The mass of a standard carving pumpkin is approximately \(12\) kg. Lavina plans to sell the pumpkins she has grown at the farmer’s market. The table she has to display the pumpkins can support \(224\) kg. If the mass of each of her pumpkins is \(12\) kg, what is the largest number of pumpkins that Lavina can put on her table?
We can make a table to calculate the total mass of various quantities of pumpkins.
| Number of Pumpkins | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Total Mass (in kg) | \(12\) | \(24\) | \(36\) | \(48\) | \(60\) | \(72\) | \(84\) | \(96\) | \(108\) | \(120\) |
| Number of Pumpkins | \(11\) | \(12\) | \(13\) | \(14\) | \(15\) | \(16\) | \(17\) | \(18\) | \(19\) |
|---|---|---|---|---|---|---|---|---|---|
| Total Mass (in kg) | \(132\) | \(144\) | \(156\) | \(168\) | \(180\) | \(192\) | \(204\) | \(216\) | \(228\) |
The total mass of \(19\) pumpkins, which is \(228\) kg, exceeds the capacity of the table. So the largest number of pumpkins Lavina can fit safely on her table is \(18\).
Having to make a table counting from \(1\) to \(19\) takes quite a bit of work. Alternatively, we could try to reduce the work by narrowing the search area. We can use easier numbers such as multiples of \(10\) to find a narrower range to check. We see that \(10 \times 12 = 120\) and \(20 \times 12 = 240\). From this we know that the answer must be between \(10\) and \(20\) pumpkins. So instead of starting our table with \(1\) pumpkin, we could start it with \(10\) pumpkins.
We might also notice that the number we are looking for (\(224\) kg) is much closer to \(240\) than \(120\). So rather than counting up, we could count down from \(240\) in a table.
| Number of Pumpkins | \(20\) | \(19\) | \(18\) |
|---|---|---|---|
| Total Mass (in kg) | \(240\) | \(228\) | \(216\) |
Again, from this result we can conclude that the largest number of pumpkins she can put on her table is \(18\).
Teacher’s Notes
In this problem we said that the pumpkins all had the same mass of \(12\) kg. However, in reality, we would not expect each of the pumpkins to have exactly the same mass. A more realistic statement would be that the mass of each pumpkin is approximately \(12\) kg, but that makes the problem a bit trickier. If we had rounded to the nearest kilogram, that means the mass of each pumpkin could be greater than or equal to \(11.5\) kg and less than \(12.5\) kg.
Let’s assume that all the pumpkins have a mass of no more than \(11.6\) kg, which is still approximately \(12\) kg. In this case we see that \(11.6 \times 19 = 220.4\) kg, so we could fit \(19\) pumpkins on the table.
Let’s assume that all the pumpkins have a mass of no less than \(12.45\) kg, which is still approximately \(12\) kg. In this case we see that \(12.45 \times 18 = 224.1\) kg, which is more than the capacity of the table.
Sometimes we need to recognize a margin of error to reflect that our physical world does not always fit into our nice and neat world of mathematical problems.