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Question
Two pipes running together can fill a cistern in `3 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.
Solution
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 1/13 min s=40/13 min s`
Part filled by one pipe in 1 min=`1/x`
Part filled by the other pipe in 1 min =`1/(x+3)`
Part filled by both pipes, running together, in 1 min =`1/x+1/(x+3)`
∴ `1/x+1/(x+3)=1/40`
⇒ `((x+3)+x)/(x(x+3))=13/40`
⇒`(2x+3)/(x^2+3a)=13/40`
⇒`13x^2+39x=80x+120`
⇒`13x^-41x-120=0`
⇒`13x^2-(65-24)x-120=0`
⇒`13x^2-65x+24x-120=0`
⇒`13x(x-5)+24(x-5)=0`
⇒`(x-5)(13x+24)=0`
⇒`x-5=0 or 13x+24=0`
⇒`x=5 or x=--24/13`
⇒`x=5` ( ∵ Speed cannot be a negative fraction)
Thus, one pipe will take 5 mins and other will take{(5+3)=8}mins to fill the cistern.
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