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DS - inequality and absolute value

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DS - inequality and absolute value

by m-sand » Sun Jan 03, 2010 5:22 am
If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > -1

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.

Can someone please explain the process of solving this ?

OA : C
Last edited by m-sand on Mon Jan 04, 2010 2:51 pm, edited 1 time in total.
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by linkinpark » Sun Jan 03, 2010 8:17 am
m-sand wrote:If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1

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.

Can someone please explain the process of solving this ?

OA : C
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by linkinpark » Sun Jan 03, 2010 8:22 am
m-sand wrote:If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1

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.

Can someone please explain the process of solving this ?

OA : C
sorry for blank post

ok from stmt x<1 could be anything consider -2, so from question 4/2 = 2 hence not less than 1, now consider 0.5, so .25/.5 = 1/2 < 1 hence stmt 1 is not sufficient

from stmt 2 x>-1 again it can be -0.5 in that case eqn will be satisfied and with x=2 it wont satisfy.

considering both -1<x<1 with x not = 0 , in this range any value will satisfy the equation to give the anser hence C
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by mehravikas » Mon Jan 04, 2010 3:59 pm
I think you should solve it using the following way -

if x is -ve: x^2 / -x < 1 -> -x < 1 or x > -1
if x is +ve: x^2 / x < 1 -> x < 1

Therefore the inequality is -1 < x < 1

Statement 1 - tells us that x is less than 1 but x could be -5 or any number less than 1 - not sufficient
Statement 2 - tells us that x is greater than -1 but it could be any number - not sufficient

Combining you get what you want :-) Therefore C
m-sand wrote:If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > -1

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.

Can someone please explain the process of solving this ?

OA : C
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