8 men, 12 women and 10 children can complete a work in 5 days, 10 days and 24 days respectively. 1 man, 1 woman and 1 child started the work. In how many days will the work get completed if the man did twice the work as woman and also twice the work as child?
If I rewrite that as 1man can complete the work in 40days in one day and similarly for woman and child.
I couldn't understand how to apply the the given condition further.
Please help me solve this problem.
A man, a woman, and a child can complete the task by themselves in $40, 120,$ and $240$ days, respectively.
If the man takes on twice the amount of work as the woman, and twice the work of the child, then the man completes half of the total task, and the woman and child each complete one quarter of the task.
Then, it gets down to who takes the longest at their part. They work simultaneously until they finish their assigned portion, then they drop out.
The man completes his half in $20$ days. The woman completes her quarter in $30$ days, and the child his/her quarter in $60$ days.
So, I think it would be done in $60$ days.
EDIT: OK, I see how they got the answer they did. I finally figured out where I had seen that kind of formula before.
There's a direct analogy between this problem and effective resistance of resistors in parallel:
$$\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$
So if the man is a $20\Omega$ resistor, the woman is a $30\Omega$ resistor, and the child a $60\Omega$ resistor, then the effective resistance is $10 \Omega.$
The analogy is this (hope you can follow it!) A lower resistance lets more electrons through per unit time than a higher resistance. For a given amount of charge, a $60\Omega$ resistor will take three times as long to let the charge through as a $20\Omega$ resistor. But with three people working at the same time (in parallel) the work gets done faster than any of them could do individually.
The question is just incredibly poorly worded. I'd change the last sentence to read: