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Joanna bought only $0.15 stamps and $0.29 stamps. How many
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26 Sep 2010, 10:48
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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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Re: Stamps
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26 Sep 2010, 10:56




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Re: Stamps
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18 Oct 2010, 01:02
I hate C traps!!The good news is that both statements do no contradict each other. So, I know that the 2nd statement provides a clue, even if it is not sufficient on its own. And, from the numbers given( 0.15, 0.29 and 4.40), I look for some sort of relationship among them. In this case, the number should be a multiple of 5 in order to give a 0 in 4.40. And, 0.15 + 0.29 =0.44. Sometimes the solution is so obvious that I cant see it even if it is staring straight at me....sigh..




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Re: Stamps
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26 Sep 2010, 11:05
C is a trap. I hate these questions, because you have to work out the possibilities. Is there a simpler way to determine if (A) only has one solution? I usually just draw a little chart and start filling it in.



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Re: Stamps
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26 Sep 2010, 11:08
I guess if there is not a quicker way, at least look for which of the numbers will be easier to work with. It is a lot easier to determine if something is divisible by 15 than 29. So start with zero 29 cent stamps and subtract it from 4.40, and see if the result is divisible by 15. If you find more than one solution, stop working and look at the next statement.



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Re: Stamps
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29 Sep 2010, 08:24
jpr200012 wrote: start with zero 29 cent stamps and subtract it from 4.40 in this particular case, I belieeve, we can't start with zero 29 cent stamps, we should start with 1, because stimulus says Jonna bought both kind of stamps.
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Re: Stamps
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29 Sep 2010, 20:57
Bunuel,im not getting why is B insufficient. we have 15x + 29y = 440 and since x=y, we have 15x + 29x = 440 then x = 10. so y= 10. Can you please explain. Thanx in advance



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Re: Stamps
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29 Sep 2010, 21:23
where 440 mentioned?? Only in statement 1. You should not use the information from statement 1 unless you are considering to combine both statement 1 and statement 2 to arrive at answer C.
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Re: Stamps
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29 Sep 2010, 21:54



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Re: Stamps
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26 Feb 2011, 11:53
Quote: So when we have equation of a type ax+by=c and we know that x and y are nonnegative integers, there can be multiple solutions possible for x and y (eg 5x+6y=60) OR just one combination (eg 15x+29y=440). Hence in some cases ax+by=c is NOT sufficient and in some cases it is sufficient. Is 5x+6y=60 a good example for this case? The only solutions to the above equation(considering only integers are acceptable; you cannot have 1.5 stamps) are x=0;y=10 (or) x=6;y=5. Unless I'm missing another solution. Don't you think 5x+10y=60 would be a better example to show multiple solutions. Just curious.



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Re: Stamps
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26 Feb 2011, 12:02
bugSniper wrote: Quote: So when we have equation of a type ax+by=c and we know that x and y are nonnegative integers, there can be multiple solutions possible for x and y (eg 5x+6y=60) OR just one combination (eg 15x+29y=440). Hence in some cases ax+by=c is NOT sufficient and in some cases it is sufficient. Is 5x+6y=60 a good example for this case? The only solutions to the above equation(considering only integers are acceptable; you cannot have 1.5 stamps) are x=0;y=10 (or) x=6;y=5. Unless I'm missing another solution. Don't you think 5x+10y=60 would be a better example to show multiple solutions. Just curious. First of all the example is not about stamps problem, it's a general example about Diophantine equations and yes, I think it's a good example as it has more than one integer solution.
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Re: Stamps
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26 Feb 2011, 12:14
Additionally if I have an equation ax+by = c; if the coefficients a,b are coprime, can I be certain that there could possibly be only one combination(other than probably a or b being 0) of a,b that would solve the equation?



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Re: Stamps
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26 Feb 2011, 12:17
Bunuel wrote: bugSniper wrote: Quote: So when we have equation of a type ax+by=c and we know that x and y are nonnegative integers, there can be multiple solutions possible for x and y (eg 5x+6y=60) OR just one combination (eg 15x+29y=440). Hence in some cases ax+by=c is NOT sufficient and in some cases it is sufficient. Is 5x+6y=60 a good example for this case? The only solutions to the above equation(considering only integers are acceptable; you cannot have 1.5 stamps) are x=0;y=10 (or) x=6;y=5. Unless I'm missing another solution. Don't you think 5x+10y=60 would be a better example to show multiple solutions. Just curious. First of all the example is not about stamps problem, it's a general example about Diophantine equations and yes, I think it's a good example as it has more than one integer solution. Fair enough. Thank you.



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Re: Stamps
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27 Feb 2011, 09:41
bugSniper wrote: Additionally if I have an equation ax+by = c; if the coefficients a,b are coprime, can I be certain that there could possibly be only one combination(other than probably a or b being 0) of a,b that would solve the equation? No, that's not generally the case. You can find very simple equations with coprime coefficients and multiple integer solutions. If you take, picking an example almost at random, 2x + 3y = 17 this will have integer solutions whenever 173y is even, so has positive integer solutions whenever y is odd (and small enough to make the equation work)  that is, it has positive integer solutions when y = 1, 3 and 5.
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Re: Stamps
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27 Feb 2011, 10:27
Ian, Please correct me. In GMAT neither statements contradict. So its good idea to take the hint from 2) statement. Solve for the value of stamps using 1) and 2) x=y and 15x + 29y = 440. Hence x=y=10 Now suspect if 1) ALONE is the "credited" answer. To prove that no other solution exists just put in random integer < 10 and random integer > 10 for x, y in the equation 15x + 29y = 440. In choosing x and y, I know for sure if x > 10 then y < 10 and vice versa. If I get more than one pair of solution, the answer is C otherwise it is A. IanStewart wrote: No, that's not generally the case. You can find very simple equations with coprime coefficients and multiple integer solutions. If you take, picking an example almost at random,
2x + 3y = 17
this will have integer solutions whenever 173y is even, so has positive integer solutions whenever y is odd (and small enough to make the equation work)  that is, it has positive integer solutions when y = 1, 3 and 5.



