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ritula
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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
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?
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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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But how did u know these boundary conditions without looking at statement 1?
lgon
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
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
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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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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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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