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Drill Rig
Task
A water well drilling rig has dug to a height of –60 feet after one full day of continuous use.
 Assuming the rig drilled at a constant rate, what was the height of the drill after 15 hours?
 If the rig has been running constantly and is currently at a height of –143.6 feet, for how long has the rig been running?
IM Commentary
The purpose of this task is to provide a context for multiplying and dividing signed rational numbers, providing a means for understanding why the signs behave the way they do when finding products. It is possible to solve this problem with or without negative numbers, depending on how the numbers are interpreted. If depths below the earth are interpreted as negative numbers (in other words, as negative height above the earth's surface), then this problem provides a good context for multiplying and dividing negative numbers. If the teacher wishes for students to use negative numbers, students can be encouraged to model the problem with a number line: the most natural way to do this is to put 0 at the surface of the earth and represent depths below the earth with negative numbers. This has been incorporated into the statement of the problem in order to encourage this approach.
This task complements the work students do with proportional relationships in grade 7 because the problem can be solved by reasoning with a proportional relationship, as shown in the first solution. For the first part of the task, students also need to make a conversion between days and hours. Because of the rate context (and signed numbers in the second solution) the teacher may wish to focus on setting up and understanding the problem, rather than on the arithmetic itself. In this case, use of calculators may be appropriate for this problem.
The information about the speed of drilling for a water rig is taken from the ''Cable tool drilling'' (section 7.4) of this Wikipedia article: http://en.wikipedia.org/wiki/Drilling_rig.
Solutions
Solution: 2 Arithmetic with positive numbers
 There are 24 hours in a day and so 15 hours is $15 \div 24 = \frac{5}{8}$ of a day. Since the rig drills 60 feet underground in a day, in $\frac{5}{8}$ of a day it will drill $$\left(\frac{5}{8} \text{ days}\right) \times \left(60 \frac{\text{feet}}{\text{day}}\right) = \frac{300}{8} \text{ feet}.$$ So the rig will have drilled $\frac{300}{8} = 37.5$ feet underground in 15 hours.
 The rig drills 60 feet underground per day so to find how long it has been running to go 143.6 feet underground, we need to calculate $$\left(143.6 \text{ feet} \right) \div \left(60 \frac{\text{feet}}{\text{day}}\right) = \frac{143.6}{60} \text{ days}.$$ This is about 2.4 days. Alternatively, it is 57.44 hours.
Solution: 1 Using a Proportional Relationship

We know that the rig drills at a constant rate, so there is a proportional relationship between the two quantities $d$, the height to which the drill has dug, and $t$, the number of days the drill runs. It drills at 60 feet per day, so we can represent this relationship with the equation: $$–60t = d$$ Since 15 hours is $\frac{15}{24} = \frac{5}{8}$ days, we can use the equation to find $d$: $$–60\cdot\frac58 = d$$ Since the depth is the same whether we think of it as a positive depth below the surface or a negative height above the surface, we can find this value by multiplying $60\cdot\frac58 = 37.5$ and then noting that the sign must be negative if we are representing positions below the surface of the earth by negative numbers. So $d = –37.5$ and the drill will be at height 37.5 feet after 15 hours.
 Likewise, if we know the drill has dug to 143.6 feet, we can use the equation again, this time to find $t$: $$60t = 143.6$$ so $t =( 143.6) \div (60)$. Since the amount of time would be the same if we were working with positive feet below the surface of the earth and a drill rate of 60 feet per day below the earth surface, we can find this value by dividing $143.6 \div 60$, which is about 2.4 days.
Solution: 3 Arithmetic with signed numbers
 We can measure the elevation below and above the ground with signed numbers, taking positive numbers to represent elevation above the ground and negative numbers to represent elevation below the ground. The rig drills below the ground at 60 feet per day so after 15 hours it will be at an elevation of $$\left(15 \text{ hours}\right) \times \left( 60 \frac{\text{feet}}{\text{day}}\right).$$ In order to calculate this, we need to convert hours to days. Since 15 hours is $\frac{15}{24} = \frac{5}{8}$ of day, after 15 hours the rig will be at an elevation of $$\left(\frac{5}{8} \text{ days} \right)\times \left(60 \frac{\text{feet}}{\text{day}}\right) = \frac{300}{8} \text{ feet}.$$ So the rig will have drilled $\frac{300}{8} = 37.5$ feet below the ground.
 Since the rig has drilled 143.6 feet under the ground, this is represented on our number line by $143.6$. It goes at a rate of 60 feet underground per day so this is $60 \frac{\text{ feet}}{\text{day}}$. So the rig has been running for $$\left(143.6 \text{ feet}\right) \div \left(60 \frac{\text{ feet}}{\text{day}}\right).$$ This is $\frac{143.6}{60}$ days. The rig has been running for a positive amount of time since it has drilled underground so this means that $\frac{143.6}{60} = \frac{143.6}{60}$ and this is the number of days the rig has been drilling. Alternatively, we will get the same answer if, instead of drilling into the ground, we are building up above the ground: so the two negative signs can be exchanged for two positive signs without changing the value of the answer.
Drill Rig
A water well drilling rig has dug to a height of –60 feet after one full day of continuous use.
 Assuming the rig drilled at a constant rate, what was the height of the drill after 15 hours?
 If the rig has been running constantly and is currently at a height of –143.6 feet, for how long has the rig been running?
Comments
Log in to commentAnn Shannon says:
almost 3 yearsI love this task and I would love it more if the commentary and solution were edited to reflect the specific language of 7. RP that talks about Proportional Relationships rather than Ratio and Proportion. I really like that "setting up and solving a proportion" is not specifically mentioned in 7.RP and I would love it if this solution and commentary did not either. If the solution path were to be developed like this: We know that the rig drills at a constant rate. (This signals that there is a proportional relationship between the two quantities "depth dug" and "digging time".) We are told that it drills at 60 feet per day. Therefore it drills at –2.5 feet per hour. From 7. RP 2(d) we will have learned how to set up the equation: –2.5h = d, where h represents number of hours and d represents depth dug below the ground in feet. We can solve, –2.5•15 = d to get –37.5 feet in 15 hours. Now, the next part is made really easy by still using –2.5h = d. When d is –143.6, we solve –2.5h = –143.6, to get h = 57.44 hours. This solution path reflects the intent of 7. RP 1 & 2 by focusing on the unit rate or constant of proportionality.
Kristin says:
almost 3 yearsThanks, Ann, this is an excellent suggestion. I've remodeled the commentary and solutions. Note that I decided to stay away from the conversion of feet per day to feet per hour, as this is a fairly sophisticated act of proportional reasoning itself and takes the focus away from the intended purpose, which is to use the context to help make sense of the sign rules for rational number arithmetic.
Let me know what you think of this version, and if you see ways to improve it.