Difference between revisions of "1998 AHSME Problems/Problem 4"

Latest revision as of 13:28, 5 July 2013

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

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

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

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 $[1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}$ .

1998 AHSME (Problems • Answer Key • Resources)
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
All AHSME Problems and Solutions

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 == Solution ==
-Note that <math>[ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c}</math>. Thus <math>[60,30,90] = [2,1,3] = [10,5,15] = \frac{2+1}{3} = 1</math>, and <math>[1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}</math>.
+Note that <math>[ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c} = [a,b,c]</math>.
+Thus <math>[60,30,90] = [2,1,3] = [10,5,15] = \frac{2+1}{3} = 1</math>, and <math>[1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}</math>.
 == See also ==
@@ Line 13: / Line 15: @@
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}