Question

It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled . How long it for each pipe to fill pool separately?

Answer 1

let larger pipe rate be x hrs/job

let smaller pipe rate be y hrs/job

A.T.Q.

x+y=1/12

4x+9y=1/2

by modifing we get

24x+24y=2

24x+54y=3

subtract and solve y

30y=1

y=1/30job/hr [rate of the smaller diameter pipe ]

smaller pipe would take 30 hrs to fill the pool

solve for x

x+[1/30]=1/12

x=[1/12]-[1/30]

x=(18/[12*13])

x=1/20 job/hr

large pipe would take 20 hrs to fill the pool

thnx :)

Answer 2

SOLUTION :  

Let the pipe of larger diameter fill the tank in x hours.

the pipe of Smaller diameter fills the tanks in (x +10) hours.

In 1 hr the part of the pool filled by the pipe of larger diameter = 1/x  

In 4 hr the part of the pool filled by the pipe of larger diameter = 4 × 1/x = 4/x

In 1 hr the part of the pool filled by the pipe of Smaller diameter = 1/(x + 10)

In 9 hr the part of the pool filled by the pipe of Smaller diameter = 9 × 1/(x + 10) = 9/(x + 10)

A.T.Q

Given : Half of the pool can be filled  

(4/x) + (9/(x+10)) = ½

[4(x + 10) + 9x] / [(x) (x + 10)] = ½

[By taking LCM]

(4x + 40 + 9x) / (x² + 10x) = ½

(13x + 40 ) / (x² + 10x) = ½

2(13x + 40 ) = (x² + 10x)  

26x + 80 = x² + 10x

x² + 10x - 26x - 80 = 0

x² -16x - 80 = 0

x² - 20x + 4x - 80 = 0

[By middle term splitting]

x(x-20) + 4(x-20) =0

(x + 4)(x - 20) = 0

(x + 4) = 0  or  (x - 20) = 0

x = - 4  or x = 20

Since, Time can't be negative , so x ≠ - 4

Therefore, x = 20

Time taken by pipe of larger diameter to fill the tank = 20 minutes.

Time taken by pipe of Smaller diameter to fill the tank = (x + 10 ) =  20 + 10 = 30 minutes.

Hence, pipe of larger diameter takes 20 minutes to fill the tank while pipe of Smaller diameter takes 30 minutes to fill the tank separately.  

HOPE THIS ANSWER WILL HELP YOU….