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12 Jul 2023, 01:30
Kudos
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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:
55%
(hard)
Question Stats:
51% (01:47) correct
49%
(01:27)
wrong
based on 96
sessions
History
Date
Time
Result
Not Attempted Yet
Does \(a = b\)?
(1) \(\frac{a}{b} = \frac{b}{a}\)
(2) \((a - b)^2 = a^2 - b^2\)
(1) \(\frac{a}{b} = \frac{b}{a}\)
(2) \((a - b)^2 = a^2 - b^2\)
12 Jul 2023, 02:24
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Stmt 1
\(\frac{a}{b} = \frac{b}{a}\)
\(\frac{a}{b}-\frac{b}{a} = 0\)
\(\frac{(a^2 - b^2)}{ab} = 0\)
\((a^2 - b^2) = 0\)
Two solutions exist, \(a = b\) or \(a = -b\) so insufficient.
Stmt 2
\((a - b)^2 = a^2 - b^2\)
\(a^2 -2ab + b^2 = a^2 - b^2\)
\(2b^2 - 2ab = 0\)
\(2b (b - a) = 0\)
Two solutions exist, \(b = 0\) or \(a = b\) so insufficient.
Stmt 1 and Stmt 2
Only common solution which exists in both conditions is \(a = b\). So sufficient.
Hence answer is C.
Hope it helped.
\(\frac{a}{b} = \frac{b}{a}\)
\(\frac{a}{b}-\frac{b}{a} = 0\)
\(\frac{(a^2 - b^2)}{ab} = 0\)
\((a^2 - b^2) = 0\)
Two solutions exist, \(a = b\) or \(a = -b\) so insufficient.
Stmt 2
\((a - b)^2 = a^2 - b^2\)
\(a^2 -2ab + b^2 = a^2 - b^2\)
\(2b^2 - 2ab = 0\)
\(2b (b - a) = 0\)
Two solutions exist, \(b = 0\) or \(a = b\) so insufficient.
Stmt 1 and Stmt 2
Only common solution which exists in both conditions is \(a = b\). So sufficient.
Hence answer is C.
Hope it helped.
Bunuel
12 Jul 2023, 14:48
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Bookmarks
(1) Only
Notice that x<>0 and b<>0
a^2=b^2 => |a|=|b|
So either a = b, or a = -b.
Not sufficient
(2) Only
a^2-2ab+b^2=a^2-b^2
-2ab+b^2=-b^2
2b^2-2ab=0
2b(b-a)=0
So either b=0 (and a can be any number) or b=a
Not sufficient.
(1)(2) Together
b cannot be 0, so b=a
Sufficient.
The answer is thus (C).
Notice that x<>0 and b<>0
a^2=b^2 => |a|=|b|
So either a = b, or a = -b.
Not sufficient
(2) Only
a^2-2ab+b^2=a^2-b^2
-2ab+b^2=-b^2
2b^2-2ab=0
2b(b-a)=0
So either b=0 (and a can be any number) or b=a
Not sufficient.
(1)(2) Together
b cannot be 0, so b=a
Sufficient.
The answer is thus (C).
