How many chocolates are owned by the person who is sitting second to the left of the person who is s...........

Logical Reasoning(SBI-PO) Puzzle Test

Directions: Study the information given below carefully and answer the questions given below. Ten people sitting in two parallel rows. Row 1 consist of A, B, C, D, E, all the person sitting in row 1 are facing south. Row 2 consist of F, G, H, I, J, all of them are facing north.

All ten people have a different number of chocolates with them. The number of chocolates they have are 6, 1, 11, 12, 9, 10, 8, 4, 2 and 7.

Neither D nor C is sitting at the extreme ends. F has an odd number of chocolates. E is sitting opposite to J who is sitting second to one of the extreme ends. J has the highest number of chocolates. H is sitting second to the left of J and is facing the one who has 6 chocolates. The one who is sitting third to the left end of the first row has a prime number of chocolates. E has more number of chocolates than I. B who is sitting at the left end has 11 number of chocolates, 9 more than the one who is sitting at the right end of the same row. A is sitting diagonally opposite to I who has even number of chocolates. D is sitting at the right side of C. The number of people sitting to the right side of G is equal to the number of people sitting at the left side of G. Addition of the number of chocolates owned by F and H makes a perfect square and that addition is less than the number of chocolates owned by G.

How many chocolates are owned by the person who is sitting second to the left of the person who is sitting opposite of G?

This questions was previously asked in

SBI PO Mains (29 Jan, 2021)

View All SBIPO Previous Year Papers

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Explanation:

Let’s break this down step by step—here’s what’s happening in the puzzle:

- Ten people, two rows, facing each other. Each person has a different number of chocolates.

- B is at the left end (row 1), has 11 chocolates. The person on the right end of row 1 has 2 chocolates, since B has 9 more chocolates than them.

- D and C can’t be at the extreme ends of row 1. D sits to the right of C.

- G is in the dead center of row 2 (because the number of people on both sides of G is equal).

- A sits diagonally opposite to I, and I must have an even number of chocolates.

- E is opposite J. J is second from one extreme end and has 12 chocolates (the highest).

- H sits second to the left of J, facing the one with 6 chocolates.

- F’s chocolates are odd, and the total chocolates with F and H make a perfect square—less than G’s.

Let’s position everyone, batch-style:

- In Row 1: B _ _ _ E (B and E at ends), so B C D A E is possible, or B D C A E, keeping in mind C and D aren’t at ends and D is to the right of C.

- G's in the middle of Row 2: F/G/H/I/J (with G at position 3).

Work out the rest:

- B = 11, E = ?. End of row 1, person with 2 chocolates.

- Assign numbers such that J=12, F and H add to 9 (perfect square; so, F=1 and H=8; or F=8 and H=1).

- I is even, so could be 2, 4, 6, 8, 10, 12 (but 12 already with J, 2 is at row 1’s right end, H has 8 or 1).

After shuffling the logic, and working through placements:

The person sitting opposite G is A (since G's in center, so A is center row 1). Opposite of G = A.

Second to the left of A is C (since row 1: B C D A E), so second left of A is C.

Now, how many chocolates does C have?

- Let's check the possible arrangements: If B = 11 (left end), E = 2 (right end), the only remaining numbers for C are 6, 4, 7, 9, and 10 (as per elimination). But as per prime number clue (person who is third left end in row 1 has a prime number), C is in that spot. The only prime number in the list is 2, 7, 11; but 2 is at the right end, 11 is with B, so C has 7 chocolates.

So here’s the answer to the question: C has 7 chocolates.

Now, scan the options:

Option 1: 12

Option 2: 6

Option 3: 7

Option 4: 9

Option 5: 11

Here’s what it all comes down to: The person sitting second to the left of the person who is sitting opposite of G has 7 chocolates.