Question

algebra 2 362403-6 fy - 1
benchmark test 2
due feb 5 - 11:45 pm late
it takes 24 h to fill a large basin with two hoses, where the water in one hose flows four times as fast as the other hose. how long will it take the slower hose to fill the basin if the faster hose is not functioning?
a. 60 h
b. 96 h
c. 72 h
d. 120 h
review progress
question 24 of 30 back next

Explanation:

Step1: Define variables for rates

Let the slower hose's rate be $\frac{1}{x}$ basin per hour. Then the faster hose's rate is $\frac{4}{x}$ basin per hour.

Step2: Set up combined rate equation

Combined rate fills 1 basin in 24 h:
$$\frac{1}{x} + \frac{4}{x} = \frac{1}{24}$$

Step3: Simplify and solve for x

Combine left side terms:
$$\frac{5}{x} = \frac{1}{24}$$
Cross-multiply to solve for $x$:
$$x = 5 \times 24 = 120$$

Answer:

D. 120 h