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Answer :
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)
\( \Rightarrow {{1}\over{x}} + {{1}\over {x - 10}} = {{8} \over {75}}\)
\( \Rightarrow {{x - 10 + x}\over{x(x - 10)}} = {{8} \over {75}}\)
\(\Rightarrow 75 (2x - 10) = 8 (x^2 - 10x)\)
\(\Rightarrow 150x – 750 = 8x^2 - 80x\)
\(\Rightarrow 8x^2 - 80x - 150x +750 = 0\)
\(\Rightarrow 4x^2 - 115x + 375 = 0\)
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.