Math Is Fun Forum

Akila.B

Member

Registered: 2007-01-14

Posts: 3

Problems involves rates and time

Can any one help me on this?

One man can do as much work in one day as a woman can do in two days. A child does one-third  of the work in a day as a woman. If a building contractor hires 39 pairs of men, women and children in the ratio 6:5:2 and pays them in all Rs1113 at the end of the days work. What must me the daily wages of a child be, if the wages are proportional to the amount of work done.

I need steps

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Problems involves rates and time

[Moved to Help Me, hopefully you'll find it. ]

First you need to work out how many of each person there are. You're told that there are 39 people in all, and that the ratio of men:women:children is 6:5:2.

So that tells you how many of each there are if there were 13 people, but as there are 39 you need to multiply by 39/13 = 3. Therefore, the ratio is 18:15:6.

Now, we're also told that women do 3 times as much work as children and that men do twice as much work as women (and so, 6 times as much as children). If the wage is proportional to the work done, then a child getting a wage of x would mean that a woman gets a wage of 3x and a man gets a wage of 6x.

From before, we worked out that there were 18 men. These all get a wage of 6x and so altogether the men get paid 18*6x = 108x.

Similarly, the women get paid 45x in total and the children get paid 6x altogether. Adding these up tells us that overall the building contractor must pay all his workers 159x.

We know that he pays out Rs1113 overall, and so from this we can work out x, the child's wage.

159x = 1113.
Divide by 159: x = 7.

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Why did the vector cross the road?
It wanted to be normal.