A tank can be filled by pipe A in 4 hours and pipe B in 6 hours. At 8:00 a.m., pipe A was opened. At what time will the tank be filled if pipe B is opened at 9:00 a.m.?

Correct — C. Take the full tank as 1 unit of work.

8:00 to 9:00, A alone fills 1⁄4, so 3⁄4 is left at 9:00
Combined rate from 9:00 = 1⁄4 + 1⁄6 = 5⁄12 per hour
Time for the rest = (3⁄4) ÷ (5⁄12) = 3⁄4 × 12⁄5 = 9⁄5 hours
9⁄5 h = 1.8 h = 1 hour 48 minutes

9:00 a.m. + 1 h 48 min = 10:48 a.m. → option (c).

Why the others are wrong

• (a)
  10:16 a.m. — 10:16 a.m. is 1 h 16 min after 9:00. At 5⁄12 per hour that fills only 19⁄36 of the tank, well short of the 3⁄4 still to go.
• (b)
  10:22 a.m. — 10:22 a.m. gives the pair 1 h 22 min, which delivers 41⁄72 — still under the 3⁄4 remaining.
• (d)
  10:18 a.m. — 10:18 a.m. allows 1 h 18 min, worth 13⁄24 of the tank. Every one of these times undercounts the 3⁄4 that A left behind.

Pipe problems run on rates, which add. A fills 1⁄4 of the tank an hour, B fills 1⁄6, so together they fill 5⁄12 an hour.

The LCM trick removes the fractions: call the tank LCM(4, 6) = 12 units. Then A is 3 units/hour and B is 2 units/hour. A puts in 3 units by 9:00, leaving 9 units at 5 units/hour, which is 9⁄5 hours — the same answer with integer arithmetic.

Key facts

• Rates add: pipes working together fill at the sum of their separate rates.
• Taking the tank as LCM(4, 6) = 12 units makes A 3 units/hour and B 2 units/hour.
• A fraction of an hour becomes minutes by multiplying by 60: 9⁄5 h × 60 = 108 min = 1 h 48 min.
• A alone would need 4 hours, so the tank filling by 10:48 a.m. (2 h 48 min) is consistent with B's help.

Common traps

• Adding 1 h 48 min to 8:00 a.m. instead of to 9:00 a.m.
• Ignoring A's one-hour head start and filling the whole tank at the combined rate.
• Reading 1.8 hours as 1 hour 8 minutes.

No directly related past PYQ was found.