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


Just a quick tool for Diophantine equations.

Lets assume the equation 4x+5y = 36. Rearrange it as x =\(\frac{(36-5y)}{4}\) or y = \(\frac{(36-4x)}{5}\). Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all we have to do is keep adding 4 to the initial value of y=0. Thus, the next value of y which will give an integral solution for x is y=4,8,12 etc. One could also subtract 4 and get subsequent values for y = -4,-8,12 etc.

Taking the second rearrangement, we can see that x=-1, we have y=8. Thus, following the same logic, the next value of x, which will lead to an integral value for y is x= 4,9,14 etc.

Now back to the given problem:

We know that 15x+29y = 440.

Now, 29y = 440-15x

or y =\(\frac{440-15x}{29}\) .Now as because the statements on GMAT don't contradict each other, we know that one of the value of x and y for the above equation is x=y=10. Thus, for the above equation, keep adding 29 to x=10 for all the successive values for getting an integral solution for y. Thus, in accordance with the given sum, we can neglect negative integral values and state that x=y=10 is the only possible solution for both x,y>0.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp [#permalink]
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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]
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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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]
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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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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

Answer: A

Cheers,
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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]
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.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamp [#permalink]
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

15x + 29y =

Stat1: She bought $4.40 worth of stamps.
15x + 29y = 440, Now, 15*29 = 435 <440, so it can have one or 2 solutions. Let's try will x=y=10. It is one solution.
Now, can you trade 29 or 15 in x and y respectively. Sufficient.

Stat2: She bought an equal number of $0.15 stamps and $0.29 stamps.
x=y, we don't know value of x and y. Insufficient.

So, I think A. :)
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