Question

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

Answer 1

Hint: We can calculate the fraction of volume filled by each water pipe in $9{\displaystyle \frac{3}{8}}$ hours, sum of which will be equal to 1 denotry full tank.
The value of the variable can be found to get the required time.

Complete step-by-step answer:
Let time taken by a smaller tap to fill the tank completely is t hrs, then volume filled in one hour = ${\displaystyle \frac{1}{t}}.$
It is given, the smaller tank takes 10 hours less. Therefore, time taken to fill the tank completely is (t-10) hours.
Volume filled in 1 hr = ${\displaystyle \frac{1}{t-10}}$
Time taken by both taps $=9{\displaystyle \frac{3}{8}}$ hours
                $={\displaystyle \frac{9\times 8+3}{8}}$
(Converting to improper fraction)
                $={\displaystyle \frac{75}{8}}\text{hrs}$
Tank filled by each tap in ${\displaystyle \frac{75}{8}}\text{ hours}$
Smaller tap $={\displaystyle \frac{1}{t}}\times {\displaystyle \frac{75}{8}}$
               $={\displaystyle \frac{75}{8t}}$
Larger tap $={\displaystyle \frac{1}{\left(t-10\right)}}\times {\displaystyle \frac{75}{8}}$
        $={\displaystyle \frac{75}{8\left(t-10\right)}}$
Sum of tanks filled by each pipe =1
Where 1 denotes completely full tank:
 ${\displaystyle \frac{75}{8t}}+{\displaystyle \frac{75}{8\left(t-10\right)}}=1$
 ${\displaystyle \frac{75}{8}}\left({\displaystyle \frac{1}{t}}+{\displaystyle \frac{1}{\left(t-10\right)}}\right)=1$
 ${\displaystyle \frac{1}{t}}+{\displaystyle \frac{1}{\left(t-10\right)}}={\displaystyle \frac{8}{75}}$
Taking LCM:
 ${\displaystyle \frac{t-10+t}{t(t-10)}}={\displaystyle \frac{8}{75}}$
$${\displaystyle \frac{2t-10}{t^2-10t}}={\displaystyle \frac{8}{75}}$$
 $(2t-10)75=8(t^2-10t)$
 $150t-750=8t^2-80t$
Rearranging:
 $8t^2-230t-750=0$
Finding roots of this quadratic equation using
 $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
Here, a = 8, b=-230, c = -750, x =t.
Substituting, we get:
 $t={\displaystyle \frac{-(-230)\pm \sqrt{{(-230)}^2-(4\times 8\times -750)}}{2\times 8}}$
 $t={\displaystyle \frac{230\pm \sqrt{28900}}{16}}$
 $t={\displaystyle \frac{230\pm 170}{16}}$
It will be divided into two cases:
 $t={\displaystyle \frac{230+170}{16}}$ $t={\displaystyle \frac{230-170}{16}}$
 $t={\displaystyle \frac{400}{16}}$ $t={\displaystyle \frac{60}{16}}$
 $t=25$ $t={\displaystyle \frac{15}{4}}$
When t =25
Time taken by smaller tank = 25 hrs
Time taken by larger tank = 25-10 = 15 hrs
When $t={\displaystyle \frac{15}{4}}$
Time taken by smaller tank $$={\displaystyle \frac{15}{4}}hrs$$
Time taken by larger tank $$={\displaystyle \frac{15}{4}}-10\text{ hrs}={\displaystyle \frac{-25}{4}}hrs$$
Time can never be negative, hence this is wrong.
Therefore, time taken by the smaller water tap to fill separately one complete tank is 25 hours and the larger tank is 15 hours.

Note: We do not use improper fractions while solving the questions and hence they are converted to mixed fractions first.Whole fraction values are taken as 1, as here, we have taken 1 for a completely filled tank.