a water tank is filled by two pipes in 48 minutes. if the larger pipe fill it alone in 40 minutes less time the smaller pipe. find the time it takes for the smaller pipe to fill the tank alone.
1/(x-40) + 1/x = 1/48
To solve this problem, let's represent the time it takes for the smaller pipe to fill the tank alone as 'x' minutes.
We are given that the larger pipe fills the tank in 40 minutes less time than the smaller pipe. Therefore, the time it takes for the larger pipe to fill the tank alone would be represented as 'x - 40' minutes.
We know that when both pipes are working together, they fill the tank in 48 minutes. This means that in 1 minute, they fill 1/48th of the tank.
Now, we can set up the equation based on the rates at which the pipes fill the tank:
1/x + 1/(x - 40) = 1/48
To solve this equation, let's find a common denominator first. In this case, the common denominator will be '48x(x-40)'.
Multiplying each term in the equation by this common denominator, we get:
48(x - 40) + 48x = x(x - 40)
Expanding and simplifying the equation, we have:
48x - 1920 + 48x = x^2 - 40x
Combining like terms, we get:
96x - 1920 = x^2 - 40x
Rearranging the equation and setting it equal to zero, we have:
x^2 - 40x - 96x + 1920 = 0
x^2 - 136x + 1920 = 0
Now, we can solve this quadratic equation. We can factor it or use the quadratic formula. Let's use the latter here:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -136, and c = 1920. Substituting these values into the quadratic formula, we get:
x = (136 ± √(136^2 - 4*1*1920)) / (2*1)
Simplifying further:
x = (136 ± √(18496 - 7680)) / 2
x = (136 ± √10776) / 2
x = (136 ± 103.8) / 2
Now, solving for both possibilities:
x1 = (136 + 103.8) / 2 ≈ 119.4 / 2 ≈ 59.7
x2 = (136 - 103.8) / 2 ≈ 32.2 / 2 ≈ 16.1
Since time cannot be negative, we discard x2.
Therefore, the time it takes for the smaller pipe to fill the tank alone is approximately 59.7 minutes.