Imagine that a clam accumulates one unit of mercury in its body from the water it lives in. How many units of mercury would a stingray accumulate if it eats 50 clams a day for 10 years, assuming no mercury is lost between the clams and stingray?

182,500 units.

Explanation:

The answer would be 182,500 units of mercury.

Assuming there is an average of 365 days in a years, 10 years is therefore equal to:

10 x 365 = 3650 days.

The stingray eats 50 clams per day. Hence, the total number of clams that would have been eaten in 10 years equals:

3650 x 50 = 182,500 clams

Each clam accumulated 1 unit of mercury in its body from the water it lives in and there is no loss of mercury between the clam and the stingray. Hence, the total number of units of mercury the stingray would have accumulated in 10 years equals:

182,500 clams x 1 unit = 182,500 units of mercury.