One interesting way to attack this problem is to start by considering what happens if we try to only buy at the cheapest price. We couldn't possibly buy more pairs than we could afford buying only the cheap pairs.

560/45 is 12 with a remainder of 20. That tells us right away that any answer larger than 12 will be too big (eliminating D and E), and also that we will need to buy at least one pair at the higher price to spend all of our money.

Now set this up algebraically:

c = # of cheap jeans

e = # of expensive jeans

560 = 45c + 65e

When the numbers are this small, it's almost always fastest to just plug in values at this point, knowing that to maximize the total number of jeans, we want the fewest expensive jeans possible. We get very lucky because as soon as we plug in 1 for e, we get 11 for c and we have our solution.

Another option would be to use the answer choices to create a second equation, and then solve the system of equations that way. We will want to have the highest answer choice that gives us an integer solution (since we can't buy fractions of jeans and we need to spend all of our money). Since we eliminated D and E, we can start with C.

560 = 45c + 65e

12 = c + e

multiple the second equation by 45:

540 = 45c + 45e

Then subtract from the first equation:

20 = 0 + 20e

1 = e

Since this represents the the highest total number in the remaining answer choices, we can just choose C and don't have to check any more answers.

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-Jordan Wolbrum

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