SOLUTION: TWO PIPES FILL UP A TANK IN 9 3/8 HRS(75/8) . IF TWO PIPE RUN SEPARATELY THE LARGER DIA TAKES 10HRS LESS THAN THE SMALLER DIA PIPE TO FILL THE TANK INDIVIDUALLY . fIND THE NO OF HR
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Question 201262: TWO PIPES FILL UP A TANK IN 9 3/8 HRS(75/8) . IF TWO PIPE RUN SEPARATELY THE LARGER DIA TAKES 10HRS LESS THAN THE SMALLER DIA PIPE TO FILL THE TANK INDIVIDUALLY . fIND THE NO OF HRS TAKE INDIVIDUAL TO FILL THE TANK ...
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
TWO PIPES FILL UP A TANK IN 9 3/8 HRS(75/8) . IF TWO PIPES RUN SEPARATELY THE LARGER DIA TAKES 10HRS LESS THAN THE SMALLER DIA PIPE TO FILL THE TANK INDIVIDUALLY . fIND THE NO OF HRS TAKE INDIVIDUAL TO FILL THE TANK ...
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Together DATA:
time = (75/8) hrs/job ; rate = (8/75) job/hr.
------------------------------------------
Larger Pipe DATA:
time = x-10 hrs/job ; rate = 1/(x-10) job/hr.
--------------------------------
Smaller Pipe DATA:
time = x hrs/job ; rate = 1/x job/hr.
---------------------------------------------
Equations:
rate + rate = together data
1/x + 1/(x-10) = 8/75
---
75(x-10) + 75x = 8x(x-10)
75x - 750 + 75x = 8x^2 - 80x
8x^2 -230x + 750 = 0
4x^2 - 115x + 375 = 0
(x-25)(4x-15) = 0
Positive realistic solution.
x = 25 hrs. (time for the smaller pipe)
x-10 = 15 hrs. (time for the larger pipe)
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Cheers,
Stan H.
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