Two pipes are connected to the same tank. When working together, they can fill the tank in 12 hrs. The larger pipe, working alone, can fill the tank in 18 hrs less time than the smaller one. How long would the smaller one take, working alone, to fill the tank?
rate of smaller pipe --- 1/x
rate of larger pipe ---- 1/(x-18) , where x > 18
combined rate = (x-18 + x)/(x(x-18)
= (2x-18)/(x^2 - 18x)
1/[(2x-18)/(x^2 - 18x)] = 12
x^2 - 18x = 24x - 216
x^2 - 42x + 216 = 0
(x - 36)(x - 6)= 0
x = 36 or x = 6
but x = 6 does not fall into my defined domain
so x = 36 <----- It would take the smaller pipe 36 hours by itself
check:
rate #1 = 1/36
rate # 2 = 1/18
combined rate = 1/36 + 1/18 = 1/12
time at combined rate = 1/(1/12) = 12
The small pipe will fill the tank in 18 hours.
Let's assume that the smaller pipe takes x hours to fill the tank on its own.
Therefore, the larger pipe would take x - 18 hours to fill the tank on its own.
To find the rate at which each pipe fills the tank, we can use the concept of "work done per unit of time."
Let the smaller pipe's rate be R1 (tank per hour) and the larger pipe's rate be R2 (tank per hour).
Working together, both pipes can fill the tank in 12 hours, so their combined rate would be:
1/12 tank per hour (R1 + R2 = 1/12).
The rate of the smaller pipe alone would be R1 = 1/x tank per hour.
The rate of the larger pipe alone would be R2 = 1/(x - 18) tank per hour.
Now, we can set up an equation using the rates:
1/x + 1/(x - 18) = 1/12
To solve this equation, we need to find a common denominator:
(x - 18 + x)/x(x - 18) = 1/12
(2x - 18)/x(x - 18) = 1/12
Cross-multiplying, we get:
12(2x - 18) = x(x - 18)
24x - 216 = x^2 - 18x
Rearranging the equation and simplifying, we have:
x^2 - 42x + 216 = 0
This quadratic equation can be factored as:
(x - 6)(x - 36) = 0
So, x = 6 or x = 36.
Since the time taken cannot be negative, the smaller pipe would take 6 hours (x = 6) to fill the tank on its own.
Therefore, the smaller pipe, working alone, would take 6 hours to fill the tank.
To solve this problem, let's assume that the smaller pipe takes X hours to fill the tank when working alone.
Now, we know that the larger pipe can fill the tank in 18 hours less time than the smaller pipe. Therefore, the larger pipe takes (X - 18) hours to fill the tank when working alone.
When the two pipes work together, they can fill the tank in 12 hours. By understanding the concept of work, we can say that the combined rate of the two pipes is equal to the sum of their individual rates.
Let's express this mathematically:
Rate of the smaller pipe when working alone = 1/X tank per hour
Rate of the larger pipe when working alone = 1/(X-18) tank per hour
Rate of the two pipes working together = 1/12 tank per hour
Using the concept of rate, we can write the equation:
1/X + 1/(X-18) = 1/12
To solve this equation, we can multiply all the terms by the least common denominator (LCD) to get rid of the denominators. The LCD in this case is 12X(X-18).
12(X-18) + 12X = X(X-18)
Distributing and simplifying,
12X - 216 + 12X = X^2 - 18X
24X - 216 = X^2 - 18X
Rearranging,
X^2 - 18X - 24X + 216 = 0
Simplifying,
X^2 - 42X + 216 = 0
To solve this quadratic equation, we can factor it or apply the quadratic formula:
(X - 6)(X - 36) = 0
This gives us two solutions: X = 6 and X = 36.
Since X represents the time taken by the smaller pipe working alone, we discard X = 36 since it would mean that the smaller pipe takes more time than the larger pipe.
Therefore, the smaller pipe takes 6 hours, working alone, to fill the tank.