Exercise: 4.3, 9. Two water taps together can fill a tank in 75/ 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 tap can separately fill the tank.

Relate Quarries: two water taps together fill a tank in 9 hours 36 minutes, two water taps together can fill a tank in 15/8 hours, two water taps together can fill a tank in 2 11/12, two water taps together fill a tank in 6 hours, two water taps together can fill a tank in 3 1/13, two water taps together can fill a tank in 100/9 hours, two water taps together can fill a tank in 3 hours 45 minutes, two pipes running together can fill a tank in 11 1/9.

Solution:

Let the time taken by smaller tap to fill the tank completely = x hours

∴ Time taken by larger tap to fill the tank completely = (x - 10) hours

Volume of tank filled by smaller tap in 1 hour = 1 / x

Volume of tank filled by larger tap in 1 hour = 1 / x-10

Given, Tank is completely filled by both taps together in $Exercise: 4.3, 9. Two water taps together can fill a tank in 75/ 8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.$ = 75 / 8 hours

Volume Of Tank filled by both taps in 1 hour= $\frac{1}{\frac{75}{8}}$ $=\frac{8}{75}$

A/Q

Volume of tank filled by smaller tap in 1 hour + Volume of tank filled by larger tap in 1 hour = Volume Of Tank filled by both taps in 1 hour

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow \frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow \frac{x-10+x}{x\left ( x-10 \right )}=\frac{8}{75}$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow \frac{2x-10}{x\left ( x-10 \right )}=\frac{8}{75}$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 75\times \left ( 2x-10 \right )=8\times x\left ( x-10 \right )$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 150x-750=8x^2-80x$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 8x^2-80x-150x+750=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 8x^2-230x+750=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 4x^2-115x+375=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 4x^2-\left ( 100+15 \right )x+375=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 4x^2-100x-15x+375=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 4x\left ( x-25 \right )-15\left ( x-25 \right )=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow \left ( 4x-15 \right )\left ( x-25 \right )=0$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow 4x=15,x=25$

$https://latex.codecogs.com/gif.image?\large \dpi{120}\Rightarrow x=\frac{15}{4},x=25$

If $https://latex.codecogs.com/gif.image?\large \dpi{120}x=\frac{15}{4}$

Then, x - 10

$https://latex.codecogs.com/gif.image?\large \dpi{120}=\frac{15}{4}-10$

$https://latex.codecogs.com/gif.image?\large \dpi{120}=\frac{15-40}{4}$

$https://latex.codecogs.com/gif.image?\large \dpi{120}=\frac{-25}{4}$ hours

Time can't be negative.

If x = 25

Then, x - 10 = 25 - 10 = 15 hours

∴ Time taken by Smaller tap to fill the tank completely = 25 hours

∴ Time taken by Larger tap to fill the tank completely = 15 hours

Exercise: 4.3, 9. Two water taps together can fill a tank in 75/ 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 tap can separately fill the tank.

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