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# pencils

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VP
Joined: 18 May 2008
Posts: 1176

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31 Jan 2009, 04:43

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SVP
Joined: 29 Aug 2007
Posts: 2420

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31 Jan 2009, 07:24
23x+21y = 130

1. If 23 cents pencil is 1, then 21 cents pencil should be 7.
2. If 23 cents pencil are 2, then 21 cents pencil should be 4.
3. If 23 cents pencil are 3, then 21 cents pencil should be 1.
4. so on.........

which makes the total value 130 cents? 2?
23x2 + 21x4 = 46 + 84 = 130
B.
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VP
Joined: 18 May 2008
Posts: 1176

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01 Feb 2009, 03:18
but in B, there are 2 variables and only one eq?dnt we require 2 eqs 2 find 2 variables? The way found the sol is hit and trial. Im very confused.Can u pls explain?
GMAT TIGER wrote:
23x+21y = 130

1. If 23 cents pencil is 1, then 21 cents pencil should be 7.
2. If 23 cents pencil are 2, then 21 cents pencil should be 4.
3. If 23 cents pencil are 3, then 21 cents pencil should be 1.
4. so on.........

which makes the total value 130 cents? 2?
23x2 + 21x4 = 46 + 84 = 130
B.
Manager
Joined: 14 Nov 2008
Posts: 182
Schools: Stanford...Wait, I will come!!!

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01 Feb 2009, 05:10
if we see the equation.. 23x+21y = 130;
then, the boundary conditions are .. 6>x> 0, 6>y>0
Also, x and y are positive integers which means x &Y cannot have fractional values.

So, its not a hit and trial..but can be calculated in few iterations.

ritula wrote:
but in B, there are 2 variables and only one eq?dnt we require 2 eqs 2 find 2 variables? The way found the sol is hit and trial. Im very confused.Can u pls explain?
GMAT TIGER wrote:
23x+21y = 130

1. If 23 cents pencil is 1, then 21 cents pencil should be 7.
2. If 23 cents pencil are 2, then 21 cents pencil should be 4.
3. If 23 cents pencil are 3, then 21 cents pencil should be 1.
4. so on.........

which makes the total value 130 cents? 2?
23x2 + 21x4 = 46 + 84 = 130
B.
VP
Joined: 18 May 2008
Posts: 1176

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01 Feb 2009, 07:29
But how did u know these boundary conditions without looking at statement 1?
lgon wrote:
if we see the equation.. 23x+21y = 130;
then, the boundary conditions are .. 6>x> 0, 6>y>0
Also, x and y are positive integers which means x &Y cannot have fractional values.

So, its not a hit and trial..but can be calculated in few iterations.

ritula wrote:
but in B, there are 2 variables and only one eq?dnt we require 2 eqs 2 find 2 variables? The way found the sol is hit and trial. Im very confused.Can u pls explain?
GMAT TIGER wrote:
23x+21y = 130

1. If 23 cents pencil is 1, then 21 cents pencil should be 7.
2. If 23 cents pencil are 2, then 21 cents pencil should be 4.
3. If 23 cents pencil are 3, then 21 cents pencil should be 1.
4. so on.........

which makes the total value 130 cents? 2?
23x2 + 21x4 = 46 + 84 = 130
B.
Director
Joined: 29 Aug 2005
Posts: 808

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01 Feb 2009, 14:55
Stmt 2: 23x+21y=130
We can infer that x and y are integers. We are after x.
For multiples of 23 and 21 to add up to even number (130), x and y need to be either both even or both odd.
You will see that only x=3 and y=3 work.
For example, x=4 and y=4 would be more than 130.
Senior Manager
Joined: 26 May 2008
Posts: 409
Schools: Kellogg Class of 2012

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01 Feb 2009, 22:23
As already explained, there are a few equations for which roots can take only one single value( like the above equation).

Yeah. It is sort of hit and trial but we can quickly make a few calculations and decide the answer

Cheers,
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--== Message from GMAT Club Team ==--

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!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: pencils &nbs [#permalink] 01 Feb 2009, 22:23
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# pencils

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