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16 Oct 2004, 17:27
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find |x+y|+|x-y|?
(1) x^2+y^2=13
(2) xy = 6
(1) x^2+y^2=13
(2) xy = 6
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16 Oct 2004, 20:08
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I guess the answer should be C (Both).
Since from both the from first and second option, there can be mutiple answers. Combining the two would give
X=3, Y=2 or Y=3,X=2
Since the absolute value is taken, both the above will give answer of |3+2| + |3-2|=|2+3|+|2-3|= 6.
Since from both the from first and second option, there can be mutiple answers. Combining the two would give
X=3, Y=2 or Y=3,X=2
Since the absolute value is taken, both the above will give answer of |3+2| + |3-2|=|2+3|+|2-3|= 6.
1. Here x and y are in all 4 quadrants (points on a circle with origin as center), so x and y will be both +ve and -ve
In all cases we can not find a value for x and y.
2. xy=6 , so x, y both either + or both -ve
Still not sufficient as x and y can take any value.
from 1 and 2,
x=2 , y=3 or
x=-2, y= -3 or
x=3, y=2 or
x=-3, y=-2
so |x+y| + |x-y|
= |2+3| + |2-3| = 6 or
= |-2-3| + |-2+3| = 6 or
= |3+2| + |3-2| = 6 or
= |-3-2| + |-3+2| = 6
so C is answer.
In all cases we can not find a value for x and y.
2. xy=6 , so x, y both either + or both -ve
Still not sufficient as x and y can take any value.
from 1 and 2,
x=2 , y=3 or
x=-2, y= -3 or
x=3, y=2 or
x=-3, y=-2
so |x+y| + |x-y|
= |2+3| + |2-3| = 6 or
= |-2-3| + |-2+3| = 6 or
= |3+2| + |3-2| = 6 or
= |-3-2| + |-3+2| = 6
so C is answer.
