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D
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If \(a > 0\), \(b > 0\) and \(c > 0\), is \(a(b - c) = 0\)?
(1) \(b - c = c - b\)
(2) \(\frac{b}{c} = \frac{c}{b}\)
(1) \(b - c = c - b\)
(2) \(\frac{b}{c} = \frac{c}{b}\)
21 Aug 2009, 19:45
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Answer is D.
a>0, b>0, c>0
for 1), b-c=c-b ==>> 2b=2c ==>>b=c
therefore a(b-c)=0, suff
for 2), b/c=c/b ==>> b^2=c^2
because b>0, c>0, ==>>b=c
therefore a(b-c)=0, suff
a>0, b>0, c>0
for 1), b-c=c-b ==>> 2b=2c ==>>b=c
therefore a(b-c)=0, suff
for 2), b/c=c/b ==>> b^2=c^2
because b>0, c>0, ==>>b=c
therefore a(b-c)=0, suff
General Discussion
25 Aug 2009, 10:32
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[quote="tejal777"]if a>0, b>0, c>0, is a(b-c)=0?
I. b-c = c-b
II. b/c = c/b
A(B-C) = 0 and as a>0 thus it is POSSIBLE ONLY WHEN B = C
from 1
b-c-c+b = 0 thus 2b = 2c thus b=c...suff
from 2
b/c = c/b all +ve thus b^2 = c^2 thus b=c....suff
D
I. b-c = c-b
II. b/c = c/b
A(B-C) = 0 and as a>0 thus it is POSSIBLE ONLY WHEN B = C
from 1
b-c-c+b = 0 thus 2b = 2c thus b=c...suff
from 2
b/c = c/b all +ve thus b^2 = c^2 thus b=c....suff
D
