Machine Production: Combined vs. Individual Work Rates

Work Rate Problems 9th-10th Grade
Problem
Working simultaneously at their respective constant rates, machines A and B produce 800 nails in x hours. Working alone at its constant rate, machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take machine B, working alone at its constant rate, to produce 800 nails?

What This Problem Teaches

  • How to set up equations using work rates (work per unit time)
  • Understanding that when machines work together, their rates add
  • Converting between rate and time using the relationship: Rate × Time = Work
  • Manipulating algebraic expressions with variables in denominators
  • Working backwards from combined performance to individual capability

Solution: Method 1 — The Rate Addition Approach

The key insight is that when machines work together, their production rates add. Let's define our unknown and work systematically.

Step 1 — Define the unknown variable

Let t = time (in hours) for machine B alone to produce 800 nails.

Step 2 — Express each machine's work rate

Work rate = Work completed ÷ Time taken

  • Machine A's rate: 800/y nails per hour
  • Machine B's rate: 800/t nails per hour
  • Combined rate: 800/x nails per hour

Step 3 — Set up the rate equation

When working together, the combined rate equals the sum of individual rates:

Combined rate = Machine A rate + Machine B rate
800/x = 800/y + 800/t

Step 4 — Solve for machine B's rate

Isolate the term containing t:

800/t = 800/x - 800/y

Step 5 — Find a common denominator

To subtract the fractions on the right side:

800/x - 800/y = 800 · (y - x)/(xy) = 800(y - x)/(xy)

So: 800/t = 800(y - x)/(xy)

Step 6 — Solve for t

Cross multiply to solve for t:

800 · xy = 800 · t · (y - x)
xy = t(y - x)
t = xy/(y - x)

Solution: Method 2 — The Unit Work Approach

Instead of tracking rates, let's think of "producing 800 nails" as 1 complete job and work with job completion rates.

Step 1 — Express rates as jobs per hour

  • Machine A completes 1 job in y hours, so its rate is 1/y jobs per hour
  • Machine B completes 1 job in t hours, so its rate is 1/t jobs per hour
  • Together they complete 1 job in x hours, so combined rate is 1/x jobs per hour

Step 2 — Set up the equation

1/x = 1/y + 1/t

Step 3 — Solve for 1/t

1/t = 1/x - 1/y = (y - x)/(xy)

Step 4 — Find t

Taking the reciprocal:

t = xy/(y - x)
Machine B alone takes xy/(y - x) hours to produce 800 nails.

Verification

Let's check our answer by substituting back into the original rate equation:

Machine A's rate:800/y nails per hour

Machine B's rate:800/(xy/(y-x)) = 800(y-x)/xy nails per hour

Combined rate:

800/y + 800(y-x)/xy = 800x/xy + 800(y-x)/xy = 800[x + (y-x)]/xy = 800y/xy = 800/x ✓

This matches the given combined rate of 800/x nails per hour, confirming our answer is correct.

Watch Out For These

✗ Mistake 1: Thinking that t = y - x
This assumes times subtract when working together, but rates add, not times. If machine A takes 6 hours alone and together they take 2 hours, machine B doesn't take 4 hours alone—it would actually take 3 hours (since 1/6 + 1/3 = 1/2).
✗ Mistake 2: Getting t = xy/(x - y) instead of xy/(y - x)
This sign error occurs when setting up the subtraction 1/x - 1/y. Remember: the combined rate (1/x) should be faster than any individual rate, so 1/x > 1/y, which means x < y.
✗ Mistake 3: Forgetting that the answer only makes sense when y > x
Physically, machine A alone (taking y hours) must be slower than both machines together (taking x hours). If y ≤ x, then machine B would need negative time, which is impossible.

The Pattern Behind This

This is a classic inverse rate combination problem. The general pattern for any two actors working together is:

If actor 1 takes time T₁ alone, actor 2 takes time T₂ alone,
and together they take time T_combined, then:

1/T_combined = 1/T₁ + 1/T₂

From this, we can derive that if we know the combined time and one individual time:

T₂ = (T₁ · T_combined)/(T₁ - T_combined)

In our problem, T₁ = y, T_combined = x, so T₂ = xy/(y - x).

This same mathematical structure appears in physics (parallel resistors), economics (production functions), and any situation where capabilities combine additively.

