Question

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

Answer 1

Hint: Assume the time taken by the tap with longer diameter to fill the tank be x hrs and the time taken by the tap with smaller diameter to fill the tank to be y hrs. Then find the time taken by both taps to fill the tank in one hour separately and together. Then the total amount of water will be equal to amount of water filled by tap with longer diameter and the amount of water filled by smaller diameter tap separately. Solve the equation formed to find the time.

Complete step-by-step answer:
Given the two taps can together fill a tank in = $1\dfrac{7}{8}$hours=$\dfrac{{15}}{8}$hours
Then the amount of water filled in the tank by the two taps in one hour=$\dfrac{1}{{\dfrac{{15}}{8}}} = \dfrac{8}{{15}}$
Now it is given that the tap with a longer diameter takes\[\;2\] hrs less than the tap with a smaller one to fill the tank separately.
Then let the time taken by the tap with longer diameter to fill the tank be x hrs.
And let the time taken by the tap with smaller diameter to fill the tank is y hrs.
According to question-
$ \Rightarrow x = y - 2$ -- (i)
Now the amount of water filled by the tap with longer diameter in one hour=$\dfrac{1}{x}$
From eq. (i) we can write,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{y - 2}}$
And the amount of water filled by the tap with smaller diameter in one hour=$\dfrac{1}{y}$
Then the total amount of water =amount of water filled by tap with longer diameter + amount of water filled by tap with smaller diameter
$ \Rightarrow \dfrac{1}{{y - 2}} + \dfrac{1}{y} = \dfrac{8}{{15}}$
On taking LCM we get,
$ \Rightarrow \dfrac{{y + y - 2}}{{{y^2} - 2y}} = \dfrac{8}{{15}}$
On cross multiplication we get,
$ \Rightarrow \left( {2y - 2} \right)15 = 8\left( {{y^2} - 2y} \right)$
On simplifying we get,
$ \Rightarrow 30y - 30 = 8{y^2} - 16y$
On separating the common terms and constant we get,
$ \Rightarrow 30y + 16y - 8{y^2} = 30$
$ \Rightarrow 46y - 8{y^2} = 30$
On taking $2$ common from both sides and cancelling it, we get-
$ \Rightarrow 23y - 4{y^2} = 15$
On multiplying negative sign both side we get,
$ \Rightarrow 4{y^2} - 23y + 15 = 0$
On factoring we get,
$ \Rightarrow 4{y^2} - 20y - 3y + 15 = 0$
$ \Rightarrow \left( {4y - 3} \right)\left( {y - 5} \right) = 0$
$ \Rightarrow 4y - 3 = 0{\text{ or y}} - 5 = 0$
$ \Rightarrow y = \dfrac{3}{4}{\text{ or }}y = 5$
On putting this value in eq. (i) we get,
For$y = \dfrac{3}{4}$ ,
$ \Rightarrow x = \dfrac{3}{4} - 2 = \dfrac{{3 - 8}}{4} = \dfrac{{ - 5}}{3}$
This is not possible as time is never negative
Then for$y = 5$ ,
$ \Rightarrow x = 5 - 2 = 3$hrs

Hence, the time taken by the tap with longer diameter to fill the tank is $3$ hrs and the time taken by tap with smaller diameter to fill the tank is $5$hrs.

Note: Here the students can also use the discriminant formula to find the factors of y which is given as-
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ For the quadratic equation of form-$a{y^2} + by + c = 0$