On dividing a number by 55, we get 28 as the remainder. On dividing the same number by 11, what is the remainder?

Correct — B. Write the number in the form the first division hands you.

N = 55k + 28 for some whole number k
11 divides 55, so 55k leaves remainder 0 against 11
28 = 11 × 2 + 6
So N = 11(5k + 2) + 6

The remainder is 6 → option (b).

Why the others are wrong

• (a)
  5 — 5 would require 28 = 11 × 2 + 5, which is 27. The true split is 28 − 22 = 6.
• (c)
  7 — 7 would require 28 = 11 × 2 + 7, which is 29. 28 − 22 = 6, so 7 overshoots by one.
• (d)
  8 — 8 is what 28 leaves against 10, not against 11. Divide by 11 and 22 goes in, leaving 6.

When one divisor is a factor of the other, the second remainder is already hiding inside the first. 55 = 5 × 11, so every multiple of 55 is also a multiple of 11 and contributes nothing new.

That leaves only the 28 to deal with, and 28 = 11 × 2 + 6.

The shortcut runs one way only. A remainder against 11 tells you nothing about the remainder against 55, because the missing piece — which multiple of 11 you landed on — is exactly what 55 would need.

Key facts

• If d divides D, then N mod d equals (N mod D) mod d.
• 55 = 5 × 11, so a remainder against 55 also settles the remainders against 5 and against 11.
• The rule does not reverse: a remainder against a factor never fixes the remainder against the multiple.

Common traps

• Dividing 55 by 11 and answering 5, which is a quotient and not a remainder
• Subtracting 11 from 28 once and stopping at 17, which is still bigger than the divisor
• Trying the reverse deduction — fixing the remainder against 55 from one against 11

No directly related past PYQ was found.