bad investment

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rahul.s: Master | Next Rank: 500 Posts; Posts: 324; Joined: Thu Dec 24, 2009 6:29 amFollowed by: 1 member; Thanked: 17 times

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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Source: Beat The GMAT — Problem Solving | Sun Feb 14, 2010 4:31 am

linkinpark: Master | Next Rank: 500 Posts; Posts: 124; Joined: Tue Dec 22, 2009 8:32 am; Location: Karjat; Thanked: 3 times

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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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

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...

R&D on GMAT algorithm:

https://www.beatthegmat.com/r-d-on-gmat- ... tml#305739

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GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sun Feb 14, 2010 8:55 am

Wow, It works....!!

Thanks Ian!

R&D on GMAT algorithm:

https://www.beatthegmat.com/r-d-on-gmat- ... tml#305739

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rahul.s: Master | Next Rank: 500 Posts; Posts: 324; Joined: Thu Dec 24, 2009 6:29 amFollowed by: 1 member; Thanked: 17 times

by rahul.s » Sun Feb 14, 2010 9:53 am

As always, thank you Ian

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zander21: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Tue Mar 29, 2011 8:08 pmFollowed by: 1 member; Thanked: 1 times

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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