Time & Work Problem: Solving 1/2 Unit of Work in 6 Days

[22990atinesh]

Homework Statement

Q. 2 Men working 3 hours/day works for 4 days to complete a work (unit of work). Calculate how many days required by 1 man working 2 hours/day to complete the 1/2 of that work ?

Homework Equations

$\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$

The Attempt at a Solution

Sol: As we know Man hour/unit work is constant. Hence $\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$
Now we can easily plug data in LHS of the above equation. But for RHS as we know, we have to calculate days required by 1 man working 2 hours/day to complete the 1/2 of that work. so we have to double the total Man hour in RHS i.e.

$\frac{2*(4*3)}{1} = \frac{2*(1*(X*2))}{1}$

$\frac{2*(4*3)}{1} = \frac{1*(X*2)}{1/2}$

$X=6 days$

Correct me, If I did something wrong.

 

[NascentOxygen]

That's the answer I got, too. ☺[/size][/color]

 

[22990atinesh]

That's the answer I got, too. ☺[/size][/color]

I'm little bit confused about formula above

 

[Vibhor]

There is no need to use the formula .You can find the answer logically as well .

 

[NascentOxygen]

I'm little bit confused about formula above

Where did it come from?

 

[22990atinesh]

As we know Man hour/unit work is constant. Hence

$\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$

If we are reducing work by half, then we have to reduce total man hours for that work by half in order to keep $\frac{MH}{W} = k$ (Ratio will be constant if and only if both numerator and denominator is multiplied by a same constant)

$\frac{2*(4*3)}{1} = \frac{1*(X*2)/2}{1/2}$

$\frac{2*(4*3)}{1} = \frac{1*(X*2)}{1}$

$X=12 days$

Please correct me. If I'm wrong

 

[HallsofIvy]

This is how I would have done it:

2 Men working 3 hours/day works for 4 days to complete a work (unit of work).

So each man works (3 hour/day)(4 days)= 12 hours to complete half the work.

Calculate how many days required by 1 man working 2 hours/day to complete the 1/2 of that work ?

At 2 hours a day, it will take 6 days to work 12 hours and do 1/2 the job.

$X=6 days$

Correct me, If I did something wrong.

Yes, that's correct- but you did entirely too much work!

 

[Ray Vickson]

Homework Statement

Q. 2 Men working 3 hours/day works for 4 days to complete a work (unit of work). Calculate how many days required by 1 man working 2 hours/day to complete the 1/2 of that work ?

Homework Equations

$\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$

The Attempt at a Solution

Sol: As we know Man hour/unit work is constant. Hence $\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$
Now we can easily plug data in LHS of the above equation. But for RHS as we know, we have to calculate days required by 1 man working 2 hours/day to complete the 1/2 of that work. so we have to double the total Man hour in RHS i.e.

$\frac{2*(4*3)}{1} = \frac{2*(1*(X*2))}{1}$

$\frac{2*(4*3)}{1} = \frac{1*(X*2)}{1/2}$

$X=6 days$

Correct me, If I did something wrong.

Not wrong, but can be made more 'elementary': the job takes 2x3x4 = 24 man-hr, so 1 man needs 12 hr to do half the job and so must work for 12/2 = 6 days.

 

[thelema418]

Labor-hours is considered a better term than man-hours. =)

Otherwise, I do a similar calculation to Vickson. I prefer to think of it geometrically. The x-axis is the number of people working; the y-axis is the number of hours worked. This creates a rectangle: the area of the rectangle is the job. When you change the number of people working on the same job, the area of the rectangle is conserved.

 

[Ray Vickson]

Labor-hours is considered a better term than man-hours. =)

Otherwise, I do a similar calculation to Vickson. I prefer to think of it geometrically. The x-axis is the number of people working; the y-axis is the number of hours worked. This creates a rectangle: the area of the rectangle is the job. When you change the number of people working on the same job, the area of the rectangle is conserved.

'Considered a better term' by whom? 'Man-hours' is absolutely standard, and 'labour (or labor) hours' is claimed in some sources to be an alternative; sometimes the horrible 'person-hours' is used instead. Anyway, in this problem, all the workers are men.

 

[thelema418]

The Bureau of Labor Statistics (United States) no longer uses "man hour" but reports "labor hour". The United Nations Development Programme recognizes the term man hour but recommends and prefers "person hour" instead. As a business manager and analyst, I utilized "labor hour" when reporting and communicating with others about workplace trends.

While it is true that the workers of this word problem are men, there is nothing that suggests the effectiveness or ineffectiveness of their work is contingent upon their "being a man" as opposed to just "being a laborer."

Utilize whatever words you wish -- but I should wish to inform the OP that there are observations we should make in the choice of words we use to mathematize this situation -- even if it is a benign word problem.

 

[22990atinesh]

Not wrong, but can be made more 'elementary': the job takes 2x3x4 = 24 man-hr, so 1 man needs 12 hr to do half the job and so must work for 12/2 = 6 days.

Thanx Ray Vickson, I got it. The total (Man) hours/ unit work is constant i.e. 24 (Man)hours/ work. So, If we have to calculate No. of days required by 1 man working 2 hours/ day to complete the 1/2 of that work. Whatever amount of (Man) hour is required by 1 man working 2 hours/ day to complete the 1/2 of that work, twice of it per unit of work will be constant. Hence

$24 = \frac{2*(1*(X*2))}{1}$

$24 = \frac{1*(X*2)}{1/2}$

$X = 6 days$