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In order to fill an empty swimming pool completely, a pipe of larger diameter alone takes 10 hour less than the time taken by the pipe of the smaller diameter alone.

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Question

In order to fill an empty swimming pool completely, a pipe of larger diameter alone takes 10 hour less than the time taken by the pipe of the smaller diameter alone. If the pipe of the larger diameter is used for 4 hours and the pipe of the smaller diameter is used for 9 hours, half of the pool is filled. In how many hours will the pipe of the larger diameter alone fill the pool.

Sum
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Solution

Let the time taken by the pipe of larger diameter to fill the pool completely be x hours and the pipe of smaller diameter be y hours.

In one hour,

Part of the pool filled by the pipe of larger diameter = `1/x`

Part of the pool filled by the pipe of smaller diameter = `1/y`

According to question,

`4/x+9/y=1/2`     .....(i)

y − x = 10     .....(ii)

Substituting the value of y from (ii) in (i), we get

`4/x + 9/(x + 10) = 1/2`

`(4(x+10)+9x)/((x+10)x)=1/2`

`(4x+40+9x)/(x^2+10x)=1/2`

`(13x+40)/(x^2+10x)=1/2`

26x + 80 = x2 + 10x

x2 − 16x − 80 = 0

x2 − 20x + 4x − 80 = 0

x(x − 20) + 4(x − 20) = 0

(x + 4)(x − 20) = 0

x = 20

Therefore, the time taken by the pipe of larger diameter to fill the pool is 20 hours.

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Chapter 6: Solving (simple) Problems (Based on Quadratic Equations) - EXERCISE 6(C) [Page 69]

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Selina Concise Mathematics [English] Class 10 ICSE
Chapter 6 Solving (simple) Problems (Based on Quadratic Equations)
EXERCISE 6(C) | Q 8. | Page 69
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