Strategy #1 – Count the Variables; Don’t do the Math!

Remember the “n equations with n variables” rule! If you have three unknowns, you’ll need three equations to solve for all three. If you only need the sum of two of those variables, however, you may only need two equations to solve! Let’s take a look at how this strategy can help us find a shortcut!

1.  A total of 2,000 t-shirts was divided among a soccer team, two baseball teams, and a track team. How many shirts did the track team receive?

(1) The track team and one of the baseball teams together received 5/7 as many shirts as the other baseball teams and the soccer team combined; and the two baseball teams each received the same number of shirts.

(2) Each baseball team received 400 fewer shirts than the soccer team and 400 more than the track team.

This is a value question for which we have four unknowns: the soccer shirts, baseball team A shirts, baseball team B shirts, and track team shirts. Statement (1) doesn’t allow us to solve for these four unknowns.

For (2), we’re given the relationship between one unknown (the soccer team) and the other three unknowns. That would allow us to choose a variable for the soccer team t-shirts, and write the other unknowns with that same variable. We don’t even have to do any math to see that this is sufficient! If we wanted to try this out algebraically, we could pick “x” for the number of shirts given out to the soccer team. Since a total of 2,000 shirts were given out, x + x + (x + 400) + (x – 400) = 2,000.  We have a linear equation with a single variable, so we know this choice is sufficient according to the “n equations, n variable” rule.

Strategy #2 – Plug in Values.

This strategy especially comes in handy for “yes or no” DS!

2. Given that A > 0 and C > 0, is (A + B) / (C + B) > A/C?

(1) B > 0

(2) A < C

For (1), we are only given that b is a positive number. Picking values is a good strategy here. If we can choose values such that we get a “yes” and choose values so that we get a “no” then we can quickly eliminate. If a = 1 and c = 1 and b = 1, then the inequality is NOT correct. However, if a = 1 and c = 2, and b = 3, we get 4/5 > 1/2, which IS correct. (2) can be proved insufficient similarly. Combined, if b is positive and a ? c, then it will continue to INCREASE the left-hand side of the inequality no matter what values we pick.