Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of ’Passing the Buck’.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediately to the right;
• Second to the left;
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as ’?’.v
Approach: Instead of building the whole circle, test each option directly against the pass moves that involve Bina and her neighbour.
Step 1: The seven children sit in a circle and the Buck moves only to a seat that is one or two places left or right of the holder. So the person immediately to Bina's right is exactly the child Bina reaches with an "immediate right" pass, and the child who reaches Bina with an "immediate left" pass.
Step 2: Take the four candidates $-$ Eshan, Chirag, Aarav, Farhan $-$ and place each, in turn, on Bina's clockwise (right) side. For each trial, check the single-step and two-step passes recorded in the game: any candidate that forces a contradiction with a recorded pass (a move that would have to skip the wrong number of seats) is eliminated.
Step 3: Only one candidate survives every recorded pass without contradiction. That child sits immediately clockwise of Bina, i.e. immediately to her right.
Answer: Eshan.
Approach: Use the equivalence "third to the left = fourth to the right" in a 7-seat circle to count the short way, then verify against the options.
Step 1: In a circle of \(7\) seats, going \(3\) seats anticlockwise (left) lands on the same person as going \(7-3 = 4\) seats clockwise (right). So "third to the left of Eshan" is the same as "fourth to the right of Eshan."
Step 2: Once the seating is fixed by the pass record, count four seats clockwise from Eshan. This shorter count avoids mistakes when wrapping around the circle.
Step 3: The seat four places clockwise from Eshan holds Chirag, matching the third-to-left count. Among the options Gaurav, Divya, Aarav and Chirag, only Chirag fits.
Answer: Chirag.
Approach: Test for uniqueness by trying to perturb each count $-$ if you can swap passes and still satisfy every clue, that type is not determined; the one you cannot perturb is the answer.
Step 1: Write the totals: the four pass-type counts add to $10$, and their signed seat-shifts ($+1,-1,+2,-2$) must move the Buck exactly from Bina's seat to Chirag's seat around the circle of $7$.
Step 2: For each pass type in turn, attempt a trade $-$ for example, replace one "second left" with a combination of others that keeps both the total of $10$ and the net displacement unchanged while respecting the recorded holders. Three of the four types admit such a trade, meaning their totals can change between valid arrangements, so they are not fixed.
Step 3: The "immediate right" count is the single type for which no valid trade exists $-$ every arrangement consistent with the start, end and recorded passes gives it the same value. Hence its total is uniquely determined.
Answer: Immediately to the right.
Approach: Use the total-receptions bookkeeping: receptions across all children must sum to $10$, so if all but one child's count can vary, the constrained one is the answer.
Step 1: Every pass creates exactly one reception, so across the $10$ rounds the children's reception counts add up to $10$. Bina starts the game (an extra hold that is not a reception) and Chirag is the final receiver, which fixes part of the picture but still leaves room for the middle children's counts to shift.
Step 2: Build the minimal and maximal possible reception counts for each candidate by routing the Buck along the different valid pass chains. Farhan, Eshan and Bina each show a range $-$ their minimum and maximum receptions differ $-$ so none of them is fixed.
Step 3: For Gaurav the minimum and maximum coincide: every valid routing forces the same number of catches on him. Since his count has zero spread, it is the only one that can be determined with certainty.
Answer: Gaurav.
Read the directions carefully and give the answer from the given options.
Who is to the immediate left of K?