Two Pumps of different sizes, working together, can empty afuel tank in 5 hours. The larger capacity pump can empty this tank in 4 hours less than the smaller one. If the larger pump isout of order, how long will it take the smaller one to do the jobalone?

Dillan Valenzuela

Dillan Valenzuela

Answered question

2022-08-03

Two Pumps of different sizes, working together, can empty afuel tank in 5 hours. The larger capacity pump can empty this tank in 4 hours less than the smaller one. If the larger pump isout of order, how long will it take the smaller one to do the jobalone?

Answer & Explanation

Kody Larsen

Kody Larsen

Beginner2022-08-04Added 11 answers

Two Pumps of different sizes, working together, can empty a fuel tank in 5 hours.
The larger capacity pump can empty this tank in 4hours less than the smaller one.
Let us assume that the smaller capacitypump can empty the tank in X hours .
Hence the larger capacity pump can empty the tank in (X -4) hours.
Now as the smaller capacity pump canempty the tank in X hours, hence we can say the smaller capacity pump in 1 hour empties 1 / X t h of the tank.
Similarly we can say that thelarger capacity pump in 1 hour empties 1 / ( X 4 ) t h of the tank
hence together in 1 hr they empty [ 1 / X + 1 / ( X 4 ) ] t h of the tank
= [(X - 4) + X]/ X(X - 4) th of the tank
= 2X - 4/X(X - 4) th of the tank
hence they empty the full tank in 1/[2X - 4/X(X - 4)]
= X(X - 4)/ (2X -4) hr
hence according to the question,
X(X - 4)/ (2X - 4) = 5
X 2 4 X = 5 ( 2 X 4 )
X 2 4 X = 10 X 20
X 2 4 X 10 X + 20 = 0
X 2 14 X + 20 = 0
Passafaromx

Passafaromx

Beginner2022-08-05Added 2 answers

Pump one = x hrs Pump two = x+4 hrs
1/x + 1/(x+4) = 1/5
( x + x + 4 ) / ( x 2 + 4 ) = ( 1 / 5 )
10 x + 20 = x 2 + 4 x
0 = x 2 6 x 20
x= 8.385
Small pump = 8.385 + 4 = 12.385 hrs

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