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07 Oct 2019, 07:28
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Difficulty:
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If r^2 + s^2 = t^2, what is the value of t ?
(1) r + s = 0
(2) 3r^2 + 2s^2 = 0
(1) r + s = 0
(2) 3r^2 + 2s^2 = 0
07 Oct 2019, 07:53
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If r^2 + s^2 = t^2, what is the value of t ?
(1) r + s = 0
Now this means r=-s, so r^2+s^2=2r^2=t^2
Thus t=+r or -r
Insufficient as we do not know the value of r or a
(2) 3r^2 + 2s^2 = 0
SUM of two NON NEGATIVE will be 0, only when both are 0.. so t=0
Sufficient
B
Posted from my mobile device
(1) r + s = 0
Now this means r=-s, so r^2+s^2=2r^2=t^2
Thus t=+r or -r
Insufficient as we do not know the value of r or a
(2) 3r^2 + 2s^2 = 0
SUM of two NON NEGATIVE will be 0, only when both are 0.. so t=0
Sufficient
B
Posted from my mobile device
07 Oct 2019, 08:20
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from statement (1), there is two possibilities,
(a) r = s = 0 ---> in this case, t = 0
(b) \(|r| = |s|\) i.e r & s have the same magnitude but opposite signs ---> in this case \(t^2 >0\) but t can be +ve or -ve
Hence, statement (1) is insufficient
from statement (2), we must conclude that r = s = 0 because the lowest value of an even power is 0 (sufficient)
(a) r = s = 0 ---> in this case, t = 0
(b) \(|r| = |s|\) i.e r & s have the same magnitude but opposite signs ---> in this case \(t^2 >0\) but t can be +ve or -ve
Hence, statement (1) is insufficient
from statement (2), we must conclude that r = s = 0 because the lowest value of an even power is 0 (sufficient)
