CAT 2023 Question Paper | Quant Slot 2

CAT Questions

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 10 : Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

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Video Explanation

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Explanatory Answer

Let A, B and C be the number of hours taken by pipes A, B and C to completely fill (or completely empty) a tank.
So the fraction of the tank filled(or emptied) by them in one hour is \( \frac { 1 } { A } , \frac { 1 } { B } , \frac { 1 } { C } \)
“Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank”
B = A – 1
“When pipes A, B and C are turned on together, the empty tank is filled in two hours”
\( \frac { 1 } { A } + \frac { 1 } { C } - \frac { 1 } { A - 1 } = \frac { 1 } { 2 } \)

“If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.”
This means, after pipe C worked for 2 hrs 15 mins (or \( \frac { 9 } { 4 } h r s \) ) and the Pipe B draining for 1 hour, the tank got filled.
\( \frac { 9 / 4 } { C } - \frac { 1 } { A - 1 } = 1 \)
\( \frac { 9 } { 4 C } - \frac { 1 } { A - 1 } = 1 \)
\( \frac { 9 } { 4 C } = 1 + \frac { 1 } { A - 1 } = \frac { A } { A - 1 } \)
\( 9 A - 9 = 4 A C \)
\( \frac { 9 } { 4 } \left( 1 - \frac { 1 } { A } \right) = C \)
\( \frac { 9 } { 4 } \left( \frac { A - 1 } { A } \right) = C \)
\( \frac { 1 } { A } + \frac { 1 } { C } - \frac { 1 } { A - 1 } = \frac { 1 } { 2 } \)
\( \frac { 1 } { A } + \frac { 4 A } { 9 ( A - 1 ) } - \frac { 9 } { 9 ( A - 1 ) } = \frac { 1 } { 2 } \)
\( \frac { 1 } { A } + \frac { 4 A - 9 } { 9 ( A - 1 ) } = \frac { 1 } { 2 } \)
\( \frac { 9 A - 9 + 4 A ^ { 2 } - 9 A } { 9 A ^ { 2 } - 9 A } = \frac { 1 } { 2 } \)
\( 8 A ^ { 2 } - 18 = 9 A ^ { 2 } - 9 A \)
\( A ^ { 2 } - 9 A + 18 = 0 \)
\( A ^ { 2 } - 6 A - 3 A - 18 = 0 \)
\( A = 6 \) or \( A = 3 \)
“Pipe A can fill the empty tank in less than five hours”
A = 3
\( C = \frac { 9 } { 4 } \left( \frac { A - 1 } { A } \right) \)
\( C = \frac { 9 } { 4 } \left( \frac { 3 - 1 } { 3 } \right) \)
\( C = \frac { 9 } { 4 } \left( \frac { 2 } { 3 } \right) = \frac { 3 } { 2 } = 1.5 \mathrm { hrs } = 90 \mathrm { mins } \)
Therefore, the time taken by pipe C to fill an empty tank is 90 min

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The answer is '90'

The question is " Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is "

Choice A is the correct answer.