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# PS:Modulus shortcut clarification

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Senior Manager
Joined: 29 Mar 2008
Posts: 345

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26 Sep 2008, 20:28
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mod(x + 1) = 2 mod(x - 1)

When one solves this with 4 different conditions, one gets the value of x=3 and x=1/3. Is there a shorter way of identifying this equation and solving it only twice (instead of trying all 4 conditions) and get the two values?

I am trying to find some time saving strategies.
I tried the 4 conditions as:
(x+1) , (x-1)
-(x+1), -(x-1)
-(x+1), (x-1)
(x+1), -(x-1)

Hope I am making this clearer.

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26 Sep 2008, 22:05
1
KUDOS
Expert's post
In many cases, the fastest way to analyze these equations is to interpret the absolute values as distances on the number line. Recall that |a-b| is the distance between a and b on the number line. So,

|x + 1| = 2 |x - 1|
|x-(-1)| = 2 |x-1|

which just says that x is twice as far from -1 as it is from 1. Clearly there will be one point between -1 and 1 which satisfies this (x = 1/3), and one point to the right of 1 (x=3).
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Joined: 29 Aug 2007
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26 Sep 2008, 23:48
1
KUDOS
leonidas wrote:
mod(x + 1) = 2 mod(x - 1)

When one solves this with 4 different conditions, one gets the value of x=3 and x=1/3. Is there a shorter way of identifying this equation and solving it only twice (instead of trying all 4 conditions) and get the two values?

I am trying to find some time saving strategies.
I tried the 4 conditions as:
(x+1) , (x-1)
-(x+1), -(x-1)
-(x+1), (x-1)
(x+1), -(x-1)

Hope I am making this clearer.

essentially these 4 conditions are in fact the same as 2 conditions:

1: if lx+1l and lx-1l both are +ve;
x+1 = 2x-2
x = 3

2: if lx+1l and lx-1l both are -ve;
-x-1 = -2x+2
x = 3

1 & 2 yeild the same value because if we change the side of 2, it becomes 1. lets do it again with if lx+1l and lx-1l both are -ve;
-x-1 = -2x+2
change the side
2x - 2 = x+1 ........ it is the same as 1.

x = 3

3: if lx+1l is +ve and lx-1l is -ve;
x+1 = -2x+2
x = 1/3

4: if lx+1l is -ve and lx-1l is +ve;
-x-1 = 2x-2
x = 1/3

the same applies to 3 and 4 too.

so do only twice i.e one for +ve and another for -ve.
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VP
Joined: 30 Jun 2008
Posts: 1018

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27 Sep 2008, 04:34
can you post the full question ?

Thanks
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Senior Manager
Joined: 29 Mar 2008
Posts: 345

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27 Sep 2008, 09:23
IanStewart- Thats one of the best ways of dealing with absolute value problems. I need more practice to think on those lines.
GMAT Tiger- I agree, I should try the equations just twice instead of all 4 conditions. This will save time.

Thanks guys....

Here is the complete problem:

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
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To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

VP
Joined: 30 Jun 2008
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27 Sep 2008, 11:30
1
KUDOS
leonidas wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

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VP
Joined: 05 Jul 2008
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27 Sep 2008, 11:38
amitdgr wrote:
leonidas wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

Squaring works or solving 2 equations works just fine.
Senior Manager
Joined: 29 Mar 2008
Posts: 345

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27 Sep 2008, 14:10
amitdgr wrote:
leonidas wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

I like this one too. So, whenever, there is an absolute vaue on either side, one of the best approaches is to square?
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

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Joined: 17 Jun 2008
Posts: 1502

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28 Sep 2008, 03:57
amitdgr wrote:
leonidas wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

Good approach. I never thought of this. This really makes calculation fool proof.

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS:Modulus shortcut clarification   [#permalink] 28 Sep 2008, 03:57
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