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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?

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.

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

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19 Dec 2012, 16:14

"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)."

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

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09 Feb 2013, 20:33

for statement 1, we get 15x + 29y = 440 as we need the value of x,so 15x = 440 -29y x = (440 - 29y)/15 It's clear from the equation that the value of y should be 5 or 10 to be divided by 15 since x is an integer. if y = 5,you cannot divide it with 15. Now try y =10, so x = 10. Statement 1 is sufficient.

Also you can see from the 2nd statement that x=y .Statement 2 itself is not sufficient as you don't know the total price.

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

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22 Mar 2013, 07:37

I'd also like to add the following -

(1) She bought $4.40 worth of stamps. - We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a "0" at the end with $0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps. - Clearly insufficient.

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

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18 Jun 2013, 18:45

I'd like to provide a demonstration for the rule above.

Let's ax+by=c...................................(1) an equation with a, b, c positive integers and GCD(a, b)=1. If (x0, y0) satisfies (1) then a(x0) + b(y0) = c Any other solution can be arrived at by varying in opposite directions x0 and y0 so the ax + by keep adding to c. a(x0 + k) + b(y0 - m) = c a(x0) + ak + b(y0) - bm = c a(x0) + b(y0) + ak - bm = c c + ak - bm = c ak = bm it follows that ak is multiple of b (and of course of a) so it is multiple of LCM(a, b)=ab since GCD(a, b) = 1. In other words ak can be written as ab*z for some integer z. The same applies to bm. So, ak = bm =ab*z So k = b*z and m = a*z. Returning back to our equation and reemplazing k and m by their expressions in terms of b and a we get: a(x0 + b*z) + b(y0 - a*z) = c Which means that the generating formula for all the solutions for (1) is: [(x0 + b*z), (y0 - a*z)] z is an integer. the first solutions generated are: etc z=-2-->(x0-2*b, y0+2*a) z=-1-->(x0-b, y0+a) z=0-->(x0, y0) z=1-->(x0+b, y0-a) z=2-->(x0+2*b, y0-2*a) etc So the minimum variation from (x0, y0) is: (x0+b, y0-a) and (x0-b, y0+a) In the case that solutions must be non-negative, then if both of the above 2 solutions is not non-negative then (x0, y0) is the only non-nergative one.

In the case that solutions must be positive, then if both of the above 2 solutions is not positive then (x0, y0) is the only positive one.

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

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10 Jan 2014, 09:46

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.

We are given: 0.15*x + 0.29*y = TC

1) Gives us TC..... And we assume this is insufficient, but that's not true. Look closely: the least common multiple between these two numbers has to be a multiple of 5, so we basically have a pretty heavy restriction for the possible combinations of the two given TC = 4.40... Also, if we take one of each, we are given: 0.15 + 0.29 = 0.44... Multiply this by 10 (which is a multiple of 5), and thus we have 4.40.. If we multiply by 5, we only get 2.20, if we multiply by 15, we get 6.60, by 20 we get 8.80.. Etc, etc... So the ONLY combination of x and y that works is x = y = 10. We have 10 of each.

2) Clearly insufficient, this only tells us that x = y, but without TC that could be any multiple of 5.

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.

Hope it helps.

Hi Bunel,

How to find the equation has more number of solutions or has single solution?

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

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13 Aug 2014, 20:15

Hi, It looks like we can use the property that a (multiple of x) + (multiple of x) = multiple of x. And from that, we can see that since 4.40 and .15 are multiples of 5, .29 has to be a multiplied by a multiple of 5 as well.

Can someone comment as to whether this property holds true for subtraction/division/multiplication?

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

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16 Aug 2015, 02:21

we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ?

we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ?

Be careful to use information from the other statement when you are evaluating each statement along in a DS question. You are correct that statement 1 is sufficient but 2 in itself (and without using the information from statement 1!) is NOT sufficient.

Statement 2 says that umber of .15$ tickets = number of .29 tickets. Now this 'equal' number can be anything from 1 to a million even. Thus we do not get 1 unique value with statement 2 alone.

