This question is based on the concept of "integral solutions". Since in such real world examples, x and y cannot be negative or fractional, usually these equations have a finite number of solutions.
After you get one solution, you will quickly know how many solutions the equation has. But getting the first set of values which satisfy the equation requires a little bit of brute force.
But the good thing here is that this is a DS question. You don't need to find the actual solution. The only thing you "need" is to establish that there is a single solution only. (Obviously, there has to be a solution since she does own $282 worth of packages.)

That can be done relatively easily.
First, check out this post for a conceptual discussion on this question type (case 2):
http://www.veritasprep.com/blog/2011/06 ... -of-thumb/

Once you understand this, the following will make sense to you.

17x + 13y = 282
If x = 0, y is 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something

Now you know that there will be only one set of values satisfying this equation. Why? Because y will be less than 17 in the first set of values which satisfy this equation. So if you want to get the next set that satisfies, you will need to subtract it by 17 which will make y negative. So in any case, there will be a unique solution to this equation.
Hence statement 1 is sufficient.

Answer (A)
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Hi MeghaP and stonecold,

Is there a shortcut-- it will depend from Q to Q, depending on the properties of the number involved..
Should you depend on a short cut-- MAY BE but do not miss out on other combinations which may as well satisfy the equation...

let me explain through this example-

NOTE- I had initially done the Q by taking values one by one, and you too when you get hang of the tables, would be able to do faster. So when now you have written about a shortcut, I can give you one in this Q..

\(17x+13y=282\)
=> \(4x+13x+13y=282\)..
\(4x+13(x+y) = 282\)...
now some info you should gather from here
1) 13(x+y) should form most of 282 as 13>4 and x+y>x...
2) (x+y) should be a multiple of 2 but not of 4---
3) if entire 282 was formed from 13(x+y), x+y <22 as 13*22>282..

so values of x+y can be 2, 6, 10, 14, 18 only..
Since we are looking for the largest value from 13(x+y), lets try with x+y =18.
\(4x+13(x+y)=4x+13*18=4x+234 = 282\)...
so \(4x= 282-234=48.. x=12\)..
satisfies all conditions written above.. OK x=12, y= 18-12=6

lets try the next possible
\(x+y = 14\)..
\(4x+13(x+y)=4x+13*14=4x+182 = 282\)...
so \(4x= 282-182=100... x=25\)..
is it possible that x= 25 and x+y=14, when both x and y are non negative.... NO
any number below this 10,8,2 can be ruled out straight way as they would make x much larger than x+y..

so our answer is that ONLY one set of number are possible..
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This is an obvious C trap question. If we combine both statements, we will get an answer.
So this problem is likely to be solved by one statement only. (With experience you can say that this statement would be A)

If we take factors of 282 we get: 2*3*47.

Statement 1
17c+13r = 282 = 6 * 47. [ where c = corn packages and r = rice packages]

Note that 17 and 13 are co-prime.

Our aim is to find such values of c and r such that 17c+13r = 47.

47 = 34+13.
c=2; r=1

We can get a unique answer.

Corn Package = 2*6 = 12 (We dont need to calculate this).

Statement 2

We know that c= 2r. But we dont know exact numbers so not sufficient.

A is the answer.

hi,
The Q puts to you a C trap..
the first statement gives you SUM of two variables..
AND the II gives you the ratio of two variables..

and there it is C, BUT more than often the SUM gives you enough INFO to be sufficient..

let the number of packages of corn be C and that of rice be R..

I. 17C + 13R = 282..
check for values of C and see if any integer value of R is possible ..
C can take values from 1 to 282/17 = 16..

