Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is:
Question:
Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is:
Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is:
Options
Answer: 90
Explanation:
Step 1: Define variables Let: Time taken by Pipe A to fill the tank = a hours Time taken by Pipe B to empty the tank = b hours Time taken by Pipe C to fill the tank = c hours Given: Pipe B empties the full tank in one hour less than A takes to fill: b = a - 1 Step 2: Write rates of pipes Pipe A fills at rate: 1/a tank per hour Pipe B empties at rate: 1/b = 1/(a-1) tank per hour Pipe C fills at rate: 1/c tank per hour When A, B, and C are on together, tank is filled in 2 hours: 2 * (1/a - 1/(a-1) + 1/c) = 1 Simplify: 1/a - 1/(a-1) + 1/c = 1/2 (Equation 1) Step 3: Translate the second scenario Pipes B and C are on together for 1 hour. Tank filled in 1 hour: 1/c - 1/(a-1) After 1 hour, B is turned off, C continues alone. Remaining fraction of tank: Remaining = 1 - (1/c - 1/(a-1)) = 1 - 1/c + 1/(a-1) = a/(a-1) - 1/c Pipe C alone fills this in 1 hour 15 minutes = 5/4 hours: (5/4) * (1/c) = a/(a-1) - 1/c Combine terms: 5/(4c) + 1/c = a/(a-1) 9/(4c) = a/(a-1) c = 9*(a-1)/(4*a) (Equation 2) Step 4: Substitute c into Equation 1 1/a - 1/(a-1) + 1/c = 1/2 1/a - 1/(a-1) + 4a/(9*(a-1)) = 1/2 Combine fractions: -1/(a-1) + 4a/(9*(a-1)) = (4a - 9)/(9*(a-1)) Equation becomes: 1/a + (4a - 9)/(9*(a-1)) = 1/2 Multiply both sides by 9a*(a-1): 9*(a-1) + a*(4a - 9) = (9a(a-1))/2 Simplify: 9a - 9 + 4a^2 - 9a = (9a^2 - 9a)/2 4a^2 - 9 = (9a^2 - 9a)/2 Multiply both sides by 2: 8a^2 - 18 = 9a^2 - 9a 0 = a^2 - 9a + 18 Step 5: Solve quadratic for a a^2 - 9a + 18 = 0 a = (9 ± sqrt(81 - 72))/2 = (9 ± 3)/2 a = 6 or 3 Given A fills in less than 5 hours, so a = 3 hours Step 6: Find c c = 9*(a-1)/(4a) = 9(3-1)/(4*3) = 18/12 = 1.5 hours Convert to minutes: 1.5 * 60 = 90 minutes
Explanation:
Step 1: Define variables Let: Time taken by Pipe A to fill the tank = a hours Time taken by Pipe B to empty the tank = b hours Time taken by Pipe C to fill the tank = c hours Given: Pipe B empties the full tank in one hour less than A takes to fill: b = a - 1 Step 2: Write rates of pipes Pipe A fills at rate: 1/a tank per hour Pipe B empties at rate: 1/b = 1/(a-1) tank per hour Pipe C fills at rate: 1/c tank per hour When A, B, and C are on together, tank is filled in 2 hours: 2 * (1/a - 1/(a-1) + 1/c) = 1 Simplify: 1/a - 1/(a-1) + 1/c = 1/2 (Equation 1) Step 3: Translate the second scenario Pipes B and C are on together for 1 hour. Tank filled in 1 hour: 1/c - 1/(a-1) After 1 hour, B is turned off, C continues alone. Remaining fraction of tank: Remaining = 1 - (1/c - 1/(a-1)) = 1 - 1/c + 1/(a-1) = a/(a-1) - 1/c Pipe C alone fills this in 1 hour 15 minutes = 5/4 hours: (5/4) * (1/c) = a/(a-1) - 1/c Combine terms: 5/(4c) + 1/c = a/(a-1) 9/(4c) = a/(a-1) c = 9*(a-1)/(4*a) (Equation 2) Step 4: Substitute c into Equation 1 1/a - 1/(a-1) + 1/c = 1/2 1/a - 1/(a-1) + 4a/(9*(a-1)) = 1/2 Combine fractions: -1/(a-1) + 4a/(9*(a-1)) = (4a - 9)/(9*(a-1)) Equation becomes: 1/a + (4a - 9)/(9*(a-1)) = 1/2 Multiply both sides by 9a*(a-1): 9*(a-1) + a*(4a - 9) = (9a(a-1))/2 Simplify: 9a - 9 + 4a^2 - 9a = (9a^2 - 9a)/2 4a^2 - 9 = (9a^2 - 9a)/2 Multiply both sides by 2: 8a^2 - 18 = 9a^2 - 9a 0 = a^2 - 9a + 18 Step 5: Solve quadratic for a a^2 - 9a + 18 = 0 a = (9 ± sqrt(81 - 72))/2 = (9 ± 3)/2 a = 6 or 3 Given A fills in less than 5 hours, so a = 3 hours Step 6: Find c c = 9*(a-1)/(4a) = 9(3-1)/(4*3) = 18/12 = 1.5 hours Convert to minutes: 1.5 * 60 = 90 minutes
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