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.

  • $\begingroup$ Do you know the correct answer? $\endgroup$ – John Sep 29 '16 at 18:25

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:

If the man works twice as hard as he usually does, how long will it take to get the work done?

  • $\begingroup$ I know the answer. But I doubt whether it's right. Your answer seems right to me. I thought so. But the answer given is 10days. Solution is as (20*30+30*60+20*60)÷(20*30*60). I can't understand how it was solved like this. $\endgroup$ – Omkar Reddy Sep 29 '16 at 18:32
  • $\begingroup$ Can the given answer be right by any chance as per the condition given? I don't think it's right. What do you say? $\endgroup$ – Omkar Reddy Sep 29 '16 at 19:14
  • $\begingroup$ I'm stumped. The expression you gave me equals $0.1$. But there are $20$s, $30$s, and $60$s in there, so maybe I'm close? LOL $\endgroup$ – John Sep 29 '16 at 20:29
  • $\begingroup$ Sorry. I meant to say (20*30*60)÷(20*30+30*60+20*60) which is 10 days. But yours is reasonable. If it's 10 days as given above, it means that they each worked for 10 days together. But it contradicts the condition which was given. Isn't it? $\endgroup$ – Omkar Reddy Sep 29 '16 at 20:37
  • $\begingroup$ And you're correct to say that man completes half of the total task, and the woman and child each complete one quarter of the task in 20, 30 and 40 days respectively. I thought anwser should be 60 days. But that last step as I mentioned confused me. I think that your answer is right. $\endgroup$ – Omkar Reddy Sep 29 '16 at 20:46

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