A large pipe can fill a tank 10 minutes faster than it takes a smaller pipe to fill the same tank. Working together, both pipes can fill the tank in 12 minutes. How long would it take the large pipe working alone to fill the tank?
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time to fill with smaller pipe --- x minutes
rate of smaller pipe = 1/x units/min
time to fill with larger pipe --- x-10 minutes
rate of larger pipe = 1/(x-10) units/min
combined rate = 1/x + 1/(x-10)
= (x-10 + x)/(x(x-10))
= (2x-10)/(x^2 - 10x)
1/[ (2x-10)/(x^2 - 10x)] = 12
(x^2 - 10x)/(2x-10) = 12
x^2 - 10x = 24x - 120
x^2 - 34x + 120 = 0
using the formula,
x = (34 ± √676)/2
= 30 minutes or 4 minutes, but x > 10
so the time to fill with the smaller pipe = 30 minutes
and the time to fill with the larger pipe is 20 minutes
check:
is the time of one more than the other equal to 10 minutes ? yes
comined rate
= 1/30 + 1/20 = 50/600 = 1/12
so time to do with combined rate = 1÷(1/12) = 12
My answer is correct.
pipe A fill the tank in 10m and pipe B can fill the tank in 15m then how long it will take to fill the tank by both A and B
To solve this problem, let's assume that the smaller pipe takes x minutes to fill the tank.
We are given that the large pipe can fill the tank 10 minutes faster than the smaller pipe. So, the large pipe would take (x - 10) minutes to fill the tank.
Now, we can set up an equation based on the information that both pipes working together can fill the tank in 12 minutes. The equation is:
1/x + 1/(x-10) = 1/12
To find the value of x, we need to solve this equation.
To do that, we can multiply both sides of the equation by 12x(x-10), which will eliminate the denominators:
12(x-10) + 12x = x(x-10)
Expanding and simplifying the equation gives us:
12x - 120 + 12x = x^2 - 10x
Combining like terms gives us a quadratic equation:
x^2 - 34x + 120 = 0
We can now solve this equation using factoring, completing the square, or the quadratic formula. In this case, we can factor the equation as follows:
(x-30)(x-4) = 0
This gives us two possible values for x: x = 30 or x = 4.
Since we are looking for the time it would take the large pipe (x-10) to fill the tank, we can disregard x = 4, as it would result in a negative time.
Therefore, the large pipe would take (30 - 10) = 20 minutes to fill the tank on its own.
So, the answer is that it would take the large pipe 20 minutes working alone to fill the tank.