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If 4/x <1/3, what is the possible range of values for x?

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If 4/x <1/3, what is the possible range of values for x? [#permalink] New post 07 Apr 2011, 04:01
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we can't change the inequality when we have -ve nd RHS +ve in reciprocal

10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

Case 1 x is +ve

x>12

Case 2 when we consider x as -ve we will have Left hand side -ve but right hand side +ve so in that case we cnt flip the inequality.
But OA is showing both x>12 nd x<12

Pls comment which condition is wrong.

thanks
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Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink] New post 07 Apr 2011, 04:39
GMATD11 wrote:
we can't change the inequality when we have -ve nd RHS +ve in reciprocal

10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

Case 1 x is +ve

x>12

Case 2 when we consider x as -ve we will have Left hand side -ve but right hand side +ve so in that case we cnt flip the inequality.
But OA is showing both x>12 nd x<12

Pls comment which condition is wrong.

thanks


4/x <1/3
12/x-1<0
(12-x)/x < 0

Means either numerator or denominators is -ve:

Case I:
If Denominator is -ve.
x<0 ------1

Numerator must be +ve
12-x > 0
-x > -12
x< 12--------------2

In equation 1 and 2, 1 is more restrictive:
x<0

Case II:
If Denominator is +ve.
x>0 ------3

Numerator must be -ve
12-x < 0
-x < -12
x > 12


In equation 3 and 4, 4 is more restrictive:
x>12

Thus;
complete Range of x:

x<0 or x>12
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Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink] New post 07 Apr 2011, 12:19
GMATD11 wrote:
we can't change the inequality when we have -ve nd RHS +ve in reciprocal


I don't understand what you mean by this?

GMATD11 wrote:
10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases


You do need to consider 2 cases, but you only need to spend any time on one of them. We know:

4/x < 1/3

This will clearly be true if x is negative, since then the left side is negative, and the right side is positive, and negative numbers are certainly smaller than positive ones. So whenever x < 0, the inequality is true.

Now for the second case: if x > 0, we can multiply both sides by x without needing to worry about reversing the inequality:

4 < x/3
12 < x

So either x < 0, or 12 < x.
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Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink] New post 07 Apr 2011, 12:38
Brilliant !

And when I know the sign of x > 0 then taking the reciprocal of left hand side Vs the right hand side will reverse the direction of the inequality

4/x < 1/3 and x > 0
or x/4 > 3 (reverse the direction)
or x > 12
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Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink] New post 29 Aug 2011, 18:13
GMATD11 wrote:
10) If 4/x <1/3, what is the possible range of values for x?


4/x < 1/3 => 12/x < 1
For the fraction 12/x to be <1 ,

Scenario1 : x has to be negative.
Scenario2 : The denominator should be greater than 12

=> x < 0 or x > 12 .

Please correct me if i am wrong.
Re: we can't change the inequality when we have LHS -ve nd RHS   [#permalink] 29 Aug 2011, 18:13
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