Two hoses together fill a pool in 2 h. If

only hose A is used, the pool fills in 3 h.
How long would it take to fill the pool if
only hose B were used?

T = 2 h.

Ta = 3 h.
Tb = ?

T = Ta*Tb/(Ta+Tb) = 2.
3*Tb/(3+Tb) = 2.
Tb = ?.

rate of hose B = 1/b

rate of hose A = 1/3
combined rate = 1/b + 1/3 = (b+3)/(3b)

so 1/( (b+3)/(3b)) = 2
3b/(b+3) = 2
3b = 2b + 6
b = 6

Just noticed this looks just like Henry's

Yes, it is I, your humble servant(LOL).

To determine how long it would take to fill the pool using only hose B, we need to first find the rate at which each hose fills the pool. Let's denote the rate at which hose A fills the pool as rateA and the rate at which hose B fills the pool as rateB.

Given that both hoses together fill the pool in 2 hours, we can say that their combined rate is 1 pool per 2 hours:
rateA + rateB = 1 pool / 2 hours

We are also given that hose A alone fills the pool in 3 hours, so we can say that its rate is 1 pool per 3 hours:
rateA = 1 pool / 3 hours

Now, we can substitute the value of rateA into the equation for their combined rate:
1 / 3 + rateB = 1 / 2

To solve for rateB, we can subtract 1/3 from both sides:
rateB = 1/2 - 1/3

To find a common denominator, we can multiply 1/2 by 3/3:
rateB = 3/6 - 2/6

Combining the fractions:
rateB = 1/6

This means that hose B fills 1/6th of the pool per hour.

To find out how long it would take to fill the pool using only hose B, we can divide 1 pool by the rate of hose B:
1 pool / (1/6 pool per hour)

Simplifying the division:
1 pool * (6/1) hour = 6 hours

Therefore, it would take 6 hours to fill the pool using only hose B.