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Two taps running together can fill a tank in $3\frac{1}{13}$ hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank ?

Answers (1)

Let the volume of the tank be $V\;m^3$

Both taps A and B take $3\frac{1}{13}$ hours to fill $V\; m^3$ .

Let tap A fill at $V_A\; m^3/hr$

And tap B fill at $V_B\; m^3/hr$

Time taken by tap A to fill the tank $=x\; hr$

Time taken by tap B to fill the tank $=(x+3)\; hr$

According to question :

$\left ( V_A+V_B \right )\times \left ( 3\frac{1}{13} \right )=V\; \; \; -(i)$

$V_A\times x=V\; \; \; -(ii)$

$V_B\times (x+3)=V\; \; \; -(iii)$

From eq (ii) $\Rightarrow V_A=\frac{V}{x}$

From eq (ii) $\Rightarrow V_B=\frac{V}{x+3}$

Putting value of $V_A$ and $V_B$ in eq (i)

$\left ( \frac{V}{x}+\frac{V}{x+3} \right )\times 3\frac{1}{13}=V$

$V\left ( \frac{1}{x}+\frac{1}{x+3} \right )\times \frac{40}{13}=V$

$\left ( \frac{x+3+x}{x(x+3)} \right )\times \frac{40}{13}=1$ (cancelling V on both sides)

$(2x+3)\times 40=13[x^2+3x]$

$80x+120=13x^2+39x$

$\Rightarrow 13x^2+39x-80x-120=0$

$\Rightarrow 13x^2-41x-120=0$

$\Rightarrow x=\frac{+41\pm \sqrt{(41)^2-4\times 13\times (-120)}}{2\times 13}$

$\Rightarrow x=\frac{41\pm \sqrt{1681+6240}}{26}$

$\Rightarrow x=\frac{41\pm \sqrt{7921}}{26}$

$\Rightarrow x=\frac{41\pm 89}{26}$

$\Rightarrow x=\frac{41+ 89}{26},\frac{41-89}{26}$

$\Rightarrow x=5,\frac{-24}{13}$

Since $x$ cannot be negative, hence $x=5$ hours.

Time taken by tap A to fill the tank $=x=5 \; hour\!s$

Time taken by tap B to fill the tank $=x+3=8 \; hour\!s$

Posted by

Ravindra Pindel

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