A top brass of a political party has to address public meetings at seven major…

2025

A top brass of a political party has to address public meetings at seven major cities — A, B, C, D, E, F and G — as part of an election campaign during a certain week. Only nine appointment slots are available that week: one on Monday, three on Tuesday, one on Wednesday, two on Thursday, and two on Friday. Two addresses must be held on Thursday and at least one on each of the other days.

Additional conditions:

  • B and E must be scheduled on the same day.

  • City A must be scheduled on Monday.

  • C cannot be scheduled on Friday.

  • G must be scheduled on Thursday.

Which of the following must be true of the appointment slots for the addresses?

  1. A.

    Three days each have exactly one public address

  2. B.

    E and B on Thursday

  3. C.

    F and G on Thursday

  4. D.

    None

Show answer & explanation

Correct answer: A

Concept

In a scheduling puzzle where each day has a fixed number of slots and only some slots are used, a statement is a 'must be true' only when it holds in EVERY valid arrangement. The reliable first step is to pin the days whose count is forced, then see how the remaining items can be spread over the flexible days.

Application

  1. Seven cities are each addressed once, so 7 addresses must be placed in the 9 available slots (Monday 1, Tuesday 3, Wednesday 1, Thursday 2, Friday 2), with every day getting at least one and Thursday getting exactly two.

  2. Monday has a single slot and needs at least one, so Monday = 1 (city A); Wednesday likewise = 1; Thursday = 2 (city G plus one more). These three days use 1 + 1 + 2 = 4 addresses.

  3. The remaining 7 − 4 = 3 addresses go to Tuesday and Friday, each needing at least one and Friday capped at two, so (Tuesday, Friday) is either (2, 1) or (1, 2). Both splits are realisable — the B–E pair, which must share a day and cannot both fit on Thursday alongside G, simply takes whichever of Tuesday or Friday carries two addresses.

  4. In both cases exactly one of Tuesday and Friday holds a single address; together with Monday and Wednesday that is exactly three days with one address each — true in every arrangement.

Cross-check

Checking the other statements:

  • 'E and B on Thursday' is impossible: one Thursday slot is already fixed by G, leaving only one free slot, so the B–E pair cannot share Thursday.

  • 'F and G on Thursday' is not forced: Thursday's second slot can go to another city instead. For instance, G and D on Thursday (with B–E on Tuesday, C on Wednesday and F on Friday) is a valid schedule, and so is G and C on Thursday — so F is only one of several possible partners.

Hence only 'three days each have exactly one public address' is guaranteed in every valid schedule; the saved answer is correct.

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