Kelvin has a part-time job picking apples. Each day, he starts with an empty container, which can hold \(960\) apples, and puts the apples that he picks into the container. Kelvin is finished his work for the day if either \(3\) hours have passed or he has filled the container.
Kelvin picks \(400\) apples on the first day, and each day steadily improves the rate at which he picks apples by picking \(21\) more apples than the day before.
After how many days of working can he fill the container in less than \(3\) hours?
Kelvin picks \(400\) apples the first day, \(421\) apples the second day, \(442\) apples the third day, and so on, increasing each day by \(21\) apples. To find the first day he can completely fill the container in less than \(3\) hours, we need to determine the number of days it takes for him to increase his productivity by \(960-400=560\) apples.
Since, \(560 \div 21 \approx 26.6\), there are \(26\) days of increase in apple picking, but where Kelvin picks an amount that is still less than \(960\) apples. That is, on day \(1\) through day \(27\), Kelvin picks less than \(960\) apples. Then on day \(28\), Kelvin picks \(960\) or more extra apples. That is, on day \(28\) and thereafter, he can fill the container in less than \(3\) hours.
Indeed, we can check that on day \(27\) Kelvin picks \(21\times 26 = 546\) extra apples, for a total of \(400 + 546 = 946\) apples. On day \(28\), Kelvin picks \(21\times 27 = 567\) extra apples, for a total of \(400 + 567= 967\) apples in \(3\) hours.