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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 ?
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags karthickhari wrote: we can solve the eqn 15X + 29Y =440 , it will have only 10 and 10 as solution, so stmt 1 is enough. If we go through stmt 2 it will also have same value it will give so cant it be solved as like 44X(x=y)=440 so x =10 ? what is wrong in this approach. Please help me out ? Be careful to use information from the other statement when you are evaluating each statement along in a DS question. You are correct that statement 1 is sufficient but 2 in itself (and without using the information from statement 1!) is NOT sufficient. Statement 2 says that umber of .15$ 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.

Hope this helps.
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 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 $0.15 stamp [#permalink] ### Show Tags 1 reto wrote: 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...

This is quite representative of a GMAT like question and thus for such questions wherein you are asked number of tickets, number of people, number of toys etc wherein only integer values can work, make sure to try to find a few sets of values for both the variables that will satisfy the given equation which in this case is 15x+29y=440.

Once you create the equation above, you can see that you could also write it as 15x=440-29y which means that 440-29y MUST be a multiple of 15 (as the other side is 15x). Thus once you start by recognizing this fact, you will see that only y=10 satisfies this. For all other values you will not get an integer value of x or get a value <0 (this is not acceptable as number of tickets can not be <0).

Hope this helps.
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 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.

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.

Bunuel,

How did you arrive at the 10-10 combination in statement 1? Trial and error?
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags manhattan187 wrote: 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.

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.

Bunuel,

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 $0.15 stamp [#permalink] ### Show Tags 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.

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.

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
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags yenh wrote: 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.

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.

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

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.

Hope this helps.
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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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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 1 BrainLab 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. 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. Intern  Joined: 08 Oct 2015 Posts: 7 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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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. 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. CEO  S Joined: 20 Mar 2014 Posts: 2560 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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dubyap wrote:
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).

Hope this helps.
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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.

Given: Joanna bought only $0.15 stamps and$0.29 stamps.
Let C = number of $0.15 stamps purchased Let E = number of$0.29 stamps purchased

Target question: What is the value of C?

Statement 1: She bought $4.40 worth of stamps We can write the equation 0.15C + 0.29E = 4.40 IMPORTANT: In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables. To determine whether this is the case here, let's examine all possible values of E. If E = 0, then the entire$4.40 was spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.40, it cannot be the case that E = 0 If E = 1, then$0.29 was spent on $0.29 stamps, leaving the remaining$4.11 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.11, it cannot be the case that E = 1 If E = 2, then$0.58 was spent on $0.29 stamps, leaving the remaining$3.82 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.82, it cannot be the case that E = 2 If E = 3, then$0.87 was spent on $0.29 stamps, leaving the remaining$3.53 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.53, it cannot be the case that E = 3 IMPORTANT: At this point, we might speed up our solution by recognizing that, in order for 0.15 to divide evenly into a number, that number must end with 5 or 0. Also recognize that, in order for the resulting value to end with a 5 or 0, E must be divisible by 5 So, from this point on, we'll just check values of E that are divisible by 5. If E = 5, then$1.45 was spent on $0.29 stamps, leaving the remaining$2.95 to be spent on $0.15 stamps. NICE!$2.95 ends with a 5. So this MIGHT work. Unfortunately, 0.15 does NOT divide evenly into 2.95. So, it cannot be the case that E = 5

Keep going!

If E = 10, then $2.90 was spent on$0.29 stamps, leaving the remaining $1.50 to be spent on$0.15 stamps. 1.50/0.15 = 10 = C. So, one possible solution is E = 10 and C = 10
If E = 15, then $4.35 was spent on$0.29 stamps, leaving the remaining $0.05 to be spent on$0.15 stamps. Doesn't work.
If E = 20, then $5.80 was spent on$0.29 stamps. Hmmm. Looks like we can stop here!

