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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
$$\dfrac{V}{x}=\dfrac{75}{8}(1+\dfrac{y}{x})=\dfrac{75}{8} (1+\dfrac{3}{5})=15$$hours
Time by pipe B=25 hours
Time by pipe A=15 hours

Mathematics

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