The first step in a DS question is to evaluate the information provided in the 2 statements individually and ONLY when you do not get a sufficient answer individually with either of the 2 statements, should you combine the statements.

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

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17 Sep 2015, 13:19

Bunuel wrote:

(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 \(y=10\).

Why is that so clear that only one integer combination fits this? Very difficult to spot...
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]

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28 Sep 2015, 11:12

Bunuel wrote:

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.

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.

How did you arrive at the 10-10 combination in statement 1? Trial and error?

1 trick to remember in such questions or similar questions to number of fruits, number of kids, number of pencils etc, the actual numbers can only be integers. You can not have fractional pencils or stamps.

Thus, based on the above question, \(15x+29y=440\) only has 1 pair of integer solution, (10,10). This is by trial and error based on the fact that both the values need to be integers.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]

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01 Oct 2015, 10:56

Bunuel wrote:

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.

Hi Bunuel, Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation.

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.

Hi Bunuel, Could you tell me how can we determine in which case ax+by=c is sufficient? We cannot test all the cases in 2 minutes In this case, we can easily see that x = y = 10 is a solution, but it seems that nothing can guarantee that there are no other solutions to this Diophantine equation.

Many thanks

Let me try to answer.

There is no other way to answer your question but to test out some pairs of values keeping in mind the values of x and y MUST be POSITIVE INTEGERS. For DS questions, you only need to get 2 values to make a statement or a combination of statements insufficient, stop, mark E and move onto next question at this point. Initially, yes this process will look extremely time consuming but once you practice a few similar question you will see that you dont need to spend a lot of time on such questions.

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

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11 Nov 2015, 04:56

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.

Pretty @!)*, let's start:

(1) 15a+29b=440 -> \(b=\frac{440-15a}{29}\) --> \(\frac{5(88-3a)}{29}\) so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here.

(2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ...

So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ?
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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.

Pretty @!)*, let's start:

(1) 15a+29b=440 -> \(b=\frac{440-15a}{29}\) --> \(\frac{5(88-3a)}{29}\) so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here.

(2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ...

So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ?

This is the most straightforward way for these questions wherein the applicable values can only be positive ingeters. Similarly, you can not have 0.5 apples or 0.33 bananas etc and hence if the question had asked about certain fruits or number of men etc , your method will still be applicable to such questions.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]

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12 Dec 2015, 14:18

Bunuel How did you systematically see that there was only one solution for the equation in statement 1? I got this answer correct but guessed. I knew it was either A or C and figured that because X was too obvious of a solution, it was A. But that leaves a lot to chance. Wondering how to best approach these diophantine eqns.

Bunuel wrote:

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.

Bunuel How did you systematically see that there was only one solution for the equation in statement 1? I got this answer correct but guessed. I knew it was either A or C and figured that because X was too obvious of a solution, it was A. But that leaves a lot to chance. Wondering how to best approach these diophantine eqns.

Let me try to answer the question. For all such questions wherein you are asked to find the number fo fruits or stamps or pencils etc, remember:

1. The number must be an integer. 2. As the number must be an integer, the minimum value of such a number is 0.

Coming back to the question at hand, from statement 1, you can generate the following equation:

0.15x+0.29y=4.40 and we need to find whether we get 1 unique value for x.

We can rewrite the given equation as x=(4.40-0.29y)/0.15 or x=(440-29y)/15 with x and y only taking integer values such that x,y \(\geq\) 0

Now, based on that information, we can see that iteratively that 440-29y can only be a factor of 15 when 440-29*y gives you a number ending with 0 or 5.

Also, a quick knowledge of what unit digits you get with a number such as A9 = 0/1/2/3/4/5/6/7/8/9 and you need to get a unit digit of 0 or 5 from the operation 440-one of the digits of obtained from 29*y, the only possible values of y can be either 0,5,10,15, 20(not needed as this will make x<0). Out of these only y=10 will give you an integer value for 'x'.

Hence the only possible value of (x,y) is (10,10).

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