A company has two types of machines, type R and type S. Operating at a constant rate, a machine of...

1. Translate the problem requirements: A type R machine completes a job in 36 hours, a type S machine completes the same job in 18 hours. We need to find how many type R machines are used when equal numbers of R and S machines complete the job together in 2 hours.
2. Determine individual machine work rates: Calculate what fraction of the job each machine type completes per hour to establish their relative speeds.
3. Set up the combined work equation: Express how the total work rate of all machines working together equals the job completion rate needed (1 job in 2 hours).
4. Solve for the number of machines: Use the combined work rate equation to find how many machines of each type are required.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in everyday terms:

• A type R machine is slower - it needs 36 hours to complete one entire job by itself
• A type S machine is faster - it needs only 18 hours to complete the same job by itself
• The company uses the same number of R machines as S machines
• Working together, all these machines complete the job in just 2 hours
• We need to find how many R machines were used

Think of it like this: if you had workers of different speeds working together on a project, the more workers you have, the faster the project gets done. We need to figure out how many of each type of worker (machine) we need.

2. Determine individual machine work rates

Let's think about how much work each machine can do per hour:

Type R machine: If it takes 36 hours to do the whole job, then in 1 hour it does \(\frac{1}{36}\) of the job. Imagine the job is like painting a house - in one hour, the R machine paints \(\frac{1}{36}\) of the house.

Type S machine: If it takes 18 hours to do the whole job, then in 1 hour it does \(\frac{1}{18}\) of the job. The S machine paints \(\frac{1}{18}\) of the house in one hour.

Notice that \(\frac{1}{18}\) is twice as big as \(\frac{1}{36}\), which makes sense because the S machine is twice as fast (18 hours vs 36 hours).

In mathematical terms:

• Rate of one R machine = \(\frac{1}{36}\) job per hour
• Rate of one S machine = \(\frac{1}{18}\) job per hour

3. Set up the combined work equation

Now let's say we use n machines of each type (n R machines and n S machines).

Think about it this way: if one R machine does \(\frac{1}{36}\) of the job per hour, then n R machines working together do \(\mathrm{n} \times \frac{1}{36} = \frac{\mathrm{n}}{36}\) of the job per hour.

Similarly, n S machines working together do \(\mathrm{n} \times \frac{1}{18} = \frac{\mathrm{n}}{18}\) of the job per hour.

When all machines work together, their work rates add up:

Total work rate = (work rate of all R machines) + (work rate of all S machines)

Total work rate = \(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18}\) jobs per hour

Since they complete the entire job in 2 hours, their combined rate must be \(\frac{1}{2}\) job per hour.

This gives us the equation: \(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18} = \frac{1}{2}\)

4. Solve for the number of machines

Let's solve \(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18} = \frac{1}{2}\) step by step.

First, let's find a common denominator for the fractions. The least common multiple of 36 and 18 is 36:

\(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18} = \frac{\mathrm{n}}{36} + \frac{2\mathrm{n}}{36} = \frac{3\mathrm{n}}{36} = \frac{\mathrm{n}}{12}\)

So our equation becomes: \(\frac{\mathrm{n}}{12} = \frac{1}{2}\)

To solve for n, multiply both sides by 12:

\(\mathrm{n} = 12 \times \frac{1}{2} = 6\)

Let's verify this makes sense: With 6 machines of each type working for 2 hours:

• 6 R machines complete: \(6 \times \frac{1}{36} \times 2 = \frac{12}{36} = \frac{1}{3}\) of the job
• 6 S machines complete: \(6 \times \frac{1}{18} \times 2 = \frac{12}{18} = \frac{2}{3}\) of the job
• Total work done: \(\frac{1}{3} + \frac{2}{3} = 1\) complete job ✓

5. Final Answer

The company used 6 machines of type R.

Looking at our answer choices: A. 3, B. 4, C. 6, D. 9, E. 12

Our answer matches choice C. 6

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "same number of each type" means: Students might think this means equal total capacity rather than equal count of machines. They could mistakenly try to balance the work contributions equally (making R machines do half the work and S machines do half the work) instead of using the same number of each machine type.

2. Confusing individual time with rate: Students often struggle to convert from "time to complete job" to "rate of work per hour." They might incorrectly use 36 and 18 directly in calculations instead of taking their reciprocals (\(\frac{1}{36}\) and \(\frac{1}{18}\)) to get the proper work rates.

3. Setting up incorrect equation for combined work: Students may forget that when multiple machines work together, their individual rates add up. They might try to average the rates or use some other incorrect combination method instead of simply adding \(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18}\).

Errors while executing the approach

1. Common denominator errors: When combining fractions \(\frac{\mathrm{n}}{36} + \frac{\mathrm{n}}{18}\), students often make mistakes finding the least common multiple. They might use 54 (36 × 18 ÷ 12) instead of 36, or incorrectly convert \(\frac{\mathrm{n}}{18}\) to \(\frac{3\mathrm{n}}{36}\) instead of \(\frac{2\mathrm{n}}{36}\).

2. Misinterpreting the time constraint: Students might set up the equation as if the total rate equals 2 (thinking 2 hours means rate = 2) instead of \(\frac{1}{2}\) (meaning the combined rate must be \(\frac{1}{2}\) job per hour to complete 1 job in 2 hours).

3. Algebraic manipulation errors: When solving \(\frac{\mathrm{n}}{12} = \frac{1}{2}\), students might incorrectly multiply or divide, getting n = 24 (multiplying by 24 instead of 12) or n = \(\frac{1}{24}\) (dividing by 12 instead of multiplying).

Errors while selecting the answer

1. Verification confusion: Students might correctly solve for n = 6 but then doubt their answer during verification. When checking that 6R + 6S machines complete the job in 2 hours, they might make arithmetic errors in the verification step and incorrectly conclude their answer is wrong.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient job size

Instead of using "1 job" as an abstract unit, let's set the total job size to 36 units of work. This number was chosen because it's the LCM of 36 and 18, ensuring all our rate calculations result in whole numbers.

Step 2: Calculate work rates per hour

• Type R machine rate: 36 units ÷ 36 hours = 1 unit per hour
• Type S machine rate: 36 units ÷ 18 hours = 2 units per hour

Step 3: Determine required total rate

To complete 36 units of work in 2 hours: \(36 ÷ 2 = 18\) units per hour needed from all machines combined

Step 4: Set up equation with equal numbers of machines

Let n = number of each type of machine
Total rate = n(1) + n(2) = 3n units per hour
We need: \(3\mathrm{n} = 18\)
Therefore: n = 6

Step 5: Verify the answer

6 type R machines: 6 × 1 = 6 units per hour
6 type S machines: 6 × 2 = 12 units per hour
Combined rate: 18 units per hour
Time to complete 36 units: \(36 ÷ 18 = 2\) hours ✓

The answer is 6 type R machines.