Question

Two water taps together can fill a tank in $9$ hours $36$ minutes. The tap of larger diameter takes $8$ hours less than the smaller one to fill the tank. Find the time in which each tap can separately fill the tank.

Answer 1

$$\begin{array}{c}\text{let the smaller diameter tap alone fill the tank in}x\text{hours...}\\ \text{and larger diameter tap fill it in}x-8\text{hours}\\ \text{so}\\ \text{when they work toghether}\end{array}$$

$$\begin{array}{c}\text{smaller tap fills}1/x\text{in one hour}\\ \text{larger tap fills}1/x-8\text{in one hour}\\ \text{toghether they fill in}\\ 1/x+1/(x-8)\text{in one hour..}\end{array}$$

$$\begin{array}{c}\text{time taken by both}=9\text{hours}36min\\ =9+36/60\text{hours}\left(\text{and by cancelling}\right)\\ =9+3/5\\ \text{both toghether fills it in}\\ \text{9/5 hours ( change into mixed fraction)}\end{array}$$

$$\begin{array}{c}=48/5\text{hours}\\ \text{there from data given}\\ 1/x+1/x-8=5/48\\ 1/x+1/x-8=5/48\\ \text{by calculating}\end{array}$$

$$\begin{array}{c}5x^2-136x+384=0\\ x=(136+\text{or}-104)/10\\ =x=24\text{or}x=16/5\\ \text{toghether.}\\ x=24\text{hours (smaller tap)}\\ \text{bigger tap}=x-8=16\text{hours}\end{array}$$

$24\text{hours whereas bigger tap takes}16hours.$