1992 AJHSME Problems/Problem 6

Problem

Suppose that [asy]  unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$a$",(1,sqrt(3)-0.2),S); label("$b$",(sqrt(3)/10,0.1),ENE); label("$c$",(2-sqrt(3)/10,0.1),WNW); [/asy] means $a+b-c$. For example, [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$5$",(1,sqrt(3)-0.2),S); label("$4$",(sqrt(3)/10,0.1),ENE); label("$6$",(2-sqrt(3)/10,0.1),WNW); [/asy] is $5+4-6 = 3$. Then the sum [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$1$",(1,sqrt(3)-0.2),S); label("$3$",(sqrt(3)/10,0.1),ENE); label("$4$",(2-sqrt(3)/10,0.1),WNW); draw((3,0)--(5,0)--(4,sqrt(3))--cycle); label("$2$",(4,sqrt(3)-0.2),S); label("$5$",(3+sqrt(3)/10,0.1),ENE); label("$6$",(5-sqrt(3)/10,0.1),WNW); label("$+$",(2.5,-0.1),N); [/asy] is

$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution

The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$

Solving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$

Since we have to add the 2 triangles the final answer is $1$, which is $\boxed{D}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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
All AJHSME/AMC 8 Problems and Solutions

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