Two water taps together can fill a tank \({9}{\dfrac{3}{8}}\)in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let time taken by tap of smaller diameter to fill the tank = x hours

Let time taken by tap of larger diameter to fill the tank = (x – 10) hours

It means that tap of smaller diameter fills \({{1} \over {x}} th\) part of tank in 1 hour.… (1)

And, tap of larger diameter fills \({{1} \over {x - 10}} th\) part of tank in 1 hour. … (2)

When two taps are used together, they fill tank in 758 hours.

In 1 hour \({{8} \over {75}} th\), they fill part of tank \({{1} \over {{{75} \over {8}}}} = {{8} \over {75}}\)… (3)
From (1), (2) and (3),
\( \Rightarrow {{1}\over{x}} + {{1}\over {x - 10}} = {{8} \over {75}}\)
\( \Rightarrow {{x - 10 + x}\over{x(x - 10)}} = {{8} \over {75}}\)
\(\Rightarrow 75 (2x - 10) = 8 (x^2 - 10x)\)
\(\Rightarrow 150x – 750 = 8x^2 - 80x\)
\(\Rightarrow 8x^2 - 80x - 150x +750 = 0\)
\(\Rightarrow 4x^2 - 115x + 375 = 0\)

Comparing equation \(4x^2 - 115x + 375 = 0\) with general equation \(ax^2 + bx + c = 0\),

We get a = 4, b = -115 and c = 375

Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow x = {{115 ± \sqrt{(-115)^2 - 4(4)(375)}} \over {(2)(4)}}\)
\(\Rightarrow x = {{115 ± \sqrt{13225 - 6000}} \over {8}} = {{115 ± \sqrt{7225}} \over {8}}\)
\(\Rightarrow x = {{115 + 85} \over {8}} , {{115 - 85} \over {8}}\)
\(\Rightarrow x = 25,3.75\)

Time taken by larger tap = x – 10 = 3.75 – 10 = -6.25 hours

Time cannot be in negative. Therefore, we ignore this value.

Time taken by larger tap = x – 10 = 25 – 10 = 15 hours

Therefore, time taken by larger tap is 15 hours and time taken by smaller tap is 25 hours.