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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 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.

Answer: A.

Hope it helps.

P.S. Pleas post one question per topic. _________________

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.

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

Show Tags

13 Oct 2012, 13:16

Expert's post

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.

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

Show Tags

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.

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. _________________

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.

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.

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