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By Allosnake (Allosnake) on Sunday, October 12, 2003 - 06:34 pm: Edit |

matH hElP aGaIN...

a boat has sprung a leak. water is coming in at a uniform rate and some has already accumulated when the leak is detected. at this point, 12 men of equal skill can pump the boat dry in 3 hours, while 5 men require 10 hours. How many men are needed to pump it dry in 2 hours?

any1?? i found its 16 men..but im not sure..or maybe 18..wut u think??

plase dun use calculus for dis

By Miseryxsignals (Miseryxsignals) on Sunday, October 12, 2003 - 07:08 pm: Edit |

i think i messed up in my reasoning, but what the heck:

12 men = 3 hours

5 men = 10 hours

you can use: y = mx + b

y = men

x = hours

i dont really know what M and B mean, im guessing M is the efficiency of the men, which should remain constant, and B corresponds to the initial water that is found.

12 = 3m + b

10 = 5m + b

2 = -2m

m = -1

10 = 5(-1) + b

b = 15

y = -x + 15

Since hours is X, you want to find the # of men (y) when x = 2

y = -(2) + 15

y = 13 men

I dont think i did it right, but it kinda seems reasonable. Can anyone check please?

By Fatjoe (Fatjoe) on Sunday, October 12, 2003 - 08:42 pm: Edit |

dunno if this is right... but here we go....

plug in the amt. of gallons... lets say 30 gallons is in the boat. well accd. to the info above if it takes 5m 10 hr for 30 gal they are pumping out 3 gal/hr.... for 12m 3hr 30 gal the rate is 10 gal. an hr. (im finding out the rate by diving the gals (30) by hrs )

Note that the rate of people to gallons is 1 to 1. an increase in 7 men led to an increase in 7 gal being pumped per hour by the men ( as it went from 3 gal/hr to 10gal/hr)

Using that notion you can figure out the answer. take the rate for what your are trying to find out and you will see that the rate to get it out in is 15 gal/hr (30 gal / 2hrs)... since that rate is 1 to 1 it requires a +5 change in people from 10-15 gal or a +12 change in people from 3 gal/10 gal... Anyway you slice it the answer is 17..

hoped this helped

By Quarky (Quarky) on Sunday, October 12, 2003 - 09:23 pm: Edit |

Here's the best way to solve it, no need for gallons or any simplification like that.

The volume of water that has already accumulated in the boat is V. The time that has already elapsed is T. Then the uniform rate of water flow in the boat is V/T.

The work that those 12 dudes perform involves removing the already present volume V and also the volume of water that flows in during the 3 hours. So the total volume of water they remove is:

V+(V/T)*3

They do it in three hours, so the rate of their work per hour is

(V+(V/T)*3)/3 = V/3+V/T

Per person, that's

**(V/3+V/T)/12**

Then the second group of 5 dudes removes this much water:

V+(V/T)*10

and their rate of work is

(V+(V/T)*10)/10 = V/10+V/T

Per person, that's **(V/10+V/T)/5**

Set the two expressions in bold equal to each other (they are equally skilled men) and solve for time T:

(V/3+V/T)/12=(V/10+V/T)/5

T=0 or T=15. zero hours makes no sense because water had already been accumulating there for some time, so T=15. That means that water has been leaking inside the boat for 15 hours before it was discovered and actions were taken.

Now plug 15 for T into one of the work rate per person equations:

(V/10+V/T)/5 = (V/10+V/15)/5 = V/30

So ONE man removes V/30 water in one hour. We need to get it done in two hours, so each man will actualyl remove 2V/30 or **V/15** water.

The total number of water that will need to be removed is:

V+(V/T)*2 = V+2V/15 = **17V/15**

Divide the two and u get the # of men:

(17V/15)/(V/15) = 17 men = final answer

Hope this helps.

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