Two water taps together fill a tank in
Question:

Two water taps together fill a tank in $1 \frac{7}{8}$ hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

 

Solution:

Let the smaller tap takes x hours to fill the tank.
Then, the larger one takes (x − 2) hours to fill the tank

Tank filled in 1 hour by smaller tap $=\frac{1}{x}$

Tank filled in 1 hour by larger tap $=\frac{1}{x-2}$

 

Tank filled in 1 hour by both the taps $=\frac{8}{15}$

According to the question,

$\frac{1}{x}+\frac{1}{x-2}=\frac{8}{15}$

$\Rightarrow \frac{(x-2)+(x)}{(x)(x-2)}=\frac{8}{15}$

$\Rightarrow \frac{2 x-2}{x^{2}-2 x}=\frac{8}{15}$

$\Rightarrow 15(2 x-2)=8\left(x^{2}-2 x\right)$

$\Rightarrow 30 x-30=8 x^{2}-16 x$

$\Rightarrow 8 x^{2}-46 x+30=0$

$\Rightarrow 4 x^{2}-23 x+15=0$

$\Rightarrow x=\frac{-(-23) \pm \sqrt{(-23)^{2}-4(4)(15)}}{2(4)}$

$\Rightarrow x=\frac{23 \pm \sqrt{529-240}}{8}$

$\Rightarrow x=\frac{23 \pm \sqrt{289}}{8}$

$\Rightarrow x=\frac{23 \pm 17}{8}$

$\Rightarrow x=\frac{23+17}{8}$ or $x=\frac{23-17}{8}$

$\Rightarrow x=5$ or $x=0.75$

But $x \neq 0.75$ $(\because(x-2)$ becomes negative)

Thus, $x=5$

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.

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