If 20 assemblers can complete a certain job in 9 ​hours, how long will the same job take if the number of assemblers is cut back to 15​?

This is a classic work problem that involves inverse proportionality. The time taken to complete a job is inversely proportional to the number of assemblers working. Here's the step-by-step solution:

### Step 1: Define the relationship

The number of assemblers \(A\) and the time \(T\) to complete the job are related by the formula:

\[
A_1 \times T_1 = A_2 \times T_2
\]

Where:
- \(A_1 = 20\) assemblers
- \(T_1 = 9\) hours
- \(A_2 = 15\) assemblers
- \(T_2 = ?\) (the time we are solving for)

### Step 2: Substitute the known values

\[
20 \times 9 = 15 \times T_2
\]

### Step 3: Solve for \(T_2\)

\[
T_2 = \frac{20 \times 9}{15} = \frac{180}{15} = 12 \, \text{hours}
\]

So, if the number of assemblers is cut back to 15, the job will take 12 hours to complete.

Would you like more details or have any questions?

Here are 5 related questions to expand this problem:
1. What if the number of assemblers is increased to 25? How long will the job take?
2. How many assemblers are needed to complete the job in 6 hours?
3. What would happen if only 10 assemblers were available? How long would the job take?
4. If 30 assemblers can complete the job in 6 hours, how long will it take for 18 assemblers?
5. How is the total work calculated in such problems?

**Tip**: In inverse proportion problems, multiplying the number of workers by the time should always give a constant value, which is the total work.

If 20 assemblers can complete a certain job in 9 hours, how long will the same job take if the number of assemblers is cut back to 15?

Discover how long it takes for 15 assemblers to complete a job when 20 assemblers can finish the same job in 9 hours. This work-time problem demonstrates inverse proportionality and the relationship between workers and time. Suitable for students in Grades 6-8.

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