Is |x-1/5| < 3/5 ?
a. -1/2 < x < 4/5
b.x > -1/7
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.
OA to follow
Inequality
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C
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Edited : C
Oh!! Silly Mistake...
Oh!! Silly Mistake...
Last edited by shankar.ashwin on Mon Sep 05, 2011 9:13 am, edited 1 time in total.
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The Answer is C
Is |x-1/5| < 3/5 ?
After solving and breaking down the fraction to decimals we can rephrase the question as
is
-0.4<x<0.8 ?
(1) -1/2 < x < 4/5
-0.5<x<0.8
x<0.8 is met but the other one says -0.5<x. What if x=-0.4. Then this does not meet the requirement of -0.4<x. Hence INSUFFICIENT.
(2)x>-1/7. x could be any value. So INSUFFICIENT
Combining we get
-0.5<x<0.8 and x>-1/7. Since -1/7 is greater than -0.5 and -0.4 we can rewrite as
-1/7<x<0.8 which automatically satisfies -0.4<x<0.8. Hence [spoiler]C[/spoiler]
Is |x-1/5| < 3/5 ?
After solving and breaking down the fraction to decimals we can rephrase the question as
is
-0.4<x<0.8 ?
(1) -1/2 < x < 4/5
-0.5<x<0.8
x<0.8 is met but the other one says -0.5<x. What if x=-0.4. Then this does not meet the requirement of -0.4<x. Hence INSUFFICIENT.
(2)x>-1/7. x could be any value. So INSUFFICIENT
Combining we get
-0.5<x<0.8 and x>-1/7. Since -1/7 is greater than -0.5 and -0.4 we can rewrite as
-1/7<x<0.8 which automatically satisfies -0.4<x<0.8. Hence [spoiler]C[/spoiler]