There is a general mathematical theory behind these kinds of equations, but it would be total overkill to learn it for the GMAT. The equations on the GMAT aren't chosen at random, so you don't need a general theory for any kind of equation; you won't be asked how many positive integer solutions there are to an equation like 11x + 14y = 491, for example. The numbers in a real GMAT question will be carefully chosen so that a test taker will have the opportunity to notice a shortcut and quickly arrive at an answer.Thouraya wrote:Hi @Ian,
So as a conclusion to what you were trying to explain above, is there a way where I can know if only ONE combination of x and y is possible for a specific total number, or whether there are MULTIPLE values which are possible? What is the best way to decide on that (if i have one equation with two unknowns).. Thank you!
If you do have a linear equation where your two unknowns represent *positive integers*, and you want to know if there is only one solution, there are only two shortcuts I can think of that might be useful. Suppose you know that someone spent a total of $2.75 on two types of fruit: apples which cost 14 cents each and oranges which cost 15 cents each, and you are asked how many pieces of fruit this person bought. You'd then write down the equation:
14x + 15y = 275
Here x and y are positive integers, and we want to find x+y. Of course we could test every possible value for x until we find one that gives an integer value for y, but we can also use one of two shortcuts. In this question, oranges and apples cost almost the same amount, which makes it very likely there will only be one solution to the equation above, since the total cost is small. You might notice that if the person buys 20 pieces of fruit, the total cost must be at least 20*14 = 280 cents, whereas if the person buys 18 pieces of fruit, the total cost must be at most 18*15 = 270 cents. So the only possibility is that the person bought 19 pieces of fruit.
Alternatively, we can use number theory. If you ever have an equation like the following, involving integers only:
14x + 15y = 275
then if two terms in your equation are divisible by some number d, the third term must also be divisible by d. Here, 15y and 275 are both divisible by 5, so 14x must be divisible by 5 as well, and thus x is divisible by 5 (to see why this is true, rewrite the equation as 14x = 275 - 15y; since 5 is a factor of the right side, it must be a factor of the left). Noticing that allows you to greatly cut down on the number of values you need to test.
If you don't think you'd spot either of these shortcuts under test conditions, don't worry about it; as Ron says above, testing values will always get you to an answer on these questions, even if it might be more time consuming. What's most important, especially for DS, is to understand why there might be precisely one solution to equations of the type discussed in this thread. That will help you to avoid mistakenly choosing answer 'C' to DS questions like the one in the original post.
