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02 Oct 2017, 23:56
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:24) correct
48%
(02:06)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
If xy = 3, what is the value of xy(x + y) ?
(1) x – y = 2
(2) x^2 + y^2 = 10
(1) x – y = 2
(2) x^2 + y^2 = 10
03 Oct 2017, 02:38
Kudos
Bookmarks
Bunuel
If xy = 3, what is the value of xy(x + y) ?
(1) x – y = 2
(2) x^2 + y^2 = 10
(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.
03 Oct 2017, 09:48
Kudos
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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
(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
