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01 Sep 2021, 08:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:19) correct
26%
(01:41)
wrong
based on 126
sessions
History
Date
Time
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Not Attempted Yet
Archit3110
Major Poster
01 Sep 2021, 08:47
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target find value of \(x(1-y)(1+y)\)
x*( 1-y^2)
#1
\(x^2=x\)
x=1 or 0
y is not know so insufficient
#2
\(y^2=x\)
y=x=1 or 0 or
y=2 , x= 4
y=3 , x=9
so on
insufficient
from 1 &2
x=0 or 1
so y = 0 or 1
but value of x*( 1-y^2)
always 0
sufficient
option C
x*( 1-y^2)
#1
\(x^2=x\)
x=1 or 0
y is not know so insufficient
#2
\(y^2=x\)
y=x=1 or 0 or
y=2 , x= 4
y=3 , x=9
so on
insufficient
from 1 &2
x=0 or 1
so y = 0 or 1
but value of x*( 1-y^2)
always 0
sufficient
option C
Bunuel
01 Sep 2021, 09:04
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What is the value of \(x(1-y)(1+y)\) ?
Stat1: \(x^2=x\)
so, x = 0 or 1. So, we don't have any fixed value of \(x(1-y)(1+y)\). Not sufficient
Stat2: \(y^2=x\)
Not sufficient. as x and y can be anything.
Combining both, if x = 0 then, y = 0 and \(x(1-y)(1+y)\) =0 and if x = 1 then y = +1 or -1, now, \(x(1-y)(1+y)\) = 0. Sufficient.
So, I think C.
Stat1: \(x^2=x\)
so, x = 0 or 1. So, we don't have any fixed value of \(x(1-y)(1+y)\). Not sufficient
Stat2: \(y^2=x\)
Not sufficient. as x and y can be anything.
Combining both, if x = 0 then, y = 0 and \(x(1-y)(1+y)\) =0 and if x = 1 then y = +1 or -1, now, \(x(1-y)(1+y)\) = 0. Sufficient.
So, I think C.
