Discussion

Explanation:

Sum of all the cards she picked in total 6 rounds = 46 + 24 + 38 + 64 + 38 + 48 = 258.

Sum if all 10 cards were picked thrice = 3 × (2 + 4 + 6 + … + 20) = 330.

∴ The sum of the 2 cards that were never picked = (330-258)/3 = 24

Now, these two cards can be (20 & 4) or (18 & 6) or (16 & 8) or (14 & 10)   … (1)

The least sum of 4 cards picked is 24. This is possible when she picks

Case 1: 2 + 4 + 6 + 12
Case 2: 2 + 4 + 8 + 10

∴ Cards 1 and 2 should definitely be picked.

From (1): The two cards never picked cannot be (20 & 4)   …(2)

The highest sum of 4 cards picked is 64. This is possible when she picks

Case 1: 10 + 16 + 18 + 20
Case 2: 12 + 14 + 18 + 20

∴ Cards 18 and 20 should definitely be picked.

From (1): The two cards never picked cannot be (18 & 6)   …(3)

From (1), (2) and (3)

The two cards never picked can be (16 & 8) or (14 & 10)

For the sum of cards to be 64 in round 4, at least one of 10 or 14 must be picked.

∴ The two cards never picked can only be (16 & 8) 

Hence, option (d).

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