Problem Solving

Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?

A. 1/3
B. 1/2
C. 2/3
D. 5/6
E. 1

Since all of the pumps are working for the same amount of time in each instance, we can use the Work = rate * time and add rates.

If A and B can fill a tank in 6/5 hour, then their rate, together, is 5/6 tank/hour rate (1 tank = rate * 6/5 hour). If you do that (convert to a rate instead of time) for all of the pump configurations, you get the following:

A + B = 5/6
A + C = 2/3
C + B = 1/2

When you solve for the variables, you get the following rates, in tank/hour:

A = 1/2
B = 1/3
C = 1/6

When the tanks work together, they will fill 1 tank/hour and therefore the answer is [E]. It's worth noting that you still need to solve the W = rate * time equation to get the answer, it just solves easily:

Work of 1 tank = 1 tank/hour * time

Time = 1 hour

Hope this helps. Good luck.

Ryan's answer is just fine. However, you are not solving this problem for credit in working it out, you are solving it to get the right answer quickly. Here is my approach:

A + B = 5/6; A + C = 2/3; B + C = 1/2

(A +B) + (A + C) + (B +C) = 5/6 + 2/3 + 1/2

2A + 2B + 2C = 5/6 + 4/6 + 3/6

2A + 2B + 2C = 12/6 or 2

(2A + 2B + 2C)/2 = 2/2

A + B + C = 1 and your answer is E.

Combined Work done by A + B + C as per given data
2(1/A+1/B+1/C)=5/6+2/3+1/2 =1/2

therefore, 1/A+1/B+1/C= 2/2=1 hr

Combined Work done by A + B + C as per given data
2(1/A+1/B+1/C)=5/6+2/3+1/2 =1/2

ksh,
Your equation is slightly mistaken. It should be
(1/2A+1/2B+1/2C)=1/2
2A + 2B + 2C = 2
2(A + B + C) =2
thus A + B + C = 1

If one does not use the reciprocals and sets up the equations simply as they are written in the question, such that: A+B = 6/5, A+C = 3/2, and B+C = 2. Is it possible to solve the problem?

nice solutions

cant understand why when you pick numbers for example

Tank = 10 and get all the respective rates

10/(6/5)= 12
10/(3/2) = 15
10/2= 5

And add 12+15+5= 32

then if 2a+2b+2c= 32

div by 2 = a+b+c= 16

then if the tank capacity was 10/16 not equals 1

dferm wrote:Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?

A. 1/3
B. 1/2
C. 2/3
D. 5/6
E. 1

Let tank = 6 liters.

Rate for A and B = w/t = 6/(6/5) = 5 liters per hour.
Rate for A and C = w/t = 6/(3/2) = 4 liters per hour.
Rate for B and C = w/t = 6/2 = 3 liters per hour.

Combining the rates above:
(A+B) + (A+C) + (B+C) = 5+4+3 = 12.
2A + 2B + 2C = 12.
A+B+C = 6 liters per hour.

Time for A+B+C = w/r = 6/6 = 1 hour.

The correct answer is E.

Mitch
What if the tank capacity is 100 ? Can u still get the 1?

luiscarlos59 wrote:Mitch
What if the tank capacity is 100 ? Can u still get the 1?

In your earlier post, you plugged in tank = 10. The division should have been performed as follows:

Rate for A+B = 10/(6/5) = 10 * 5/6 = 50/6 = 25/3.
Rate for A+C = 10/(3/2) = 10 * 2/3 = 20/3.
Rate for B+C = 10/2 = 5.

Combining the rates, we get:
(A+B) + (A+C) + (B+C) = 50/6 + 20/3 + 5.
2A + 2B + 2C = 25/3 + 20/3 + 5.
2A + 2B + 2C = 20.
A+B+C = 10.

Time for A+B+C = w/r = 10/10 = 1 hour.

Same answer, but the math is messier.
Although we can plug in any value for the tank, we should plug in a number that is easily divided by the given times (6/5, 3/2 and 2).
To divide by 6/5 we multiply its reciprocal 5/6; to divide by 3/2 we multiply by its reciprocal 2/3.
Thus, a good value for the tank will be a multiple of 6.

RT= D Works AMAZINGLY FASTER THAN any other approach (IMO :))

(Ra+Rb) *6/5 =1 => (Ra+Rb)=5/6 -->I
(Ra+Rc) *3/2 =1 => (Ra+Rc)=4/6 (converted 2/3 to 4/6 for easy calculation) -->II
(Ra+Rb) *2 =1 => (Rb+Rc)=3/6 (converted 1/2 to 3/6 for easy calculation) -->III

ADDING I, II, III

2(Ra+Rb+Rc) = 12/6 =2
(Ra+Rb+Rc)*1 = 1

Time = 1 hour , E

dferm wrote:Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?

A. 1/3
B. 1/2
C. 2/3
D. 5/6
E. 1

I ran into this problem today and after reviewing all the postings I still have an issue.
The question is
"How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?" and not "how much of the tank, the three pomps working simultaneously, can fill up in an hour?", so why my approach wasn't good?

A+B=1hr12min(6/5) 1 tank
A+C=1hr30min(3/2) 1 tank
B+C=2hr 1 tank
-----------------------------------
2(A+B+C)=4hr42min 3 tanks
A+B+C=2hr21min 3 tanks
A+B+C=47min 1 tank

I do not understand where is my mistake.

Anurag@Gurome wrote:Hi crisro!

If you can complete an work in 2 hours and I can complete it in 3 hours, then are we going to take (2 + 3) = 5 hours to complete the work?

Certainly NO.
If we work together, we will complete it in less than 2 hours.

To find how much time we will take we need to see how much of the work we are going to complete in unit time. You can do 1/2 of the work in one hour and I can do 1/3 of the work. Hence, working together, we will complete (1/2 + 1/3) = 5/6 of the work in one hour.

Therefore, to complete the work we will take 6/5 hours = (1 + 1/5) hours = 1 hour 12 minutes

I hope it will clear your confusion.

Thanks, It took me all night long to figure out that my approach was wrong:(