2002 AMC 10A Problem 16 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10A Problem 16

Below is the professionally curated solution for Problem 16 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 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 substitution

Difficulty rating: 1330

16.

If $a+1=b+2=c+3=d+4=a+b+c+d+5,$ then $a+b+c+d$ is

$-5$

$-\dfrac{10}{3}$

$-\dfrac{7}{3}$

$\dfrac{5}{3}$

$5$

Solution:

Let the common value be $k.$ Then $a=k-1,$ $b=k-2,$ $c=k-3,$ $d=k-4,$ so $a+b+c+d=4k-10.$

Since $a+b+c+d+5=k,$ we get $4k-10+5=k,$ so $3k=5$ and $k=\dfrac{5}{3}.$ Then $a+b+c+d=k-5=\dfrac{5}{3}-5=-\dfrac{10}{3}.$

Thus, the correct answer is B.

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