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26 Sep 2010, 11:48
17
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75% (hard)

Question Stats:

53% (01:48) correct 47% (01:43) wrong based on 2023 sessions

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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: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 26 Sep 2010, 11:56 26 58 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.
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 16 Jun 2013, 17:32 11 20 Here is the method to never fail to answer correctly Diophantine-equations-related Data Sufficiency problems. 1. First, be sure that the 2 variables must be non-negative integers or positive integers and that each statement provides a linear equation relating the 2 variables. Furthermore, be sure that the 2 equations are not equivalent (2x+3y=20 and 6x+9y=60 are equivalent) and are reduced to the form: ax + by = c whith integral coefficients and constant term such that GCF (a,b)=1. 2. Find an initial Solution: "Take advantage" of the fact that statements never contradict each other and thus system of equations constructed with both statements have always at least one solution. So resolve the system of equations. 3. Unicity: Once you arrive to a solution, say (x0, y0), go back to the first statement alone, for example, and check the unicity of the solution using only that statement by applying the test below. In case the solution is unique, statement 2 is superfluous and statement 1 is sufficient. The answer is A or D. In case the solution is not unique the answer is B, C or E. Apply the test on statement (2). And update your answer. If there is more than one solution using each statement alone then the answer is C. ------------------------------------------------------------------------------------------------------------------------------------------------- Now here is the rule that indicates whether or not a non-negative integer solution is unique to an equation: Suppose the equation be: ax+by=c (reduced with a, b, c positive integers. i.e. GCF(a,b)=1) If (x0-b)<0 AND (y0-a)<0 then there is no other non-negative integer solution than (x0, y0) and the corresponding statement is sufficent. If (x0-b)>=0 OR (y0-a)>=0 then other non-negative integers solutions exist and the statement is not sufficient. If the variables must be positive the test is: If (x0 - b)<=0 AND (y0 - a)<=0 then there is no other positive integer solution than (x0, y0) and the corresponding statement is sufficent. If (x0 - b)>0 OR (y0 - a)>0 then other positive integers solutions exist and the statement is not sufficient. Note: The test is to subtract each coefficient from the solution found for the opposite variable. -------------------------------------------------------------------------------------------------------------------------------------------------- Let's apply this to a real GMAT problem: A man buys some juice boxes. The boxes are from two different brands, A and B. How many boxes of brand A did the man buy if he bought$5.29 worth of boxes?
(1) The price of brand A box is $0.81 and the price of brand B box is$0.31
(2) The total amount of boxes is 9

Variables here must be positive integers -number of juice boxes- since it is suggested that some juice boxes are from brand A and the rest from brand B.
1. The equations provided are:
(1) 0.81A + 0.31B = 5.29.
(2) A + B = 9
Which are reduced to:
(1) 81A + 31B = 529
(2) A + B = 9,
which is a system of reduced, linear, non-equivalent equations.

2. Find an initial Solution:
(1) 81A + 31B = 529. GCF(31, 81)=1.
(2) A + B = 9

Mutliplying (2) by 31 and subtracting it from (1) we get:
50A=250 so A=5 and B=4.
An initial solution is (5, 4)

3. Unicity:
Unicity for statement (1):
81(5) + 31(4) = 529
Since
(5 - 31) <=0 AND (4 - 81)<=0 then there no other positive solution than (5, 4) so statement (1) is sufficient.

Unicity for statement (2):
It is obvious that statement (2) alone is not sufficient but the test is still applicable.
1(5) + 1(4) = 9
Since
(5 - 1) >0 OR (4 - 1) > 0 then there are other positive solutions than (5, 4) so statement (2) is not sufficient.

Hope this helps.
##### General Discussion
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Joined: 30 May 2010
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26 Sep 2010, 12:08
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.
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29 Sep 2010, 21:57
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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29 Sep 2010, 22:54
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. _________________ Manager Joined: 06 Apr 2010 Posts: 62 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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18 Oct 2010, 02:02
8
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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26 Feb 2011, 13:02
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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27 Feb 2011, 10:41
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, 22:42
35
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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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 07 Feb 2013, 12:56 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 Retired Moderator Joined: 05 Jul 2006 Posts: 1384 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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08 Feb 2013, 12:21
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
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Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 23 Mar 2013, 00:47 1 5 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. _________________ Intern Joined: 09 Apr 2013 Posts: 2 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamp  [#permalink]

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18 Jun 2013, 18:45
1
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.

CQFD