i solved this task as the following
100=100(1+X)(1-Y) --> y=x/(1+X)
and at the end i couldnt see my answer among the answer choises, hence thought i had solved the task incorrectly and chose (c) 2x/(1 + 2x) which seemed more or less similar to mine. My question is when dealing with percents is it better to choose x/100 or just x? eg if i read "The value of an investment increases by x% during January and decreases by y% during February" is it better to write 100*(1+x)(1-y) or 100(1+x/100)(1-y/100) i understand that they are similar in meaning but if i went with the second version id have higher chance to pick the right answer because the denominator in my answer and that in (E) would be the same
Responding to a pm:
Whether to use x/100 or x is not a choice. It depends entirely on the question.
Note that the question here says: "The value of an investment increases by x% during January" You are given that it increases by x% i.e. x per cent i.e. x per 100. So it increases by x/100. So the new value becomes Old value + Old value * (x/100). So we get the new value by multiplying the old value by (1 + x/100) When you solve for x, you might get a value similar to 10 or 20 etc. So the % increase is 10% or 20%.
We cannot choose to use (1 + x) instead.
Similarly, if the question says that the % increase in value is x.... when you solve for x, you will get something similar to 0.1 or 0.2 which is 10% or 20%.
Depending on the question, you need to use either (1 + x/100) or (1 + x). _________________
Re: Nice question on percentage [#permalink]
29 Oct 2013, 09:22
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: \(I=I(1+\frac{x}{100})(1-\frac{y}{100})\) --> \(100^2=(100+x)(100-y)\)
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?
gmatclubot
Re: Nice question on percentage
[#permalink]
29 Oct 2013, 09:22
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