Question
Download Solution PDFThree pipes A, B and C can fill a cistern in 204 hours. After working together for 68 hours, C is closed and A and B filled the remaining part in 204 hours. Find the number of hours taken by C alone to fill the cistern.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Three pipes A, B and C can fill a cistern = 204 hours
Calculation:
(A + B + C) × 204 = (A + B + C) × 68 + (A + B) × 204
⇒ C × 204 = (A + B + C) × 68
⇒ C × 3 = (A + B + C) × 1
⇒ C : (A + B + C) = 1 : 3
Total work = (A + B + C) × 204 = 3 × 204 = 612 unit
Time is taken by C = Total work / Efficiency of C
⇒ 612 / 1
⇒ 612 hours
The number of hours required by C alone to fill the cistern is 612 hr
∴ The correct answer is option (3).
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