Solve Work Rate Problem: 6 Men Colouring 36m Cloth in x Days

[NotaMathPerson]

if 4 men can colour 48 m of cloth in 2 days, in how many days can 6 men colour a 36 m cloth?

Can you give me a general method on solving this problem? Thanks!

 

[MarkFL]

I would solve this by computing how many m. of cloth 1 man colors in 1 day...

 

[NotaMathPerson]

I would solve this by computing how many m. of cloth 1 man colors in 1 day...

I found that one man can do 6m of cloth a day. By 48/(8) =6

If one man can do 6m a day then 6 men can do 36m in one day.

But I want some formula or something that would generalize the process. I having difficulty doing that.

 

[NotaMathPerson]

Hello guys!

As I was searching on the net for a general formula to solve this problem I stubled upon this

M1*H1*D1*E1*W2=M2*H2*D2*E2*W1

Where M = # of men; D = # of days; H = # of hours ;W = # something done; E = efficiency.

Now if I will relate it to the problem above

M1 = 4 men, W1 = 48 m, D1 = 2 days.
M2 = 6, W2 = 36m, D2 = ? days.

From here it is only a matter plugging in the given to get 1 day as an answer.

Now my questions are

1. What do we call the value when we multiply out both sides of M1*H1*D1*E1*W2=M2*H2*D2*E2*W1(what is the unit or name associated to it?)

2. Why the W2 is on the left hand side? Same with W1 that is on right hand side of the eqn.

3. Why do we equate M1*H1*D1*E1*W2=M2*H2*D2*E2*W1?

Please I need to learn this. Kindly explain. Thanks!

 

[MarkFL]

The amount of work that gets done can be assumed to be proportional to the number of men working and the length of time they work:

\(\displaystyle W_n=kM_nt_n\)

Thus, given two groups of men working (presumably all at the same rate), we may state:

\(\displaystyle k=\frac{W_1}{M_1t_1}=\frac{W_2}{M_2t_2}\implies W_1M_2t_2=W_2M_1t_1\)

Both sides would have units of "man-days of work."

From this, we may write:

\(\displaystyle t_2=\frac{W_2M_1t_1}{W_1M_2}\)

Plugging in the values from the original problem, we find:

\(\displaystyle t_2=\frac{36\cdot4\cdot2}{48\cdot6}=1\)

 

[NotaMathPerson]

The amount of work that gets done can be assumed to be proportional to the number of men working and the length of time they work:

\(\displaystyle W_n=kM_nt_n\)

Thus, given two groups of men working (presumably all at the same rate), we may state:

\(\displaystyle k=\frac{W_1}{M_1t_1}=\frac{W_2}{M_2t_2}\implies W_1M_2t_2=W_2M_1t_1\)

Both sides would have units of "man-days of work."

From this, we may write:

\(\displaystyle t_2=\frac{W_2M_1t_1}{W_1M_2}\)

Plugging in the values from the original problem, we find:

\(\displaystyle t_2=\frac{36\cdot4\cdot2}{48\cdot6}=1\)

Hello!

What does the efficiency mean in the formula?

 

[MarkFL]

Hello!

What does the efficiency mean in the formula?

I would say efficiency is related to the amount of work done per man-time...so let's put efficiency $E$ into the equation:

\(\displaystyle W_n=kE_nM_nt_n\)

Dimensional analysis shows that $k$ is now dimensionless (before, $k$ included efficiency)...and so we can now state:

\(\displaystyle k=\frac{W_1}{E_1M_1t_1}=\frac{W_2}{E_2M_2t_2}\implies W_1E_2M_2t_2=W_2E_1M_1t_1\)

 

[Marcelo Arevalo]

Can it be tripple ratio?

4:48:2 = 6:36:x
men to meter ratio defined;
12:2 = 6:x
mean & extreme
x = 12/12 = 1
1 day is the answer

is it possible??