Question

If two pipes function simultaneously, a reservoir will be filled in 12hrs. First pipe fills the reservoir 10hrs faster than the second pipe. How many hours will the second pipe take to fill the reservoir

Answer 1

let the time taken by smaller pipe be x hr
therefore time taken by large pipe!=x +10
by applying unitary method u will get following :::::
1/x +1/x +10=1/12
on solving u will get
x +x +10/x sq. +10x =1/12
solve it......
u will get
x=40hr
and larger =50
hopw this may help u!!!!!
mark as brainliest!!!

Answer 2

SOLUTION :  

Let the faster pipe fill the reservoir in x h.

Then, the  slower pipe the reservoir in (x + 10) h

In 1 hour the faster pipe fills the portion of the reservoir : 1/x  

In 12 hour the faster pipe fills the portion of the reservoir : 12 ×  1/x  = 12/x

In 1 hour the slower pipe fills the portion of the reservoir : 1/(x + 10)

In 12 hour the faster pipe fills the portion of the reservoir : 12 × 1/(x + 10) = 12/(x +10)

A.T.Q  

12/x + 12/(x +10) = 1

12(1/x + 1/(x + 10) ) = 1

1/x + 1/(x + 10) = 1/12

(x + 10 + x ) / [x(x + 10)] = 1/12  

[By taking LCM]

2x +10 /(x² + 10x) = 1/12  

x² + 10x = 12(2x +10)

x² + 10x = 24x + 120

x² + 10x - 24x - 120 = 0

x² - 14x - 120 = 0

x² + 6x - 20x - 120 = 0

[By splitting middle term]

x(x + 6) - 20(x + 6) = 0

(x - 20) (x + 6) = 0

(x - 20) or  (x + 6) = 0

x = 20  or  x = - 6

Since, time cannot be negative, so x ≠ - 6  

Therefore, x = 20  

The faster pipe takes 20 hours to fill the reservoir

Hence, the slower pipe (second) takes (x + 20) = 30 hours to fill the reservoir.

HOPE THIS ANSWER WILL HELP YOU…...