Question

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

Answer 1

Hint: To solve this question we need to know the concept of Time and Work and fraction. In this question we have to change the mixed fraction into unlike fraction. We then equate the sum of the one hour work done by both the taps to the value given in the question. Reciprocal of the time taken to do the work is the part of the whole work done.

Complete step by step solution:
The question asks us to find the total time taken by the two taps separately to fill the complete tank. The conditions given are that the two taps together fill up the tank in $9\dfrac{3}{8}$ hours and the tap of larger diameter takes $10$ hours less than the smaller one .
Let us start solving by considering the fact that the smaller diameter tap fills the water in the tank in $x$ hours.
Now the part of tank filled by the smaller tap in one hour is =$\dfrac{1}{x}$
Consider the time taken by the larger tap to fill the tank be $\left( x-10 \right)$ hours.
Now the part of tank filled by the larger tap in one hour is =$\dfrac{1}{\left( x-10 \right)}$
According to the question both the tap takes $9\dfrac{3}{8}$ hours to fill the tank completely.
According to the question, time taken by $B$to complete the work is $40$ days. We need to convert the mixed fraction $9\dfrac{3}{8}$into a fraction.
$\Rightarrow ~9\dfrac{3}{8}$ hours
$\Rightarrow \dfrac{9\times 8+3}{8}$ hours
$\Rightarrow \dfrac{72+3}{8}$ hours
$\Rightarrow \dfrac{75}{8}$ hours
Part of the water filled by both the taps in an hour is =$\dfrac{8}{75}$
Calculating the number of hours required by the taps to fill the tank is:
$\Rightarrow \dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}$
On finding the L.C.M of $x$ and $\left( x-10 \right)$ we get:
$\Rightarrow \dfrac{x-10+x}{x\left( x-10 \right)}=\dfrac{8}{75}$
$\Rightarrow \dfrac{2x-10}{x\left( x-10 \right)}=\dfrac{8}{75}$
On forming the quadratic equation we get:
$\Rightarrow 75\left( 2x-10 \right)=8\times x\left( x-10 \right)$
$\Rightarrow 150x-750=8{{x}^{2}}-80x$
$\Rightarrow 8{{x}^{2}}-230x+750=0$
Dividing each term by $2$ we get:
$\Rightarrow 4{{x}^{2}}-115x+375=0$
On middle term factoring the above equation:
$\Rightarrow 4{{x}^{2}}-100x-15x+375=0$
$\Rightarrow 4x\left( x-25 \right)-15\left( x-25 \right)=0$
$\Rightarrow \left( 4x-15 \right)\left( x-25 \right)=0$
Now analysing the values of $x$ we get:
Case1: When $x=\dfrac{15}{4}$ hours , then the time taken by larger tap will be $x-10$ which is
$\Rightarrow \dfrac{15}{4}-10$
$\Rightarrow \dfrac{15-40}{4}$
$\Rightarrow \dfrac{-25}{4}$
Since time cannot be negative so $x=\dfrac{15}{4}$ is not valid.
Case 2: When $x=25$ hours , then the time taken by larger tap will be $x-10$ which is
$\Rightarrow 25-10$
$\Rightarrow 15$hours
Since time is positive so $x=25$ is valid.
$\therefore $ The time at which smaller and larger taps can separately fill the tank is $25$ hours and $15$ hours respectively.

Note: While solving this question always keep in mind that \[Time\text{ }=\text{ }\dfrac{Work\text{ }done}{Efficiency}\].To solve these type of question we need to find the work done in one hour. To recheck the answer we can see that the time taken by both the taps together to complete a work is less than the time taken by any one of the taps to fill the tank. So through the answer it is clear that time taken by both the tap is $\dfrac{75}{8}$ hours while time taken to fill the tap individually is $25$ hours and $15$ hours.