MFaulkner wrote:Work can be done by 8 men and 10 women in 25 days. The same work can be done by 10 children and 5 women in 25 days.

In how many days would it take 2 children and 3 men to complete the same work?

Hello MFaulkner, in this problem we could formulate only two equations with three variables. For the number of variables exceeding the number of equations we must assign parameter to the third variable. Depending on the type of parameter used, we attain various degrees of a precision for the represented values. Let's start

x= men
y=women
z=children

one-day work - men and women, 8x+10y=1/25 Equation (1); children and women, 10z+5y=1/25 Equation (2)

Take 2 times the Equation (2), 20z+10y=2/25
then, subtract Equation (1) from Equation (2) after multiplication
20z+10y - 8x-10y = 2/25-1/25
20z-8x=1/25, 500z-200x=1

Work completed is equal to (500z-200x) or 1

Time required for the completion of work by children and men (2z+3x)
(500z-200x)/(2z+3x)=Time (if not integer, add [+] 0.5)

To find time, we must assign the parametric condition to a denominator (2z+3x) which must Not Be equal (=) to 0.
It can not be negative too, since the work pieces completed by children (z) and men (x) are positive values. Hence, we formulate the inequality |2z| + |3x| > 0

Let's solve the system of equation and one inequality for now:
{ 500z-200x=1
{ |2z| + |3x| > 0

We must consider only z>0 and x>0 (see the parametric conditions above)
{500z-200x=1
{2z => 3x
{3x => 2z

let's focus on 2z=3x, since this condition is present in both inequalities z=3x/2

500*(3x/2)-200x=1, 750x-200x=1, 550x=1, x=1/550
z=3/(550*2)=3/1100

Now plug in the values for x (1/550) and z (3/1100)
3x+2z, 3*(1/550)+2*(3/1100) = 3/550+6/1100 = (6+6)/1100 = 12/1100 = 3/275

The work completed is equal to 500z-200x=1
Time=Work completed/(3x+2z), 1/(3/275)=275/3

2 children and 3 men should complete the work at 275/3 (@ if not integer, add [+] 0.5) 275/3+1/2=553/6 ~ 92 days

I bet your answer is different. It's natural, since we have three variables and only two defined equations.

Night Reader,

Thank you! That was an extremely thorough answer & you were very helpful in making your reasoning clear.

Only question I have now is why did you use 'hours' as the unit of time at the end? Shouldn't this be days? 91 days?

MFaulkner wrote:Night Reader,

Thank you! That was an extremely thorough answer & you were very helpful in making your reasoning clear.

Only question I have now is why did you use 'hours' as the unit of time at the end? Shouldn't this be days? 91 days?

Since a unit of measurement in the problem is a day, I have edited the solution.