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Re: The value of an investment increases by x% during January and [#permalink]
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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.
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Re: The value of an investment increases by x% during January and [#permalink]
There was a formula for successive percents: is there a way to solve it using the same.

Net percent = a+ b + ab/100

where a is the initial percent and b is the next percent change.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/02 ... e-changes/

Please help!
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mbafall2011 wrote:
There was a formula for successive percents: is there a way to solve it using the same.

Net percent = a+ b + ab/100

where a is the initial percent and b is the next percent change.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/02 ... e-changes/

Please help!


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)
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Re: The value of an investment increases by x% during January and [#permalink]
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VeritasPrepKarishma wrote:
mbafall2011 wrote:
There was a formula for successive percents: is there a way to solve it using the same.

Net percent = a+ b + ab/100

where a is the initial percent and b is the next percent change.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/02 ... e-changes/

Please help!


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)

thanks.
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Re: The value of an investment increases by x% during January and [#permalink]
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144144 wrote:

can someone please explain me how 100x/(100+x) == 100 - 10,000/(100+x)

thanks.


\(100 - \frac{10,000}{(100+x)} = \frac{(10,000 + 100x - 10,000)}{(100+x)}\) = \(\frac{100x}{(100+x)}\)
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Re: The value of an investment increases by x% during January and [#permalink]
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VeritasPrepKarishma wrote:
144144 wrote:

can someone please explain me how 100x/(100+x) == 100 - 10,000/(100+x)

thanks.


\(100 - \frac{10,000}{(100+x)} = \frac{(10,000 + 100x - 10,000)}{(100+x)}\) = \(\frac{100x}{(100+x)}\)


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.
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catennacio wrote:
VeritasPrepKarishma wrote:
144144 wrote:

can someone please explain me how 100x/(100+x) == 100 - 10,000/(100+x)

thanks.


\(100 - \frac{10,000}{(100+x)} = \frac{(10,000 + 100x - 10,000)}{(100+x)}\) = \(\frac{100x}{(100+x)}\)


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.


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

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).
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Re: The value of an investment increases by x% during January and [#permalink]
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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 and [#permalink]
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 \(y=100-\frac{10,000}{100+x}\).
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Re: The value of an investment increases by x% during January and [#permalink]
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 \(y=100-\frac{10,000}{100+x}\).



\(y=\frac{100x}{100+x}\)

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 and [#permalink]
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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 ?

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)


Multiplying by 5/4 = a 25% increase
Multiplying by 4/5 = a 20% decrease

Multiplying by 2/1 = a 100% increase
Multiplying by 1/2 = a 50% decrease

Multiplying by 4/1 = a 300% increase
Multiplying by 1/4 = a 75% decrease

For the investment to be same at the end of February, it must be multiplied by RECIPROCALS -- the equivalent of multiplying by a FACTOR OF 1 -- so that the value of the investment does not change.
Case 1: 5/4 * 4/5 --> a 25% increase followed by a 20% decrease --> x=25, y=20
Case 2: 2/1 * 1/2 --> a 100% increase followed by a 50% decrease --> x=100, y=50
Case 3: 4/1 * 1/4 --> a 300% increase followed by a 75% decrease --> x=300, y=75

What is the value of y in terms of x?
Any of the cases above can be used to determine the correct answer.
In Case 1, plugging x=25 into the correct answer will yield 20 (the value of y).
In Case 2, plugging x=100 into the correct answer will yield 50 (the value of y).
In Case 3, plugging x=300 into the correct answer will yield 75 (the value of y).

Only E is viable.
If we consider Case 3 and plug x=300 into E, we get:
\(100 - \frac{10000}{400} = 100 - \frac{100}{4} = 100 - 25 = 75\)

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

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.


VeritasKarishma can you pls explain how you got this expression (10/110) *100= 100/11 %

thanks! :)
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Re: The value of an investment increases by x% during January and [#permalink]
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dave13 wrote:
Quote:

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.


VeritasKarishma can you pls explain how you got this expression (10/110) *100= 100/11 %

thanks! :)



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?

You need to reduce 110 by 10 to make it 100 again.
10 is what percentage of 110?

It is (10/110) *100
which is equal to 100/11 %

So you need to reduce 110 by 100/11% to make it 100 again.
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Re: The value of an investment increases by x% during January and [#permalink]
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).

I didn’t got the last step…
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