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705-805 (Hard)|   Algebra|      
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Bunuel
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If xy = 3, what is the value of xy(x + y) ? 02 Oct 2017, 23:56
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E
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Difficulty:

75% (hard)

Question Stats:

52% (02:24) correct 48% (02:05) wrong based on 123 sessions

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If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10
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E
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fmik7894
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If xy = 3, what is the value of xy(x + y) ? 03 Oct 2017, 02:38
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Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10
Show more

I took the conventional way.

We have x=3/y and we need the value of 3(x + y).

S1. x – y = 2
Substitute the equation \(x=3/y\) in the above equation to get \(y^2+2y-3=0\)
You will get y=1 and x=3 ; y=-3 and x=-1
Both the sets give you different value for 3(x+y)
Not sufficient

S2. \(x^2 + y^2 = 10\)
Again substitute \(x=3/y\) to obtain \(y^4 - 10y^2 + 9=0\)
The sets you obtain are:
x=3 and y=1
x=-3 and y=-1
x=1 and y=3
x=-1 and y=-3
The above values give you different answers for 3(x+y)
Not sufficient

S1 and S2 combined
Only value common to both condition is x=-1 and y=-3
This gives you -12 as the value of 3(x+y)
Sufficient

Answer : C

I know there has to be a faster way. Please suggest.
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If xy = 3, what is the value of xy(x + y) ? 03 Oct 2017, 09:48
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Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10
Show more

we need to calculate the value of \(3(x+y)\). Hence need to know the value of \((x+y)\)

Statement 1: \((x+y)^2=(x-y)^2+4xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Statement 2: \((x+y)^2=x^2+y^2+2xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Combining 1 & 2 we get \(x+y=±4\). Hence no unique value. Insufficient

Option E
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If xy = 3, what is the value of xy(x + y) ? 03 Oct 2017, 09:53
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fmik7894
Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10

I took the conventional way.

We have x=3/y and we need the value of 3(x + y).

S1. x – y = 2
Substitute the equation \(x=3/y\) in the above equation to get \(y^2+2y-3=0\)
You will get y=1 and x=3 ; y=-3 and x=-1
Both the sets give you different value for 3(x+y)
Not sufficient

S2. \(x^2 + y^2 = 10\)
Again substitute \(x=3/y\) to obtain \(y^4 - 10y^2 + 9=0\)
The sets you obtain are:
x=3 and y=1
x=-3 and y=-1
x=1 and y=3
x=-1 and y=-3
The above values give you different answers for 3(x+y)
Not sufficient

S1 and S2 combined
Only value common to both condition is x=-1 and y=-3
This gives you -12 as the value of 3(x+y)
Sufficient

Answer : C

I know there has to be a faster way. Please suggest.
Show more

Hi fmik7894

Apart from being a lengthy method, you have ignored one common value. See highlighted section. as there is no unique solution, IMO answer should be E
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If xy = 3, what is the value of xy(x + y) ? 06 Oct 2017, 00:33
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Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10
Show more

(1) x – y = 2

Let x =-1 , y=-3......... x – y = 2............3 (-4) = -12

Let x= 3 , y= 1......... ..x – y = 2............3 (4) = 12

Two values

Insufficient

(2) x^2 + y^2 = 10

Use same examples above

Insufficient

Combine 1 &b2

Use same examples above

No clear soution

Insufficient

Answer: E
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If xy = 3, what is the value of xy(x + y) ? 15 Jul 2018, 01:27
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niks18
Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10

we need to calculate the value of \(3(x+y)\). Hence need to know the value of \((x+y)\)

Statement 1: \((x+y)^2=(x-y)^2+4xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Statement 2: \((x+y)^2=x^2+y^2+2xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Combining 1 & 2 we get \(x+y=±4\). Hence no unique value. Insufficient

Option E
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Not able to understand the highlighted part. Please elaborate.
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If xy = 3, what is the value of xy(x + y) ? 15 Jul 2018, 07:22
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AkshdeepS
niks18
Bunuel
If xy = 3, what is the value of xy(x + y) ?

(1) x – y = 2
(2) x^2 + y^2 = 10

we need to calculate the value of \(3(x+y)\). Hence need to know the value of \((x+y)\)

Statement 1: \((x+y)^2=(x-y)^2+4xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Statement 2: \((x+y)^2=x^2+y^2+2xy\), substitute the values to get

\((x+y)^2=16\) or \(x+y=±4\). Hence no unique value. Insufficient

Combining 1 & 2 we get \(x+y=±4\). Hence no unique value. Insufficient

Option E

Not able to understand the highlighted part. Please elaborate.
Show more

Hi AkshdeepS

\((x+y)^2=(x-y)^2+4xy\), basically this is a helpful formula. Expand both sides of the equation and you will get the relation.

\((x+y)^2=x^2+y^2+2xy\) and

\((x-y)^2+4xy=x^2+y^2-2xy+4xy=x^2+y^2+2xy\)

why I used this formula because you need to find x+y and the statement gives you x-y. Hence I need a formula that connects the two.
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If xy = 3, what is the value of xy(x + y) ? 17 Sep 2019, 05:54
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