Question

1. Pipe A alone can fill a reservoir in 12 min. Pipe B can fill it in only 8 min. How long would it take both pipes to full the reservoir?
Show your Solution​

Answer 1

Answer:

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.

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Answer 2

Step-by-step explanation:

Answer:

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.

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