Linear equation

Question

Two water pipes can fill an empty tank in 10hr. If the larger pipe can fill the tank in 4.5hr less time than is required by smaller pipe, how long will it take each pipe alone to fill the tank?

Answer 1

Recognize that this is a Work-Rate problem.

A=rt where

r=rate

t=time

A=work accomplished

 

Let:

P_L=x..........time (hours) for large pipe to fill tank

P_S=x+4.5...time (hours) for small pipe to fill tank

P_L+P_S=10...time (hours) for both pipes to fill tank

1/P_L=1/x....rate of work done by large pipe when both pipes are filling

1/P_S=1/(x+4.5)...rate of work done by small pipe when both pipes are

                          filling

 

 

Set up a table for the formula for rt=A:

 

Pipe     Rate(r)   *   Time(t) =  Work Accomplished(A)

---------------------------------------------------------------------------------

P_L        1/x               10            10/x................large pipe portion of work

P_S         1/(x+4.5)      10            10/(x+4.5)......small pipe portion of work

 

So, add the work accomplished values together to get the whole job...

10/x + 10/(x+4.5) = 1......each pipe's contribution to one "whole" job.

 

Multiply all terms in equation by "x(x+4.5)" to eliminate denominators

      and simplify to get....

x²-15.5x-45=0

 

Solve by above equation for "x" by factoring, using quadratic

      formula, etc. get...

x=18

 

Therefore:

P_L=18 hours.................. takes large pipe 18hrs to fill tank working alone

P_S=18+4.5=22.5 hours...takes small pipe 22.5hrs to fill tank working

                                     alone

 

Note:  Can also get percentage of each pipe's contribution to the

              "one whole job" (decimal form) by replacing x=18 into

              the "Work Accomplished (A)" terms in the above table...

P_L=.56

P_s=.44

Added together, get 1.0 or "one whole job."

 

Make sure any values solved are checked with given info in the

              question.  Check your work!