Question

♡HOLA

Three pipe X, Y. and Z are fitted to a tank. For any pipe, the rate of filling is the same as that
of the rate of emptying. The rates of filling of X, Y. and Z are in the ratio 2:3:4. X alone
can fill the tank in 5 h. Find the time taken in hours, to fill the tank if X is used as an emptying
pipe, whereas the other two are used as filling pipes.


pls. solve it asap.....
​

Answer 1

Answer:

The time required is 2 hours.

Step-by-step explanation:

Given :

Ratio of rates filling as well as emptying the tank - 2 : 3 : 4

X takes = 5 hours to fill the tank

To find :

Time required to fill the tank when X is emptying and Y and Z are used filling

Solution :

Let the 'a' be the common proportion,

Rate of filling/emptying of Pipe X = 2a litres per hour

Rate of filling/emptying of Pipe Y = 3a litres per hour

Rate of filling/emptying of Pipe Z = 4a litres per hour

X alone takes 5 hours to fill the tank, the capacity of the tank =

$\sf{\longrightarrow} \: \dfrac{Capacity \: of \: tank}{2a} = 5 \: hours \\ \\ \sf{\longrightarrow} Capacity \: of \: tank = 2a \times 5 \\ \\ \sf{\longrightarrow} \: Capacity \: of \: tank = 10a \: litres$

Capacity = 10a litres

$\rule{300}{1.5}$

$\textsf{\underline{\underline{According to the question -}}}$

Pipe X is emptying the tank whereas Pipe Y and Pipe Z are filling the tank.

The filling pipe does a positive work (+) where as the tank which empties or leaks does a negative work (–).

Add the rates of each pipe to get the rate of filling the tank per hour in the given condition.

$\tt{\longrightarrow} \: 3a + 4a + ( - 2a) \\ \\ \sf{\longrightarrow} \:7a + ( - 2a) \\ \\ \sf{\longrightarrow} \:5a$

Rate of filling the tank = 5a litres per hour

$\rule{300}{1.5}$

★ $\textsf{\underline{\underline{Time required to fill the tank - }}}$

Capacity of the tank = 10a litres

$\boxed{\sf{Time \: required = \frac{Capacity \: of \: the \: tank}{Rate \: of \: filling}}}$

Divide the capacity by rate =

$\sf{\longrightarrow} \: \dfrac{10a}{5a} \\ \\ \sf{\longrightarrow} \:2$

Time required = 2 hours

$\therefore$ The time required is 2 hours.

Answer 2

$\huge \boxed{ \fcolorbox{cyan}{red}{Answer : }}$

●Given:

☆Three pipe X, Y. and Z are fitted to a tank. For any pipe, the rate of filling is the same as that

of the rate of emptying.

☆The rates of filling of X, Y. and Z are in the ratio 2:3:4

$\sf{so \: x \: alone \: can \: fill \: the \: tank \: in \: 5hours}$

$\sf{we \: have \: to \: find \: the \: tank \: required \: when \:x \: is \: emptying}$

$\sf{y \: and \: z \: are \: used \: filling}$

$\rm{let \: take \: the \: a}$

●$\sf{emptying \: of \: x = 2a}$

●$\sf{emptying \: of \: y = 3a}$

●$\sf{emptying \: of \: y = 4a}$

●$\tt{so \: this \: is \: litre \: per \: hour}$

___________________________

●$\sf{X \: alone \: takes \: 5 hours \: to \: fill \: the \: tank}$

●$\tt{capacity \: of \: the \: tank = }$

◍$\sf{ \frac{capacity \: of \: tank}{2a} = 5hr}$

◍$\sf{capacity \: of \: tank = 2a \times 5}$

❃$\sf{capacity \: of \: tank = 10a \: liter}$

❃$\bf{ \huge{ \boxed{ \red{ \tt{capacity \: of \: tank \: 10a \: liter \: }}}}}$

so then

❃X is used as an emptying pipe, whereas the other two are used as filling pipes.

☆so we have to add the rates of pipes to get the rate of filling the tank per hour

◇$\rm{3a + 4a + ( - 2a)}$

◇$\rm{7a + ( - 2a)}$

◇$\rm{5a}$

◇$\tt{rate \: of \: filling \: the \: tank = 5a \: liter \: pr \: hr}$

now time required to fill the tank

$\sf{tank \: of \: capacity = 10a \: liters}$

so

$\tt{time \: required = \frac{capacity \: of \: tank}{rate \: of \: filling}}$

$\sf{then}$

$\rm{ \frac{5a}{10a}}$

$\sf{2}$

$\sf{ \huge \boxed{ \pink{ \sf{time \: required = 2hours : }}}}$