1991 AMC 8 Problem 7

Question

The value of

\tiny\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}

is closest to

10{,}000{,}000

100{,}000{,}000

1{,}000{,}000{,}000

10{,}000{,}000{,}000

100{,}000{,}000{,}000

The numerator shares a common factor of

487{,}000

Round each number to a single leading digit before dividing, for example

487{,}000 \approx 500{,}000

and

(19{,}367)(0.05) \approx 1000

Factor the numerator:

487{,}000

\cdot\, (12{,}027{,}300 + 9{,}621{,}001).

Rounding to leading digits, this is about

500{,}000

\times (10{,}000{,}000 + 10{,}000{,}000)

= 500{,}000 \times 2

\times 10^{7}.

The denominator is about

20{,}000 \times 0.05 = 1000.

So the value is roughly

\dfrac{500{,}000 \times 2 \times 10^{7}}{1000}

= 10^{10}

= 10{,}000{,}000{,}000.

Thus, the correct answer is D .

Problem 7 in other years