# Solve this prob!

• Apr 14th 2010, 02:21 AM
jgv115
Solve this prob!
Bob calculated that his 8 professional workers would take 15 hours each to pack a few parcels.

If all the workers work at the same rate, how long would it take 12 workers to pack all the parcels.

I somehow got 10. I dunno if it is right.

Can someone not necessarily tell me the answer but point me in the right direction?
• Apr 14th 2010, 02:36 AM
pickslides
Quote:

Originally Posted by jgv115
Bob calculated that his 8 professional workers would take 15 hours each to pack a few parcels.

If all the workers work at the same rate, how long would it take 12 workers to pack all the parcels.

I somehow got 10. I dunno if it is right.

Can someone not necessarily tell me the answer but point me in the right direction?

Solve...

$\displaystyle \frac{15}{8}=\frac{x}{12}$
• Apr 14th 2010, 02:42 AM
jgv115

$\displaystyle x = 22.5$

22.5 hours...

The number of hours should be lower than 15 hours because there are more people working on it...
• Apr 14th 2010, 02:49 AM
sa-ri-ga-ma
Quote:

Originally Posted by jgv115
Bob calculated that his 8 professional workers would take 15 hours each to pack a few parcels.
If all the workers work at the same rate, how long would it take 12 workers to pack all the parcels.

Total number of man hours needed to complete the packing is 8*15.
To complete the same work, 12 workers take x hours each. So
8*15 = 12*x.
Solve for x.
• Apr 14th 2010, 02:51 AM
pickslides
Yes, I should read more closely.

You are saying 10 hours because you have 50% more workers (4 extra out of a new total of 12 is one third) and therefore time taken should be cut by one third.
• Apr 14th 2010, 02:54 AM
jgv115
mm ok.

In that case:

12 workers are joined by 8 new workers. New workers work at half the rate of normal workers. How long would it take to pack the boxes?

Here's what I did: 8 new workers is equivalent to 4 "normal" workers.

So $\displaystyle 15*8 = 16 * x$

$\displaystyle 120 = 16x$

$\displaystyle x= 7.5$

Is that all good?
• Apr 14th 2010, 03:02 AM
sa-ri-ga-ma
Yes. You are right.
• Apr 14th 2010, 03:27 AM
jgv115
(remember that training workers that half the time of normal workers)

Some normal workers and 6 training workers would have taken 12 hours to complete the job. How many normal workers were there?

We know the job is 120.

Let x be the number of normal workers:

$\displaystyle 120= (x+3) * 12$

$\displaystyle x= 7$

Is this all good?

Last one:

If after 2.5 hours, two normal workers and all training workers had to leave. How much longer would the remaining workers take to finish packing than if no one had left.

So basically I worked out a rate.

In 12 hours, 10 normal workers (7 normal + 6 training) would complete 120. So if they were given only 2.5 hours then...

$\displaystyle \frac {120}{12} = \frac{x}{2.5}$ x = how much work was done.

$\displaystyle x= 25$

So the remaining work is 120-25 which is 95. There are 5 (7-2) normal workers left and no training workers (all left)

$\displaystyle 95 = 5x$ where x = the number of hours it takes

$\displaystyle x= 19$

So altogether the workers took 21.5 hours to complete the task when some workers left.

It took 12 hours for 10 workers to do the job.

So 21.5 - 12 = 9.5

So it would take 9.5 hours longer.

Is this correct? :S
• May 2nd 2010, 04:39 AM
yatfungyuen
Yes, I think so. I got the same answer as you