So, there is only one possible scenario that meets the given conditions.
So, it MUST be the case that E = 10 and C = 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: She bought an equal number of $0.15 stamps and$0.29 stamps.
We have no idea how much money Joanna spent on stamps.
As such, there are infinitely many scenarios that satisfy statement 2. Here are two:
Case a: She bought 3 $0.29 stamps and 3$0.15 stamps. In this case, the answer to the target question is C = 3
Case b: She bought 8 $0.29 stamps and 8$0.15 stamps. In this case, the answer to the target question is C = 8
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 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.

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:
http://gmatclub.com/forum/eunice-sold-s ... 09602.html
http://gmatclub.com/forum/martha-bought ... 00204.html
http://gmatclub.com/forum/a-rental-car- ... 05682.html
http://gmatclub.com/forum/joe-bought-on ... 06212.html
http://gmatclub.com/forum/a-certain-fru ... 01966.html
http://gmatclub.com/forum/joanna-bought ... 01743.html
http://gmatclub.com/forum/at-an-amuseme ... 26814.html
http://gmatclub.com/forum/collections-c ... 81062.html

Hope it helps.

Bunuel,

How are you sure that only one combination possible in equation 1.....Is it Trial error or any method to find....I always struggle in optimizing values... Please help!
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags Bought$0.15 and $0.29. How many$0.15?

Can be 0.15, 0.30, 0.45, 0.60, 0.75, 0.90, 1.05, 1.20, 1.35, 1.50, 1.65..

1. Bought $4.40 --> 0.15A + 0.29B = 4.40 The amount of B need to be resulted in cost of x.x0, x.x5 B=10 --> 2.9 --> (4.4-2.9)/0.15 =10 B=20 --> 5.8 cant (coz >4.4) Since the only possibility is: 10*0.29=2.9 10*0.15=1.5 (2.9+1.5=4.4) Therefore 1 is Sufficient. 2. Sane amount of 0.15 & 0.29. Can be anything --> Therefore 2 is inSufficient. Cheers, Eyal Manager  S Joined: 10 Jun 2014 Posts: 82 Location: India Concentration: Operations, Finance WE: Manufacturing and Production (Energy and Utilities) Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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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. For more on this type of questions check: http://gmatclub.com/forum/eunice-sold-s ... 09602.html http://gmatclub.com/forum/martha-bought ... 00204.html http://gmatclub.com/forum/a-rental-car- ... 05682.html http://gmatclub.com/forum/joe-bought-on ... 06212.html http://gmatclub.com/forum/a-certain-fru ... 01966.html http://gmatclub.com/forum/joanna-bought ... 01743.html http://gmatclub.com/forum/at-an-amuseme ... 26814.html http://gmatclub.com/forum/collections-c ... 81062.html Hope it helps. <Only one integer combination of xx and yy is possible to satisfy> .....How to arrive this ? through lengthy try of combinations or is there any shortcut process to arrive this Senior Manager  G Joined: 04 Aug 2010 Posts: 496 Schools: Dartmouth College Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

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 = the number of 15-cent stamps and y = the number of 29-cent stamps.

Statement 2:
x=y
Here, x and y can be any positive integral values.
INSUFFICIENT.

Statement 1:
15x+29y = 440

Solve for x and y when x=y, as required in Statement 2.
Substituting x=y into 15x+29y=440, we get:
15x + 29x = 440
44x = 440
x=10, implying that y=10

Thus, one solution for 15x+29y=440 is as follows:
x=10, y=10
In accordance with the rule discussed above, the value of x may be altered only in INCREMENTS OF 29 (the coefficient for y), while the value of y may be altered only in INCREMENTS OF 15 (the coefficient for x).
Not possible:
If x increases by 29 and y decreases by 15, then y will be negative.
If x decreases by 29 and y increases by 15, then x will be negative.
Implication:
The only nonnegative integral solution for 15x+29y=440 is x=10 and y=10.
SUFFICIENT.

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Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 15x+29y=440 29y=2*a mod 3 440= 2 mod 3 15x=0 mod3 2a=2 a=1 y=3k+1 Also, 29y=440-15x y must be a multiple of 5 y can be 10, 25 40, 55, 70....so on We can clearly see at y= 25, 40,55, 70 or bigger values, x will be negative. 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. Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp   [#permalink] 20 Sep 2019, 12:44

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