rajatvmittal wrote:Is x2+y2>3z? (2 is square and z is multiplication)
a)(x+y)2=9z, and (x−y)2 = z (2 is square)
b) z=0
OA C
Statement 1:
Adding together (x+y)² = 9z and (x-y)² = z, we get:
(x+y)² + (x-y)² = 9z+z
(x² + 2xy + y²) + (x² - 2xy + y²) = 10z
2(x²+y²) = 10z
x²+y² = 5z.
If x²+y²=0 and z=0, then x²+y²=3z.
If x²+y²=10 and z=2, then x²+y²>3z.
INSUFFICIENT.
Statement 2:
If x²+y²=0 and z=0, then x²+y²=3z.
If x²+y²=10 and z=0, then x²+y²>3z.
INSUFFICIENT.
Statements combined:
Since x²+y² = 5z and z=0, x²+y²=0.
Thus, it is not true that x²+y²>3z.
SUFFICIENT.
The correct answer is
C.
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