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## kaplan 800 question

This topic has 1 expert reply and 3 member replies
kandelaki Senior | Next Rank: 100 Posts
Joined
05 Feb 2007
Posted:
37 messages

#### kaplan 800 question

Thu Feb 22, 2007 10:54 pm
if -1/2 and -1/4

which is the minimum value of xy^2 ?

choices
-1/75
-1/50
-1/48
-1/32
-1/16

the explanation says that the minimum value means farthest from 0. thats true , yes. however i believe farthest from 0 we achive by taking lowest possibilies of x and y ( that is 1/3 form x and 1/5 from y) and get -1/75 as answerr
but the official answer they took was 1/2 and 1/4 and got -1/32
i am all confused here

kandelaki Senior | Next Rank: 100 Posts
Joined
05 Feb 2007
Posted:
37 messages
Tue Feb 27, 2007 2:47 am
Oh, now i get it! thanks Stacey!!

### GMAT/MBA Expert

Stacey Koprince GMAT Instructor
Joined
27 Dec 2006
Posted:
2228 messages
Followed by:
681 members
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GMAT Score:
780
Mon Feb 26, 2007 11:00 pm
Also, just FYI - this might be what is messing you up:

-75 is further from zero than -32.

BUT -1/75 is closer to zero than -1/32.

Think about two fractions that are much easier to understand: -1/2 and -1/3. Which one is closer to zero? -1/3. And that has the larger number in the denominator.

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banona Senior | Next Rank: 100 Posts
Joined
30 Dec 2006
Posted:
38 messages
1
Fri Feb 23, 2007 2:56 pm
Hi,
Personnally, when manipulating inequatlities of negative numbers I convet them to inequalties of positive numbers to make them regular and easy to treat;

Fo this one,I would take the positive number (-x) which would be 1/3 < -x < 1/2

I would consider y^2 as the same of (-y)^2 and get,
1/25 < y^2 < 1/16 as 1/5 <-y < 1/4

then, the product -xy^2 as a positive number, would fit the following inequality : 1/75 < -xy^2 < 1/32

Now, it's easy to get xy^2 inequality as follows:

-1/32 < xy^2< -1/75

it's then obvious that the minimum value of xy^2 is -1/32

I hope I am right

_________________
I appologize for my Frenchy-English.
I am working on it.

jayhawk2001 Community Manager
Joined
28 Jan 2007
Posted:
789 messages
Followed by:
1 members
30
Sat Feb 24, 2007 8:54 am
For min and max values question, it is important to know the sign of
the end-result.

In this case, we know that the xy^2 will be negative. So, minimum
value of xy^2 will have to be a number which is farther away from zero.
(A simple example would be to compare -1 and -3, -3 is the minimum since it is farther away from 0 compared to -1.)

So, min (xy^2) is in essence min (x) * max (y^2) which yields
-1/2 * 1/16 = -1/32. Please note that we can split min (xy^2) as
above only because we already know what the sign of the
end result will be.

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