Question 1095128: three men can individually do a job in 5,6 and 7 hours respectively . what fractions of the entire job is done by the three men together in 1 hour ?
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = quantity
the quantity is equal to 1 job.
therefore rate * time = 1
for the first person, the equation becomes rate * 5 = 1
rate for the first person is 1/5 of the job in 1 hour.
for the second person, the equation becomes rate * 6 = 1
rate for the second person is 1/6 of the job in 1 hour.,
for the third person, the equation becomes rate * 7 = 1
rate for the third person is 1/7 of the job in 1 hour.
when all 3 persons work together, their rates are additive.
their combined rate is (1/5 + 1/6 + 1/7) of the job in 1 hour.
place (1/5 + 1/6 + 1/7) under the common denominator of 210 and the combined rate becomes (42 + 35 + 30) / 210 = 107/210 of the job in 1 hour.
their combined rate is 107/210 of the job in 1 hour.
that's your solution.
to determine how long it takes to complete the job, you get:
r * t = q becomes 107/210 * t = 1
solve for t to get t = 1 / (107/210) = 1 * 210/107 = 210/107 hours.
the first person can do the job in 5 hours, therefore his rate is 1/5 of the job in 1 hour.
multiply that by 210/107 hours and he completes 1/5 * 210/107 = 42/107 of the job in 210/107 hours.
the second person can do the job in 6 hours, therefore his rate is 1/6 of the job in 1 hour.
multiply that by 210/107 hours and he completes 1/6 * 210/107 = 35/107 of the job in 210/107 hours.
the third person can do the job in 7 hours, therefore his rate is 1/7 of the job in 1 hour.
multiply that by 210/107 hours and he completes 1/7 * 210/107 = 30/107 of the job in 210/107 hours.
add up the contribution of each person and you get 42/107 + 35/107 + 30/107 = 107/107 of the job in 210/107 hours.
107/107 is equal to 1, therefore the whole job is completed in 210/107 hours.
210/107 hours is equal to 1.962616822 hours in decimal form.
Answer by MathTherapy(10556)  (Show Source):
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