Two pipes can fill a tank in 9 hours. If the bigger of the pipes has a rate of 1 ½ times of the smaller, how long would it take each pipe to fill the tank if opened alone?
(3 / 2s) + (1 / s) = 1 / 9
27/2 + 9 = s = 45/2
b = 2/3 s = 15
To solve this problem, we can assign variables to represent the rates of the two pipes. Let's say the rate of the smaller pipe is r (in tanks per hour). Then, the rate of the larger pipe is 1 ½ times the rate of the smaller pipe, which is 1.5r (in tanks per hour).
We are also given that both pipes together can fill the tank in 9 hours. This means that the combined rate of the two pipes is equal to 1 tank per 9 hours, or 1/9 tanks per hour.
Using the rates, we can set up an equation:
r + 1.5r = 1/9
Combining like terms, we have:
2.5r = 1/9
To isolate r, we divide both sides of the equation by 2.5:
r = (1/9) ÷ 2.5
r = 1/22.5
r = 2/45
Now, we know the rate of the smaller pipe is 2/45 tanks per hour. To find out how long it would take for the smaller pipe to fill the tank alone, we take the reciprocal of the rate:
1 / (2/45)
Simplifying, we get:
1 ÷ (2/45) = 45/2 = 22.5 hours
Therefore, it would take the smaller pipe 22.5 hours to fill the tank if opened alone.
Since the rate of the larger pipe is 1.5 times that of the smaller pipe, its rate is 1.5 * (2/45) = 3/45 tanks per hour. To find out how long it would take for the larger pipe to fill the tank alone, we take the reciprocal of the rate:
1 / (3/45)
Simplifying, we get:
1 ÷ (3/45) = 45/3 = 15 hours
Therefore, it would take the larger pipe 15 hours to fill the tank if opened alone.