  Question

Two water taps together can fill a tank in $$9 \dfrac{3}{8}$$ hours. The tap of larger diameter takes $$10$$ hours less than smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution

Let $$x$$ be speed of water from pipe with layer diameter and $$y$$ be speed of water from other pipe.here speed refer to the volume flowing per hour.$$V$$ be volume of tank.Time taken when both pipes are used=$$\dfrac{V}{x+y}=9 \dfrac{3}{8}=\dfrac{75}{8}$$When pipe with larger diameter is used,time=$$\dfrac{V}{x}$$when other pipe is uses,time=$$\dfrac{V}{y}$$Now,$$\dfrac{V}{y}+10=\dfrac{V}{x}$$$$\dfrac{V}{y}-\dfrac{V}{x}=10$$ ..(1)and $$\dfrac{V}{x+y}=\dfrac{75}{8}$$$$8V=75(x+y)$$$$8\dfrac{V}{x}=75(1+\dfrac{y}{x})$$$$\dfrac{V}{x}=\dfrac{75}{8}(1+\dfrac{y}{x})$$..(2)similarly $$\dfrac{V}{y}=\dfrac{75}{8} (1+\dfrac{x}{y})$$ ..(3)from (1),(2),(3)$$\dfrac{75}{8} (\dfrac{y}{x}-\dfrac{x}{y})=10$$$$\dfrac{y}{x}-\dfrac{x}{y}=\dfrac{80}{75}=\dfrac{16}{15}$$let $$\dfrac{y}{x}=t$$$$t-\dfrac{1}{t}=\dfrac{16}{15}$$$$15t^2-15=16t$$$$15t^2-16t-15=0$$$$t=-\dfrac{5}{3}$$ or $$t=+0.6$$$$t \neq -\dfrac{5}{3} \neq \dfrac{y}{x}$$$$\dfrac{V}{y}=\dfrac{75}{8} (1+\dfrac{x}{y})=\dfrac{75}{8} (1+\dfrac{5}{3})=25$$ hours$$\dfrac{V}{x}=\dfrac{75}{8}(1+\dfrac{y}{x})=\dfrac{75}{8} (1+\dfrac{3}{5})=15$$hoursTime by pipe B=25 hoursTime by pipe A=15 hoursMathematics

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