M39-75

Question

Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours. Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours. Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank. What fraction of the tank was filled by Pump A?

A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)

A.  \frac{1}{24} 
B.  \frac{1}{8} 
C.  \frac{3}{8} 
D.  \frac{1}{2} 
E.  \frac{2}{3}

Bunuel · Answer

Official Solution:

Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours. Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours. Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank. What fraction of the tank was filled by Pump A?

A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)

A.  \frac{1}{24} 
B.  \frac{1}{8} 
C.  \frac{3}{8} 
D.  \frac{1}{2} 
E.  \frac{2}{3}

Let the time when all three pumps were working together be  t  hours. Then:
 
Pump A worked for  t + 4  hours and filled  \frac{1}{12} * (t + 4)  of the tank;
 
Pump B worked for  t + 2  hours and filled  \frac{1}{5} * (t + 2)  of the tank;
 
Pump C worked for  t  hours and filled  \frac{1}{4} * t  of the tank;
 
Hence, we'd have:
   
 \frac{1}{12} * (t + 4) + \frac{1}{5} * (t + 2) + \frac{1}{4} * t = 1 
   
Multiply the entire equation by 60 to eliminate fractions:
   
 5(t + 4) +12(t + 2) + 15t= 60  
 
 32t + 44= 60  
 
 t=\frac{1}{2}  
   
Therefore, Pump A filled  \frac{1}{12} * (t + 4)=\frac{1}{12} * (\frac{1}{2} + 4)=\frac{3}{8}  of the tank.

Answer: C

julianaabreu · Answer

I did not quite understand the solution. i did not understand t +4, t+2

Bunuel · Answer

Timeline
 
 First 2 hours: only A works. 
 Next 2 hours: A and B work. 
 Last t hours: A, B, C all work.

So:
 
 Pump A works in all three parts: 2 + 2 + t = t + 4 hours. 
 Pump B works only in parts 2 and 3: 2 + t = t + 2 hours. 
 Pump C works only in part 3: t hours.

That is where t + 4 and t + 2 come from.

Adit_ · Answer

I have a small doubt not regard to the problem but with regard to the full tests. I often see problems rated as "Medium" and when I click the question page (this page in this case) it shows me 655-705 which is clearly Hard level. Which one is the true level of the problem? And are the full tests actually a bit harsh in that aspect marking a few hard questions to be medium?

Bunuel · Answer

The two difficulty labels are based on different data pools, so they do not always match.

For the tests, the difficulty is based on how students perform on the questions within the tests themselves. On the forum, PS, DS, etc, the difficulty is based on student performance over time in those separate pools.

Because the underlying attempt data are different, a question can appear as Medium in one place and Hard in another.

M39-75

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