Question 54·Hard·Ratios, Rates, Proportional Relationships, and Units
A water tank is initially full. Pump A can drain the full water tank in 10 hours when it works alone at a constant rate. Pump B also drains water at a constant rate, but its draining time for the full tank is unknown.
Both pumps are turned on at the same time. After exactly 3 hours, Pump B is turned off, and Pump A continues working by itself. The tank becomes completely empty 6 hours after Pump B is turned off.
Working alone at its constant rate, how many hours would Pump B need to drain the full tank?
(Express the answer as an integer)
For work and rate problems, always start by treating the entire job (here, the full tank) as 1 unit, then convert each time into a rate using rate = work ÷ time. For multi-stage situations (some time with two pumps, some with one), compute how much of the job is done in each stage using work = rate × time, and remember that the total work across all stages must add up to 1. Write an equation that reflects the correct stage(s) for the unknown rate, then solve for the variable—this is usually faster and safer than trying to reason with proportions or averages in your head.
Hints
Turn times into rates
Think of the whole tank as 1 unit. If Pump A empties the tank in 10 hours, what is Pump A’s rate (fraction of the tank per hour)? Then represent Pump B’s unknown rate using a variable.
Use the last 6 hours of Pump A working alone
During the final 6 hours, only Pump A is running. How much of the tank does Pump A empty in those 6 hours at its constant rate? What fraction of the tank must have been emptied before those 6 hours?
Focus on the first 3 hours with both pumps
In the first 3 hours, both pumps are on. Write an expression for how much of the tank they empty together in 3 hours in terms of Pump A’s rate and Pump B’s rate.
Set up and solve the equation
Set your expression for the amount emptied in the first 3 hours equal to the fraction of the tank that had to be emptied in those 3 hours. Then solve the equation for Pump B’s time to empty the whole tank alone.
Desmos Guide
Express the amount emptied in the first 3 hours
In Desmos, enter the function:
f(x) = 3*(1/10 + 1/x)
This represents the fraction of the tank emptied during the first 3 hours when both pumps are on, in terms of Pump B’s time to empty a full tank alone.
Graph the target fraction and find the intersection
From the problem, the first 3 hours must empty of the tank. In Desmos, on a new line enter:
y = 2/5
Also enter:
y = f(x)
Zoom or adjust the viewing window so you can see where the two graphs intersect for positive . The x-coordinate of that intersection is the number of hours Pump B would need to empty the full tank alone.
Step-by-step Explanation
Define the rates of the pumps
Treat the full tank as 1 unit of work.
- Pump A empties the tank in 10 hours, so its rate is tank per hour.
- Let be the number of hours Pump B needs to empty the full tank when working alone. Then B’s rate is tank per hour.
Use the last 6 hours with Pump A alone
After Pump B is turned off, Pump A works alone for 6 more hours until the tank is empty.
- In those 6 hours, Pump A empties
of the tank.
- That means the first 3 hours (when both pumps were on) must have emptied the remaining part of the tank:
So together, in the first 3 hours, the pumps emptied of the tank.
Write an equation for the first 3 hours with both pumps
When both pumps are on, their combined rate is
In 3 hours, they empty
of the tank.
From Step 2, we know this must equal , so the equation is
Solve the equation for Pump B’s time
Start from
Distribute the 3:
Rewrite as to match denominators:
Subtract from both sides:
Now solve for :
So, working alone at its constant rate, Pump B would need 30 hours to drain the full tank.