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Help on 2 DS problems

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Help on 2 DS problems

by crackgmat007 » Mon Apr 27, 2009 5:07 pm
Can someone help me on the below problems pls:

A. If X^3*Y=24, what is the value of (X^3*Y^3-X^2*Y^2)
1. X^2*Y^2=36
2. X^3*Y^2=72

OA - B, IMO - D

B. Which expression is larger -> 1/(5-x) or x/5
1. x<8
2. x>-8

OA - E, IMO - B

My bad, rectified the expression in problem # B, the first term is not negative.
Last edited by crackgmat007 on Mon Apr 27, 2009 6:32 pm, edited 1 time in total.
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by cramya » Mon Apr 27, 2009 5:41 pm
A. If X^3*Y=24, what is the value of (X^3*Y^3-X^2*Y^2)
1. X^2*Y^2=36
2. X^3*Y^2=72
Its not D because

To find x^3y^3 - x^2y^2

= x^y^2 (xy-1)

Stmt I

We know x^2y^2=36 but xy can be -6 or 6

36(-6-1) and 36(6-1)

2 different answer hence INSUFF

Hope this helps!

Regards,
CR
Last edited by cramya on Mon Apr 27, 2009 5:56 pm, edited 1 time in total.
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by cramya » Mon Apr 27, 2009 5:50 pm
Good problem.
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Re: Help on 2 DS problems

by Vemuri » Mon Apr 27, 2009 6:04 pm
crackgmat007 wrote: A. If X^3*Y=24, what is the value of (X^3*Y^3-X^2*Y^2)
1. X^2*Y^2=36
2. X^3*Y^2=72

OA - B, IMO - D
The question stem states that X^3*Y=24. This tells us that X & Y are either both positive or both negative.

Stmt 1: Given, X^2*Y^2=36. From this we can derive X*Y=+-6. But, since X & Y are either both positive or negative, X*Y=6. From this lets try to evaluate the expression X^3*Y^3-X^2*Y^2. This can be reorganized as:
X^2*Y^2(X*Y) - X^2-Y^2 ==> 36*6-36 ==> 36*5 ==> 180 (Sufficient)

Stmt 2: Given X^3*Y^2=72. We also know from the question stem that X^3*Y=24. Dividing both, we will get the value of Y=3. If we know Y, we can determine X=8. Hence, we can solve the expression. Sufficient.

My answer is D. I am interested in understanding why Stmt 1 is not sufficient. Will wait to see some responses on this.
crackgmat007 wrote: B. Which expression is larger - 1/(5-x) or x/5
1. x<8
2. x>-8

OA - E, IMO - B
Stmt 1: As long as x >=0, x/5 is larger than -1/(5-x). But, when x<0, -1/(x-5) is larger than x/5. So, this statement is insufficient.

Stmt 2: Same explanation as in Stmt 1.

Both combined, will result in the same explanation & so we cannot determine which expression is larger for sure.

Hence, answer should be E
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Reectified the problem

by crackgmat007 » Mon Apr 27, 2009 6:35 pm
@ Vemui
There was a minor error in the problem. Actually the first term in the second problem is not negative. Pls take a look again and let me know if your answer changes. Thanks.
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Re: Reectified the problem

by Vemuri » Mon Apr 27, 2009 8:03 pm
crackgmat007 wrote:@ Vemui
There was a minor error in the problem. Actually the first term in the second problem is not negative. Pls take a look again and let me know if your answer changes. Thanks.
Thanks for correcting the problem, but it does not change the answer. Both the statements are still not sufficient to determine which is greater. So, the answer is E.
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by gmat740 » Mon Apr 27, 2009 10:40 pm
A. If X^3*Y=24, what is the value of (X^3*Y^3-X^2*Y^2)
1. X^2*Y^2=36
2. X^3*Y^2=72
given X^3*Y=24..........(1)
we need to find: (X^3*Y^3-X^2*Y^2)
it can be written as x^2*y^2(xy - 1)

(I) x^2*y^2 = 36.........(2)

so divide (1) by (2)
we get x/y = 2/3, but we need to find xy and not x/y

some may argue over here that :
x^2*y^2 = 36 => (xy) =6

but friends, it may also mean (xy) = -6

remember x^2 = 4 means x= +2 and -2

so using (I) we cannot get one answer


(II)

x^3*y^2 =72
(x^3*y)*y = 72 => 24y = 72 => y=3 and so we can get the value of x also and we can easily arrive at the answer

Hope this helps
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by Vemuri » Mon Apr 27, 2009 11:55 pm
gmat740 wrote: given X^3*Y=24..........(1)
we need to find: (X^3*Y^3-X^2*Y^2)
it can be written as x^2*y^2(xy - 1)

(I) x^2*y^2 = 36.........(2)

so divide (1) by (2)
we get x/y = 2/3, but we need to find xy and not x/y

some may argue over here that :
x^2*y^2 = 36 => (xy) =6

but friends, it may also mean (xy) = -6

remember x^2 = 4 means x= +2 and -2

so using (I) we cannot get one answer
Well, but the question stem provides us with X^3*Y=24. This is possible if X & Y are either both positive or both negative. So, XY will also yield a positive result.
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by gmat740 » Tue Apr 28, 2009 4:08 am
Well that's a good point...thanks for pointing this out Vemuri....
given X^3*Y=24..........(1)
we need to find: (X^3*Y^3-X^2*Y^2)
it can be written as x^2*y^2(xy - 1)
So,
if you say xy =+6
so,
6*x^2 =24
so,
x^2=4
x = +2 and -2

We must have only 1 value of x

Although this is not required for calculation,but to solve the ambiguity

So you actually get 2 different values of x

This is actually not a solving but an explanation why (I) is incorrect.
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