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multiplying by unknown variables in inequalities

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multiplying by unknown variables in inequalities [#permalink] New post 06 Oct 2011, 10:36
In the MGMAT book, it mentions that when multiplying by unknown variables, take both positive/negative scenarios into account

4/x<1/3?

1. 12<x

2. little confused here. do I take x to be -x?
4/-x<1/3
12<-x
-12>x?

?????
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Manager
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Re: multiplying by unknown variables in inequalities [#permalink] New post 06 Oct 2011, 12:50
Its not the value of x you need to put as negative.

When you substitute a sample value in an equation you need to check with both positive an negative values

for ur question

4/x<1/3?

1. 12<x

if above is the statement 1 then no need to substitute a negative value as they have given x>12

means take some number more than 12 (take 13)

then eqn will become

4/13<1/3
==>12/39<13/39 then given eqn is correct

1 is sufficient
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Manager
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Re: multiplying by unknown variables in inequalities [#permalink] New post 06 Oct 2011, 12:54
For the below question

inequality-121619.html

i tried to substitute x = -x and the method was confusing but when i assumed x as 2 and -2 it was easy to identify whether the statement is enough or not
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Re: multiplying by unknown variables in inequalities [#permalink] New post 07 Oct 2011, 02:34
Expert's post
386390 wrote:
In the MGMAT book, it mentions that when multiplying by unknown variables, take both positive/negative scenarios into account

4/x<1/3?

1. 12<x

2. little confused here. do I take x to be -x?
4/-x<1/3
12<-x
-12>x?

?????


Best way to deal with inequalities is by keeping 0 on the right hand side.

Is 4/x < 1/3?
Is 4/x - 1/3 < 0?
Is (12 - x)/3x < 0?
Is (x - 12)/3x > 0

When will (x-12)/3x be positive?

It will be positive when x > 12 or when x < 0 (How? You can find it in a moment. Check out this thread: inequalities-trick-91482.html?hilit=trick%20inequalities )

Statement 1 is sufficient.
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Re: multiplying by unknown variables in inequalities   [#permalink] 07 Oct 2011, 02:34
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