Lesson 17

If you were babysitting, would you rather 

• Charge $5 for the first hour and $8 for each additional hour?

Or 

• Charge $15 for the first hour and $6 for each additional hour? 

Explain your reasoning. 

The amount of water in two tanks every 5 minutes is shown in the table.

time (minutes)	tank 1 (liters)	tank 2 (liters)
0	25	1000
5	175	900
10	325	800
15	475	700
20	625	600
25	775	500
30	925	400
35	1075	300
40	1225	200
45	1375	100
50	1525	0

1. Describe what is happening in each tank. Either draw a picture, say it verbally, or write a few sentences.
2. Use the table to estimate when the tanks will have the same amount of water.


3. The amount of water (in liters) in tank 1 after \(t\) minutes is \(30t + 25\). The amount of water (in liters) in tank 2 after \(t\) minutes is \( \text-20t + 1000\). Find the time when the amount of water will be equal.

1. What is the height of each elevator at this time?
2. How long would it take each elevator to reach ground level at this time?
3. If the two elevators travel toward one another, at what height do they pass each other? How long would it take?
4. If you are on an underground parking level 14 meters below ground, which elevator would reach you first?

Imagine a full 1,500 liter water tank that springs a leak, losing 2 liters per minute. We could represent the number of liters left in the tank with the expression \(\text-2x+1,\!500\), where \(x\) represents the number of minutes the tank has been leaking.

Now imagine at the same time, a second tank has 300 liters and is being filled at a rate of 6 liters per minute. We could represent the amount of water in liters in this second tank with the expression \(6x+300\), where \(x\) represents the number of minutes that have passed.

Since one tank is losing water and the other is gaining water, at some point they will have the same amount of water—but when? Asking when the two tanks have the same number of liters is the same as asking when \(\text-2x+1,\!500\) (the number of liters in the first tank after \(x\) minutes) is equal to \(6x+300\) (the number of liters in the second tank after \(x\) minutes),

\(\text-2x+1,\!500=6x+300.\)

Solving for \(x\) gives us \(x=150\) minutes. So after 150 minutes, the number of liters of the first tank is equal to the number of liters of the second tank. But how much water is actually in each tank at that time? Since both tanks have the same number of liters after 150 minutes, we could substitute \(x=150\) minutes into either expression.

Using the expression for the first tank, we get \(\text-2(150)+1,\!500\) which is equal to \(\text-300+1,\!500\), or 1,200 liters.

If we use the expression for the second tank, we get \(6(150)+300\), or just \(900+300\), which is also 1,200 liters. That means that after 150 minutes, each tank has 1,200 liters.