Two water pipes together can fill a tank in9/3/8 the pipe of larger di

1.	Two water pipes together can fill a tank in9/3/8 the pipe of larger diameter takes 10 hours less than the smaller one to fill the tank separately find the time in which each pipe can fill the tank separately
Answer» Answer: Answer: Let the tap with SMALLER diameter fills the tank alone in X HOURS Let the tap with larger diameter fills the tank alone in (x – 10) hours. In 1 hour, the tap with a smaller diameter can FILL 1/x part of the tank. In 1 hour, the tap with larger diameter can fill 1/(x – 10) part of the tank. The tank is filled up in 75/8 hours. Thus, in 1 hour the taps fill 8/75 part of the tank. 1/x + 1/(x-10) = 8/75 (x-10) + x / x(x-10) = 8/75 2x – 10/x(x-10) = 8/75 75 (2x-10) = 8(x2-10x) by cross multiplication 150x – 750 = 8x2 – 80x 8x2 − 230x + 750 = 0 4x2−115x + 375 = 0 4x2 − 100x −15x + 375 = 0 4x(x−25)−15(x−25) = 0 (4x−15)(x−25) = 0 4x−15 = 0 or x – 25 = 0 x = 15/4 or x = 25 Case 1: When x = 15/4 Then x – 10 = 15/4 – 10 ⇒ 15-40/4 ⇒ -25/4 Time can never be negative, so x = 15/4 is not possible. Case 2: When x = 25 then x – 10 = 25 – 10 = 15 ∴ The tap of smaller diameter can separately fill the tank in 25 hours, and the time taken by the larger tap to fill the tank = (25 – 10) = 15 hours.

Two water pipes together can fill a tank in9/3/8 the pipe of larger diameter takes 10 hours less than the smaller one to fill the tank separately find the time in which each pipe can fill the tank separately

Answer»

Answer:

Answer:

Let the tap with SMALLER diameter fills the tank alone in X HOURS

Let the tap with larger diameter fills the tank alone in (x – 10) hours.

In 1 hour, the tap with a smaller diameter can FILL 1/x part of the tank.

In 1 hour, the tap with larger diameter can fill 1/(x – 10) part of the tank.

The tank is filled up in 75/8 hours.

Thus, in 1 hour the taps fill 8/75 part of the tank.

1/x + 1/(x-10) = 8/75

(x-10) + x / x(x-10) = 8/75

2x – 10/x(x-10) = 8/75

75 (2x-10) = 8(x2-10x) by cross multiplication

150x – 750 = 8x2 – 80x

8x2 − 230x + 750 = 0

4x2−115x + 375 = 0

4x2 − 100x −15x + 375 = 0

4x(x−25)−15(x−25) = 0

(4x−15)(x−25) = 0

4x−15 = 0 or x – 25 = 0

x = 15/4 or x = 25

Case 1: When x = 15/4

Then x – 10 = 15/4 – 10

⇒ 15-40/4

⇒ -25/4

Time can never be negative, so x = 15/4 is not possible.

Case 2: When x = 25 then

x – 10 = 25 – 10 = 15

∴ The tap of smaller diameter can separately fill the tank in 25 hours, and the time taken by the larger tap to fill the tank = (25 – 10) = 15 hours.