IIM B UG (B.Sc-Hons.)- Sample Paper 2026 · Logical Reasoning · Set 1: Rubbing out numbers on a board (Q10–Q13)

The full set and all 4 questions (Q5–Q8) from IIM B UG (B.Sc-Hons.)- Sample Paper 2026, with answers and solutions.

Directions: Study the following information carefully and answer the questions that follow.

The numbers 1, 2, 3, and so on up to 20 are written on a board. A move is made as follows: choose any two of the numbers currently on the board, rub out both of them, and then write down one new number equal to the larger of the two minus the smaller.

Every move therefore leaves the board with one number fewer than before. Moves continue in this way until only a single number is left.

Question 5 · Quant Based Puzzles and Optimisation

Which one of the following CANNOT be the number finally left?
  1. A
    7
  2. B
    2
  3. C
    20
  4. D
    0
Pick an option, then check.
Answer & solution

Answer: Option A

Watch what a move does to the total of all the numbers on the board.

Two numbers a and b are replaced by their difference, so the total falls by a + b − |a − b|, which equals twice the smaller of the two.

That change is always even, so the total keeps the same odd-or-even character from start to finish.

The starting total is 1 + 2 + … + 20 = 210, which is even.

So the single number left at the end must be even.

Of the options, 7 is the only odd one, so 7 cannot be the final number.

Question 6 · Quant Based Puzzles and Optimisation

Suppose instead of 1 to 20 the numbers 1 to 21 are written on the board at the start, and the same moves are made until one number is left. Which one of the following could be the number finally left?
  1. A
    10
  2. B
    21
  3. C
    4
  4. D
    0
Pick an option, then check.
Answer & solution

Answer: Option B

The same rule applies: each move changes the total by an even amount, so the total keeps its odd-or-even character.

The starting total is now 1 + 2 + … + 21 = 231, which is odd.

So the number finally left must be odd, which rules out 10, 4 and 0.

That leaves 21, and 21 really is reachable.

Set 21 aside and work on 1 to 20 alone: pair them as (1, 2), (3, 4), … , (19, 20), and each pair leaves 1, giving ten 1s.

Pair those ten 1s to leave five 0s, then combine the 0s to leave a single 0.

Finally take 21 and 0, whose difference is 21.

So 21 is the only option that could be the number left.

Question 7 · Quant Based Puzzles and Optimisation

As the moves are carried out, which one of the following never changes?
  1. A
    Whether the count of odd numbers is even or odd
  2. B
    The largest number on the board
  3. C
    How many different numbers appear on the board at once
  4. D
    The total of the numbers on the board
Pick an option, then check.
Answer & solution

Answer: Option A

Test each option against a single move, which replaces two numbers by their difference.

Take the count of odd numbers and check the three possible cases.

Two odd numbers: their difference is even, so two odds leave and none arrives — the count drops by 2 and its odd-or-even character is unchanged.

Two even numbers: their difference is even, so the count of odds does not move at all.

One odd and one even: their difference is odd, so one odd leaves and one odd arrives — again no change.

In every case the count of odd numbers keeps the same odd-or-even character, so option (1) never changes.

The largest number can fall, for example 20 and 19 leave 1, so (2) is wrong.

The board shrinks by one number each move and repeats appear, so (3) changes.

The total falls with almost every move, so (4) changes.

Question 8 · Quant Based Puzzles and Optimisation

Suppose instead that the numbers written on the board at the start are 3, 6, 9, and so on up to 60, that is the first twenty multiples of 3. The same moves as in the passage are then made until one number is left. Which one of the following CANNOT be the number finally left?
  1. A
    0
  2. B
    45
  3. C
    6
  4. D
    30
Pick an option, then check.
Answer & solution

Answer: Option B

Two things are preserved here, and both must hold for the final number.

First, every number on the board starts as a multiple of 3, and the difference of two multiples of 3 is again a multiple of 3. So the final number is a multiple of 3.

Second, the total keeps its odd-or-even character, as before. The starting total is 3 + 6 + … + 60 = 3 × 210 = 630, which is even, so the final number must be even.

Being both a multiple of 3 and even means the final number must be a multiple of 6.

Checking the options: 0, 6 and 30 are all multiples of 6, but 45 is odd.

So 45 cannot be the number finally left.

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