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CAT 2023 Slot 2 — QA Question 8

Pipes & CisternsEasy

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

Answer & solution

  • A

    60

  • 90

  • C

    75

  • D

    120

Solution

Easy

Work in rates (tank/hour). Let A, C take AA and CC hours; the drain B takes A−1A-1 hours. Write one equation for "all three together fill in 2 h" and another for "B for 1 h, then C for 1141\tfrac14 h fills it". Eliminate CC to get a quadratic in AA, pick the root <5\lt 5, then back-solve CC.

Rates are positive for fill pipes A, C and negative for drain B. Speeds: A→1AA\to\frac1A, B→−1A−1B\to-\frac{1}{A-1}, C→1CC\to\frac1C. Note 1 h 15 min=541\text{ h }15\text{ min}=\frac54 h, so C runs 1+54=941+\frac54=\frac94 h in Case 2 while B runs only the first 11 h.

1

Case 1 — all three for 2 hours fill the tank:

1A−1A−1+1C=12...(1)\frac1A-\frac{1}{A-1}+\frac1C=\frac12\qquad\text{...(1)}
2

Case 2 — B for 1 h (then off), C for 94\tfrac94 h total fills the tank:

−1A−1⋅1+1C⋅94=1...(2)-\frac{1}{A-1}\cdot 1+\frac1C\cdot\frac94=1\qquad\text{...(2)}
3

Eliminate CC: compute 94×(1)−(2)\frac94\times(1)-(2) to cancel the 1C\frac1C terms:

94A−94(A−1)+1A−1=98−1⇒ 94A−54(A−1)=18⇒ 18A−10A−1=1\begin{aligned} &\frac{9}{4A}-\frac{9}{4(A-1)}+\frac{1}{A-1}=\frac98-1\\ &\Rightarrow\ \frac{9}{4A}-\frac{5}{4(A-1)}=\frac18\\ &\Rightarrow\ \frac{18}{A}-\frac{10}{A-1}=1 \end{aligned}
4

Solve the quadratic (multiply through by A(A−1)A(A-1)):

18(A−1)−10A=A(A−1)⇒ 8A−18=A2−A⇒ A2−9A+18=0⇒ (A−3)(A−6)=0 ⇒ A=3 or 6\begin{aligned} &18(A-1)-10A=A(A-1)\\ &\Rightarrow\ 8A-18=A^2-A\\ &\Rightarrow\ A^2-9A+18=0\\ &\Rightarrow\ (A-3)(A-6)=0\ \Rightarrow\ A=3\ \text{or}\ 6 \end{aligned}

Since A fills in less than 5 hours, take A=3A=3.

5

Back-solve CC from (1) with A=3A=3, so A−1=2A-1=2:

13−12+1C=12⇒ 1C=12+12−13=23⇒ C=32 h=90 minutes\begin{aligned} &\frac13-\frac12+\frac1C=\frac12\\ &\Rightarrow\ \frac1C=\frac12+\frac12-\frac13=\frac23\\ &\Rightarrow\ C=\frac32\ \text{h}=90\ \text{minutes} \end{aligned}

9090 minutes — option (b).