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Home » Two pipes running together can fill a cistern in minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Two pipes running together can fill a cistern in 3 \frac{1}{13} minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
Two pipes running together can fill a cistern in 3 \frac{1}{13} minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Let one pipe fills the cistern in x mins.

Therefore, the other pipe will fill the cistern in (x+3) mins.

Time taken by both, running together, to fill the cistern =3 \frac{1}{13} \min s=\frac{40}{13} \min s

Part filled by one pipe in 1 \min =\frac{1}{x}

Part filled by the other pipe in 1 \mathrm{~min}=\frac{1}{x+3}

Part filled by both pipes, running together, in 1 \mathrm{~min}=\frac{1}{x}+\frac{1}{x+3}

\begin{array}{l} \therefore \frac{1}{x}+\frac{1}{x+3}=\frac{1}{\frac{40}{13}} \\ \Rightarrow \frac{(x+3)+x}{x(x+3)}=\frac{13}{40} \\ \Rightarrow \frac{2 x+3}{x^{2}+3 a}=\frac{13}{40} \\ \Rightarrow 13 x^{2}+39 x=80 x+120 \\ \Rightarrow 13 x^{2}-41 x-120=0 \\ \Rightarrow 13 x^{2}-(65-24) x-120=0 \\ \Rightarrow 13 x^{2}-65 x+24 x-120=0 \\ \Rightarrow 13 x(x-5)+24(x-5)=0 \\ \Rightarrow(x-5)(13 x+24)=0 \\ \Rightarrow x-5=0 \text { or } 13 x+24=0 \\ \Rightarrow x=5 \text { or } x=\frac{-24}{13} \\ \Rightarrow x=5 \quad(\because \text { Speed cannot be a negative fraction) } \end{array}

Thus, one pipe will take 5 mins and other will take \{(5+3)=8\} mins to fill the cistern.

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