Two pipes together can fill a tank in 20/3 hours if the pipe of smaller diameter takes 3 hours more than the pipe of larger diameter to fill the tank separately find the time taken by each pipe to fill the tank separately
Answers
Answer :
Let,
A pipe of big diameter is Y, and another smaller diameter pipe is X
According to the question,
Two pipes fill a tank together in 20/3 hours
So,
X + Y =20/3 ..........(1)
The pipes of small diameter take more 3 hours more than small diameter of pipe
There for X = Y + 3
X - Y = 3 ...........(2)
sum of equation (1) and (2)
X + Y = 20/3.....(1)
X - Y = 3..........(2)
2X = 20/3 +3
2X = 29/3
X = 29/ 6 hours
Put the value of X in to equation (1)
29/ 6 + Y = 20/ 3
Y = 20/ 3 - 29/ 6
Y = 11/ 6 hours
It means pipes of small diameter take time to fill tank = 11/ 6 hours
and pipes of big diameter take time to fill tank = 29/ 6 hours
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Step-by-step explanation:
Given Two pipes together can fill a tank in 20/3 hours . If pipe of smaller diameter takes 3 hours more than the pipe of larger diameter to fill the tank separately, find the time taken by pipe to fill the tank
- Let pipe A and pipe B fill the tank in p hours and p + 3 hours.
- In 1 hour pipe A , 1/p part of the tank will be full
- Also in 1 hour for pipe B, 1/p + 3 part of the tank will be full.
- So it will be 1/p + 1/p+ 3
- = p + 3 + p / p(p + 3)
- = 2 p + 3 / p(p + 3)
- Therefore 2p + 3 / p(p + 3) part of the tank will be full in 1 hour
- So 1 full tank will fill in 1 / (2 p + 3) / p(p + 3)
- = p(p + 3) / 2p + 3
- Now p(p + 3) / 2p + 3 = 20 / 3
- Or 3p (p + 3) = 40 p + 60
- Or 3p^2 – 31 p – 60 = 0
- Now p = - b + - √b^2 – 4ac / 2a
- = - (- 31) + - √(-31)^2 + - 4 (3)(-60) / 2(3)
- = 31 + - √961 + 720 / 6
- = 31 + - √1681 / 6
- = 31 + - 41 / 6
- = 72 / 6 or - 10 / 6
- = 12 hours
So the time taken by the pipe to fill the tank will be 12 hours.
Reference link will be
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