Question

Two pipes running together can fill a tank in $11\frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Answer 1

SOLUTION :  

Given : Two water taps together can fill a tank in 11 1/9  = 100/9 minutes

Let the faster pipe fill the tank  in ‘x’ minutes

Smaller pipe fills the tank in (x + 5) minutes

In 1 minute the faster pipe fills the portion of the tank : 1/x  

In 100/9 minutes  the faster pipe fills the portion of the tank : 100/9 ×  1/x  = 100/9x

In 1 minute the Smaller pipe fills the portion of the tank : 1/(x + 5)

In 100/9 minutes the Smaller pipe fills the portion of the tank : 100/9 × 1/(x + 5) = 100/9(x + 5)

A.T.Q

100/9x + 100/9(x + 5) = 1

100/9[1/x + 1/(x + 5)] = 1

1/x + 1/(x + 5) = 9/100

(x + 5 + x)/ [x(x + 5)] = 9/100

[By taking LCM]

(2x + 5)/(x² + 5x) = 9/100

100 (2x + 5) = 9(x² + 5x)  

200x + 500 = 9x² + 45x

9x² + 45x - 200x - 500 = 0

9x² - 155x - 500 = 0

9x² - 180x + 25x  - 500 = 0

[By middle term splitting]

9x (x - 20) + 25 (x - 20) = 0

(9x + 25) (x - 20) = 0

(9x + 25) = 0  or  (x - 20) = 0

x = -25/9 or  x = 20

Since, Time can't be negative , so x ≠ - 25/9

Therefore, x = 20

Time taken by faster pipe to fill the tank = 20 minutes.

Time taken by slower pipe to fill the tank = (x + 5 ) =  20 + 5 = 25 minutes.

Hence, faster pipe takes 20 minutes to fill the tank while slower pipe takes 25 minutes to fill the tank.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Let the faster pipe fill the tank  in ‘x’ minutes

Smaller pipe fills the tank in (x + 5) minutes

In 1 minute the faster pipe fills the portion of the tank : 1/x  

In 100/9 minutes  the faster pipe fills the portion of the tank : 100/9 ×  1/x  = 100/9x

In 1 minute the Smaller pipe fills the portion of the tank : 1/(x + 5)

In 100/9 minutes the Smaller pipe fills the portion of the tank : 100/9 × 1/(x + 5) = 100/9(x + 5)

A.T.Q

100/9x + 100/9(x + 5) = 1

100/9[1/x + 1/(x + 5)] = 1

1/x + 1/(x + 5) = 9/100

(x + 5 + x)/ [x(x + 5)] = 9/100

[By taking LCM]

(2x + 5)/(x² + 5x) = 9/100

100 (2x + 5) = 9(x² + 5x)  

200x + 500 = 9x² + 45x

9x² + 45x - 200x - 500 = 0

9x² - 155x - 500 = 0

9x² - 180x + 25x  - 500 = 0

[By middle term splitting]

9x (x - 20) + 25 (x - 20) = 0

(9x + 25) (x - 20) = 0

(9x + 25) = 0  or  (x - 20) = 0

x = -25/9 or  x = 20

Since, Time can't be negative , so x ≠ - 25/9

Therefore, x = 20

Time taken by faster pipe to fill the tank = 20 minutes.

Time taken by slower pipe to fill the tank = (x + 5 ) =  20 + 5 = 25 minutes.

Hence, faster pipe takes 20 minutes to fill the tank while slower pipe takes 25 minutes to fill the tank.