1955 AMC 12 Problem 38

Attempt Problem 38 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

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38.

Four positive integers are given. Select any three of these integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers 29,29, 23,23, 2121 and 1717 are obtained. One of the original integers is:

1919

2121

2323

2929

1717

Answer: B
Concepts:linear systemaveragessum invariant
Difficulty rating: 1840
Small Hint:

If the original sum is SS and the singled-out integer is x,x, the result is S+2x3\frac{S+2x}{3}

Big Hint:

Sum all four reported results to determine SS

Solution:

If the original integers sum to S,S, the result associated with singled-out integer xx is x+Sx3=S+2x3. x+\frac{S-x}{3}=\frac{S+2x}{3}. Summing all four reported results counts the total as 2S,2S, so S=29+23+21+172=45. S=\frac{29+23+21+17}{2}=45. The result 2929 therefore comes from x=3(29)452=21.x=\dfrac{3(29)-45}{2}=21.

Thus, one original integer is 21,21, and the correct answer is B.

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