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 $[ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c} = [a,b,c]$.

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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