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inequalities

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inequalities [#permalink]

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New post 04 Jun 2009, 18:12
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1. If 4/x < 1/3 what is the possible range of values for x?



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


Any shorter way of attempting these two questions? Thanks

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Re: inequalities [#permalink]

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New post 04 Jun 2009, 20:33
skim wrote:
1. If 4/x < 1/3 what is the possible range of values for x?
2. If 4/x < -1/3 what is the possible range of values for x?

Any shorter way of attempting these two questions? Thanks


1. 4/x < 1/3
4/x could be -ve or +ve so x could be also +ve or -ve i.e. x has 2 values.
4/x - 1/3 < 0
(12 - x ) / 3x < 0
12 - x < 0
12 < x .........i.

Also x is:
x < 0 ...... ii


2. 4/x < -1/3
4/x can only -ve so x should also be -ve i.e x has one value.
4/x + 1/3 < 0
(12 + x)/3x < 0
x < -12..................i
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GT

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Re: inequalities [#permalink]

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New post 05 Jun 2009, 02:10
4/x < 1/3
4/x - 1/3 < 0
(12 - x ) / 3x < 0
12 - x < 0
12 < x

Also
x < 0


4/x < -1/3

4/x + 1/3 < 0
(12 + x)/3x < 0
x < -12
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Re: inequalities [#permalink]

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New post 08 Jun 2009, 06:38
Thanks, GT and amolsk11.

Reference is made to the first question. Although I understand:

12 < x .........i.

I don't understand:

Also x is:
x < 0 ...... ii


Would appreciate clarification to the above. Thanks.

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Re: inequalities [#permalink]

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New post 08 Jun 2009, 15:07
In inequalities like these I find at first the limits of range, when x=y. In this case it will be:
1. 4/x = 1/3 i.e. when x=12. Then, if you want to lessen 4/x, you should INCREASE x, as x is denominator (try picking numbers, such as 16 and 8 for example and you will see the right way). Thus, the range of possible values are x>12.

2. IMO, for the 2nd question the answer is -12 < x < 0
If 4/x < -1/3 what is the possible range of values for x?
First of all, x is only negative, i.e. x<0. Otherwise, 4/x will be positive and >-1/3
Then, one of the margins for values of x is when 4/x = -1/3, i.e. when x=-12. If we want 4/x to become less, we should INCREASE the denominator, try again picking some numbers such as -8 (-4/8 < -1/3) and -16 (-4/16 > -1/3). But we can increase x only till 0. Thus, the range of solutions is -12 > x >0.

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Re: inequalities [#permalink]

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New post 08 Jun 2009, 20:37
Hi Gmat Tiger and amolsk11,
Can you pls. clarify my doubt regarding your solution for the second inequality.

You guys arrived at
(12+X)/3X <0
Q1. Can we cross multiply without considering the sign in the inequality. I generallt refrain myself from multiplying or diving an inequality.
Q2. If we know that x is -ve why are we not inverting the sign of the inequality ?
Q3. If we pretend that we are not bothered of the sign then i think cross multiplyinh is not allowed across inequalities.

Pls. help

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Re: inequalities [#permalink]

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New post 09 Jun 2009, 01:29
The OA for the second inequality question is -12 < x < 0

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Re: inequalities   [#permalink] 09 Jun 2009, 01:29
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