Question:

Seven people, A, B, C, D, E, F and G, want to go boating. Only two boats are available, and their seating must follow these rules:
(I) A will go in the same boat as E.
(II) F cannot go in the same boat as C, unless D also joins them.
(III) Neither B nor C can be placed in the same boat as G.
(IV) A single boat can carry a maximum of four people.

If E is in the same boat as F, which of the following is the complete and accurate list of people in the other boat?

Show Hint

Work out who can safely become the fourth passenger with A, E and F without breaking the G avoidance rule or the F-C-D rule.
Updated On: Jul 14, 2026
  • F and E
  • G and A
  • D and A
  • C, D and B
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The Correct Option is D

Solution and Explanation

Step 1: Lock in the fixed pair.
Rule I keeps A and E together always. Since E sits with F here, the boat already has A, E and F, three people so far.

Step 2: Test adding C to this boat.
If C joined A, E, F, rule II would force D in too, since F and C cannot share a boat alone. That gives five people, A, E, F, C, D, which breaks the four person cap in rule IV. So C cannot join this boat.

Step 3: Test adding B or D alone.
If B alone joined without C, the remaining three, C, D, G, would end up in the other boat together, but rule III bans C from sitting with G, so that fails. If D alone joined, the remaining three, B, C, G, would sit together, again breaking rule III since B and C cannot sit with G. Both attempts fail.

Step 4: Add G instead.
If G joins A, E, F, the boat is full at four people. The leftover three, B, C and D, go to the other boat, and this boat has no G, so rule III is satisfied for both B and C. No other rule is broken either.

Final Answer:
The only seating that satisfies every rule places A, E, F, G in one boat and leaves C, D and B in the other. \[ \boxed{\text{C, D and B}} \]
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