Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours (i.e. $\dfrac{75}{8}$ hours). The tap of larger diameter takes $10$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Step-by-Step Solution
Key Concept: Let smaller tap take $x$ hours, larger tap take $x - 10$ hours. In 1 hour, work done $= \dfrac{1}{x} + \dfrac{1}{x-10} = \dfrac{8}{75}$.
Let time taken by smaller tap $= x$ hours. Time by larger tap $= x - 10$ hours. [1.0 Mark]
In 1 hour, portion filled by smaller tap $= 1/x$, by larger tap $= 1/(x-10)$. Total filled together in 1 hour $= \dfrac{1}{75/8} = \dfrac{8}{75}$. [1.0 Mark]
Equation: $\dfrac{1}{x} + \dfrac{1}{x-10} = \dfrac{8}{75} \Rightarrow \dfrac{x - 10 + x}{x(x - 10)} = \dfrac{8}{75} \Rightarrow \dfrac{2x - 10}{x^2 - 10x} = \dfrac{8}{75}$. [1.0 Mark]
$75(2x - 10) = 8(x^2 - 10x) \Rightarrow 150x - 750 = 8x^2 - 80x \Rightarrow 8x^2 - 230x + 750 = 0$. [1.0 Mark]
Divide by 2: $4x^2 - 115x + 375 = 0 \Rightarrow 4x^2 - 100x - 15x + 375 = 0 \Rightarrow 4x(x - 25) - 15(x - 25) = 0 \Rightarrow (4x - 15)(x - 25) = 0$.
$x = 25$ or $x = 15/4 = 3.75$.
If $x = 3.75$, larger tap time $x - 10 = -6.25$ (negative, impossible). So $x = 25$ hours.
Smaller tap takes $25$ hours, larger tap takes $15$ hours. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Letting smaller tap time $= x$, larger tap $= x - 10$: 1.0 Mark
Forming rate of work equation $\dfrac{1}{x} + \dfrac{1}{x-10} = \dfrac{8}{75}$: 1.0 Mark
Simplifying to quadratic $4x^2 - 115x + 375 = 0$: 1.0 Mark
Solving quadratic for $x = 25$ (rejecting $3.75$): 1.0 Mark
Final times: Smaller tap $= 25$ hrs, Larger tap $= 15$ hrs: 1.0 Mark
Correct Answer: