Comprehension

Read the information given below to answer the questions.

A, B, C, D, E, F, G and H want to have a dinner on a round table and they have worked out the following seating arrangements.
(a) A will sit beside C.
(b) H will sit beside A.
(c) C will sit beside E.
(d) F will sit beside H.
(e) E will sit beside G.
(f) D will sit beside F.
(g) G will sit beside B.
(h) B will sit beside D.

Question: 1

Which of the following is wrong?

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When many “beside” conditions appear, grow a single open chain by forcing second neighbors; it usually closes into one unique circle. For left/right in round tables with everyone facing the center: right = anti-clockwise, left = clockwise.
Updated On: Jul 15, 2026
  • A will be to the immediate right of C
  • D will be to the immediate left of B
  • E will be to the immediate right of A
  • F will be to the immediate left of D
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The Correct Option is C

Approach Solution - 1

Step 1: Build the chain using “beside” constraints
A must be adjacent to C and H (a), (b). So A’s two neighbors are exactly \{C, H\}.
From (c), C must also sit beside E. Since C already uses one side for A, the other neighbor of C is E. Thus the local order is either A–C–E or E–C–A.
From (e), E sits beside G. E already sits next to C, therefore E’s other neighbor is G. Extend: A–C–E–G (in some direction).
From (g), G sits beside B. Hence G’s other neighbor is B. Extend: A–C–E–G–B.
From (h), B sits beside D. Therefore B’s other neighbor is D. Extend: A–C–E–G–B–D.
From (f), D sits beside F, so D’s other neighbor is F. Extend: …–B–D–F.
From (d), F sits beside H, thus F’s other neighbor is H. Because H must also be next to A (b), closing the circle gives the unique clockwise order:
\[ A – C – E – G – B – D – F – H – (back to A) \] Step 2: Fix “left” and “right” around a round table
In circular seating, everyone faces the center unless stated otherwise. A person’s immediate right is the seat anti-clockwise from them; the immediate left is the seat clockwise from them. (Stand at a seat and face the center—your right-hand neighbor sits anti-clockwise.)
Step 3: Test each option on the arrangement

(A) A to the immediate right of C: For C, the anti-clockwise neighbor is A (order … A–C–E …). Hence true.
(B) D to the immediate left of B: For B, the clockwise neighbor (left) is D (… G–B–D …). Hence true.
(C) E to the immediate right of A: For A, the anti-clockwise neighbor (right) is H (… F–H–A–C …), not E. Hence false.
(D) F to the immediate left of D: For D, the clockwise neighbor (left) is F (… B–D–F …). Hence true. Conclusion: Only statement (C) is wrong.
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Approach Solution -2

Using the seating loop A - C - E - G - B - D - F - H (back to A), with everyone facing the centre so that a clockwise step lands on a person's immediate left and an anti-clockwise step lands on their immediate right, let's test each statement:

  1. A will be to the immediate right of C: Moving anti-clockwise from C (backwards along the loop) lands on A, since the loop reads ...A-C-E.... Since anti-clockwise gives the right-hand neighbour, A is indeed to the immediate right of C. This statement is true.
  2. D will be to the immediate left of B: Moving clockwise (forward) from B along the loop ...G-B-D... lands on D. Since clockwise gives the left-hand neighbour, D is indeed to the immediate left of B. This statement is true.
  3. E will be to the immediate right of A: A's two neighbours in the loop are C (forward, its left) and H (backward, its right), since the loop closes as ...F-H-A-C.... E is nowhere adjacent to A at all; E sits two places away from A on the C side. So A's immediate right is H, not E, making this statement false.
  4. F will be to the immediate left of D: Moving clockwise (forward) from D along the loop ...B-D-F... lands on F. Since clockwise gives the left-hand neighbour, F is indeed to the immediate left of D. This statement is true.

Three of the four statements check out against the loop, and only the claim about E and A fails, since A's actual right-hand neighbour is H, not E.

Therefore, the correct answer is E will be to the immediate right of A.

