12 Days of Christmas GMAT Competition - Day 2: Jingle, an elf

Let's assume that the distances covered be x, y and z.

\(x \leq y \leq z\)

\(x + y + z = 1080\) --- Equation 1

Ques: \(\frac{x + y + z }{3} > y\)

\(x + y + z > 3y\)

\(x + z > 2y\)

From equation 1

\(1080 - y > 2y\)

\(3y < 1080\)

\(y < 360\)

Hence the second largest distance should be less 360.

Statement 2

(2) On one of the days, Jingle covered 360 miles.

Let's assume that the smallest distance covered is 360. In that case, the other distances should also be 360 for the sum to be 1080. In this case the mean = median. The answer is No.

If the second largest distance is 360, then the other two distances must sum to 720, and the mean is equal to median. The answer is No.

The largest distance is 360, then the other distances should also be 360 for the sum to be 1080. In this case the mean = median. The answer is No.

The statement is sufficient.

Statement 1

(1) The range of the distances Jingle covered over these days was equal to his daily average (arithmetic mean) distance.

\(z - x = \frac{x + y + z }{3}\)

\(3z - 3x = x + y + z\)

\(2z - 4x = y\)

With this equation we cannot determine anything useful.

We can multiple possible values depending on the values chosen for x, y and z.

Hence this statement alone is not sufficient to answer the question.

IMO B