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1991 AJHSME Problems/Problem 7

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Problem

The value of \frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)} is closest to

\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qqu...

Solution

We can make the approximations \begin{align*}487,000 &\approx 500,000 \\ 12,027,300 &\approx 12,000,000 \\9,621,001 &\approx 10,000,000 \\19,367...

Using these instead of the original numbers for an estimate, we have \begin{align*}\frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \...

See Also

1991 AJHSME (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
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
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