SOLUTION: A tap fills a tank in 60 mins, another tap fills the same tank in 25 mins, and a third tap fills the same tank in 15 mins. How much time will those take if they run at the same tim

Algebra ->  Proportions -> SOLUTION: A tap fills a tank in 60 mins, another tap fills the same tank in 25 mins, and a third tap fills the same tank in 15 mins. How much time will those take if they run at the same tim      Log On


   



Question 1134694: A tap fills a tank in 60 mins, another tap fills the same tank in 25 mins, and a third tap fills the same tank in 15 mins. How much time will those take if they run at the same time to fill this tank?
Found 3 solutions by ikleyn, josmiceli, greenestamps:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
First tap fills  1%2F60  of the tank volume per minute.


The second tap fills  1%2F25  of the tank volume per minute.


The third tap fills  1%2F15  of the tank volume per minute.


Working together, the three taps fill  1%2F60+%2B+1%2F25+%2B+1%2F15 = 5%2F300+%2B+12%2F300+%2B+20%2F300 = %285%2B12%2B20%29%2F300 = 37%2F300 of the tank volume per minute.


So, the filling process will require 300%2F37 = 8.11 minutes, if all three taps run at the same time.

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It is a standard and typical joint work problem.

There is a wide variety of similar solved joint-work problems with detailed explanations in this site.  See the lessons
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive


Read them and get be trained in solving joint-work problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems"  of the section  "Word problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Add their rates of filling, which is:
[ how much of the tank is filled ] / [ time to fill that much ]
+1%2F1+ means 1 tank filled in 1 hr
Let +t+ = time in hrs to fill the tank with all 3 open
+1%2F60+%2B+1%2F25+%2B+1%2F15+=+1%2Ft+ ( in minutes )
+1%2F%2860%2F60%29+%2B+1%2F%2825%2F60%29+%2B+1%2F%2815%2F60%29+=+1%2Ft+ ( in hrs )
+5%2F5+%2B+1%2F%285%2F12%29+%2B+1%2F%283%2F12%29+=+1%2Ft+
+1+%2B+12%2F5+%2B+4+=+1%2Ft+
Multiply both sides by +5t+
+5t+%2B+12t+%2B+20t+=+5+
+37t+=+5+
+t+=+5%2F37+ hrs
or
+t+=+%285%2F37%29%2A60+ min
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check:
+1+%2B+12%2F5+%2B+4+=+1%2F%285%2F37%29+
+5+%2B+12%2F5+=+37%2F5+
+25+%2B+12+=+37+
OK

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


In the clearly presented solution from tutor @ikleyn, the problem is solved by the standard algebraic method -- looking at the fraction of the job each tap does in one minute, and solving an equation using the least common denominator of those fractions.

Here is an alternative method that many students prefer, since it avoids using fractions in the calculations.

The individual times for the three taps to fill the tank are 60, 25, and 15 minutes.

(1) Find the least common MULTIPLE of those times -- 300.

(2) In 300 minutes...
(a) the first tap could fill the tank 300/60 = 5 times;
(b) the second tap could fill the tank 300/25 = 12 times; and
(c) the third tap could fill the tank 300/15 = 20 times.

So in 300 minutes, the three taps could fill the tank 5+12+20 = 37 times. That means the time required to fill the one tank is 300/37 minutes.