Math, asked by akulshrestha15, 9 months ago

two water taps together can fill a tank in 9whole 3/8 hours . the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately . find the time in which each tap can separately fill te tank ( please iska sahi answer dena ye class 10th ncert ka question hai . chapter 4 ex 4.3 question no 9)​

Answers

Answered by Garvit951
2

Answer:

Step-by-step explanation:

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(x – 10)

As given, the tank can be filled in 9 3/8 = 75/8 hours by both the pipes together.

Therefore,

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) = 8x2 – 80x

⇒ 150x – 750 = 8x2 – 80x

⇒ 8x2 – 230x +750 = 0

⇒ 8x2 – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

Answered by Anonymous
0

Answer:

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