Pipe X can fill a tank in 60 hours while pipe Y can fill the tank in 72 hours. Both pipes are opened together for 20 hours. How much of the tank is left empty ?

Correct — D. The four options are printed as pictures of fractions. Work out the empty part, then match it.

X fills 1⁄60 of the tank per hour, Y fills 1⁄72 per hour.
LCM of 60 and 72 is 360, so together = 6⁄360 + 5⁄360 = 11⁄360 per hour.

In 20 hours they fill 20 × 11⁄360 = 220⁄360 = 11⁄18.

Left empty = 1 − 11⁄18 = 7⁄18, which is the fraction shown in option (d).

Why the others are wrong

• (a)
  Option (a) shows 3⁄8. That would be the empty part only if the pair filled 5⁄8 in 20 hours, a combined rate of 1⁄32 per hour. Their real rate, 11⁄360, is a shade slower, so slightly more than 3⁄8 is left.
• (b)
  Option (b) shows 1⁄8, which needs 7⁄8 of the tank filled in 20 hours — a combined rate of 7⁄160 per hour. The pair manage only 11⁄360, about seven-tenths of that, so far more than 1⁄8 stays empty.
• (c)
  Option (c) shows 243⁄360. The denominator is the right LCM but the numerator is not: 20 hours fills 220⁄360, so the empty part is 360 − 220 = 140⁄360, which reduces to 7⁄18.

Pipe questions are rate questions. A pipe that fills a tank in t hours contributes 1⁄t of the tank per hour, and simultaneous pipes add their rates.

Take an LCM denominator early — 360 here — and the addition stops involving decimals.

The last line matters as much as the arithmetic: the question asks how much is left empty, so the filled fraction has to be subtracted from 1.

Key facts

• A pipe filling in t hours has a rate of 1⁄t of the tank per hour.
• 1⁄60 + 1⁄72 = 11⁄360 of the tank per hour.
• In 20 hours the pair fills 220⁄360, which reduces to 11⁄18.
• The empty share is 1 − 11⁄18 = 7⁄18, roughly 38.9% of the tank.

Common traps

• Answering 11⁄18, the part filled, when the question asks for the part left empty
• Adding the times, 60 + 72, instead of adding the rates
• Losing the LCM step and adding 1⁄60 and 1⁄72 as though the denominators matched

No directly related past PYQ was found.