Question

Two pipes running together can fill a cistern in 3 {1/13} min. If one pipe takes 3 min more than the other to fill it, then find the time in which each pipe would fill the cistern.

Answer 1

Answer:

In 3 1/13= 40/13 min the slower pipe filled the cistern= 40/13 × (1/x+3) = 40/13(x+3). Hence, Faster pipe takes 5 min to fill the cistern while slower pipe takes (x+3) = 5+3= 8 min to fill the cistern.

Answer 2

$\sf \: SOLUTION : -$

Let faster pipe takes x min to fill the cistern.

Then, slower pipe will take ( x + 3 ) min to fill cistern.

$\sf \: Since, \: portion \: of \: the \: cistern \: filled \: by \: the \: faster \: pipe \: in \: 1 \: min = \frac{1}{x} .$

$\therefore \: \sf \: Portion \: of \: the \: cistern \: filled \: by \: the \: faster \: pipe \: in \: 3 \frac{1}{13} min$

$\sf = 3 \frac{1}{13} \times \frac{1}{x} = \frac{40}{13x}$

$\sf \: Similarly, \: portion \: of \: the \: cistern \: filled \: by \: slower \: pipe$

$\sf \: in \: 3 \frac{1}{13} \: min = \frac{40}{13} \times \frac{1}{(x + 3)} = \frac{40}{13(x + 3)}$

According to the question,

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

$\implies \: \sf \: \frac{40}{13} \bigg [ \frac{1}{x} + \frac{1}{x + 3} \bigg] = 1$

$\implies \: \sf \: 40 \bigg[ \frac{x + 3 + x}{x(x + 3) } \bigg ] = 13 \implies \: 40(2x + 3) = 13x(x + 3)$

$\implies \: \sf \: 80x + 120 = 13 {x}^{2} + 39x \implies \: 13 {x}^{2} + 39x - 80x - 120 = 0$

$\implies \: \sf \: 13 {x}^{2} - 41x - 120 = 0$

$\implies \: \sf \: 13 {x}^{2} - 65x + 24x - 120 = 0 \: \: \: \: \: \: [by \: factorisation]$

$\implies \: \sf \: 13x(x - 5) + 24(x - 5) = 0$

$\implies \: \sf \: (x - 5)(13x + 24) = 0$

$\implies \: \sf \: x - 5 = 0 \: or \: 13x + 24 = 0$

$\implies \: \sf \: x = 5 \: or \: x = \frac{ - 24}{13}$

$\because \: \sf \: Time \: cannot \: be \: negative.$

$\therefore \: \sf \: x = 5$

Hence, faster pipe takes 5 min to fill the cistern while slower pipe takes ( 5 + 3 ) = 8 min to fill the cistern.