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# if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18

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if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink]

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16 Jun 2010, 10:41
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can someone help with the below DS problems.

1st problem
-----------------
if xy=-6 what is xy(x+y)
1.x-y = 5
2.xy^2 = 18.

2nd problem
-----------
can the +ve integer "p" be expressed as the product of two integers, each of which is greater than 1?
1. 31<p<37
2. p is odd.
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Re: NUMBERS !!!! [#permalink]

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16 Jun 2010, 11:08
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1. If xy=-6, what is xy(x+y)?

As given that $$xy=-6$$, then we should be able to determine only the value of $$x+y$$.

(1) $$x-y = 5$$ --> $$x=y+5$$ --> $$(y+5)y=-6$$ --> solving for $$y$$ gives $$y=-3$$ or $$y=-2$$ --> if $$y=-3$$, then $$x=2$$ and $$x+y=-1$$ but if $$y=-2$$, thren $$x=3$$ and $$x+y=1$$. Two values for $$x+y$$. Not sufficient.

(2) $$xy^2 = 18$$ --> $$(xy)*y=18$$ --> as $$xy=-6$$ --> $$-6y=18$$ --> $$y=-3$$ and $$x=2$$ --> $$x+y=-1$$. Sufficient.

2. Can the positive integer p be expressed as the product of two integers, each of which is greater than 1?

If positive integer p cannot be expressed as the product of two integers >1 it would mean that p is a prime number. So, basically question asks is p prime?

(1) 31<p<37 --> between these numbers there is no prime. Hence ANY integer from these range CAN be expresses as the product of two numbers. Sufficient.

(2) p is odd --> odd numbers can be primes as well as non-primes. Not sufficient.

Hope it helps.

P.S. Pleas post one question per topic.
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Re: NUMBERS !!!! [#permalink]

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16 Jun 2010, 11:37
Many thanks Bunuel !!!
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Re: NUMBERS !!!! [#permalink]

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13 Oct 2012, 12:57
Thanks for the explanation!
Can this also be solved as:

xy=-6 => the question amounts to -6(x+y) = ?

a) x-y=5.. => x = y+5 => -6(2y+5) = ?

Y can take any value.. hence, insufficient.

b) sufficient because as y=-3 and x =2, -6(x+y) can be determined.

Hence B.

Is the reasoning for Statement 1 above being insufficient, correct?

Bunuel wrote:
1. If xy=-6, what is xy(x+y)?

As given that $$xy=-6$$, then we should be able to determine only the value of $$x+y$$.

(1) $$x-y = 5$$ --> $$x=y+5$$ --> $$(y+5)y=-6$$ --> solving for $$y$$ gives $$y=-3$$ or $$y=-2$$ --> if $$y=-3$$, then $$x=2$$ and $$x+y=-1$$ but if $$y=-2$$, thren $$x=3$$ and $$x+y=1$$. Two values for $$x+y$$. Not sufficient.

(2) $$xy^2 = 18$$ --> $$(xy)*y=18$$ --> as $$xy=-6$$ --> $$-6y=18$$ --> $$y=-3$$ and $$x=2$$ --> $$x+y=-1$$. Sufficient.

2. Can the positive integer p be expressed as the product of two integers, each of which is greater than 1?

If positive integer p can not be expressed as the product of two integers >1 it would mean that p is a prime number. So, basically question asks is p prime?

(1) 31<p<37 --> between these numbers there is no prime. Hence ANY integer from these range CAN be expresses as the product of two numbers. Sufficient.

(2) p is odd --> odd numbers can be primes as well as non-primes. Not sufficient.

Hope it helps.

P.S. Pleas post one question per topic.
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if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink]

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13 Oct 2012, 13:16
prep wrote:
Thanks for the explanation!
Can this also be solved as:

xy=-6 => the question amounts to -6(x+y) = ?

a) x-y=5.. => x = y+5 => -6(2y+5) = ?

Y can take any value.. hence, insufficient.

b) sufficient because as y=-3 and x =2, -6(x+y) can be determined.

Hence B.

Is the reasoning for Statement 1 above being insufficient, correct?

