Two hoses are connected to a swimming pool. Working together, they can fill the pool in 4 hr. The larger hose, working alone, can fill the pool in 6 hr less time than the smaller one. How long would it take the smaller one, woking alone, to fill the pool?

Please use the work equation

The large one can fill in x hours,

then the small one in (x+6) hours.
During each hour, working together, the two hoses can fill 1/x+1/(x+6) of the pool, or (2x+6)/(x(x+6)).

Since using both hoses, it fills the pool in 4 hours, each hour fills 1/4 of the pool, therefore:

1/x + 1/(x+6) = 1/4

(2x+6)/(x(x+6)) = 1/4
4(2x+6)=x(x+6)
x² -2x -24 = 0
Solve for x (reject the negative root).
The large one will fill in x hours alone, and the small one, x+6 hours.

Let's assume the time it takes for the smaller hose to fill the pool alone is x hours.

According to the given information, the larger hose can fill the pool in x - 6 hours.

The rate of work of the smaller hose is 1/x of the pool per hour.

The rate of work of the larger hose is 1/(x-6) of the pool per hour.

Together, they can fill the pool in 4 hours, so their combined rate of work is 1/4 of the pool per hour.

According to the work equation, the combined rate of work is equal to the sum of the individual rates of work:

1/x + 1/(x-6) = 1/4

To solve this equation, we need to find a common denominator:

(4(x-6) + 4x)/(x(x-6)) = 1/4

(4x - 24 + 4x)/(x^2 - 6x) = 1/4

(8x - 24)/(x^2 - 6x) = 1/4

Cross multiplying:

4(8x - 24) = x^2 - 6x

32x - 96 = x^2 - 6x

Rearranging the equation to solve for x:

x^2 - 38x + 96 = 0

Factoring the quadratic equation:

(x - 24)(x - 4) = 0

Solving for x, we get two values: x = 24 and x = 4.

Since the smaller hose cannot fill the pool in less time than the larger hose, we can discard the x = 4 solution.

Therefore, the time it takes for the smaller hose to fill the pool alone is 24 hours.

To solve this problem, we can use the concept of work. The work equation states that work done = rate of work × time. In this case, the work is to fill the entire pool.

Let's assume that the rate of work for the larger hose is R1 (pools per hour) and the rate of work for the smaller hose is R2 (pools per hour).

From the given information, we can create the following equations:
1. (R1 + R2) × 4 = 1 (since it takes both hoses 4 hours to fill the pool)
2. R1 × (t - 6) = 1 (since the larger hose takes 6 hours less than the smaller hose to fill the pool)

Now, let's solve these equations to find the values of R1 and t (time taken by the smaller hose to fill the pool):

From equation 2: R1 × (t - 6) = 1
Dividing both sides by (t - 6): R1 = 1 / (t - 6)

Substituting this value of R1 into equation 1: (1 / (t - 6) + R2) × 4 = 1
Simplifying: 4 / (t - 6) + 4R2 = 1
Rearranging: R2 = (1 - 4 / (t - 6)) / 4
Simplifying further: R2 = ((t - 6) - 4) / 4(t - 6)
R2 = (t - 10) / 4(t - 6)

Now, substitute this value of R2 into equation 1:
(1 / (t - 6) + (t - 10) / 4(t - 6)) × 4 = 1
Simplify: 1 + (t - 10) / (t - 6) = 1 / 4
Rearrange: (t - 10) / (t - 6) = 1/4 - 1
Simplify: (t - 10) / (t - 6) = -3/4

Cross-multiply: 4(t - 10) = -3(t - 6)
Simplify: 4t - 40 = -3t + 18
Rearrange: 7t = 58
Divide both sides by 7: t = 8.29

Therefore, it would take the smaller hose, working alone, approximately 8.29 hours to fill the pool.