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# Joanna bought only $0.15 stamps and$0.29 stamps. How many

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Joanna bought only $0.15 stamps and$0.29 stamps. How many [#permalink]

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05 Feb 2005, 21:14
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Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.
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05 Feb 2005, 22:22
I got (D). Both are sufficient.
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05 Feb 2005, 22:26
C looks obvious but, A alone is sufficient.

15x + 29y = 440, condition x, y are integers.

y can only be 10.
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05 Feb 2005, 22:36
mantha wrote:
C looks obvious but, A alone is sufficient.

15x + 29y = 440, condition x, y are integers.

y can only be 10.

Can you explain how you know that y can only be 10? I was learning towards that answer but wasn't sure how to tell if there'd be any other values taht would leave integers for both variables.
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05 Feb 2005, 22:48
In order for the sum to be a factor of 10, each must be a factor of 5 or each must be a factor of 10.

It is easier to check the multiples for 29.

29*5 = 145, in which case 15 is not an integer from 15x + 29y = 440.
29*10 = 290, => x = 10, a possibility
29*15 = 435, again x cannot be an integer.
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05 Feb 2005, 23:17
A is sufficient to answer the question. in this case, we can use trial and error approach.

i suspect B is sufficient.
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06 Feb 2005, 01:13
YAA, OA is D

BOTH ARE SUFFICIENT
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06 Feb 2005, 01:25
I agree A is sufficient but it means you have to make all the calculation to be sure there is no other match between the 2 kind of stamps to reach 4.40 It seems quite a long way to me. There might be a shortcut...i think the shortest way might be the one explained by Mantha

DLMD, I am sorry but I dont see how D could be the OA. There is no way to find the solution with only statement B. It could be 2, 3, 15, 100 or any other number of stamps because the only information you have is that you have the same number of stamps for each category...
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06 Feb 2005, 05:09
This is a classic trap Q

I get D too
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06 Feb 2005, 06:46
HongHu wrote:
15x+29x=440, x=10. sufficient

hmm........
how did you get .15x+.29x=4.40 from statement ii? pls explain..........
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06 Feb 2005, 07:30
rxs0005 wrote:
This is a classic trap Q

I get D too

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06 Feb 2005, 09:20
I don't see how statement 2 by itself is sufficient. All you know that that statement is that there are the same number of .15 stamps as .29. Couldn't that number be anything? For example, there could be one .15 and one .29 or there could be a million .15 and a million .29.
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06 Feb 2005, 09:21
banerjeea_98 wrote:
"C"....

My bad, didn't even check the state 1 properly, I think OA shud be "A" instead of "D".
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06 Feb 2005, 11:14
Antmavel wrote:
I agree A is sufficient but it means you have to make all the calculation to be sure there is no other match between the 2 kind of stamps to reach 4.40 It seems quite a long way to me. There might be a shortcut...i think the shortest way might be the one explained by Mantha

DLMD, I am sorry but I dont see how D could be the OA. There is no way to find the solution with only statement B. It could be 2, 3, 15, 100 or any other number of stamps because the only information you have is that you have the same number of stamps for each category...

Antmavel, you are right, there is no way to get the answer just from statement 2.

Actually, the OA i have is A, not D
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06 Feb 2005, 21:52
Hmmm obviously I was hullusinating. There's something to be said about getting enough sleep for one's abilities of critical reasoning ...
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06 Feb 2005, 23:37
HongHu wrote:
Hmmm obviously I was hullusinating. There's something to be said about getting enough sleep for one's abilities of critical reasoning ...

I thought you don't have to sleep
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06 Feb 2005, 23:58
Sigh.
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07 Feb 2005, 06:15
DLMD wrote:
HongHu wrote:
Hmmm obviously I was hullusinating. There's something to be said about getting enough sleep for one's abilities of critical reasoning ...

I thought you don't have to sleep

good one...Honghu is everywhere....
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