Step-by-Step Solution

Topic: Work, Time & Efficiency

1. Establish One-Hour Work Rates

• • Work done by A in 1 hour = 1/10
• • Work done by B in 1 hour = 1/12
• • Work done by C in 1 hour = -1/15 (Negative because it empties)

(I) PIPES A AND B TOGETHER

Combined Rate (A + B) = (1/10) + (1/12)

LCM of 10 and 12 is 60.

Rate = (6 + 5) / 60 = 11 / 60

Time Taken: Reciprocal of work rate.

(II) PIPES A AND C TOGETHER

Combined Rate (A + C) = (1/10) – (1/15)

LCM of 10 and 15 is 30.

Rate = (3 – 2) / 30 = 1 / 30

(III)(A) ALL THREE PIPES (A, B, C)

Combined Rate = (1/10) + (1/12) – (1/15)

LCM of 10, 12, 15 is 60.

Rate = (6 + 5 – 4) / 60 = 7 / 60

(III)(B) FINDING WHEN PIPE B WAS TURNED OFF

Let pipe B be open for t hours.

Pipe A was open for the entire duration (6 hours).

Total Work Equation: (Work by A) + (Work by B) = 1 (Full Tank)

(1/10 × 6) + (1/12 × t) = 1

Solve for t:

6/10 + t/12 = 1

3/5 + t/12 = 1

t/12 = 1 – 3/5

t/12 = 2/5

t = (2 × 12) / 5 = 24 / 5