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M12 Q32

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M12 Q32 [#permalink] New post 07 Dec 2008, 09:32
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If x^3*y^4 = 2000, what is y?

1. x is an integer
2. y is an integer

[Reveal] Spoiler: OA
E

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Re: M12 Q32 [#permalink] New post 07 Dec 2008, 14:23
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echizen wrote:
x^3*y^4 = 2000 . What is y?

1. x is an integer
2. y is an integer


Statement 1: x is an integer:
if x is an integer, y could or could not be an integer. for ex:

if x = 1, y = -/+(2000^(1/4))
if x = 2, y = 250^(1/4)
if x = 5, y = 2 or -2
if x = 2000^(1/3), y = 1 or -1

Statement 2: y is an integer

if y = 1 or -1, x = (2000^1/3)
if y = 2 or -2, x = 125^(1/3)
if y = 5, x = (16/5)^1/3

from 1 and 2: y = + or - 2.

Therefore, E.

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Re: M12 Q32 [#permalink] New post 07 Dec 2008, 16:02
Thanks for your reply this helps :)
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Re: M12 Q32 [#permalink] New post 11 Jun 2010, 22:13
Agree with E.

Statement 1: Insufficient.
Statement 2: Insuffiecient.

Combining 1 & 2: 5^3 x 2^4 =2000

Now, (-2)^4 = (2)^4= 16... thats a good catch here, + 2 & -2 raised to a +ve exponent yield the same positive result. Hence, y does not have definite value, it could be either +2 or -2.

Hence, E.
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Re: M12 Q32 [#permalink] New post 11 Jun 2010, 23:46
I will go with option E as there are two set of values satisfying the above eqn..

x = 5 & y = 2
x = 5 & y = -2.

So option E
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Re: M12 Q32 [#permalink] New post 13 Aug 2010, 01:01
Its E.

For ex: 5 and 2 solved for X and Y resp.
But Y cannot be determined it can be +2 or -2!
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Re: M12 Q32 [#permalink] New post 09 Sep 2010, 05:19
St1: x is an integer - insufficient, coz value of y can be anything, right from an integer to a fractional value...( the only fact that is derived here is x will be definitely positive)
St2: y is an integer - insufficient, x can take any value.

St1 & St2 - x should be positive, y can be positive or negative..

Therefore, E
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Re: M12 Q32 [#permalink] New post 09 Sep 2010, 17:10
E

came up with one solution using (1)
2000 = 1000 * 2 = 10^3 * (sqrt(sqrt(2)) ) ^ 4

similar solution can be obtained for (2)

however, many questions remain unanswered. or simply put, if you plug in multiple values, you may or may not get 2000 as the product.

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Re: M12 Q32 [#permalink] New post 09 Sep 2010, 19:28
Ans E: X must be +ve but Y can be +ve or -ve

2000=2*2*2*2*5*5*5 or 2000=-2*-2*-2*-2*5*5*5

so,E

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Re: M12 Q32 [#permalink] New post 10 Sep 2010, 08:00
Thanks all... this seemed like a slam-dunk E..... due to no "exclusions" or narrowing of scope with the statements 1 or 2. (eg: x>0, y>0)...
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Re: M12 Q32 [#permalink] New post 10 Sep 2010, 20:58
my opinion as follows.

x^3*y^4=2000
(x*y)^3*y=2*10^3

thus, x*y must produce a multiple of 10.
(x*y)^3=k*10^3
k*y=2

if k is larger than 2, then y cannot be an integer.

if y was a fraction, then x might be a fraction. (8/3 and 30/8, the two produce 10)
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Re: M12 Q32 [#permalink] New post 14 Sep 2010, 21:43
Answer is E.

The catch point here is to pick numbers of different varieties in order to try to deny and support a statement.. Since both are integers, each can be +ve or -ve.

Therefore, y has +2 or -2.

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Re: M12 Q32 [#permalink] New post 15 Jun 2011, 13:18
What is the fastest way to determine that x and y are 5 and +/- 2?

Or would an easier way to answer this question be simply to note that because y is raised to an even exponent it will have to have an +/- answer and therefore the answer will be E?

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Re: M12 Q32 [#permalink] New post 15 Jun 2011, 16:54
Its E!

Missed considering the negative part..Silly of me.. :|

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Re: M12 Q32 [#permalink] New post 16 Jun 2011, 18:25
In the DS section -- its useful to take a moment to just think about what concept the GMAT is likely trying to test. This makes it much easier to solve the problem than just using the brute force method.
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Re: M12 Q32 [#permalink] New post 29 Oct 2011, 19:00
I chose E, but for two diff reasons:

1) X and Y can be +ve or -ve
2) Is "star" (e.g. *) defined as multiplication? I thought multiplication was "x"?

I figured since you don't know what function "*" is, you cannot solve the problem anyway. I encountered problems like these on Kaplan Advanced.
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Re: M12 Q32 [#permalink] New post 19 Jun 2012, 06:08
Straight E. The key point is to remember that even exponents mask the sign of the base number. The rest of the solution is as illustriously noted by other posters on this thread.

Cheers.

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Re: M12 Q32 [#permalink] New post 19 Jun 2013, 04:17
Expert's post
echizen wrote:
If x^3*y^4 = 2000, what is y?

1. x is an integer
2. y is an integer

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions


If x^3*y^4 = 2000, what is y?

(1) x is an integer.
(2) y is an integer.

Each statement alone is insufficient. When considering together we have: x^3*y^4 = 5^3*2^4=2000, but since y is in even power it can be 2 as well as -2. Not sufficient.

Answer: E.

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Re: M12 Q32 [#permalink] New post 19 Jun 2013, 13:21
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So I am surmizing, that to get the answer for this

1) Do a prime factorization of 2000. 2 raised to 4 and 5 raised to 3

2) Realize that you have a cube and power of four in the multiplication. The integer raised to the power of four can either be a negative or a positive.

So you cant really determine if y is -2 or +2.

Answer is E
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Re: M12 Q32 [#permalink] New post 05 Jun 2014, 21:02
Hi Bunuel,
I have a question here,
so when you considered both the statement and got x^3*y^4 = 5^3*2^4=2000.
i want to know how you reached there? did you try plugging in x=1 ,2, 3 , 4 and 5 and y =1 , 2 etc?just trying to understand the thought process . when i tried to take both the statements i could not do that .


Bunuel wrote:
echizen wrote:
If x^3*y^4 = 2000, what is y?

1. x is an integer
2. y is an integer

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions


If x^3*y^4 = 2000, what is y?

(1) x is an integer.
(2) y is an integer.

Each statement alone is insufficient. When considering together we have: x^3*y^4 = 5^3*2^4=2000, but since y is in even power it can be 2 as well as -2. Not sufficient.

Answer: E.

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Re: M12 Q32   [#permalink] 05 Jun 2014, 21:02
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