Question

Two water taps together can fill a tank in 1 {7/8 } hrs. The tap with longer diameter takes 2 hrs less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer 1

$\boxed{ \boxed{ \bf {\pink{SOLUTION }}}}$

$\rm \: Let \: the \: time \: taken \: by \: smaller \: tap \: to \: fill \: tank \: completely \: = x \: hrs$

$\rm \: So, \: volume \: of \: tank \: filled \: by \: smaller \: tap \: in \: 1 \: hr = \frac{1}{x}$

$\rm \: Volume \: of \: tank \: filled \: by \: larger \: tap \: in \: 1 \: hr = \frac{1}{x - 2}$

$\rm \: Now, \: time \: taken \: by \: both \: taps \: to \: fill = 1 \frac{7}{8} = \frac{15}{8} \: hrs$

$\rm \: Tank \: filled \: by \: smaller \: tap \: in \: \frac{15}{8}hrs = \frac{1}{x} \times \frac{15}{8} = \frac{15}{8x}$

$\rm \: Tank \: filled \: by \: larger \: tap \: in \: \frac{15}{8}hrs = \frac{1}{x - 2}\times \frac{15}{8} = \frac{15}{8(x - 2)}$

$\rm \: Therefore \: \frac{15}{8x} + \frac{15}{8(x - 2)} = 1 \implies \: \frac{15}{8} \bigg[ \frac{1}{x} + \frac{1}{x - 2} \bigg] = 1$

$\implies \: \rm \: \frac{2(x - 1)}{ {x}^{2} - 2x} = \frac{8}{15}$

$\implies \: \rm \: 15(x - 1) = 4( {x}^{2} - 2x)$

$\implies \: \rm \: 23x = 4 {x}^{2} + 15$

$\implies \: \rm \: 4x {}^{2} - 23x + 15 = 0$

$\rm \: x = \frac{ - ( - 23) \pm \sqrt{( - 23) {}^{2} - 4.4.15} }{2.4}$

$\rm \: x = \frac{23 \pm \: 17}{8}$

$\boxed{ \rm \red{ • Taking \: Positive \: Sign}}$

$\rm \: x = \frac{23 + 17}{8}$

$\rm \: x = 5$

$\boxed{ \rm \red{• Taking \: Negative \: Sign}}$

$\rm \: x = \frac{23 - 17}{8}$

$\rm \: x = \frac{3}{4}$

$\rm \: Taking \: x = 5$

$\rm \: Time \: taken \: by \: smaller \: tap = 5 \: hrs$

$\rm \: Time \: taken \: by \: large \: tap = x - 2 = 5 - 2 = 3 \: hrs$

$\rm \: Taking \: x = \frac{3}{4}$

$\rm \: Time \: taken \: by \: smaller \: tap = \frac{3}{4} \: hr$

$\rm \: Time \: taken \: by \: large \: tap = x - 2$

$\rm = \frac{3}{4} - 2 = \frac{ - 5}{4}, \: its \: not \: solution.$