Question

It can take 12hrs to fill 2 swimming pool pipes....if the pipe of the larger diameter is used for 4hrs and the smaller one is used for 9hrs only half the pool can be filled...how long would it take to fill the pool seperatley

Answer 1

Let the time taken to fill the tank with the larger diameter be ⇒ 'x' hours

So therefore, the smaller diameter will fill the tank in ⇒ $(x+10)$ hours.

With this information, we know that-

The portion of the pool that will be filled in 1 hour by the pipe of the larger diameter would be ⇒ $\frac{1}{x}$

The portion of the pool that will be filled in 1 hour by the pipe of the smaller diameter would be ⇒ $4\times \frac{1}{x} =\frac{4}{x}$

So we can say-

The portion of the pool filled by the pipe of the smaller diameter in 9 hours would be ⇒ $9\times1\div (x +10) = 9\div (x+10)$

Now lets answer the question with the provided information.

According to the question, half of the pool can be filled  

$\frac{4}{x}+(9\div(x+10))=\frac{1}{2}$

$[4(x+10)+9x]\div [(x)(x+10)]=\frac{1}{2}$

$(4x+40+9x)\div (x^2+10x)=\frac{1}{2}$

$(13x+40)\div (x^2+10x)=\frac{1}{2}$

$2(13x+40)=(x^2+10x)$

$26x+80=x^2+10x$

$x^2+10x-26x-80=0$

$x^2-16x-80=0$

$x^2 - 20x + 4x - 80 = 0$

$x(x-20) + 4(x-20) =0$

$(x + 4)(x - 20) = 0$

$(x + 4) = 0 \ or\ (x - 20) = 0$

$x = - 4\ or \ x = 20$

We know that time cannot be represented in negative form. So here the value of 'x' is not (-4)

∴ x = 20

Time taken by pipe of larger diameter to fill the tank ⇒ 20 min

Time taken by pipe of Smaller diameter to fill the tank ⇒30 min

∴ The pipe with the larger diameter takes 20 minutes to fill the tank and the pipe with the smaller diameter takes 30 minutes to fill the tank.

Answer 2

$\huge \purple {Answer}$

Let the time taken to fill the tank with larger diameter= x hrs

Therefore, the smaller diameter will fill the tank= (x+10)(x+10) hrs

We know that:-

Portion of the pool that will be filled in 1 hour by the pipe of larger diameter= $\frac {1}{x}$

Portion of the pool that will be filled in 1 hour by the pipe of the smaller diameter= 4×$\frac{1}{x}$ =$\frac{4}{x}$

Therefore, Portion of the pool filled by the pipe of the smaller diameter in 9 hours= 9×1÷ (x +10) =9÷(x+10)

According to Statement:-

$\frac{4}{x}$+(9÷(x+10))= $\frac {1}{2}$

[4(x+10)+9x]÷[(x)(x+10)]=$\frac{1}{2}$

(4x+40+9x) ÷ (x²+10x)=$\frac{1}{2}$

(13x+40)÷(x²+10x)=$\frac{1}{2}$

2(13x+40)=(x​²+10x)

26x+80=x²+10x

x²+10x-26x-80=0

x²-16x-80=0

x² - 20x + 4x - 80 = 0

x(x-20) + 4(x-20) =0

(x + 4)(x - 20) =0

(x + 4) = 0 or (x - 20) =0

x = - 4 or x = 20 

We know that time can never be negative.

$\boxed {x = 20}$

Therefore, Time taken by pipe of larger diameter to fill the tank= 20 minutes

Time taken by pipe of Smaller diameter to fill the tank= 30 minutes

∴ The pipe with the larger diameter takes 20 minutes and the pipe with the smaller diameter takes 30 minutes to fill the tank.