Question

It can take 12 hours to fill the swimming pool using 2 pipes. if the pipe of larger diameter is used for 3 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool can be filled. how long would it take for each pipe to fill the tank separately pls experts answer this one

Answer 1

$\sf\large\underline{Let:-}$

$\sf{\implies Time\:taken\:by\:_{(small\:pipe)}=x}$

$\sf{\implies Time\:taken\:by\:_{(large\:pipe)}=y}$

$\sf\large\underline{To\: Find:-}$

$\sf{\implies Time\:taken\: separately\:by\:_{(each\:pipe)}=?}$

$\sf\large\underline{Solution:-}$

To calculate the time taken to fill the pool separately at first we have to set up equation as per the given clue in the question then solve the equation. After solving the equation you got the time taken to fill the pool separately.

$\sf\large\underline{Calculation\:for\:1st\:equation:-}$

$\sf{\implies Time\:taken\:in\:an\:hour\:_{(small\:pipe)}=\frac{1}{x}}$

$\sf{\implies Time\:taken\:in\:an\:hour\:_{(large\:pipe)}=\frac{1}{y}}$

$\sf{\implies Time\:taken\:_{(small\:pipe+large\:pipe)}=\frac{1}{12}}$

$\tt{\implies \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}}$

$\tt{\implies \dfrac{y+x}{xy}=\dfrac{1}{12}}$

$\tt{\implies 12y+12x=xy-----(i)}$

$\sf\large\underline{Calculation\:for\:2nd\:equation:-}$

$\sf{\implies Time\:taken\:in\:an\:hour\:_{(small\:pipe)}=\frac{9}{x}}$

$\sf{\implies Time\:taken\:in\:an\:hour\:_{(large\:pipe)}=\frac{3}{y}}$

$\tt{\implies \dfrac{9}{x}+\dfrac{3}{y}=\dfrac{1}{2}}$

$\tt{\implies \dfrac{9y+3x}{xy}=\dfrac{1}{2}}$

$\tt{\implies 18y+6x=xy-----(ii)}$

• In eq (i) multiple by 18 and (ii) by 12 then subract:-]

$\tt{\implies 216y+216x=18xy}$

$\tt{\implies 216y+72x=12xy}$

• By solving we get here:-]

$\tt{\implies 144x=6xy}$

$\tt{\implies 6y=144}$

$\tt{\implies y=24}$

• Putting the value of y=24 in eq (i):-]

$\tt{\implies 12x+12y=xy}$

$\tt{\implies 12x+12*24=24*x}$

$\tt{\implies 12x+288=24x}$

$\tt{\implies 24x-12x=288}$

$\tt{\implies 12x=288}$

$\tt{\implies x=24}$

$\sf\large{Hence,}$

$\sf{\implies Time\:taken\:by\:_{(small\:pipe)}=24\:hours}$

$\sf{\implies Time\:taken\:by\:_{(large\:pipe)}=24\:hours}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

It takes 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 3 hours  and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled. How long would each pipe  take to fill the swimming pool?

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

It would take 24 hours for each pipe to fill the tank separately

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

Let the pipe with larger diameter be "L" and the one with smaller diameter

be "S"

And also :

Let pipe "L" work at a rate of x hours and pipe "S" work at a rate of y hours

So we can say that :

In 1 hour pipe L can fill the pool = $\sf{\dfrac{1}{x}}$

In 1 hour pipe S can fill the pool = $\sf{\dfrac{1}{y}}$

It is given that if both pipes are together then they take 12 hours to fill the pool

If the are together then the equation can be formulated as in 1 hour they fill the pool:

$\sf{\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12} \quad \ldots \ldots(1)}$

Now, in 3 hour pipe L can fill the pool is =

$\sf{\dfrac{1}{x}\times 3=\dfrac{3}{x}}$

So in 9 hour pipe S can fill the pool is =

$\sf{\dfrac{1}{9}\times 9=\dfrac{9}{y}}$

Then the new equation can be formulated as they fill half of the pool the equation can be written as:

$\sf{\dfrac{4}{\mathrm{x}}+\dfrac{9}{\mathrm{y}}=\dfrac{1}{2} \quad \ldots \ldots(2)}$

Now Let $\dfrac{1}{\mathrm{x}}=\mathrm{p}$ and $\frac{1}{\mathrm{y}}=\mathrm{q}$   $.......(3)$

Putting in equation (1) and (2), than the equations can be written as:

$\sf{p+q=\dfrac{1}{12}}$

Multiplying both sides by 12 :

12p + 12q = 1  $.......(4)$

And :

$\sf{3p+9q=\dfrac{1}{2}}$

Multiplying both sides by 2 :

6p + 18q = 1 $.......(5)$

_________________________

Now, solve for values of p and q

$\begin{bmatrix}\sf{12p+12q=1\\\\ \sf{6p+18q=1}\end{bmatrix}$

_________________________

Step 1 : $\text{Isolate p for 12 p+12 q=1}$

$12p+12q=1$

$\implies12p+12q-12q=1-12q$

$\implies12p=1-12q$

$\implies\displaystyle\frac{12p}{12}=\frac{1}{12}-\frac{12q}{12}$

$\sf{p=\dfrac{1-12q}{12}}$

Step 2 : $\sf{\mathrm{Subsititute\:}p=\dfrac{1-12q}{12}\:\:in\:6p+18q=1}$

$\begin{bmatrix}6\cdot \dfrac{1-12q}{12}+18q=1\end{bmatrix}$

$\implies\begin{bmatrix}\dfrac{1+24q}{2}=1\end{bmatrix}$

Step 3 : $\sf{Isolate\:\: q \:\:for \:\dfrac{1+24 q}{2}=1}$

$\mathrm{For\:}p=\dfrac{1-12q}{12}$

$\mathrm{Subsititute\:}\boxed{\sf{q=\dfrac{1}{24}}}$

$\sf{p=\dfrac{1-12\cdot \tfrac{1}{24}}{12}}$

$\boxed{\sf{p=\dfrac{1}{24}}}$

Hence,

$\boxed{\sf{p=\dfrac{1}{24},\:q=\dfrac{1}{24}}}$

_________________________

From (3) we can say that x = 24 and y =24 Hence the pipe "L" would take 24 hours and the pipe "S" would take 24 hours to fill the pool separately.