Question

Two pipes running together can fill a tank in 11 1/9 minutes .if 1pipe takes 5min more than the other to fill the tank seperately. Find the time in which each pipe would fill the tank seperately.

Answer 1

let the time taken by 1 pipe a be x
then pipe taken by 2nd pipe b =x+5
time taken by both pipe together 11 1/9 or 100/9
1/a+1/b=1/t
1/x+1/(x+5)=9/100
(x+5+x)/x(x+5)=100/9
100(2x+5)=9(x2+5x)
200x+500=9x2+45x
9x2-155x-500=0
9x2-180x+25x-500=0
9x(x-20)+25(x-20)=0
(x-20)(9x+25)=0
since time cannot be negative
x=20min
x+5=25min

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the time taken by faster pipe be x minutes.

And the time taken by slower pipe be (x + 5) minutes.

Tank filled by faster pipe in 1 minute = 1/x

Tank filled by faster pipe in 1 minute = 1/(x + 5)

According to the Question,

⇒ 1/x + 1/(x + 5) = 9/100

⇒ (x + 5 + x)/x(x + 5) = 9/100

⇒ (2x + 5)/(x² + 5x) = 9/100

⇒ 200x + 500 = 9x² + 45x

⇒ 9x² + 45x - 200x - 500 = 0

⇒ 9x² - 155x - 500 = 0

⇒ 9x² - 180x + 25x  - 500 = 0

⇒ 9x(x - 20) + 25(x - 20) = 0

⇒ (9x + 25) (x - 20) = 0

⇒ x = - 25/9, 20 (As x can't be negative)

⇒ x = 20 minutes.

Faster Pipe = x = 20 minutes

Slower Pipe = x + 5 = 20 + 5 = 25 minutes.