.
You are given that that first (the faster) pump works as effectively as three slower pumps.
Therefore, you may think that 4 = 3+1 slower pumps, actually, work, and these 4 pumps fill the tank in 12 minutes.
Hence, one single slower pump can fill the tank in 4*12 = 48 minutes, working alone.
Next, since the faster pump is 3 times as fast, it needs only 48/3 = 16 minutes to fill the tank, working alone.
Solved.
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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.