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04 Oct 2019, 07:41
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C
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A changing system in x and y is such that y never equals 0 and x always equals a constant K. What is the constant K equal to?
(1) x + y equals either 3 or 5.
(2) x – y equals either 3 or 5.
(1) x + y equals either 3 or 5.
(2) x – y equals either 3 or 5.
Source: Nova GMAT
07 Oct 2019, 01:13
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1 alone is not sufficient
2 alone is not sufficient
when we consider 1 and 2 together
Case 1. x+y=5 and x-y=3
then x=4 and y=1
Case 2. x+y=3 and x-y=5
then x=4 and y=-1
so both cases x=k=4
Option C
2 alone is not sufficient
when we consider 1 and 2 together
Case 1. x+y=5 and x-y=3
then x=4 and y=1
Case 2. x+y=3 and x-y=5
then x=4 and y=-1
so both cases x=k=4
Option C
07 Oct 2019, 08:35
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SajjadAhmad
A changing system in x and y is such that y never equals 0 and x always equals a constant K. What is the constant K equal to?
(1) x + y equals either 3 or 5.
(2) x – y equals either 3 or 5.
(1) x + y equals either 3 or 5.
(2) x – y equals either 3 or 5.
Source: Nova GMAT
(1) x + y equals either 3 or 5.
x=1 y=2 ( x+y =3) or x=1 y=4 ( x+y =5)
x=2 y=1 ( x+y =3) or x=2 y=3 ( x+y =5) As we can have different x values INSUFF .
( Many such combo's possible just taking one example )
( NOTE : x has to be constant , y can change but \(\neq\)0 )
(2) x - y equals either 3 or 5.
x=6 y=3 ( x - y=3) or x=6 y=1 ( x - y=5)
x=7 y=4 ( x - y=3) or x=7 y=2 ( x - y=5)
As we can have different x values INSUFF .
1+2
x+y =3
x - y=3
This combo is not possible as this will lead to y being 0
x+y =5
x - y=5
This combo is also NOT possible as this will lead to y being 0
x+y =3
x - y=5
Possible x=4 and y=-1
x+y =5
x - y=3
Possible x=4 and y= 1
Since we are getting a constant value for x this is our K,SUFF.
Ans-C
