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

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Manager
Joined: 06 Apr 2010
Posts: 117
Joanna bought only $0.15 stamps and$0.29 stamps. How many  [#permalink]

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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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Joined: 02 Sep 2009
Posts: 51280

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26 Sep 2010, 10:56
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49
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.

Let $$x$$ be the # of $0.15 stamps and $$y$$ the # of$0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$

(1) She bought $4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient. (2) She bought an equal number of$0.15 stamps and $0.29 stamps --> $$x=y$$. Not sufficient. Answer: A. So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative 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. For more on this type of questions check: eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html collections-confused-need-a-help-81062.html Hope it helps. _________________ ##### Most Helpful Community Reply Manager Joined: 06 Apr 2010 Posts: 69 Re: Stamps [#permalink] ### Show Tags 18 Oct 2010, 01:02 7 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.. ##### General Discussion Manager Joined: 30 May 2010 Posts: 176 Re: Stamps [#permalink] ### Show Tags 26 Sep 2010, 11:05 3 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. Manager Joined: 30 May 2010 Posts: 176 Re: Stamps [#permalink] ### Show Tags 26 Sep 2010, 11:08 1 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. Manager Joined: 21 Oct 2007 Posts: 180 GRE 1: Q780 V540 Re: Stamps [#permalink] ### Show Tags 29 Sep 2010, 08:24 3 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. _________________ If you like my post, please press Kudos+1 Manager Joined: 13 Aug 2010 Posts: 158 Re: Stamps [#permalink] ### Show Tags 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 Senior Manager Status: Upset about the verbal score - SC, CR and RC are going to be my friend Joined: 30 Jun 2010 Posts: 291 Re: Stamps [#permalink] ### Show Tags 29 Sep 2010, 21:23 5 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. _________________ My gmat story MGMAT1 - 630 Q44V32 MGMAT2 - 650 Q41V38 MGMAT3 - 680 Q44V37 GMATPrep1 - 660 Q49V31 Knewton1 - 550 Q40V27 Math Expert Joined: 02 Sep 2009 Posts: 51280 Re: Stamps [#permalink] ### Show Tags 29 Sep 2010, 21:54 1 prab wrote: 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 As noted above by Dreamy you can not use info from statement (1) to solve statement (2), so for (2) we don't know that total$4.40 were spent.
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26 Feb 2011, 11:53
1
Quote:
So when we have equation of a type ax+by=c and we know that x and y are non-negative 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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Joined: 02 Sep 2009
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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 non-negative 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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26 Feb 2011, 12:14
Additionally if I have an equation ax+by = c; if the coefficients a,b are co-prime, 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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Joined: 23 Jan 2011
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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 non-negative 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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Joined: 24 Jun 2008
Posts: 1322

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27 Feb 2011, 09:41
1
bugSniper wrote:
Additionally if I have an equation ax+by = c; if the coefficients a,b are co-prime, 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 17-3y 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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27 Feb 2011, 10:27
1
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 17-3y 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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Joined: 24 Jun 2008
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27 Feb 2011, 21:42
32
6
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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Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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27 Feb 2011, 23:15
1
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).
Intern
Joined: 02 Aug 2012
Posts: 15
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many  [#permalink]

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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...
Intern
Joined: 17 Dec 2012
Posts: 4
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many  [#permalink]

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07 Feb 2013, 11:56
1
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  [#permalink]

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08 Feb 2013, 11:21
3
1
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 [#permalink] 08 Feb 2013, 11:21

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

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