Question

Three taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr, respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank?
(a) 8
(b) 9
(c) 10
(d) 11

Answer 1

Hint: To solve this problem, we need to know the basics of unitary method. We will try to use the analogy of speed, distance and time to solve this problem $\left( Speed=\dfrac{Dis\tan ce}{Time} \right)$ . In this problem, speed is analogous to volumetric flow rate (amount of volume per hours), distance is analogous to amount of volume and time is analogous to number of hours. Now, we can proceed with this question.

Complete step-by-step solution -
We first start by analysing each tank. Since, A takes 10 hr to independently fill the tank. Thus, let’s assume volume of tank as V and let the volumetric flow rate of tank A be a. Thus, we have,
a = $\dfrac{Volume}{time}$ (using the speed, distance and time analogy on this problem) --
a = $\dfrac{V}{10}$ --(1)
Now, B takes 20 hr to independently fill the tank. Let the volumetric flow rate of tank B be b. Thus, we have,
b = $\dfrac{Volume}{time}$ (using the speed, distance and time analogy on this problem)
b = $\dfrac{V}{20}$ -- (2)
Now, C takes 30 hr to independently fill the tank. Let the volumetric flow rate of tank C be c. Thus, we have,
c = $\dfrac{Volume}{time}$ (using the speed, distance and time analogy on this problem)
c = $\dfrac{V}{30}$ -- (3)
Since, exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour, for first 3 hours, we have A and B, B and C and finally A and C operating thus the amount of tank filled in first 3 hours (let’s assume V’ amount of volume filled in first 3 hours). We have,
V’ = $\left( \dfrac{V}{10}+\dfrac{V}{20} \right)$ $\times $ 1 + $\left( \dfrac{V}{20}+\dfrac{V}{30} \right)$ $\times $ 1 + \[\left( \dfrac{V}{10}+\dfrac{V}{30} \right)\] $\times $ 1
We get this by using the results of (1), (2) and (3). Since, if A and B work together, their volumetric flow rates would be added similar to the speed (using the analogy). We do the same for B and C, and A and C.
V’ = $\dfrac{3V}{20}$ + $\dfrac{5V}{60}$ + $\dfrac{4V}{30}$
V’ = $\dfrac{22V}{60}$
Thus, in 3 hours, $\dfrac{22V}{60}$ of the tank was filled. The remaining tank is $V-\dfrac{22V}{60}=\dfrac{38V}{60}$ . Now, we want to find the minimum time to fill the tank, thus we need to fill the tank by the pair which will fill the tanks the fastest. We can see from (1), (2) and (3) that A and B have the highest volumetric flow rates. Thus, for minimum time, the remaining tank should be filled by A and B. The volumetric flow rate of A and B combined is $\left( \dfrac{V}{10}+\dfrac{V}{20} \right)$ = $\dfrac{3V}{20}$ . Thus, for remaining tank, we have the
Time = $\dfrac{Volume}{Volumetric\text{ flow rate}}$
Time = $\dfrac{\dfrac{38V}{60}}{\dfrac{3V}{20}}$ (Since, the remaining volume is $\dfrac{38V}{60}$ and combined volumetric flow rate is $\dfrac{3V}{20}$ )
Time = $\dfrac{38}{9}$ hours
Time = 4.22 hours
Thus, minimum time to fill would be the next largest integer (that is 5). Thus, total time is 3 + 5 = 8 hours.
Hence, the correct answer is (a) 8 hours.

Note: In this problem, we have used the unitary method disguised in the form of speed, distance and time analogy. One can also approach this problem by using the original form of unitary method. However, due to the complexity of the problem, it would take much more time to solve the problem using this technique.