Question

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

Answer 1

Hint: To solve this problem, we should know the concept of volume filled vs hours problems. A tap or combination of taps can fill the tank in x hours means that the tap or combination of taps can fill $\dfrac{1}{x}$ portion of the tank in 1 hour. Let the smaller tap fill the tank in $x$ hours. The larger one fills the tank in $x-10$ hours and we can write the portions done by them individually in 1 hour. From the above statement, we can write that both the tanks fill the $\dfrac{8}{75}$ portion of the tank in 1 hour. We know that the sum of individual works in one hour will be equal to the work done by both of them simultaneously in one hour. Using the above values, we can find the required times.

Complete step by step answer:
We are given that two taps fill the tank in $9\dfrac{3}{8}=\dfrac{72+3}{8}=\dfrac{75}{8}$ hours. We know that a tap or combination of taps can fill the tank in x hours means that the tap or combination of taps can fill $\dfrac{1}{x}$ portion of the tank in 1 hour. Using this, we can write that both the taps fill $\dfrac{8}{75}$ portion of the tank in 1 hour working simultaneously.
Let us consider that the smaller tap fills the tank in $x$ hours. From the question, we can write that the larger one fills the tank in $x-10$ hours. From the above statement, we can write that
Smaller tap fills the $\dfrac{1}{x}$ portion of the tank in 1 hour.
Larger tap fills the $\dfrac{1}{x-10}$ portion of the tank in one hour.
We know that the sum of individual works in one hour will be equal to the work done by both of them simultaneously in one hour. Using this, we can write that
$\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}$
Taking L.C.M and solving, we get
$\begin{align}
  & \dfrac{x-10+x}{x\left( x-10 \right)}=\dfrac{8}{75} \\
 & \dfrac{2x-10}{x\left( x-10 \right)}=\dfrac{8}{75} \\
\end{align}$
Cross multiplying gives
$\begin{align}
  & 75\left( 2x-10 \right)=8\left( {{x}^{2}}-10x \right) \\
 & 8{{x}^{2}}-80x-150x+750=0 \\
 & 8{{x}^{2}}-230x+750=0 \\
 & 4{{x}^{2}}-115x+375=0 \\
 & 4{{x}^{2}}-100x-15x+375=0 \\
 & 4x\left( x-25 \right)-15\left( x-25 \right)=0 \\
 & \left( x-25 \right)\left( 4x-15 \right)=0 \\
 & x=25,\dfrac{15}{4} \\
\end{align}$
As both the values $x,x-10$ should be positive, we can take the value of x to be $x=25$
The larger tap takes $x-10=25-10=15$ hours to fill the tank individually.

$\therefore $ The times required for smaller tap and larger tap to fill the tank separately are $25\text{ and 15}$ hours respectively.

Note: Some students make a mistake while concluding the answer. Instead of writing the answer as 25,15 hours they write that 25, 35 hours. So, it is always a good practice to check back the answer that we get in these types of questions. Checking the solution, we get $\dfrac{1}{25}+\dfrac{1}{15}=\dfrac{15+25}{15\times 25}=\dfrac{40}{15\times 25}=\dfrac{8}{75}$ which is the initial assumed value. Like this, we should verify the answer that we get with the values given in the question.