Bunuel wrote:
1. If xy=-6, what is xy(x+y)?

As given that $$xy=-6$$, then we should be able to determine only the value of $$x+y$$.

(1) $$x-y = 5$$ --> $$x=y+5$$ --> $$(y+5)y=-6$$ --> solving for $$y$$ gives $$y=-3$$ or $$y=-2$$ --> if $$y=-3$$, then $$x=2$$ and $$x+y=-1$$ but if $$y=-2$$, thren $$x=3$$ and $$x+y=1$$. Two values for $$x+y$$. Not sufficient.

(2) $$xy^2 = 18$$ --> $$(xy)*y=18$$ --> as $$xy=-6$$ --> $$-6y=18$$ --> $$y=-3$$ and $$x=2$$ --> $$x+y=-1$$. Sufficient.

2. Can the positive integer p be expressed as the product of two integers, each of which is greater than 1?

If positive integer p can not be expressed as the product of two integers >1 it would mean that p is a prime number. So, basically question asks is p prime?

(1) 31<p<37 --> between these numbers there is no prime. Hence ANY integer from these range CAN be expresses as the product of two numbers. Sufficient.

(2) p is odd --> odd numbers can be primes as well as non-primes. Not sufficient.

Hope it helps.

P.S. Pleas post one question per topic.

Not quite. As you can see from my post: xy=-6 and x-y=5 has two sets of solutions: x=2 and y=-3 OR x=3 and y=-2. So, saying that we cannot determine the value of -6(2y+5) because y can take ANY value is not correct.

Hope it's clear.
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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink]

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14 Oct 2012, 11:07
Bunuel wrote:
Not quite. As you can see from my post: xy=-6 and x-y=5 has two sets of solutions: x=2 and y=-3 OR x=3 and y=-2. So, saying that we cannot determine the value of -6(2y+5) because y can take ANY value is not correct.

Hope it's clear.

Thanks! I totally understand your solution and the fact that a quadratic gets two roots and hence two solutions.
But for a moment consider y takes 0. (the question has no restriction on y)
thus -6(5) = -30
Now, if y takes 2, then -6(9) = -54... and so on.
Hence, there isn't a distinct value the equation xy(x+y) is arriving at, given the conditions. Thus, insufficient. Not sure if this is the way to approach it though!

Another way could be: -6(x+y) = z (some constant) and then solve x-y=5 , simultaneously. So, three unknowns and two equations. Hence no solutions.
Please let me know your thoughts on this.
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Re: NUMBERS !!!! [#permalink]

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14 Oct 2012, 21:51
prep wrote:
Thanks for the explanation!
Can this also be solved as:

xy=-6 => the question amounts to -6(x+y) = ?

a) x-y=5.. => x = y+5 => -6(2y+5) = ?

Y can take any value.. hence, insufficient.

Is the reasoning for Statement 1 above being insufficient, correct?

Responding to a pm:

"xy=-6 => the question amounts to -6(x+y) = ?"
This is fine. You have replaced one part fo the question but don't forget that we still have some info on the other part i.e. on x and y. We know that xy = -6 so x = -6/y

Statement 1 tells us x-y=5.. => x = y+5. Fine.

The questions becomes: -6(2y+5) = ?

What are the restrictions on y?
-6/y = y + 5
-6 = y^2 + 5y
y = -2 or -3

Mind you, y cannot be 0 because xy must be -6 and x-y must be 5.
y can take NO value other than -2 or -3.

Hence, it is incorrect to say that y can take any value and hence not sufficient.

Say, statement 2 were different and it gave you 2 values for y: -2 or 5
If you use your logic, you would say that the answers is E since you cannot get a unique value for y. But in that case, the answer would have been C since -2 would be the common value out of the two statements.

