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Joe bought only twenty cent stamps and thirty cent stamps.

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Joe bought only twenty cent stamps and thirty cent stamps. [#permalink]

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New post 16 Jun 2010, 13:01
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C
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Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.
(2) Joe bought a total of $2.50 worth of stamps.

Just want to make sure my reasoning is correct on this one - we can only use integers values for the numbers stamps, correct?

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Re: Joe bought only twenty cent stamps and thirty cent stamps. [#permalink]

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New post 16 Jun 2010, 13:50
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I am not sure if this is the most accurate solution, but here's my explanation as to why the answer is C.

Looking at the first statement:

20 cent stamps: > 8

This is insufficient, so we can eliminate answer choices A and D

Now looking at second statement, we see that \(0.20a + 0.30 b = 2.5\)

But there are many possible values that will satisfy this condition, and hence this statement is insufficient by itself. So we can eliminate B.

Combining the two statements, we know that \(a > 8\), so let us assume \(a = x + 8\) where x can take the values of 1, 2, 3 and so on (Number of stamps can only be integers; you can't buy half a stamp :D)

Substituting \(a = x+8\) into the equation we had from statement 2 we get

0.20a + 0.30b = 2.5

0.20(x+8) + 0.30b = 2.5

0.20x + 1.60 + 0.30b = 2.5

0.20x + 0.30b = 2.5 - 1.6 = 0.9

Now, we know that x can take the values of 1, 2 and 3 and so on

For x = 1 b = \(\frac {0.7}{0.3}\) - Eliminate (Not an integer)

For x = 2 b = \(\frac {0.5}{0.3}\) - Eliminate (Not an integer)

For x = 3 b = 1 - This is an integer, that would give rise to the answer being a = 11 and b = 1. But to be sure, we can check one or two more values.

For x = 4, b = \(\frac{0.1}{0.3}\)

For x = 5, b becomes negative. So we can stop here.

Hence, on combining the statements we see it's sufficient to answer the question.

Hope this helps. I wonder if there are other shorter methods to figure this out?

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Re: Joe bought only twenty cent stamps and thirty cent stamps. [#permalink]

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New post 16 Jun 2010, 14:05
thanks for the answer above.... my solution is similar but doesnt use equations....

I assumed 9 *20cents stamps => 1.80 => 70 cents leftover (but this is not a integer value for the # of 30 cent stamps!)

so i moved to 10*20 cent stamps => 2.00 with 50 cents leftover for calculating # of 30 cent stamps used and yet again, we see that we dont get integer value for the # of 30 cent stamps....ie: 50/30=> 1.67 30c stamps....

hence the answer is 11 20cent stamps yielding 1 30cent stamp ...11,1 is the pair and good answer

hence C.

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Re: Joe bought only twenty cent stamps and thirty cent stamps. [#permalink]

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New post 16 Jun 2010, 15:27
Thanks guys! Good reasoning for both responses. Pran's solution is a little faster but both work.
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Re: Joe bought only twenty cent stamps and thirty cent stamps. [#permalink]

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New post 11 Aug 2017, 03:14
Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps --> clearly insufficient.

(2) Joe bought a total of $2.50 worth of stamps --> \(2x+3y=25\): as \(x\) and \(y\) must be an integers we must check whether this equation has unique solution (for more on this check below links) --> \(2x=25-3y\), so 25 minus multiple of 3 must be multiple of 2, following pairs of (x,y) are possible: (2, 7), (5, 5), (8, 3), (11, 1). Not sufficient.

(1)+(2) As from (1) \(x>8\) then from (2) only one pair is left: \(x=11\) and \(y=1\). Sufficient.

Answer: C.


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Re: Joe bought only twenty cent stamps and thirty cent stamps.   [#permalink] 11 Aug 2017, 03:14
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