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pencils

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VP
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New post 31 Jan 2009, 04:43
pls explain ur answer

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New post 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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New post 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.
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Re: pencils  [#permalink]

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New post 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.
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New post 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.
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Re: pencils  [#permalink]

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New post 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.
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Re: pencils  [#permalink]

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New post 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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Re: pencils &nbs [#permalink] 01 Feb 2009, 22:23
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