Problem Solving

A student responded to all of the 22 questions on a test and received a score of 63.5. If the scores were derived by adding 3.5 points for each correct answer and deducting 1 point for each incorrect answer, how many questions did the student answer incorrectly?

(A) 3
(B) 4
(C) 15
(D) 18
(E) 20

OA A

heshamelaziry wrote:A student responded to all of the 22 questions on a test and received a score of 63.5. If the scores were derived by adding 3.5 points for each correct answer and deducting 1 point for each incorrect answer, how many questions did the student answer incorrectly?

(A) 3
(B) 4
(C) 15
(D) 18
(E) 20

If you're an algebra person, you can derive the equation from the word problem. Good news! If you're not an algebra person, we can backsolve very quickly as well.

As always, when backsolving we want to start by applying logic/common sense. The student scored 63.5. If the student had gotten a lot wrong, there's no way to achieve that score. Accordingly, choices C, D and E don't make any sense. (Remember, the question asks how many answers the student got WRONG.)

So, right away we know that either A or B is correct. Worse case, 50/50 guess.

Since there's only two choices, we can start with either one... let's try (B) 4.

If the student got 4 wrong, the student got 18 right, so the score is:

18(3.5) - 4(1).

At this point we're done - we see that this will generate and integer score, and we want 63.5, so there's no way that B could be correct. Choose (A)!

Of course, if we had noticed this trend during our "common sense/logic" stage of the process, we would have realized that to get a non-integer score, the students needs to get an odd number right/wrong and eliminated B, D and E right off the bad.

So, just by common sense we can eliminate B, C, D and E - we love it when that happens.

* * *

If we were to set up the equation, we'd have gotten:

3.5C - 1(22-C) = 63.5

in which C is the number of correct choices and (22-C) is the number of incorrect choices. We can solve for C and then (22-C) to get the right answer.

Of course, we can also set up:

3.5(22-I) - 1(I) = 63.5

in which I is the number of incorrect choices and solve for I directly - the only downside to the second approach is that the math may be more annoying, but it saves us a step.

One other point - this question really stresses the value of Step 4 of the Kaplan Method for Problem Solving: always double check the question. A lot of test takers would solve for the number of correct responses by mistake, not realizing that isn't what the question is asking for. On the real GMAT "19" almost certainly would have been one of the choices, just to punish those sloppy test takers.