Question

Two pipes together can fill a water tank is 6 hrs and 40min. Find the time each will take to fill the tank if one of the two pipes can fill it in 3hrs less than the other.

Answer 1

$\mathfrak{\huge{\underline {Answer:}}}$

$\sf{Let\:the\:time \:taken\:by\:one\:tap\:be}$ = x hours

(Let this tap be = $\sf{tap_{1}}$)

$\sf{The\:time \:taken\:by\:the\:other\:tap}$ = x + 3 hours

(Let this tap be = $\sf{tap_{2}}$)

Next step :-

$\sf{One\:hour\:work\:of\:tap_{1} }$ = $\frac{1}{x}$

$\sf{One\:hour\:work\:of\:tap_{2} }$ = $\frac{1}{x+3}$

Now, the total time is given to be = 6 hours and 40 minutes = $\frac{20}{3}$ hours

$\sf{The\:one\:hour\:work\:together}$ = $\frac{3}{20}$

We can write that:-

=》 $\sf{\frac{1}{x} + \frac{1}{x + 3} = \frac{3}{20}}$

Solve this formed equation further

=》 $\sf{\frac{2x + 3}{x^{2} + 3x} = \frac{3}{20}}$

Cross multiply the equation further

=》 $\sf{40x + 60 = 3x^{2} + 9x}$

=》 $\sf{3x^{2} - 31x - 60 = 0}$

Use the quadratic formula for further calculations

$\mathfrak{Quadratic \:Equation}$ = $\tt{\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

Here, the values of the variable will be :-

a = 3

b = (-31)

c = (-60)

=》 $\tt{\frac{31 \pm \sqrt{(-31)^{2} - 4(3)(-60)}}{2(3)}}$

=》 $\tt{\frac{31 \pm \sqrt{961 - 720}}{6}}$

=》 $\tt{\frac{31 \pm \sqrt{1681}}{6}}$

=》 $\tt{\frac{31 \pm 41}}{6}}$

=》 x = $\tt{\frac{31 + 41}{6}}$, $\tt{\frac{31- 41}{6}}$

=》 x = $\tt{\frac{72}{6} , \frac{-10}{6}}$

=》 x = $\tt{12, \frac{-5}{3}}$

Since Time > 0, the only accepted value is 12.

Hence, time taken by $\sf{tap_{1}}$ = 12 hours

Time taken by another tap = 12 + 3 = 15 hours

Answer : $\sf{\huge{12\:hours,\:15\:hours}}$

Answer 2

Solution:-

Let the time taken to fill the tank by smaller tap be "x" hours.

=> Time taken to fill the tank by the larger tank = ( x+3) hours.

Now,

Tank filled in 1 hour by smaller tap = 1/x

& Tank filled in 1 hour by Larger tap = 1/(x+3).

Time taken by both the tapes to fill the tank = 6 hour 40 minute.

=> 6 + 40/60 = 20/3 hours.

A.T.Q.

$= > \frac{1}{x} + \frac{1}{x + 3} = \frac{3}{20} \\ \\ = > \frac{x + 3 + x}{x(x + 3)} = \frac{3}{20} \\ \\ = > 20(3 + 2x) = 3( {x}^{2} + 3x) \\ \\ = > 60 + 40x = 3 {x}^{2} + 9x \\ \\ = > 3 {x}^{2} + 9x - 40x - 60 = 0 \\ \\ = > 3 {x}^{2} - 31x - 60 = 0 \\ \\ = > 3 {x}^{2} - (36 - 5)x - 60 = 0 \\ \\ = > 3 {x}^{2} - 36x + 5x - 60 = 0 \\ \\ = > 3x(x - 12) + 5(x - 12) = 0 \\ \\ = > (x - 12)(3x + 5) = 0 \\ \\ = > x = 12 \: \: \: \: and \: \: \: \: \: x = \frac{ - 5}{3} (neglect)$

Hence,

Time Taken by the smaller tap to fill the tank = 12 hours.

& Time taken by the larger tap to fill the tank = 12 +3 = 15 hours.