Pipe X can fill a tank in 9 hours and Pipe Y can fill it in 21 hours. If they are opened on alternate hours and Pipe X is opened first, in how many hours shall the tank be full?

Correct — D. Give the tank the capacity LCM(9, 21) = 63 units.

X fills 63 ⁄ 9 = 7 units/hour
Y fills 63 ⁄ 21 = 3 units/hour
One X-then-Y cycle of 2 hours = 10 units

6 cycles = 12 hours = 60 units, leaving 3 units
Hour 13 belongs to X, at 7 units/hour
3 ⁄ 7 hour finishes the tank

Total = 12 + 3⁄7 = 12 3⁄7 hours → option (d)

Why the others are wrong

• (a)
  10 3⁄7 carries the right fractional tail but too few cycles. Five cycles fill only 50 of the 63 units, leaving 13 — more than one X hour can clear.
• (b)
  11 3⁄7 stops at 57 units: five cycles plus one X hour. Six units remain, and the next hour belongs to Y at 3 units/hour, so the tank is not finished within that hour.
• (c)
  9 3⁄7 is exactly three hours short. Nine hours of alternation reaches 47 of the 63 units — four cycles plus one X hour — with 16 units still to fill.

Alternate-hour pipe problems are counted in cycles, never by merging the two pipes into one combined rate. Give the tank the LCM of the two times as its capacity and both rates come out whole.

Here the tank is 63 units, X does 7 an hour, Y does 3, and one full X-then-Y cycle does 10.

Divide 63 by 10 for the complete cycles, then finish the remainder with whichever pipe is due next. Order matters: with X first the leftover is cleared at 7 units an hour, so the tail is 3⁄7 of an hour rather than a whole one.

Key facts

• LCM(9, 21) = 63, so X fills 7 units an hour and Y fills 3.
• One two-hour X-then-Y cycle fills 10 units.
• Six cycles take 12 hours and fill 60 units, leaving 3 units for X to finish in 3⁄7 hour.

Common traps

• Averaging the two rates into a single pipe, which throws away the alternation.
• Rounding the final part-hour up to a whole hour.
• Beginning the cycle with Y when the stem opens with X.

No directly related past PYQ was found.