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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi burgess

To do problems like this, vary one thing at a time.

If the number of days is held constant and you reduce the time from 7 to 6 hours, that will need more laborers (as they've got less time). So multiply by 7/6.

If the number of days is now allowed to change from 18 to 30 days, that will need less laborers (they've got more days). So multiply by 18/30.

So the number of laborers required will now be:

30 x 7/6 x 18/30 =

Alternatively, invent a unit to describe what has to be done. Let's call it the person/hour/day unit.

To start with there are 30 x 7 x 18 person/hour/days of work to do.

But they work 6 hours in a day, so 30 x 7 x 18 divided by 6 person/days to complete the job.

But they have 30 days to do it, so (30 x 7 x 18)/6 divided by 30 people will be needed.

As you can see, it comes to the same calculation either way.

Hope that helps,

Bob

ps. More on this at http://www.mathsisfun.com/algebra/direc … ional.html

*Last edited by bob bundy (2014-07-01 22:43:09)*

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Thank you much for explaining in detail!!!

bob bundy wrote:

hi burgess

To do problems like this, vary one thing at a time.

If the number of days is held constant and you reduce the time from 7 to 6 hours, that will need more laborers (as they've got less time). So multiply by 7/6.

If the number of days is now allowed to change from 18 to 30 days, that will need less laborers (they've got more days). So multiply by 18/30.

So the number of laborers required will now be:

30 x 7/6 x 18/30 =

Alternatively, invent a unit to describe what has to be done. Let's call it the person/hour/day unit.

To start with there are 30 x 7 x 18 person/hour/days of work to do.

But they work 6 hours in a day, so 30 x 7 x 18 divided by 6 person/days to complete the job.

But they have 30 days to do it, so (30 x 7 x 18)/6 divided by 30 people will be needed.

As you can see, it comes to the same calculation either way.

Hope that helps,

Bob

ps. More on this at http://www.mathsisfun.com/algebra/direc … ional.html

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