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bad investment

by rahul.s » Sun Feb 14, 2010 4:31 am
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) / 10000
(E) 100 - (10000 / 100 + x)

OA: E
Source: MGMAT
Level: 700 - 800

could someone please explain it by plugging in numbers as well as algebraically?
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by linkinpark » Sun Feb 14, 2010 4:46 am
let n be actual invst at Jan beginning
at Jan end it'll be = n + n * x/100
and in Feb end n+n*x/100 - y/100(n + n*x/100) which is = n

now n = n+n*x/100(1-y/100)
n= n(1+x/100)*(1-y/100)

1/(1+x/100) = 100-y/100
100/(100+x) = 100-y/100
10000/(100 +x) = 100-y

so y = 100 - 10000/(100 +x)

in problems such as this one, I prefer using variables and setting up equations directly instead of backsolving, saves time.
530->480->580
when posting a question don't post OA(even masked) before some discussion.
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by gmatmachoman » Sun Feb 14, 2010 8:26 am
LOL..
I got :
100 *x/(100+x) which is same as 100 - 10000/(100 +x)


When I saw my answer, it was not resembling any of the answer choices given and made my head swirl ard...Dont get carried away... :D :D
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by Ian Stewart » Sun Feb 14, 2010 8:46 am
I'd likely do the question algebraically, as was done above, but there are certainly other options, some quite fast. If we are careful to choose extremely simple numbers, for x and y, we can determine which answer choice describes the relationship between x and y. Say our investment is worth $100 to begin with, then doubles in value to $200, then drops back to $100. Then we have a 100% increase followed by a 50% decrease; that is, x = 100 and y = 50. We can now plug x = 100 into each answer choice until we find one that is equal to 50; only E is remotely close.

Note that if, by bad luck, two different answer choices both gave us 50, we would need to try a different set of numbers to work out which of those two choices was correct - one reason why it can be preferable just to do things algebraically.
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by gmatmachoman » Sun Feb 14, 2010 8:55 am
Wow, It works....!!

Thanks Ian!
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by rahul.s » Sun Feb 14, 2010 9:53 am
As always, thank you Ian :)
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by zander21 » Tue Aug 30, 2011 8:30 pm
Ian Stewart wrote:I'd likely do the question algebraically, as was done above, but there are certainly other options, some quite fast. If we are careful to choose extremely simple numbers, for x and y, we can determine which answer choice describes the relationship between x and y. Say our investment is worth $100 to begin with, then doubles in value to $200, then drops back to $100. Then we have a 100% increase followed by a 50% decrease; that is, x = 100 and y = 50. We can now plug x = 100 into each answer choice until we find one that is equal to 50; only E is remotely close.

Note that if, by bad luck, two different answer choices both gave us 50, we would need to try a different set of numbers to work out which of those two choices was correct - one reason why it can be preferable just to do things algebraically.
This is the perfect explanation and by far by fastest, easiest way to answer. Thanks.
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