Question

One tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.
(a) 15 hrs and 10 hrs
(b) 12 hrs and 13 hrs
(c) 21 hrs and 4 hrs
(d) 11 hrs and 14 hrs

Answer 1

Hint: For solving this problem we need to consider that the smaller tap takes \['x'\] hours and the bigger tap takes \['y'\] hours to fill the tank of height \['H'\] alone. Then we apply the given two conditions to mathematical equations and solve them for getting the values of \['x,y'\]. While applying the conditions we need to keep in mind that times of individual taps should not be added to get time for filling tanks together. We need to add the height not time.

Complete step-by-step solution
Let us assume that height of the tank as \['H'\]
Let us assume that the smaller tap takes \['y'\] hours to fill the tank alone.
Let us assume that a bigger tap takes \['x'\] hours to fill the tank alone.
We are given that a smaller tap takes 5 hours more than a bigger tap to fill the tank alone.
Applying this condition mathematically we get
\[\Rightarrow y=x+5....equation(i)\]
We know that a bigger tap takes \['x'\] hours to fill a tank of height \['H'\] alone.
So, we can write that the bigger tap fills \[\dfrac{H}{x}\] height in 1 hour.
Similarly, we know that a smaller tap takes \['y'\] hours to fill the tank of height \['H'\] alone.
So, we can write that the smaller tap fills \[\dfrac{H}{y}\] height in 1 hour.
Let us assume that \['h'\] is the height filled by both taps together.
Since we are not allowed to add the times we can add heights that both taps cover in one hour.
So, we can write
\[\begin{align}
  & \Rightarrow h=\dfrac{H}{x}+\dfrac{H}{y} \\
 & \Rightarrow h=H\left( \dfrac{1}{x}+\dfrac{1}{y} \right) \\
\end{align}\]
We are given that both taps together take 6 hours to fill the tank.
We know that \['h'\] is the height filled by both taps together we can write from the given condition that
\[\Rightarrow h=\dfrac{H}{6}\]
Now, by equating both the values of \['h'\] we get
\[\begin{align}
  & \Rightarrow H\left( \dfrac{1}{x}+\dfrac{1}{y} \right)=\dfrac{H}{6} \\
 & \Rightarrow xy=6\left( x+y \right).....equation(ii) \\
\end{align}\]
By substituting the value of \['y'\] we got from equation (i) in equation (ii) we get
\[\begin{align}
  & \Rightarrow x\left( x+5 \right)=6\left( x+x+5 \right) \\
 & \Rightarrow {{x}^{2}}+5x=12x+30 \\
 & \Rightarrow {{x}^{2}}-7x-30=0 \\
\end{align}\]
Here, we got the quadratic equation in \['x'\].
By solving it we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-10x+3x-30=0 \\
 & \Rightarrow x\left( x-10 \right)+3\left( x-10 \right)=0 \\
 & \Rightarrow \left( x-10 \right)\left( x+3 \right)=0 \\
 & \Rightarrow x=10,-3 \\
\end{align}\]
Here, the variable \['x'\] represents time which cannot be negative. So, \[x=10\].
By substituting \[x=10\] in equation (i) we get
\[\begin{align}
  & \Rightarrow y=x+5 \\
 & \Rightarrow y=10+5=15 \\
\end{align}\]
Therefore the taps take 10 hrs and 15 hrs to fill the tank alone. So, option (a) is the correct answer.

Note: Students will make mistakes in taking the first condition, that is they will consider the total time of 6 hours as the sum of time taken by both taps together that is \[x+y=6\] which is wrong. We are not allowed to add the times when the situations are taken together. Nut instead we can add work done in small change in time as we considered 1 hour in the solution.