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Approach Solution -3

With everyone seated in the fixed clockwise loop A-C-E-G-B-D-F-H (back to A), it helps to assign each seat a position number, 1 through 8, in this exact clockwise order: A=1, C=2, E=3, G=4, B=5, D=6, F=7, H=8, with position 8 followed by position 1 again. Facing the centre, a person's immediate left is the next position number clockwise, adding 1 and wrapping 8 back to 1, and their immediate right is the previous position number, subtracting 1 and wrapping 1 back to 8. Let's check each statement using this numbering:

  1. A will be to the immediate right of C: C is position 2, so C's right-hand neighbour is position \( 2 - 1 = 1 \), which is A. This statement is true.
  2. D will be to the immediate left of B: B is position 5, so B's left-hand neighbour is position \( 5 + 1 = 6 \), which is D. This statement is true.
  3. E will be to the immediate right of A: A is position 1, so A's right-hand neighbour is position \( 1 - 1 \), which wraps around to position 8, that is, H, not E, which is position 3. This statement is false.
  4. F will be to the immediate left of D: D is position 6, so D's left-hand neighbour is position \( 6 + 1 = 7 \), which is F. This statement is true.

Working through the position numbers confirms that only the claim about E being to the right of A fails; A's actual right-hand neighbour is H.

Therefore, the correct answer is E will be to the immediate right of A.

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Question: 2

Which of the following is correct?

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Fix one validated circular order and stick to a consistent left/right rule (facing centre: right = anti-clockwise, left = clockwise). Then check each option directly on the ring.
Updated On: Jul 15, 2026
  • B will be to the immediate left of D
  • H will be to the immediate right of A
  • C will be to the immediate right of F
  • B will be to the immediate left of H
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The Correct Option is B

Approach Solution - 1

Recap of the unique circular order (from Q186 constraints)
Working through the “beside” conditions yields the only consistent clockwise seating:
\[ A – C – E – G – B – D – F – H – (back to A) \] Left/Right convention on round tables
All face the centre. A person’s immediate right is the seat anti-clockwise from them; the immediate left is the seat clockwise from them. Test each option against the arrangement

(A) B to the immediate left of D? For D, the clockwise neighbor (left) is F (order … B–D–F …). Not B. False.
(B) H to the immediate right of A? For A, the anti-clockwise neighbor (right) is H (… F–H–A–C …). True.
(C) C to the immediate right of F? For F, the anti-clockwise neighbor (right) is D (… D–F–H …). Not C. False.
(D) B to the immediate left of H? For H, the clockwise neighbor (left) is A (… F–H–A …). Not B. False. Conclusion: Only statement (B) is correct.
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Approach Solution -2

Using the same seating loop A - C - E - G - B - D - F - H (back to A), with clockwise steps giving a person's immediate left and anti-clockwise steps giving their immediate right, let's test each statement:

  1. B will be to the immediate left of D: Moving clockwise (forward) from D along the loop ...B-D-F... lands on F, not B, so D's immediate left is F. This statement is false.
  2. H will be to the immediate right of A: The loop closes as ...F-H-A-C..., so moving anti-clockwise (backward) from A lands on H. Since anti-clockwise gives the right-hand neighbour, H is indeed to the immediate right of A. This statement is true.
  3. C will be to the immediate right of F: Moving anti-clockwise (backward) from F along the loop ...D-F-H... lands on D, not C, so F's immediate right is D. This statement is false.
  4. B will be to the immediate left of H: Moving clockwise (forward) from H along the loop ...H-A-C... lands on A, not B, so H's immediate left is A. This statement is false.

Only the statement about H sitting to the immediate right of A matches the seating loop.

Therefore, the correct answer is H will be to the immediate right of A.

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Approach Solution -3

Using the same position-numbering system, A=1, C=2, E=3, G=4, B=5, D=6, F=7, H=8 clockwise, with left neighbour at position+1 and right neighbour at position-1, wrapping around the 8 seats, let's check each statement:

  1. B will be to the immediate left of D: D is position 6, so D's left neighbour is position \( 6 + 1 = 7 \), which is F, not B, which is position 5. This statement is false.
  2. H will be to the immediate right of A: A is position 1, so A's right neighbour is position \( 1 - 1 \), wrapping around to position 8, which is H. This statement is true.
  3. C will be to the immediate right of F: F is position 7, so F's right neighbour is position \( 7 - 1 = 6 \), which is D, not C, which is position 2. This statement is false.
  4. B will be to the immediate left of H: H is position 8, so H's left neighbour is position \( 8 + 1 \), wrapping around to position 1, which is A, not B. This statement is false.

Working through the position numbers shows that only the claim about H sitting to the right of A holds up.

