MathDB

29

Part of 1950 AMC 12/AHSME

Problems(1)

AHSME 1950- part 2

Source:

7/20/2008
A manufacturer built a machine which will address 500500 envelopes in 88 minutes. He wishes to build another machine so that when both are operating together they will address 500500 envelopes in 22 minutes. The equation used to find how many minutes xx it would require the second machine to address 500500 envelopes alone is:
<spanclass=′latex−bold′>(A)</span> 8−x=2<spanclass=′latex−bold′>(B)</span> 18+1x=12<spanclass=′latex−bold′>(C)</span> 5008+500x=500<spanclass=′latex−bold′>(D)</span> x2+x8=1<spanclass=′latex−bold′>(E)</span> None of these answers<span class='latex-bold'>(A)</span>\ 8-x=2 \qquad <span class='latex-bold'>(B)</span>\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2} \qquad <span class='latex-bold'>(C)</span>\ \dfrac{500}{8}+\dfrac{500}{x}=500 \qquad <span class='latex-bold'>(D)</span>\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\ <span class='latex-bold'>(E)</span>\ \text{None of these answers}