Real Applications

  • Manufacturing: Determining individual machine capacity when you know combined production line output
  • Computing: Calculating individual processor speeds when parallel processing performance is known
  • Team productivity: Estimating how long a team member would take solo based on team performance data

What If?

1
Different Work Amount
Working together, machines A and B produce 1200 nails in x hours. Working alone, machine A produces 1200 nails in y hours. In terms of x and y, how many hours does it take machine B to produce 1200 nails alone?
Step 1 — Set up with new work amount

Let t = time for machine B alone to produce 1200 nails. The work amount doesn't change the relationship structure.

Step 2 — Express rates

Machine A: 1200/y nails/hour, Machine B: 1200/t nails/hour, Combined: 1200/x nails/hour

Step 3 — Rate equation

1200/x = 1200/y + 1200/t

Step 4 — Solve for t

Divide by 1200: 1/x = 1/y + 1/t, so 1/t = 1/x - 1/y = (y-x)/(xy)

Answer

t = xy/(y-x) hours — the same formula regardless of work amount!

2
Three Machines
Machines A, B, and C together produce 800 nails in 2 hours. Machine A alone takes 8 hours, machine B alone takes 6 hours. How long would machine C alone take for 800 nails?
Step 1 — Find individual rates

A: 800/8 = 100 nails/hour, B: 800/6 = 133.33 nails/hour, Combined: 800/2 = 400 nails/hour

Step 2 — Find C's rate

C's rate = Combined rate - A's rate - B's rate = 400 - 100 - 133.33 = 166.67 nails/hour

Step 3 — Find C's time

Time = Work ÷ Rate = 800 ÷ 166.67 = 4.8 hours

Verification

Check: 100 + 133.33 + 166.67 = 400 ✓

Answer

Machine C alone takes 4.8 hours (4 hours 48 minutes)

3
Reverse Problem
Machine A produces 800 nails in 6 hours alone. Machine B produces 800 nails in 4 hours alone. How many hours do they take working together to produce 800 nails?
Step 1 — Find individual rates

Machine A: 800/6 = 133.33 nails/hour, Machine B: 800/4 = 200 nails/hour

Step 2 — Find combined rate

Combined rate = 133.33 + 200 = 333.33 nails/hour

Step 3 — Find combined time

Time = Work ÷ Rate = 800 ÷ 333.33 = 2.4 hours

Alternative using formula

Using 1/t = 1/6 + 1/4 = 2/12 + 3/12 = 5/12, so t = 12/5 = 2.4 hours

Answer

Working together: 2.4 hours (2 hours 24 minutes)

4
Staggered Start
Machine A works alone for 1 hour, then machine B joins. Together they finish producing 800 nails in a total of 3 hours from when A started. Machine A alone takes 5 hours for 800 nails. How long would B alone take for 800 nails?
Step 1 — Analyze the timeline

A works alone for 1 hour, then both work for 2 more hours (total 3 hours)

Step 2 — Calculate A's production

A's rate: 800/5 = 160 nails/hour. In 3 hours, A produces: 160 × 3 = 480 nails

Step 3 — Calculate B's production

B works for only 2 hours and produces: 800 - 480 = 320 nails

Step 4 — Find B's rate and time

B's rate: 320 ÷ 2 = 160 nails/hour. Time for 800 nails: 800 ÷ 160 = 5 hours

Answer

Machine B alone takes 5 hours — same as machine A!

Frequently Asked Questions

How do you combine work rates when machines work together? +
When machines work together, their rates add. If machine A produces 800 nails in y hours (rate = 800/y per hour) and together they produce 800 nails in x hours (combined rate = 800/x per hour), then machine B's rate = 800/x - 800/y per hour.
What's the formula for finding individual time from combined work rate? +
If the combined rate is C and one machine's rate is R₁, then the other machine's rate is R₂ = C - R₁. To find time, use Time = Work ÷ Rate. In this problem, machine B's time = 800 ÷ (800/x - 800/y) = xy/(y-x).
Why can't you just subtract the times x and y directly? +
Because rates add, not times. If one machine takes 4 hours and together they take 2 hours, the other machine doesn't take 2 hours alone—it takes 4 hours (since 1/4 + 1/4 = 1/2, so together they have rate 1/2 jobs per hour).
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-08

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