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Re: Stamps
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27 Feb 2011, 21:42
gmat1220 wrote: Ian, Please correct me. In GMAT neither statements contradict. So its good idea to take the hint from 2) statement. Solve for the value of stamps using 1) and 2) x=y and 15x + 29y = 440. Hence x=y=10
Yes, since the Statements never contradict each other, you can be sure from Statement 2 that there must be one solution where x=y, even when you only use Statement 1 alone. The only question then is whether there might be a second solution. gmat1220 wrote: Now suspect if 1) ALONE is the "credited" answer. To prove that no other solution exists just put in random integer < 10 and random integer > 10 for x, y in the equation 15x + 29y = 440. In choosing x and y, I know for sure if x > 10 then y < 10 and vice versa. If I get more than one pair of solution, the answer is C otherwise it is A.
No, I would not just haphazardly plug in all conceivable values of y here to see which work; that would take a long time. We have an equation involving positive integers: 15x + 29y = 440 Now, two of the numbers (15 and 440) are multiples of 5. That guarantees that the third number, 29y, is also a multiple of 5, and so y must be a multiple of 5 (if it is not immediately clear that 29y needs to be a multiple of 5 here, you can rewrite the equation as 29y = 440  15x = 5(88  3x), from which we can see that 29y is equal to a multiple of 5). Doing this you greatly cut down on the number of values you need to test; you now only need to check y= 5, 10 and 15 (since if y = 20, the sum is too large).
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Re: Stamps
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27 Feb 2011, 23:15
Ian Your explanation almost blows me away Such a profound explanation about factors. I am lovin it ! Please have my kudos ! IanStewart wrote: gmat1220 wrote: Ian, Please correct me. In GMAT neither statements contradict. So its good idea to take the hint from 2) statement. Solve for the value of stamps using 1) and 2) x=y and 15x + 29y = 440. Hence x=y=10
Yes, since the Statements never contradict each other, you can be sure from Statement 2 that there must be one solution where x=y, even when you only use Statement 1 alone. The only question then is whether there might be a second solution. gmat1220 wrote: Now suspect if 1) ALONE is the "credited" answer. To prove that no other solution exists just put in random integer < 10 and random integer > 10 for x, y in the equation 15x + 29y = 440. In choosing x and y, I know for sure if x > 10 then y < 10 and vice versa. If I get more than one pair of solution, the answer is C otherwise it is A.
No, I would not just haphazardly plug in all conceivable values of y here to see which work; that would take a long time. We have an equation involving positive integers: 15x + 29y = 440 Now, two of the numbers (15 and 440) are multiples of 5. That guarantees that the third number, 29y, is also a multiple of 5, and so y must be a multiple of 5 (if it is not immediately clear that 29y needs to be a multiple of 5 here, you can rewrite the equation as 29y = 440  15x = 5(88  3x), from which we can see that 29y is equal to a multiple of 5). Doing this you greatly cut down on the number of values you need to test; you now only need to check y= 5, 10 and 15 (since if y = 20, the sum is too large).



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many
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19 Dec 2012, 15:14
"Now, two of the numbers (15 and 440) are multiples of 5. That guarantees that the third number, 29y, is also a multiple of 5, and so y must be a multiple of 5 (if it is not immediately clear that 29y needs to be a multiple of 5 here, you can rewrite the equation as 29y = 440  15x = 5(88  3x), from which we can see that 29y is equal to a multiple of 5). Doing this you greatly cut down on the number of values you need to test; you now only need to check y= 5, 10 and 15 (since if y = 20, the sum is too large)."
Thank you for this explanation. Wow...



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many
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07 Feb 2013, 11:56
Hi, just saw this very useful information in a MGMAT explanation. In order to prove that no other pair exists, you could figure out what number of stamps are required to do a TRADE between the $0.15 and $0.29 stamps. You would need to trade 29 of the $0.15 stamps against 15 of the $0.29 stamps. Hence you need at least either 30 of the $0.15 stamps or 16 of the $0.29 stamps to be able to do a trade, because according to the statment Joanna buys at least one of each stamp. To further illustrate this, let's assume Joanna bought $8.80 worth of stamps. Then she could have bought 20 of each of the stamps. (20 * $0.15) + (20 * $0.29) = $8.80 Furthermore you could trade 15 of the $0.29 stamps against 29 of the $0.15 stamps. [(20 + 29) * $0.15] + [(20  15) * $0.29] = $8.80 Since the amount of $4.40 limits the number of stamps to 10 each, there is no trade possible and therefore you don't need to do further tests. Thanks to Tim from MGMAT



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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many
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08 Feb 2013, 11:21
udaymathapati wrote: 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. Key is to realise that 0.15+0.29 = 0.44 and thus from statement one u can buy 10 combinations of the 0.44 stamps




Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many &nbs
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