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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] New post 16 Jun 2010, 09:41
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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] New post 16 Jun 2010, 10:08
Expert's post
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 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.
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Re: NUMBERS !!!! [#permalink] New post 16 Jun 2010, 10:37
Many thanks Bunuel !!!
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Re: NUMBERS !!!! [#permalink] New post 13 Oct 2012, 11:57
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.

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.
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if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] New post 13 Oct 2012, 12:16
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.

Hope it's clear.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] New post 14 Oct 2012, 10:07
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] New post 14 Oct 2012, 20:51
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.

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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Re: NUMBERS !!!! [#permalink] New post 14 Oct 2012, 20:54
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.
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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] New post 14 Oct 2012, 20:55
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.
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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] New post 14 Oct 2012, 21:27
Expert's post
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)
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Re: if xy=-6 what is xy(x+y) 1.x-y = 5 2.xy^2 = 18 [#permalink] New post 14 Oct 2012, 22:28
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, 22:28
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