2012 AMC 10A Problem 8

Below is the professionally curated solution for Problem 8 of the 2012 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10A 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:system of equations

Difficulty rating: 1020

8.

The sums of three whole numbers taken in pairs are 12,17,12, 17, and 19.19. What is the middle number?

44

55

66

77

88

Solution:

Let the three numbers be a,b,ca, b, c where a<b<c.a \lt b \lt c. None of them are equal, since all three sums are different.

Then a+b=12,a+c=17, a + b = 12, a + c = 17, and b+c=19. b + c = 19.

Adding all three equations together gives us 2(a+b+c)=48. 2(a + b + c) = 48.

Then a+b+c=24, a + b + c = 24, from which we can subtract a+c=17, a + c = 17, to get b=7.b = 7.

Thus, D is the correct answer.

Problem 8 in Other Years