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1998 AHSME Problems/Problem 4

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Problem

Define [a,b,c] to mean \frac {a+b}c, where c \neq 0. What is the value of

\left[[60,30,90],[2,1,3],[10,5,15]\right]?

\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.5 \qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ }1.5 \qquad \mathrm{(E) \ }2

Solution

Note that [ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c}. Thus [60,30,90] = [2,1,3] = [10,5,15] = \frac{2+1}{3} = 1, and [1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}.

See also

1998 AHSME (Problems)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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