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2002 AMC 10A Problem 16

Problem 16 of 25IntermediateAlgebra

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

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Solution

Let the common value be k.k. Then a=k−1,a=k-1, b=k−2,b=k-2, c=k−3,c=k-3, d=k−4,d=k-4, so a+b+c+d=4k−10.a+b+c+d=4k-10. Since a+b+c+d+5=k,a+b+c+d+5=k, we get 4k−10+5=k,4k-10+5=k, so 3k=53k=5 and k=53.k=\dfrac{5}{3}. Then a+b+c+d=k−5a+b+c+d=k-5 =53−5=−103.=\dfrac{5}{3}-5=-\dfrac{10}{3}. Thus, the correct answer is B.

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Concepts: system of equations · substitution

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