Two taps, having different rates of flow are used to fill a large water tank. If tap A is used on its own it will take 5 hours longer to fill the tank that it would take tap B to fill it on its own. Together, the taps would fill the tap in 6 hours. Assuming that the taps are running at full capacity

Oh and the qn is How long will it take tap A to fill the tank


how long will it take tap B to fill the tank
help

Tap B = X hrs.

Tap A = (X + 5) hrs.

1/X + 1/(X + 5) = 1/6,
Multiply both sides by common denominator, x(x + 5):
x + 5 + x = x(x + 5) / 6,
Multiply both sides by 6:
6x + 30 + 6x = x(x + 5),
12x + 30 = x^2 + 5x,
Combine like-terms and set Eq = to 0:
-x^2 + 7x + 30 = 0,
Multiply both sides by -1,:
x^2 + 7x + 30 = 0,
Factor Eq:
(x + 3) (x - 10) = 0,
x + 3 = 0, x = -3,
x - 10 = 0, x = 10.
Select positive value:
x = 10 hrs = tap B HRS,
X + 5 = 10 + 5 = 15hrs = 15 = Tap A hrs.

To solve the problem, let's denote the rate of flow of tap A as a (in tanks per hour) and the rate of flow of tap B as b (in tanks per hour).

According to the given information:
- If tap A is used on its own, it takes 5 hours longer to fill the tank than tap B on its own. This can be represented as: 1/a = 1/b + 5.
- When both taps are used together, they can fill the tank in 6 hours. This can be represented as: 1/a + 1/b = 1/6.

Now, let's solve these equations to find the values of a and b.

From the first equation, we can rearrange it as: 1/a - 1/b = 5. Multiplying both sides by ab, we get: b - a = 5b.
Similarly, from the second equation, we can rearrange it as: b + a = 6b.

Adding the two equations together, we get: (b - a) + (b + a) = 5b + 6b.
Simplifying the equation, we have: 2b = 11b.
Dividing both sides by 11b, we get: 2 = 11.

This implies that there is no solution to the system of equations. Therefore, there is no rate of flow for taps A and B that satisfies the given conditions.