Word problem help please!!

Question

A Chinese restaurant has a large goldfish pond. Suppose that an inlet pipe and a hose together can fill the pond in 3 hours. The inlet pipe alone can complete the job in one hour less time than the hose alone. Find the time that the hose can complete the job alone and the time that the inlet pipe can complete the job alone.

Answer 1

The key to problems like this is that rates add.   

Let p = the filling rate for the pipe and h = the filling rate for the hose.

Then  h + p = 1/3        (that is  - a rate of one third of a pool per hour)

    This equation can be written as   h = 1/3 - p

The time required for the pipe working alone is  1/p  

The time required for the hose working alone is  1/h

So 1/h = 1 + 1/p.  This can be rearranged as h = p/(1 + p)

  Combining these two equations results in   1/3 - p = p/(1 + p)

   This last equation can be converted to a standard form quadratic equation and solved using the quadratic formula.

The result is  p = (1/6)sqrt(37) - 5/6 =  .1804642.

  h can be found by substitution back into  h = 1/3 - p  to get h = 7/6 - (1/6)sqrt(37) =  .15287

  Finally, the time the hose can complete the job alone is 1/h = 6.541  hrs

and the time the pipe can complete the job alone is 1/p = 5.541 hrs

 

 

Answer 2

Not for the weak at heart....

You can use other variables to represent the times and rates of speed, but they are not nearly as descriptive.

I tried to make it as clean and neat as I can. Is there an equation editor on here?

I also documented each step so you can follow along with each step, one at a time.

 

R-pipe : the rate at which the pipe fills the pond
T-pipe : the time it takes the pipe to fill the pond

R-hose: the rate at which the hose fills the pond
T-hose: the time it takes the hose to fill the pond

1 = R-pipe * T-pipe
1 = R-hose*(T-Pipe+1) where T-hose = T-pipe + 1 <-- it takes the hose an hour longer

1 = (R-hose + R-pipe)* 3 <--- they fill it in 3 hours

Solving the first two equations for R-hose and R-pipe,respectively, then
plugging them into the third equation

1 = [(1/(T-pipe+1)) + (1/T-pipe) ]* 3

 

                1                       1

1 =  [  --------------  +     ---------- ]  * 3   <-- writes it like this

            T-pipe+1             T-pipe 

           T-pipe + (T-pipe + 1)
1 = ---------------------------- * 3 <--- common denominator is (T-pipe+1)*T-pipe
             (T-pipe + 1)*T-pipe

               2*T-pipe + 1
1 = -------------------------- * 3 <--- combines like terms in numerator
           (T-pipe + 1)*T-pipe


1                  2*T-pipe + 1
------ = ----------------------- <-- divides both sides by 3
3            (T-pipe + 1)*T-pipe

(T-pipe + 1)*T-pipe = 3 * (2*T-pipe + 1) <--- cross multiplies

(T-pipe)^2 + T-pipe = 6*T-pipe + 3 <--- distributive property, both sides

(T-pipe)^2 + T-pipe - 6*T-pipe - 3 = 0 <--- moves everything to left side

(T-pipe)^2 - 5*T-pipe - 3 = 0 <--- combines like terms

 

This is a quadratic equation in terms of t-pipe. It MUST be solved

by quadratic formula because it cannot be factored.

T-pipe = [-(-5) +OR- square-root( (-5)^2 - 4*1*(-3)) ]/(2*1) <--- quadratic in terms of T-pipe with A=1, B=-5, C=-3

= [ 5 +OR- square-root( 25 - - 12) ]/2    <---  the square of (-5) is 25;  multiplication before subtraction

= [ 5 +OR- square-root( 25 + 12) ]/2   <-- definition of subtraction: 25 - -12 = 25 + 12 = 37

= [ 5 +OR- square-root(37) ]/2

So there APPEARS to be two values for T-pipe.
1) [5 + square-root(37)]/2 = 5.541381265419 or just over 5 and 1/2 hours

2) [5 - square-root(37)]/2 = -0.541381265419. Since time cannot have a negative
value, this answer is disqualified and rejected.

T-hose = T-pipe + 1 = 5.541381265419 + 1 = 6.541381265419 or just over six and 1/2 hours.

The rates of speed for both  them is just the reciprocal of their times.
So the Pipe fills 1/5.541381265419 = 0.18046 of the pond in an hour.
The host fills 1/6.541381265419 = .152873 of the pond in an hour.
Combined, they fill the pond at a rate of 0.18046 + 0.152873 = 0.3333.... = 0.3.

            This is 1/3 of an hour or 20 minutes.

Therefore, working together, it will take them 3 hours to fill the pond, which proves the answers are
correct.

 

In conclusion, the answers are that it takes the pipe just over 5 and 1/2 hours if working alone

and the hose will need just over 6 and 1/2 hours working alone.

The exact answers are highlighted in bold above. These answers are PROVEN to be correct.

Answer 3

Hi Harley.

 

Rate problems can be really confusing...

 

No matter if the pond is filled by the inlet pipe alone or the hose alone or with them both, all three methods will be putting the same amount of water into the pond.  So let's just say that the pond has 100 gallons.  We can make this anything, as long as it is the same for all three problems.

 

If the time it takes for the pipe to fill the pond is "p" and the time it takes for the hose to fill the pond is "h", then the times we need are found by rate = volume/time, so if we are looking for times, we will need volume/rate.

 

timeBOTH = 3 hours

rateBOTH = 100 gallons/3 hours or 33.3333 gallons per hour

 

timePIPEalone = p   (one hour less than the hose alone)

ratePIPEalone = 100/p  gallons per hour

 

timeHOSEalone = p + 1

rateHOSEalone = 100/(p+1)  gallons per hour

 

 

If we add the ratePIPEalone (in gal/hr) + rateHOSEalone (in gal/hr), this should equal the rateBOTH (in gal/hr).

 

So...

 

100/p  +  100/(p+1)  =  100/3

 

left side is the addition of fractions, so both fractions need a common denominator of "p(p+1)".

 

(p +1) (100)                    (p) (100)                    100

-------------      +      ---------------------  =        -------

(p + 1) (p)                       (p) (p+1)                     3

 

 

This becomes

 

100p + 100 + 100p                    100

------------------------      =    ----------------------

    p² + p                                         3

 

 

 

Cross multiply and simplify:

 

3(200p + 100)  =  100 (p² + p)

 

 

Distribute on both sides:

600p + 300  =  100 p²  + 100p

 

 

Move everything to the right side:

 0  =  100 p² - 500p  - 300

 

 

Divide everything on both sides by 100 (just to make simplifying easier....)

 

 

0 = p² -5p - 3

 

 

 

Since this does not factor easily, you will have to use the quadratic formula to get

 

5 ±√[25 - (4)(1)(-3)]

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

           2(1)

 

5±√[25+12]

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

        2

 

 

5 ±√37

-----------   =   +5.542  or  -0.541

    2

 

 

So it looks like the pipe alone can fill the pond in 5.542 hours,  the hose alone can fill the pipe in 6.542 hours.

 

 

Hope this helps.  :-)

 

Vicki