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CAT 2021 Slot 2 — QA Question 18

Pipes & CisternsEasy

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is:

Answer & solution

  • A

    140

  • B

    120

  • 144

  • D

    264

Solution

Easy

Let aa be A's fill rate and bb be B's drain rate (units of tank per hour). Write the total work for both scenarios; the tank is full (1 tankful) in each, so equate the two expressions to relate aa and bb. Then the time for A alone is tanka\dfrac{\text{tank}}{a}.

Both scenarios start filling at 2 pm. Hours that A runs and hours that B drains:

ScenarioA runs (hrs)B drains (hrs)Net work
A at 2, B at 3, full at 10878a−7b8a-7b
A at 2, B at 4, full at 6424a−2b4a-2b
1

Equate the two (each equals one full tank).

8a−7b=4a−2b⇒ 4a=5b ⇒ b=45a\begin{aligned} &8a-7b=4a-2b\\ &\Rightarrow\ 4a=5b\ \Rightarrow\ b=\tfrac{4}{5}a \end{aligned}
2

Find the tank size in terms of aa. Use scenario 2 (=1=1 tank) and substitute b=45ab=\tfrac45 a.

Tank=4a−2b=4a−2⋅45a=4a−85a=125a\begin{aligned} &\text{Tank}=4a-2b=4a-2\cdot\tfrac{4}{5}a=4a-\tfrac{8}{5}a=\tfrac{12}{5}a \end{aligned}
3

Time for A alone. Divide tank by rate aa.

t=Tanka=125aa=125 hr=144 min\begin{aligned} &t=\frac{\text{Tank}}{a}=\frac{\tfrac{12}{5}a}{a}=\frac{12}{5}\text{ hr}=144\text{ min} \end{aligned}
144 minutes144\text{ minutes}
CAT 2021 Slot 2 QA Q18: Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A — Solution | TheCATExam