Three pipes can fill a water tank constantly.The time taken for the first two pipes working simultaneously to fill the tank is equivalent to time taken by the third pipe to fill it alone.If the second pipe fills the tank in 10 hours faster than the first pipe and 8 hours slower than the third pipe.Then find out the time required by the first pipe is:

Question

Three pipes can fill a water tank constantly.The time taken for the first two pipes working simultaneously to fill the tank is equivalent to time taken by the third pipe to fill it alone.If the second pipe fills the tank in 10 hours faster than the first pipe and 8 hours slower than the third pipe.Then find out the time required by the first pipe is:

Answer 1

T1*T2/(T1+T2) = T3.

T1*(T1-10)/(T1+(T1-10)) = T1-18.

(T1^2-10T1)/(2T1-10) = T1-18.
Multiply by 2T1-10:
T1^2-10T1 = 2T1^2-36T1-10T1+180
-T1^2 + 36T1 - 180 = 0
T1^2 - 36T1 + 180 = 0.
-6*-30 = 180, -6 + -30 = -36 = B.
(T1-6)(T1-30) = 0.

T1-6 = 0, T1 = 6 h.

T1-30 = 0, T1 = 30 Hours = Solution.

Thirty hours was selected because T2 cannot be 10 hours less than six hours.

Answer 2

To solve this problem, let's assume that the first pipe takes 'x' hours to fill the tank alone.

Given that the second pipe takes 10 hours less than the first pipe, it will take (x - 10) hours to fill the tank alone.

Similarly, given that the second pipe takes 8 hours slower than the third pipe, the third pipe will take (x - 10 + 8) = (x - 2) hours to fill the tank alone.

Now, we know that the time taken by the first two pipes working simultaneously is equivalent to the time taken by the third pipe alone.

Therefore, the time taken by the first two pipes is (1/x) + (1/(x - 10)) and the time taken by the third pipe is 1/(x - 2).

According to the problem, the time taken by the first two pipes equals the time taken by the third pipe. So, we can set up the following equation:

(1/x) + (1/(x - 10)) = 1/(x - 2)

To solve this equation, we can cross-multiply:

(x - 10)(x - 2) + x(x - 2) = x(x - 10)

Simplifying the equation:

(x - 10)(x - 2) + x(x - 2) = x(x - 10)
(x^2 - 12x + 20) + (x^2 - 2x) = x^2 - 10x
2x^2 - 14x + 20 = x^2 - 10x

Rearranging the equation to solve for x:

2x^2 - x^2 - 14x + 10x - 20 = 0
x^2 - 4x - 20 = 0

This is a quadratic equation, which can be factored:

(x - 6)(x + 4) = 0

Setting each factor equal to zero:

x - 6 = 0 or x + 4 = 0

Solving for x:

x = 6 or x = -4

Since time cannot be negative, the solution is x = 6.

Therefore, the time required by the first pipe to fill the tank alone is 6 hours.