Only C as 12 and R as 6 is possible..
SUff

II. She has twice as many packages of corn as of rice
C = 2R..
various answers possible..

A
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Hey Chetan I have a doubt in this Question..
I marked the correct answer as i thought in my head that 13 and 17 are prime so the common factor would repeat after 13*17 hence only one value satisfy.
Now my question is the process of putting the integers as 1,2,3,4...takes a loooooong time..
Is there any other shorter way to tackle this question .
regards
S.C.S.A
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I have come across a lot of such questions to recognize the trap. But I do not know the mathematical way to conclude that only one combination will fit in. For example, in this case C has to be 12 and R has to be 6. How can we rule out other possibilities? Please suggest. Thank you!

I hate such questions because, as in statement 1, you must be sure that such equation doesn't have unique solution. And with such numbers..How come should i guess about 12 and 6? Picking numbers is time-consuming
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One thing I find convenient in these kind of questions is to get help from the other statement that is given as a trap. For example, for this one '(2) She has twice as many packages of corn as of rice' statement should help you pick out the numbers efficiently since you would be choosing twice as as many corn packages as of rice. C = 12 R =6 becomes easy to find

Hi,
it may not be advisable here, unless you are very careful...
say 12 and 6 you have found same way..
BUT there is another solution for the EQUATION, then statement I will not be sufficient whereas you will mark it as sufficient..
so you have to ensure that there are no further solutions apart from 12 and 6..
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That is certainly a good idea, provided it is used carefully. I wrote a post on this concept sometime back:
http://www.veritasprep.com/blog/2014/07 ... t-part-ii/
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there is one similar question in og 2015 but it is more easy because we can prove that there is only one set of solution in case of A.

is this problem from gmatprep? i doubt that gmat force us to make too complex accounting

Note that this is a 700+ level question. This concept can be tested in GMAT. If you use some tricks, you can solve the question very easily with few calculations.
Check out my discussion on this question above. Also, knuckleduster talks about a strategy that can be used here to solve it very quickly. I have also given a link above where that strategy has been discussed in detail.
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Here's my simple solution. Since 17 & 13 are both primes (always look out for prime numbers), we can use prime factorization.

282 - divide by 2 -> 141 divide by 3--> 47 (which breaks into 17 *2 + 13(1))

So (17(2)+13(1) * 6) = $282.... there's only one way to get to $282. 17*12 + 13*6. other methods won't work so sufficient.

Should be out of this question in 60 seconds if you're quick.
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VeritasKarishma, Bunuel

To clarify, is the quick, "lightbulb" method to solve this question recognizing that the LCM of 13 and 17 is found only once between 0 and 282? Or is it that because we can rule out any solution of Rice being greater than 17 and can therefore deduce that the solution must exist with Rice<17 ---> and therefore, if we subtract 17 from this number (to find the next solution) it will give us a negative and thus no additional solution?

I understand that both ways that I have outlined are based on the same concept, so my question is can we just go straight to the "LCM occurring only once test" or do we have to rule out a few potential values first?

Thanks in advance!

LCM of 13 and 17 doesn't have much to do with it since you are not multiplying 13 with 17 but instead 13 with some number and 17 with another.

e.g. 17x + 13y = 277
x = 1, y = 20 satisfies this.
Next solution will be x = 14, y = 3
Both valid solutions though LCM of 13 and 17 occurs only once in 277 too.
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Why do we need to assume it's an integral solution?
We can very well assume that the package quantities are not integers.

Generally, on the GMAT questions, "INTEGERS" are mentioned specifically in the question stem.

We are told that "Pam owns an inventory of [i]unopened packages of corn and rice...[/i]" so we can deduce that the number of packages must be integers.
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this co-prime won't always work. for example 3x+5y=42>> 3x+5y=6x7. it'll work with 6(x=2), but not with 7, given that we are only looking for integers >=0. but then if x=2 is a solution, you could multiple by 7 and have x=14,y=0 as a starting point, and then go down by the LCM approach which mean 3x-15 and 5y+15. so x=14,9,4 and y=0,3,6. however it's a bit easier to solve for a smaller number like 42 and not so easy for something larger where lots of iterations will just take time

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