## Task

According to an article in Runners' World magazine:

On average the human body is more than 50 percent water [by weight]. Runners and other endurance athletes average around 60 percent. This equals about 120 soda cans' worth of water in a 160-pound runner!

Investigate their calculation. Approximately how many soda cansâ€™ worth of water are in the body of a 160-pound runner? What unprovided information do you need to answer this question?

## IM Commentary

This task provides students with an opportunity to engage in Standard for Mathematical Practice 6, attending to precision. It intentionally omits some relevant information -- namely, that a typical soda can holds 12 oz of fluid, that a pound is equivalent to 16 dry ounces, and that an ounce of water weighs approximately 1.04 dry ounces (at the temperature of the human body) -- in the interest of having students discover that these are relevant quantities. The incompleteness of the problem statement makes the task more amenable to having students do work in groups.

## Solution

According to the $60\%$ figure, a 160-pound runner contains about $.6\times 160=96$ pounds of water, or $96\times 16=1536$ ounces. In turn, 1536 ounces of water occupies a volume of $1536\text{ oz}\cdot\frac{1\text{ fl. oz}}{1.043\text{ oz}}\approx 1473$ fluid ounces. Finally, dividing these 1473 ounces into 12-ounce cans will require apprix $1473\div 12\approx 123$ cans.

Alternatively, we can write down a single equation encapsulating all of the above steps, using units to guide our way from the runners weight in pounds to the number of soda cans: $$ 160\text{ lbs}\times\frac{0.6\text{ lbs water}}{\text{ lb }}\times \frac{16\text{ oz}}{\text{ lb water}}\times\frac{1\text{ fl. oz}}{1.043\text{ oz}}\times\frac{1\text{ can}}{12\text{ fl. oz}}\approx 123 \text{ cans.} $$

Given the approximations made in the problem (most notably, the very approximate 60%), it is reasonable to report an answer only to the nearest multiple of 10, and report a final answer of $\boxed{120\text{ cans,}}$ leading us to concur with the magazine's calculation.

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