Therefore, the correct answer is H will be to the immediate right of A.

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Question: 3

A and F will become neighbours if:

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When two people are separated by exactly one fixed neighbour in a circular arrangement, they can become adjacent only if that common neighbour moves.
Updated On: Jul 15, 2026
  • B agrees to change her sitting position
  • C agrees to change her sitting position
  • G agrees to change her sitting position
  • H agrees to change her sitting position
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The Correct Option is D

Approach Solution - 1

Established clockwise order (from Q186): A – C – E – G – B – D – F – H – (back to A).
Here, A’s neighbours are C and H; F’s neighbours are D and H. The only person sitting between A and F is H.
Therefore, unless H vacates that seat (i.e., H changes position), A and F cannot become adjacent. Changing the seats of B, C, or G does not remove H from between A and F. Hence only option (D) works.
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Approach Solution -2

The seating loop is A - C - E - G - B - D - F - H (back to A). A and F are not currently adjacent; going one way around the table from A we pass C, E, G, B, D before reaching F (five people in between), and going the other way we pass only H before reaching F. So the short path between A and F has exactly one person sitting in it: H. Let's test each option:

  1. B agrees to change her sitting position: B sits between G and D, nowhere near the short A-H-F stretch, so moving B does not bring A and F any closer together.
  2. C agrees to change her sitting position: C sits between A and E, on the long side of the table relative to F, so moving C does not affect the short A-H-F stretch either.
  3. G agrees to change her sitting position: G sits between E and B, also on the long side away from the A-F gap, so moving G leaves the short path unaffected.
  4. H agrees to change her sitting position: H is the one and only person occupying the seat directly between A and F on the short side of the table. If H moves away from that seat, A and F would become directly adjacent.

Since H is the sole obstruction between A and F, only a change in H's position can make them neighbours.

Therefore, the correct answer is H agrees to change her sitting position.

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Approach Solution -3

A and F are not currently neighbours, so the question asks whose seat would need to change to bring them together. A useful way to see this is to write out each person's current two neighbours as a small list, then look for anyone who appears in BOTH A's neighbour list and F's neighbour list, since such a shared person is effectively the single bridge sitting between them:

  1. B: B's neighbours are G and D. B does not appear in A's neighbour list, C and H, or in F's neighbour list, D and H, so moving B has no bearing on the seat directly between A and F.
  2. C: C's neighbours are A and E. While C is one of A's neighbours, C does not appear anywhere in F's neighbour list, so moving C affects only A's other side, not the side facing F.
  3. G: G's neighbours are E and B. G appears in neither A's nor F's neighbour list, so moving G does nothing to close the gap between them.
  4. H: H's neighbours are F and A, meaning H is the one person who appears in BOTH A's neighbour list and F's neighbour list at once. H is the shared bridge sitting directly between them.

Since H is the only person common to both A's and F's immediate neighbour lists, only a change in H's seat can remove the single person separating A from F.

Therefore, the correct answer is H agrees to change her sitting position.

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Question: 4

During sitting:

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For $n=8$ in a round table, the person directly opposite is found by moving $+4$ seats (mod 8).
Updated On: Jul 15, 2026
  • A will be directly facing C
  • B will be directly facing C
  • A will be directly facing B
  • B will be directly facing D
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The Correct Option is C

Approach Solution - 1

With 8 seats, opposite seats are 4 places apart. Using the fixed order:
Indices (clockwise) $0$ A, $1$ C, $2$ E, $3$ G, $4$ B, $5$ D, $6$ F, $7$ H.
Opposites: A↔B (0↔4), C↔D (1↔5), E↔F (2↔6), G↔H (3↔7).
Thus A faces B, C faces D, etc. Only statement (C) matches this. Statements (A), (B), and (D) contradict the opposite pairs.
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Approach Solution -2

With eight people seated evenly around a round table, the person directly opposite anyone is exactly four seats away, counted in either direction, since \( 8 \div 2 = 4 \). Using the loop A - C - E - G - B - D - F - H (back to A), let's count four steps from A in each direction and check each option:

  1. A will be directly facing C: C is only one seat away from A along the loop, not four seats away in either direction, so C cannot be directly opposite A.
  2. B will be directly facing C: Counting four steps clockwise from C, C-E-G-B, is only 3 steps to B, not 4, so B is not exactly opposite C; C's true opposite is D (counting C-E-G-B-D is 4 steps).
  3. A will be directly facing B: Counting four steps clockwise from A, A-C-E-G-B, is exactly four steps, landing on B. Counting four steps the other way (anti-clockwise) from A, A-H-F-D-B, is also exactly four steps, landing on B again. Both directions agree that B sits directly opposite A.
  4. B will be directly facing D: Counting four steps from B, B-D-F-H-A, lands on A in four steps, not D, so B's true opposite is A, not D.

