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# Joanna bought only $0.15 stamps and$0.29 stamps

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

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19 Aug 2006, 00:15
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Hi guys is there a better way to attack these types of question outside of trial and error?

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.

In this case i assumed that the only way to satisfy condition one is if the 29 cent stamp would be either 5 or 10 so as to open the possibility that 440-29x is divisible by 15. Is this a valid assumption?

But what if the question were to be something like this

Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 cent pencil, how many 23 cent pencil did Martha buy?

1) Martha bought a total of 6 pencils
2) The total value of the pencils bought was 130
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19 Aug 2006, 00:58
Q1: A
Here is my approach:

Total = 15x+29y

St1: 15x+29y = 440
this means 29y must be a multiple of 5. So y could be 5 or 10
If y = 10 then x = 10
if y = 5 then x = 295/15 which is not an integer.
So x = 10 and y = 10 : SUFF

St2: x = y: INSUFF

Q2: B

St1: x +y = 6: INSUFF

St2: 21x+23y = 135
y could vary from 1 to 5
When y =
1. 5 then x = 0
2. 4 then x = 43/21
3. 3 then x = 66/21
4. 2 then x = 89/21
5. 1 then x = 112/21
SUFF
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19 Aug 2006, 01:03
Thanks So in question 2 you really have to plug in values.
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19 Aug 2006, 11:46
ps_dahiya wrote:
Q1: A
Here is my approach:

Total = 15x+29y

St1: 15x+29y = 440
this means 29y must be a multiple of 5. So y could be 5 or 10
If y = 10 then x = 10
if y = 5 then x = 295/15 which is not an integer.
So x = 10 and y = 10 : SUFF

St2: x = y: INSUFF

Q2: B

St1: x +y = 6: INSUFF

St2: 21x+23y = 135
y could vary from 1 to 5
When y =
1. 5 then x = 0 21x + 115 = 135; x=20/21. Thus answer should be E
2. 4 then x = 43/21
3. 3 then x = 66/21
4. 2 then x = 89/21
5. 1 then x = 112/21
SUFF

My approach to the first problem was
from 1, 15x + 29y = 440
from 2, x = y
combining both, we have 44x=440;x=10 and y=10.
Hence (C).
Could you please explain what is wrong with this logic pls? Also is (E) the correct answer for problem 2 ? (my stmt in red in the quote above)
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19 Aug 2006, 13:03
ps_dahiya wrote:
Q1: A
Here is my approach:

Total = 15x+29y

St1: 15x+29y = 440
this means 29y must be a multiple of 5. So y could be 5 or 10
If y = 10 then x = 10
if y = 5 then x = 295/15 which is not an integer.
So x = 10 and y = 10 : SUFF

St2: x = y: INSUFF

Q2: B

St1: x +y = 6: INSUFF

St2: 21x+23y = 135
y could vary from 1 to 5
When y =
1. 5 then x = 0
2. 4 then x = 43/21
3. 3 then x = 66/21
4. 2 then x = 89/21
5. 1 then x = 112/21
SUFF

Careful, ps_dahiya, you have been gifted with a fine math mind, but the last part needs to be reworked due to a careless error!

Obviously 130 is not a multiple of either 21 or 23, and 130=21(6)+4

i.e. 6 21cent pencils only would fall 4cents short of 1.30, so 'trade' two of them for 23cent pencils. No other comination will total $1.30 CEO Joined: 20 Nov 2005 Posts: 2911 Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008 Followers: 24 Kudos [?]: 273 [0], given: 0 ### Show Tags 19 Aug 2006, 20:48 kevincan wrote: ps_dahiya wrote: Q1: A Here is my approach: Total = 15x+29y St1: 15x+29y = 440 this means 29y must be a multiple of 5. So y could be 5 or 10 If y = 10 then x = 10 if y = 5 then x = 295/15 which is not an integer. So x = 10 and y = 10 : SUFF St2: x = y: INSUFF Q2: B St1: x +y = 6: INSUFF St2: 21x+23y = 135 y could vary from 1 to 5 When y = 1. 5 then x = 0 2. 4 then x = 43/21 3. 3 then x = 66/21 4. 2 then x = 89/21 5. 1 then x = 112/21 SUFF Careful, ps_dahiya, you have been gifted with a fine math mind, but the last part needs to be reworked due to a careless error! Obviously 130 is not a multiple of either 21 or 23, and 130=21(6)+4 i.e. 6 21cent pencils only would fall 4cents short of 1.30, so 'trade' two of them for 23cent pencils. No other comination will total$1.30

Thanks Kevin. These silly mistakes are a big problem for me. Could you please suggest, how can I avoid these? Should I read the question twice or should I read the solution twice or something else????
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20 Aug 2006, 00:29
IF you are very fast, then do what I did in the school, do each question twice. People rarely make the same silly mistake twice
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20 Aug 2006, 00:35
kevincan wrote:
IF you are very fast, then do what I did in the school, do each question twice. People rarely make the same silly mistake twice

I got what you are saying but I am not very fast. Just to give you an idea, I generally complete the quant section 5-10 minutes early with a minimum score of 49. So it is not feasible to complete each question twice. Any other suggestions????
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20 Aug 2006, 00:35
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