Question

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. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Suppose the faster pipe takes $x$ minutes to fill the cistern. Therefore, the slower pipe will take $(x+3)$ minutes to fill the cistern.

$\therefore$ Portion of the cistern filled by the faster pipe in one minute $=\frac{1}{x}$

$\Rightarrow$ Portion of the cistern filled by the faster pipe in $\frac{40}{13}$ minutes $=\frac{1}{x}\times \frac{40}{13}=\frac{40}{13x}$

Similarly,

Portion of the cistern filled by the slower pipe in $\frac{40}{13}$ minutes.

$=\frac{1}{x+3}\times \frac{40}{13}=\frac{40}{13(x+3)}$

It is given that the cistern is filled in $\frac{40}{13}$ minutes

$\therefore \frac{40}{13x}+\frac{40}{13(x+3)}=1$

$\Rightarrow \frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}$

$\frac{x+3+x}{x(x+3)}=\frac{13}{40}$

$\Rightarrow 40(2x+3)=13x(x+3)$

$\Rightarrow 80x+120=13x^2+39x$

$\Rightarrow 13x^2-41x-120=0$

$\Rightarrow 13x^2-65x+24x-120$

$\Rightarrow 13x(x-5)+24(x-5)=0$

$\Rightarrow (x-5)(13x+24)=0$

$\Rightarrow x-5=0or,13x+24=0$

$\Rightarrow x=5or,x=\frac{-24}{13}\Rightarrow x=5$ [since x >0]

Hence, the faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 minutes to fill the cistern