SOLUTION: If it takes 3men 56hours to dig a hole 4mX6mX5m, and two of the men work twice
as fast as the third,find the number of hours that will take the two faster men to dig this hole to
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as fast as the third,find the number of hours that will take the two faster men to dig this hole to
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Question 941565: If it takes 3men 56hours to dig a hole 4mX6mX5m, and two of the men work twice
as fast as the third,find the number of hours that will take the two faster men to dig this hole together. ( answer is 70hours - show your solution ) Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! If it takes 3men 56hours to dig a hole 4mX6mX5m, and
two of the men work twice as fast as the third,
let t = time required by the faster man to do the job alone
then
2t = time required by the slower man alone
let the completed job = 1
:
A shared work equation for the three men + + = 1
multiply equation by 2t, cancel the denominators and you have:
56 + 2(56) + 2(56) = 2t
56 + 112 + 112
280 = 2t
t = 280/2
t = 140 hrs alone, the faster guy
:
find the number of hours that will take the two faster men to dig this hole together.
let x = no. hrs the two working together + = 1
multiply by 140
x + x = 140
2x = 140
x = 70 hrs, the two faster guys working
