In a computer game, there are builders and destroyers. Together there are 20 of them. Some of them try to build a wall around a castle while the rest try to demolish it. Each of the builders can build the wall alone in 15 hours while any of the destroyers can demolish it in 10 hours. If all 20 builders and destroyers are made active when there is no wall and the wall get built in 3 hours, how many of them are destroyers?
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Solving the Computer Game Builder Destroyer Problem
This problem involves understanding work rates and how they combine when some entities are building (positive rate) and others are demolishing (negative rate).
Let's break down the problem:
- Total number of individuals (builders + destroyers) = 20
- Time for one builder to build the wall alone = 15 hours
- Time for one destroyer to demolish the wall alone = 10 hours
- Time taken for all 20 together to build the wall = 3 hours
Understanding Work Rates
The work rate is the amount of work done per unit of time. In this case, the 'work' is building or demolishing one wall.
- The rate of one builder is \( \frac{1}{15} \) of the wall per hour.
- The rate of one destroyer is \( \frac{1}{10} \) of the wall per hour. Since destroyers demolish, their contribution is negative. So, the rate of one destroyer is \( -\frac{1}{10} \) of the wall per hour.
Setting Up the Equations
Let \( b \) be the number of builders and \( d \) be the number of destroyers.
We know the total number of individuals is 20:
\( b + d = 20 \quad (Equation\ 1) \)
The combined work rate of all 20 individuals is the sum of the rates of all builders and all destroyers.
- Total rate of \( b \) builders = \( b \times \frac{1}{15} = \frac{b}{15} \)
- Total rate of \( d \) destroyers = \( d \times \left(-\frac{1}{10}\right) = -\frac{d}{10} \)
The net combined rate is \( \frac{b}{15} - \frac{d}{10} \). This is the fraction of the wall built per hour when all 20 are active.
We are told that the wall is built in 3 hours by all 20 working together. The total work done is 1 (representing one wall). The formula Work = Rate × Time applies here.
\( \left(\frac{b}{15} - \frac{d}{10}\right) \times 3 = 1 \quad (Equation\ 2) \)
Solving for the Number of Destroyers
We have two equations with two variables, \( b \) and \( d \). We want to find \( d \).
From Equation 1, we can express \( b \) in terms of \( d \):
\( b = 20 - d \)
Substitute this expression for \( b \) into Equation 2:
\( \left(\frac{20 - d}{15} - \frac{d}{10}\right) \times 3 = 1 \)
Divide both sides by 3:
\( \frac{20 - d}{15} - \frac{d}{10} = \frac{1}{3} \)
To eliminate the denominators (15, 10, and 3), we find their least common multiple, which is 30. Multiply every term by 30:
\( 30 \times \frac{20 - d}{15} - 30 \times \frac{d}{10} = 30 \times \frac{1}{3} \)
\( 2(20 - d) - 3d = 10 \)
Distribute and simplify:
\( 40 - 2d - 3d = 10 \)
\( 40 - 5d = 10 \)
Subtract 40 from both sides:
\( -5d = 10 - 40 \)
\( -5d = -30 \)
Divide by -5:
\( d = \frac{-30}{-5} \)
\( d = 6 \)
So, there are 6 destroyers.
If there are 6 destroyers, then the number of builders is \( b = 20 - 6 = 14 \).
Verification
Let's check if 14 builders and 6 destroyers can build the wall in 3 hours.
- Rate of 14 builders = \( 14 \times \frac{1}{15} = \frac{14}{15} \)
- Rate of 6 destroyers = \( 6 \times \left(-\frac{1}{10}\right) = -\frac{6}{10} = -\frac{3}{5} \)
Combined rate = \( \frac{14}{15} - \frac{3}{5} \)
Find a common denominator (15):
\( \frac{14}{15} - \frac{3 \times 3}{5 \times 3} = \frac{14}{15} - \frac{9}{15} = \frac{14 - 9}{15} = \frac{5}{15} = \frac{1}{3} \)
The combined rate is \( \frac{1}{3} \) of the wall per hour.
Time taken = Work / Rate = \( 1 / \left(\frac{1}{3}\right) = 1 \times 3 = 3 \) hours.
This matches the information given in the problem. Therefore, the number of destroyers is correct.
| Item | Description | Value/Expression |
|---|---|---|
| Total individuals | Builders + Destroyers | 20 |
| Builder Rate | Work per hour per builder | \( \frac{1}{15} \) |
| Destroyer Rate | Work per hour per destroyer (demolish) | \( -\frac{1}{10} \) |
| Time to build wall | All working together | 3 hours |
| Number of Builders | Let it be | \( b \) |
| Number of Destroyers | Let it be | \( d \) |
| Total People Equation | \( b + d = 20 \) | \( b = 20 - d \) |
| Combined Rate Equation | \( \left(\frac{b}{15} - \frac{d}{10}\right) \times 3 = 1 \) | \( \frac{b}{15} - \frac{d}{10} = \frac{1}{3} \) |
Conclusion
Based on the calculations, the number of destroyers is 6.
Revision Table: Key Concepts Review
| Concept | Explanation | How it applies here |
|---|---|---|
| Work Rate | The amount of work done per unit of time. It's usually represented as \( \frac{1}{\text{Time Taken}} \). | Builder rate \( = \frac{1}{15} \), Destroyer rate \( = -\frac{1}{10} \). |
| Combined Work Rate | The sum of individual work rates. If some entities undo work, their rates are negative. | Net rate \( = (\text{Number of Builders} \times \text{Builder Rate}) + (\text{Number of Destroyers} \times \text{Destroyer Rate}) \). |
| Work Done Formula | Work = Rate \( \times \) Time. If the total work is building one item, Work = 1. | \( (\text{Net Combined Rate}) \times 3\ \text{hours} = 1\ \text{wall} \). |
| Solving System of Equations | Using two or more equations with multiple variables to find the values of the variables. Substitution or elimination methods can be used. | We used substitution, solving for \( b \) in the first equation and plugging it into the second. |
Additional Information on Work Problems
Work problems often involve calculating rates and combining them. Here are some common scenarios and tips:
- Individuals Working Together: If two people can complete a job in \( T_1 \) and \( T_2 \) hours respectively, their combined rate is \( \frac{1}{T_1} + \frac{1}{T_2} \). The time they take together is \( \frac{1}{\text{Combined Rate}} \).
- Working Against Each Other: Like in this problem, if one entity builds and another destroys, subtract the rates. The net rate is positive if the building rate is greater than the demolishing rate, leading to completion of the positive work.
- Multiple People/Machines: If there are \( N \) identical workers, and one worker takes \( T \) time, the rate of one worker is \( \frac{1}{T} \). The rate of \( N \) workers is \( N \times \frac{1}{T} \).
- Setting up Equations: Always define your variables clearly (e.g., number of builders, number of destroyers). Write down the total number equation and the work/rate/time equation based on the information given.
These types of word problems require careful reading to identify the rates and how they interact, then setting up algebraic equations to solve for the unknown quantity.
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