Pipes A & B can fill a storage tank in 8hours and 12 hours, respectively . With the tank empty ,pipe A was turned on at noon, and then pipe B was turned on at 1:30 P.M.At what time was the tank full. Please help me!!

Let V be the tank volume. Pipe A fills at (1/8 V) per hour and pipe B fills at (1/12) V per hour. Let t be the elapsed time after 1:30. At that time, the volume filled (x) is

x = (1.5 + t) * (1/8) V + t (1/12) V = V

0.1875 V + 0.125 V t + 0.0833 V t = V
0.1875 + 0.2083 t = 1
Solve for t, which will be in hours. I get 3.90, or 3 hours 54 minutes after 1:30, when Pipe B was turned on

Check my algebra; I don't guarantee it

Isn't it t-1.5 though? Because since pipe b is 1.5 hours behind, and x is the number of hours for pipe a

To solve the problem, we can use the concept of rates and work.

First, let's find the individual rates at which pipe A and pipe B can fill the tank:

Pipe A's rate = 1/8 of the tank volume per hour
Pipe B's rate = 1/12 of the tank volume per hour

Since both pipes work together to fill the tank, we can add their rates to find the combined rate:

Combined rate = Pipe A's rate + Pipe B's rate
= 1/8 + 1/12
= 3/24 + 2/24
= 5/24 of the tank volume per hour

Now, let's calculate the time it takes to fill the tank.

From noon to 1:30 PM, or 1.5 hours, only pipe A is running. So, during this time, pipe A fills:
Volume filled by pipe A = Pipe A's rate * time
= (1/8) * 1.5
= 3/16 of the tank volume

At 1:30 PM, both pipes A and B are running. We need to find the additional time it takes for the tank to get full. Let's represent this time as 't' in hours.

During this additional time (t), pipe A fills:
Additional volume filled by pipe A = Pipe A's rate * t
= (1/8) * t
= t/8 of the tank volume

During this additional time (t), pipe B fills:
Volume filled by pipe B = Pipe B's rate * t
= (1/12) * t
= t/12 of the tank volume

The sum of the volumes filled by pipe A and pipe B must be equal to the total volume of the tank. So, we can set up the equation:

3/16 + t/8 + t/12 = 1

To solve this equation, we can find a common denominator of 48:

(9/48) + (6t/48) + (4t/48) = 1

Combining the terms, we get:

(9 + 6t + 4t)/48 = 1

Simplifying further:

9 + 10t = 48

10t = 48 - 9
10t = 39

t = 39/10
t ≈ 3.9

So, the additional time it takes to fill the tank is approximately 3.9 hours.

To find the time at which the tank is full, we add the additional time (3.9 hours) to the time when pipe B was turned on (1:30 PM), which gives us:

Time when the tank is full = 1:30 PM + 3.9 hours
= 5:24 PM

Therefore, the tank is full at approximately 5:24 PM.