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CAT 2018 Slot 1 — QA Question 4

Pipes & CisternsEasy

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Answer & solution

Correct answer: 10

Answer: 10

Solution

Easy

Let each filling pipe add aa and each draining pipe remove bb (litres/hour). The same tank volume fills in 6 h with 66 fillers and 55 drainers, and in 60 h with 55 fillers and 66 drainers. Equate the two volumes to relate aa and bb, then find the time for 22 fillers and 11 drainer.

1

Two expressions for the tank volume. Net rate ×\times time, in both scenarios.

V=6(6a−5b)V=60(5a−6b)\begin{aligned} &V = 6(6a - 5b)\\ &V = 60(5a - 6b) \end{aligned}
2

Relate aa and bb. Equate the two volumes.

6(6a−5b)=60(5a−6b)⇒ 6a−5b=50a−60b(divide by 6)⇒ 55b=44a⇒ a=54b\begin{aligned} &6(6a-5b) = 60(5a-6b)\\ &\Rightarrow\ 6a - 5b = 50a - 60b \quad\text{(divide by 6)}\\ &\Rightarrow\ 55b = 44a\\ &\Rightarrow\ a = \tfrac54 b \end{aligned}
3

Time with 2 fillers and 1 drainer. Let it take mm hours; its volume equals VV.

m(2a−b)=6(6a−5b)⇒ m(2⋅54b−b)=6(6⋅54b−5b)(use a=54b)⇒ m(64b)=6(104b)\begin{aligned} &m(2a - b) = 6(6a - 5b)\\ &\Rightarrow\ m\left(2\cdot\tfrac54 b - b\right) = 6\left(6\cdot\tfrac54 b - 5b\right) \quad\text{(use }a=\tfrac54 b\text{)}\\ &\Rightarrow\ m\left(\tfrac{6}{4}b\right) = 6\left(\tfrac{10}{4}b\right) \end{aligned}
4

Solve for mm.

64m=604⇒ m=10 hours\begin{aligned} &\tfrac{6}{4}m = \tfrac{60}{4}\\ &\Rightarrow\ m = 10 \text{ hours} \end{aligned}
10 hours\boxed{10 \text{ hours}}

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CAT 2018 Slot 1 QA Q4: A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate — Solution | TheCATExam