Problem A and Solution

Piling Wood

The Taylor family heats their home with wood. They recently got a large delivery of cut logs which were left in a pile on their driveway. Janelle and Alphonso carry logs from the pile and stack them in the woodshed. In one trip, Janelle can carry \(4\) logs at a time and Alphonso can carry \(3\) logs at a time. They each take the same amount of time to carry logs from the pile, stack them in the woodshed, and walk back to the pile.

If they each take as many logs as they can on each trip from the pile to the woodshed, how many logs have Janelle and Alphonso stacked after making \(10\) trips from the pile to the woodshed?

When Janelle and Alphonso started, there were \(200\) logs in the pile. How many trips from the pile to the woodshed do they have to take in order to move all the logs?

It takes approximately \(30\) seconds for Janelle and Alphonso to carry logs from the pile, stack them in the woodshed, and return to the pile. Approximately how many minutes does it take them to stack all of the logs in part (b)?

On the last trip, Janelle and Alphonso carry the same number of logs. If they each take as many logs as they can on their earlier trips, how many logs do they each carry on the last trip?

In total, Janelle and Alphonso carry \(4 + 3 = 7\) logs per trip from the log pile to the woodshed. This means they carry \(7 \times 10 = 70\) logs in \(10\) trips.

To figure out how many trips it takes to move all the logs, we could skip count by \(7\)s until we get to or pass \(200\): \[7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, \ldots\]

Then we would count how many \(7\)s it takes to get past \(200\) and we would see that it is \(29\).

Alternatively, we know from part (a) that after \(10\) trips, they will have moved \(70\) logs. So, after \(20\) trips they will have moved \(140\) logs, and after \(30\) trips they will have moved \(210\) logs. If we count backwards from \(210\), we can determine that after \(29\) trips they will have moved \(203\) logs, and after \(28\) trips they will have moved \(196\) logs. Therefore, we can conclude that it takes \(29\) trips to move all the logs.

Another way to calculate this is to divide \(200\) by \(7\) to obtain \(28\) remainder \(4\). Since there is a remainder after \(28\) trips, then it will take one more trip to move all the logs. Therefore, we can conclude that it takes \(29\) trips to move all the logs.

Since each trip takes approximately \(30\) seconds, then \(2\) trips take approximately \(1\) minute. This means that \(28\) trips would take approximately \(14\) minutes. Therefore, \(29\) trips would take approximately \(14\) and a half minutes.

After \(28\) trips, they have moved \(28 \times 7 = 196\) logs. This means there are \(200 - 196 = 4\) logs left for their last trip. Since Janelle and Alphonso carry the same number of logs on their last trip, then they each must carry \(2\) logs.