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Problem of the Week
Problem A and Solution
Cleaning Calculation

Problem

A housekeeper is responsible for cleaning all the rooms on one floor of a hotel. The floor has \(16\) regular rooms and \(5\) suites. Regular rooms take \(20\) minutes each to clean. Suites take \(30\) minutes each to clean.

  1. How long does it take to clean all the rooms on the floor of the hotel?

  2. If the housekeeper starts cleaning at \(10\):\(00\) a.m. and does not take a break, at what time is the job finished?

Solution

  1. We can calculate the time it takes to clean the regular rooms by skip counting by \(20\):

    \(20\), \(40\), \(60\), \(80\), \(100\), \(120\), \(140\), \(160\), \(180\), \(200\), \(220\), \(240\), \(260\), \(280\), \(300\), \(320\)

    We see that it takes a total of \(320\) minutes to clean the regular rooms.

    We can calculate the time it takes to clean the suites by skip counting by \(30\):

    \(30\), \(60\), \(90\), \(120\), \(150\)

    We see that it takes a total of \(150\) minutes to clean the suites. We add these two numbers together to see that it takes a total of \(320 + 150 = 470\) minutes to clean the entire floor.

    From this total number of minutes, we could calculate the result in hours and minutes. However, we could also recognize that when skip counting by \(20\), we get to \(60\) minutes after three \(20\)-minute intervals. This means it takes an hour to clean three regular rooms. We could count a bit differently:

    \(20\) min, \(40\) min, \(1\) hour, \(80\) min, \(100\) min, \(2\) hours, \(140\) min, \(160\) min, \(3\) hours, \(200\) min, \(220\) min, \(4\) hours, \(260\) min, \(280\) mins, \(5\) hours, \(320\) mins

    We can see that \(320\) minutes is equal to \(5\) hours and \(20\) minutes.

    Similarly, we get to \(60\) minutes after two \(30\)-minute intervals. With similar counting we see that \(150\) minutes is equal to \(2\) hours and \(30\) minutes.

    Now we can find the total time taken as: \[5~\text{hours} + 2~\text{hours} + 20~\text{minutes} + 30~\text{minutes}\] which is equal to \(7\) hours and \(50\) minutes.

  2. We can use a timeline to determine the time that is \(7\) hours and \(50\) minutes after \(10\):\(00\) a.m.

    The hours 10 am, 11 am, 12 pm, 1 pm, 2 pm, 3 pm, 4 pm, 5 pm, and 6 pm are plotted on a time line with 5 tick marks between each hour. An arrow is drawn from 10 am ending at the fifth tick mark between 5 pm and 6 pm with the length of 7 hours and 50 minutes indicated.

    Therefore, the housekeeper will finish the job at \(5\):\(50\) p.m.

Teacher’s Notes

Although we often think of rates as being related to distance and time, we can describe this problem in terms of rates. The units we use when describing rates is:

(some quantity)/(some unit of time)

For example, a car’s speed may be measured in km/hr or a computer’s download speed may be measured in MBits/sec.

When working with rates involving distance and time, we can consider this formula that shows the relationship between rate, distance, and time: \[\text{rate} = \dfrac{\text{distance}}{\text{time}}\]

Given any two of the values, we can use the formula to calculate the third. For example, if we know the rate and time, we can calculate the distance. Often we rearrange the formula to make it easier to calculate the missing value, where the missing value is isolated on one side of the equals sign. \[\text{rate} \times \text{time} = \text{distance}~~~~~~~\text{and}~~~~~~~\text{time} = \dfrac{\text{distance}}{\text{rate}}\]

For this problem, we can consider the following formula for the rate of cleaning: \[\text{rate} = \dfrac{\text{rooms}}{\text{time}}\]

For regular rooms, we know the rate is \(\dfrac{1~\text{room}}{20~\text{min}}\) and the number of rooms is \(16\).

We can rearrange the formula to isolate the time: \[\text{time} = \dfrac{\text{rooms}}{\text{rate}}\]

Now we can substitute the actual numbers into this formula: \[\text{time} = \dfrac{16~\text{rooms}}{\frac{1~\text{room}}{20~\text{mins}}}\]

To calculate this result we must divide by a fraction, which we can do by inverting the fraction and multiplying like this:

\[\text{time} = 16~\cancel{\text{rooms}} \times \dfrac{20~\text{mins}}{1~\cancel{\text{room}}}\] \[\text{time} = 320~\text{mins}\]