IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Concept:Use the relation: $\text{Efficiency} = \frac{1}{\text{Time}}$ for each pipe, and compare the final times obtained for Quantity A and Quantity B.Explanation:Quantity A:Let the time taken by pipe A be $x$ hours. Then pipe B takes $(x + 7)$ hours.Pipes C and D together fill the tank in $3.5$ hours, so their combined efficiency is $\frac{1}{3.5} = \frac{2}{7}$ per hour.All four pipes together fill the tank in $2$ hours, so their combined efficiency is $\frac{1}{2}$ per hour.Hence, efficiency of A and B together is:$$\frac{1}{2} - \frac{2}{7} = \frac{3}{14}$$So, $\frac{1}{x} + \frac{1}{x+7} = \frac{3}{14}$.Solving: $\frac{2x+7}{x(x+7)} = \frac{3}{14}$$14(2x+7) = 3x(x+7)$$28x + 98 = 3x^2 + 21x$$3x^2 - 7x - 98 = 0$$(3x+14)(x-7) = 0$Thus, $x = 7$ hours. So, Quantity A = $7$ hours.Quantity B:Time ratio $A : B = 3 : 2$ and $B : C = 1 : 3$, so $A : B : C = 3 : 2 : 6$.Let their times be $3k$, $2k$, and $6k$ hours respectively.Pipe C is an outlet pipe, so its efficiency is negative.Net efficiency of A, B, and C working together:$\frac{1}{3k} + \frac{1}{2k} - \frac{1}{6k} = \frac{2 + 3 - 1}{6k} = \frac{4}{6k} = \frac{2}{3k}$They fill the tank in $7.5$ hours, so $\frac{2}{3k} = \frac{1}{7.5}$$\Rightarrow 3k = 15$, hence $k = 5$.Therefore, time taken by pipe A alone $= 3k = 15$ hours.So, Quantity B = $15$ hours.Since $7 < 15$, Quantity A is less than Quantity B.Answer:Quantity A $(7)$ < Quantity B $(15)$.$\therefore$ The correct option is B. Quantity A < Quantity B.