Bigger learning in the end m = 2g
m + g = 4h
=> 3g = 4h
w + 2g = 2h
2m + 2g = 2h
=> w = 2m = 4g
w + m + g
= 4g + 2g + g
= 7g
3g -> 4h
7g -> 12/7 h
Doubt - I am always confused as to for hours when to use 1/h vs just h as done in this solution
This is a very important distinction.
There are two valid languages for work problems.
Language 1: Rate language (uses 1/h)
If a worker takes h hours to do 1 job, then:
Rate = 1/h
Example:
Man + Girl take 4 hours.
Then:
M + G = 1/4
jobs per hour.
This is the algebraic/rate approach.
Language 2: Time language (uses just h)
Instead of converting to rates, you compare groups that finish the same job in the same time.
Example:
Woman + Man → 2 hours
Woman + 2 Girls → 2 hours
Since both groups take exactly 2 hours, they must have equal total productivity.
So:
1m = 2g
No 1/2 appears anywhere.
You're reasoning with equivalent workers, not rates.
Why the solution never writes 1/4
When it says:
1m + 1g → 4h
it is not saying:
m + g = 4
It is simply recording a fact:
"This group needs 4 hours."
Then it replaces m by 2g:
3g → 4h
meaning:
"Three girls need 4 hours."
Still no rates.
When MUST you use 1/h?
Use 1/h when you want to:
- Add rates
- Subtract rates
- Solve equations
Example:
A takes 6 hours.
B takes 12 hours.
You cannot write:
6 + 12
Instead:
A = 1/6
B = 1/12
A + B = 1/4
So together they take 4 hours.
Whenever workers are being combined mathematically, use rates.
When can you avoid 1/h?
When you're only making equivalence statements:
- Same work
- Same time
- Therefore same productivity
Example:
W + M → 2h
W + 2G → 2h
Same job. Same time.
Therefore:
M = 2G
No rates needed.
Quick rule
Ask yourself:
"Am I adding/subtracting rates?"
Yes → use rates (1/h).
No → if I'm only comparing groups that finish the same work in the same time, use worker-equivalence.
In this problem, the elegant solution never needed to add rates, so it stayed entirely in the worker-equivalence world and never converted hours into 1/h.
That's why you saw 4h instead of 1/4.