Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 17:19

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The value of an investment increases by x% during January

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 38858
Followers: 7728

Kudos [?]: 106047 [0], given: 11607

Re: Nice question on percentage [#permalink]

### Show Tags

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: $$I=I(1+\frac{x}{100})(1-\frac{y}{100})$$ --> $$100^2=(100+x)(100-y)$$

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?

$$1=(1+\frac{x}{100})(1-\frac{y}{100})$$;

$$1=(\frac{100+x}{100})(\frac{100-y}{100})$$;

$$100*100=(100+x)(100-y)$$.

Hope it's clear.
_________________
Intern
Joined: 30 Jul 2011
Posts: 11
Location: United States
Concentration: Finance, Economics
GMAT Date: 05-31-2024
Followers: 1

Kudos [?]: 6 [0], given: 22

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

### Show Tags

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.
Senior Manager
Joined: 15 Aug 2013
Posts: 314
Followers: 0

Kudos [?]: 63 [0], given: 23

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

### Show Tags

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 $$y=100-\frac{10,000}{100+x}$$.

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 $$y=100-\frac{10,000}{100+x}$$.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1857
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 51

Kudos [?]: 2168 [0], given: 193

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

### Show Tags

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 $$y=100-\frac{10,000}{100+x}$$.

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 $$y=100-\frac{10,000}{100+x}$$.

$$y = 100 - \frac{10000}{100+x}$$

$$y = \frac{100 * 100 + 100x - 100000}{100+x}$$

$$y = \frac{100x}{100+x}$$
_________________

Kindly press "+1 Kudos" to appreciate

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1857
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 51

Kudos [?]: 2168 [0], given: 193

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 $$y=100-\frac{10,000}{100+x}$$.

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

$$y = 100 + \frac{100x}{100+x} - 100$$

$$y = 100 + \frac{100x - 10000 -100x}{100+x}$$

$$y = 100 - \frac{10000}{100+x}$$
_________________

Kindly press "+1 Kudos" to appreciate

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15431
Followers: 649

Kudos [?]: 207 [0], given: 0

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

### Show Tags

15 Aug 2015, 19:42
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15431
Followers: 649

Kudos [?]: 207 [0], given: 0

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

### Show Tags

31 Oct 2016, 03:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 03 Jan 2017
Posts: 204
Followers: 0

Kudos [?]: 2 [0], given: 4

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

### Show Tags

21 Mar 2017, 15:43
I tried to pick smart numbers and then to test the answers

so: 100 * x->(100-y)*100*x
we need y

let's x=25
125->125*0,8
hence y=20%
if we test the cases, E works well
Re: The value of an investment increases by x% during January   [#permalink] 21 Mar 2017, 15:43

Go to page   Previous    1   2   [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
The value of a particular investment increased by 5% the first year 2 26 Apr 2017, 11:33
1 The product of x and y is a constant. If the value of x is increased b 5 04 Feb 2016, 23:20
10 Money invested at x%, compounded annually, triples in value in approxi 9 18 Mar 2017, 01:46
8 The value of land increases by a% during June and decreases 6 07 Sep 2016, 20:30
14 The value of investment Q increased by q percent from t 7 07 Apr 2016, 19:46
Display posts from previous: Sort by