Task
This headline appeared in a newspaper.
Every day 7% of Americans eat at Giantburger restaurants

Decide whether this headline is true using the following information.
 There are about $8\times 10^3$ Giantburger restaurants in America.
 Each restaurant serves on average $2.5\times10^3$ people every day.
 There are about $3\times10^8$ Americans.
Explain your reasons and show clearly how you figured it out.
Solution
If there are about $8\times 10^3$ Giantburger restaurants in America and each restaurant serves about $2.5\times10^3$ people every day, then about
$$8\times 10^3 \cdot 2.5\times10^3 = 20 \times 10^6 = 2 \times 10^7$$
people eat at a Giantburger restaurant every day.
Since there are about $3\times10^8$ Americans, the percent of Americans who eat at a Giantburger restaurant every day can be computed by dividing the number of restaurant patrons by the total number of people:
$$2 \times 10^7\div 3\times10^8 = \frac23 \times 10^{1}$$
Since $$\frac23 \times 10^{1} = \frac23 \times \frac{1}{10} = \frac{2}{30} = \frac{1}{15} = 0.06\overline{6},$$ our estimate is that $6\frac23 \%$ of Americans eat a Giantburger restaurant every day, which is reasonably close to the claim in the newspaper.
Comments
Log in to commenth. patton says:
6 monthsJust to offer another solution path that my students chose: Find 7% of 3 x 10^8 > (7 x 10^2) x (3 x 10^8) = 21 x (10^6). So the claim is that 21 million Americans eat at Giantburger. The actual is (8 x 10^3) x (2.5 x 10^3) = 20 x 10^6. So 20 million Americans actually eat at Giantburger. A comparison of the actual to the claim will provide a solution. The actual is 1 million less, which opens up the task up for nice discussion about whether being 1 million off is a reasonable claim. When is it reasonable to be 1 million off? When is it unacceptable to be 1 million off?