Mr. Frank has 1.03 kilograms of fertilizer for the plants in his nursery. He wants every plant to get 95 mg of fertilizer 4 times each year. What is the number of plants he could fertilize with that amount? How much fertilizer will he have left over?

Step-by-step explanation:

The amount each plant requires in a year is 4 × 0.095 g = 0.38 g. The amount on hand is 1.03 kg = 1,030 g.

Then the number of plants Mr. Frank can fertilize (for one year) is ...

  (1030 g)/(0.38 g/plant) = 2710 20/38 plants

That is 2710 plants with (20/38)×(0.38 g) = 0.20 g left over.

Comment on the solution

The usual consideration of "significant figures" in a problem like this would say there is no leftover. The amount of fertilizer is taken to be ±.005 kilograms, equivalent to ±13 plants. The amount per plant is taken to be ±.5 mg, a possible error of 0.53%, so another ±14 plants. In other words, we don't know the number of plants that can be fertilized any closer than about ±27.

Also, it seems quite unlikely that Mr Frank could make 2700 measurements of 95 milligrams with such precision that he would have exactly 200 mg left over--even if he actually started with exactly 1,030,000 mg of fertilizer.

Note, too, that the number of "plants" reported here is actually "plant·years." That is, Mr. Frank could fertilize 1355 plants for 2 years, or 542 plants for 5 years.

1/4 because there is 4 grades.

stops on an even number.