2016 AMC 12A Problem 4

Below is the professionally curated solution for Problem 4 of the 2016 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:meanmedian (data)mode

Difficulty rating: 1100

4.

The mean, median, and mode of the 77 data values 60,100,x,40,50,200,9060, 100, x, 40, 50, 200, 90 are all equal to x.x. What is the value of x?x?

5050

6060

7575

9090

100100

Solution:

The mean condition gives 60+100+x+40+50+200+907=540+x7=x, \dfrac{60+100+x+40+50+200+90}{7}=\dfrac{540+x}{7}=x, so 540+x=7x540+x=7x and x=90.x=90.

In nondecreasing order the data are 40,50,60,90,90,100,200,40, 50, 60, 90, 90, 100, 200, so the median is 9090 and the mode is 90,90, as required.

Thus, the correct answer is D.

Problem 4 in Other Years