Counting four seats around the loop in either direction from A consistently lands on B, confirming they sit directly across from each other.

Therefore, the correct answer is A will be directly facing B.

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Approach Solution -3

On a round table seating exactly eight people, being "directly facing" someone means there are exactly three people sitting between the pair on each side, three plus the two people themselves plus three more equals all eight seats. Using the loop A-C-E-G-B-D-F-H (back to A), let's check each option by counting people on both sides:

  1. A will be directly facing C: Going from A to C the short way passes zero people in between, they are immediate neighbours, not opposite. This statement is false.
  2. B will be directly facing C: Going from C to B one way, C-E-G-B, passes two people, E and G, in between, not three, so they are not directly opposite. This statement is false.
  3. A will be directly facing B: Going from A to B one way, A-C-E-G-B, passes three people, C, E, G, in between, and going the other way, A-H-F-D-B, also passes exactly three people, H, F, D, in between. Both directions agree on a three-person gap, confirming they are directly opposite.
  4. B will be directly facing D: B and D are actually named as neighbours in clue (h), "B will sit beside D," so they have zero people between them, not three, meaning they cannot possibly be opposite each other.

Counting three people on both sides between A and B confirms they sit directly across the table from each other, while the other pairs fail this count.

Therefore, the correct answer is A will be directly facing B.

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Question: 5

H will be sitting between:

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Once the circle is fixed, read “between” as “the two immediate neighbours of the person” and simply list their adjacent seats.
Updated On: Jul 15, 2026
  • C and B
  • A and F
  • D and C
  • E and G
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The Correct Option is B

Approach Solution - 1

From the fixed arrangement A – C – E – G – B – D – F – H – A, the neighbours of H are F (on one side) and A (on the other). Hence H sits between A and F. Other pairs listed do not flank H in the circle.
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Approach Solution -2

The question asks which two people H sits between. Instead of reconstructing the entire seating order, we can answer this directly from just two of the original clues. Clue (b) states "H will sit beside A", and clue (d) states "F will sit beside H". Since a round table gives every person exactly two immediate neighbours, and these two clues already name both of H's neighbours directly, no further reconstruction is needed to answer this particular question. Let's still check each option against this:

  1. C and B: Neither C nor B is mentioned as sitting beside H in any of the original clues, so this pairing does not match H's actual neighbours.
  2. A and F: Clue (b) directly places A beside H, and clue (d) directly places F beside H. Both of H's neighbours are accounted for exactly by this pair.
  3. D and C: Neither D nor C is linked to H in the original clues; D sits beside B and F, while C sits beside A and E, so this option is incorrect.
  4. E and G: Neither E nor G is linked to H at all in the clues; they sit beside each other and beside C and B respectively, unrelated to H's seat.

Since the original clues directly state that H sits beside both A and F, and no other person is ever linked to H, H's two neighbours are confirmed as A and F.

Therefore, the correct answer is A and F.

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Approach Solution -3

Using the completed seating loop A-C-E-G-B-D-F-H, back to A, we can find H's neighbours simply by locating H in this ring and reading off the person immediately before and immediately after it. Let's check each option against the ring directly:

  1. C and B: In the ring, C sits between A and E, while B sits between G and D, neither one is positioned next to H at all, so this pairing does not match H's actual neighbours.
  2. A and F: The ring reads ...D-F-H-A-C..., so the person immediately before H is F, and the person immediately after H is A. Both A and F sit directly on either side of H in the ring.
  3. D and C: D sits between B and F in the ring, and C sits between A and E, neither of them is adjacent to H, so this pairing is incorrect.
  4. E and G: E sits between C and G, and G sits between E and B, neither is anywhere near H's position in the ring.

Reading the ring directly around H's position shows its two immediate neighbours are F on one side and A on the other, matching only the second option.

Therefore, the correct answer is A and F.

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