Asked by Ronald
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
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
Answers
Answered by
Jeff
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
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