Hence, it is important to understand and utilize all the constraints given.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 15 Oct 2011 Posts: 41 Followers: 0 Kudos [?]: 7 [0], given: 37 Re: NUMBERS !!!! [#permalink] ### Show Tags 14 Oct 2012, 21:54 Thanks a lot! That makes it clear VeritasPrepKarishma wrote: prep wrote: Thanks for the explanation! Can this also be solved as: xy=-6 => the question amounts to -6(x+y) = ? a) x-y=5.. => x = y+5 => -6(2y+5) = ? Y can take any value.. hence, insufficient. Is the reasoning for Statement 1 above being insufficient, correct? Responding to a pm: "xy=-6 => the question amounts to -6(x+y) = ?" This is fine. You have replaced one part fo the question but don't forget that we still have some info on the other part i.e. on x and y. We know that xy = -6 so x = -6/y Statement 1 tells us x-y=5.. => x = y+5. Fine. The questions becomes: -6(2y+5) = ? What are the restrictions on y? -6/y = y + 5 -6 = y^2 + 5y y = 2 or 3 Mind you, y cannot be 0 because xy must be -6 and x-y must be 5. y can take NO value other than 2 or 3. Hence, it is incorrect to say that y can take any value and hence not sufficient. Say, statement 2 were different and it gave you 2 values for y: 2 or 5 If you use your logic, you would say that the answers is E since you cannot get a unique value for y. But in that case, the answer would have been C since 2 would be the common value out of the two statements. Hence, it is important to understand and utilize all the constraints given. Intern Joined: 15 Oct 2011 Posts: 41 Followers: 0 Kudos [?]: 7 [0], given: 37 Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] ### Show Tags 14 Oct 2012, 21:55 Thanks a lot for the explanations!! Bunuel wrote: prep wrote: Thanks for the explanation! Can this also be solved as: xy=-6 => the question amounts to -6(x+y) = ? a) x-y=5.. => x = y+5 => -6(2y+5) = ? Y can take any value.. hence, insufficient. b) sufficient because as y=-3 and x =2, -6(x+y) can be determined. Hence B. Is the reasoning for Statement 1 above being insufficient, correct? Bunuel wrote: 1. If xy=-6, what is xy(x+y)? As given that $$xy=-6$$, then we should be able to determine only the value of $$x+y$$. (1) $$x-y = 5$$ --> $$x=y+5$$ --> $$(y+5)y=-6$$ --> solving for $$y$$ gives $$y=-3$$ or $$y=-2$$ --> if $$y=-3$$, then $$x=2$$ and $$x+y=-1$$ but if $$y=-2$$, thren $$x=3$$ and $$x+y=1$$. Two values for $$x+y$$. Not sufficient. (2) $$xy^2 = 18$$ --> $$(xy)*y=18$$ --> as $$xy=-6$$ --> $$-6y=18$$ --> $$y=-3$$ and $$x=2$$ --> $$x+y=-1$$. Sufficient. Answer: B. 2. Can the positive integer p be expressed as the product of two integers, each of which is greater than 1? If positive integer p can not be expressed as the product of two integers >1 it would mean that p is a prime number. So, basically question asks is p prime? (1) 31<p<37 --> between these numbers there is no prime. Hence ANY integer from these range CAN be expresses as the product of two numbers. Sufficient. (2) p is odd --> odd numbers can be primes as well as non-primes. Not sufficient. Answer: A. Hope it helps. P.S. Pleas post one question per topic. Not quite. As you can see from my post: xy=-6 and x-y=5 has two sets of solutions: x=2 and y=-3 OR x=3 and y=-2. So, saying that we cannot determine the value of -6(2y+5) because y can take ANY value is not correct. Hope it's clear. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2291 Kudos [?]: 15146 [0], given: 224 Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] ### Show Tags 14 Oct 2012, 22:27 warya75 wrote: can someone help with the below DS problems. 1st problem ----------------- if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18. You can use some identities of algebra to solve it too. 1. x-y = 5 (x + y)^2 = (x - y)^2 + 4xy = 5^2 + 4(-6) = 1 So (x+y) can be 1 or -1. Not sufficient. 2. xy^2 = 18 (-6)y = 18 So y = -3 which gives x = 2 Now you have a unique value for xy(x+y) so sufficient. Answer (B) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink]

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14 Oct 2012, 23:28
Thanks a lot nice sharing of data sufficiency questions.
Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18   [#permalink] 14 Oct 2012, 23:28
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# if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18

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