Question 503580
A 90,000 gallon water tank can be filled in two hours by opening valve A alone and in two and a half hours by opening valve B alone. It can be emptied in three hours by opening valve C alone. How long will it take to fill the tank under each of the following conditions? If an answer is not whole hours, include hours, minutes and seconds in your answer.
<pre>
To avoid so many fractions, let's use minutes:

A tank can be filled in 120 minutes by opening valve A alone and in 150 
minutes by opening valve B alone. It can be emptied in 180 minutes by 
opening valve C alone. 

Make this chart. Fill in 1's for the number of tanks filled, and -1 for C,
since to empty a tank is mathematically the sames as "filling -1 tanks".
Fill in the times required as stated in the problem: 


                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         
       B only                        1            150         
       C only                       -1            180        
     


Now fill in the rates in tanks/hour by dividing the number of tanks
filled by the the number of hours


                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180

Now we'll bring in condition "a":  Fill in 1 for the number of tanks filled
x for the number of minutes required and fill in the rate by dividing
number of tanks filled by minutes.

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
  a. A&C only                        1             x            1/x                                              

We use

{{{(matrix(2,1,"A's",rate))+(matrix(2,1,"C's",rate))=(matrix(4,1,combined,rate, 
for, "A&C"))}}}  {{{1/120 + (-1/180)=1/x}}}, answer 360 minutes = 6 hours
--------------------------

Now we'll bring in condition "b":  Fill in 1 for the number of tanks filled
x for the number of minutes required and fill in the rate by dividing
number of tanks filled by minutes.

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
  b. A&B only                        1             x            1/x                                              

We use

{{{(matrix(2,1,"A's",rate))+(matrix(2,1,"B's",rate))=(matrix(4,1,combined,rate, 
for, "A&B"))}}}  {{{1/120 + (1/150)=1/x}}}, answer 66 2/3 minutes or
1 hour, 6 minutes, 40 seconds  

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

Now we'll bring in condition "c":  Fill in 1 for the number of tanks filled
x for the number of minutes required and fill in the rate by dividing
number of tanks filled by minutes.

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
  c. B&C only                        1             x            1/x                                              

We use

{{{(matrix(2,1,"B's",rate))+(matrix(2,1,"C's",rate))=(matrix(4,1,combined,rate, 
for, "A&C"))}}}  {{{1/150 + (-1/180)=1/x}}}, answer 900 minutes = 15 hours
--------------------------

Now we'll bring in condition "d", which is in two parts.
For the first 20 minutes fill in A's rate, 1/120 and 20 for the time in minutes:
Then fill in the "tanks filled" by multiplying rate by the time, getting 
20/120 or 1/6 of a tank filled.  Then put x for the number of minutes after the
first 20 minutes, and the sum of A's and B's rates 1/120 + 1/150 = 3/200.
Then to find the number of tanks filled we multiple the rate by the time,
and get (3/200)x  

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/1
d1. A only for 20 min              1/6            20          1/120 
d2. A&B only after 20 minutes    (3/200)x          x          3/200

We use

{{{(matrix(8,1,
part,
of,
tank,
filled,
in,
first, 
20,
minutes))+

(matrix(5,1,
part, 
of, 
tank, 
filled,
afterward))=

(matrix(2,1,
1,
tank))}}}  {{{1/6 + expr(3/200)x=1}}}, answer 500/9 minutes 
= 55 minutes 33 1/3 seconds.
Adding on the first 20 minutes, 75 minutes, 33 1/3 seconds or

1 hour, 15 minutes, 33 1/3 seconds. 


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

Now we'll bring in condition "e", which is also in two parts.
For the first 20 minutes fill in A&C's rate, 1/120-1/180 = 1/72, and
20 minutes for the time 
Then fill in the "tanks filled" by multiplying rate by the time, getting 
20/120 or 1/6 of a tank filled.  Then put x for the number of minutes after the
first 20 minutes, and the sum of A's,B's, and C's rates 
1/120 + 1/150 - 1/180 = 17/1800.
Then to find the number of tanks filled we multiple the rate by the time,
and get (3/200)x  

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
e1. A&C only for 20 minutes        5/18            20         1/72 
e2. A&B&C after 20 minutes    (17/1800)x            x        17/1800  

We use

{{{(matrix(8,1,
part,
of,
tank,
filled,
in,
first, 
20,
minutes))+

(matrix(5,1,
part, 
of, 
tank, 
filled,
afterward))=

(matrix(2,1,
1,
tank))}}}  {{{5/18 + expr(17/1800)x=1}}}, answer 1300/17 minutes 
= 76 minutes 28 4/17 seconds.
Adding on the first 20 minutes, 96 minutes, 28 4/17 seconds or

1 hour, 36 minutes, 28 4/17 seconds. 


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

Now we'll bring in condition "f", which is also in two parts.
For the first 20 minutes fill in As rate, 1/120, and
20 minutes for the time 
Then fill in the "tanks filled" by multiplying rate by the time, getting 
20/120 or 1/6 of a tank filled.  Then put x for the number of minutes after the
first 20 minutes, and the sum of A's,B's, and C's rates 
1/120 + 1/150 - 1/180 = 17/1800.
Then to find the number of tanks filled we multiple the rate by the time,
and get (3/200)x  

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
f1. A for 20 minutes                1/6            20          1/120 
f2. A&B&C after 20 minutes    (17/1800)x            x        17/1800

 We use

{{{(matrix(8,1,
part,
of,
tank,
filled,
in,
first, 
20,
minutes))+

(matrix(5,1,
part, 
of, 
tank, 
filled,
afterward))=

(matrix(2,1,
1,
tank))}}}  {{{1/6 + expr(17/1800)x=1}}}, answer 1500/17 minutes 
= 88 minutes 14 2/17 seconds.
Adding on the first 20 minutes, 108 minutes, 14 2/17 seconds or

1 hour, 48 minutes, 14 2/17 seconds. 


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

Now we'll bring in condition "g", which is also in three parts.
For the first 20 minutes fill in As rate, 1/120, and
20 minutes for the time 
Then fill in the "tanks filled" by multiplying rate by the time, getting 
20/120 or 1/6 of a tank filled.  
For the next 20 minutes fill in A&B's rate, 1/120+1/150 = 3/200, and
20 minutes for the time 
Then fill in the "tanks filled" by multiplying rate by the time, getting 
60/200 = 3/10 of a tank filled.


Then put x for the number of minutes after the
first 40 minutes, and the sum of A's and B's rates 
1/120 + 1/150 = 3/200.
Then to find the number of tanks filled we multiple the rate by the time,
and get x/120  

                               Number of      Time in       Rate in
                               Tanks filled     minutes       tanks/hour
       A only                        1            120         1/120
       B only                        1            150         1/150
       C only                       -1            180        -1/180
g1. A only for 20 minutes           1/6            20         1/120 
g2. A&B after 20 minutes           3/10            20         3/200
g3. A only after 40 minutes       x/120             x         1/120



f1. A for 20 minutes                1/6            20          1/120 
f2. A&B&C after 20 minutes    (17/1800)x            x        17/1800

 We use

{{{(matrix(8,1,
part,
of,
tank,
filled,
in,
first, 
20,
minutes))+

(matrix(8,1,
part,
of,
tank,
filled,
in,
second, 
20,
minutes))+



(matrix(5,1,
part, 
of, 
tank, 
filled,
afterward))=

(matrix(2,1,
1,
tank))}}}  {{{1/6+3/10+x/120=1}}}, 64 minutes
Adding on the first 40 minutes, 104 minutes or

1 hour, 44 minutes. 


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

You do the last one!

Edwin</pre>