In a college library, four different business newspapers — Economic Times,…
2025
In a college library, four different business newspapers — Economic Times, Business Standard, Business Line, and Financial Express — are available. All students visit the library regularly, but 20% of them do not read any business newspaper. The four newspapers, in the order given above, are read by 230, 180, 180, and 220 students respectively. For any two of the four newspapers, exactly 20 students read exactly those two newspapers. 30 students read all four newspapers, and no student reads exactly three of the four newspapers.
Suppose every student in the college — including those who currently read no newspaper — starts reading one additional newspaper (from the four listed above) that they are not reading at present; a student who already reads all four newspapers has none left to add and is therefore unaffected. What is the least number of students who then read all four newspapers?
- A.
45
- B.
90
- C.
50
- D.
30
Attempted by 5 students.
Show answer & explanation
Correct answer: D
Concept: When a group of people is classified only by how many of several categories they fall into (0, exactly-1, exactly-2, ... exactly-n), two identities always hold: (a) each category's total headcount (e.g. total readers of a single newspaper) equals the sum, over every ‘exactly-k’ group, of k times that group's size, since a person in the exactly-k group is counted once for each of the k categories they belong to; and (b) the number of people belonging to at least one category equals the sum of the sizes of all the exactly-k groups. Once the exactly-k group sizes are known, applying a fixed rule that shifts every person's category by a known amount (here, ‘each person now belongs to exactly one more category than before, choosing one they were not already in’) tells you exactly how the group sizes move.
Add the four individual newspaper readership figures: 230 + 180 + 180 + 220 = 810. By identity (a) above, this total equals 1×(exactly-one readers) + 2×(exactly-two readers) + 3×(exactly-three readers) + 4×(exactly-four readers).
The problem gives the exactly-three-readers count as 0 and the exactly-four-readers count as 30.
The problem also gives, for every one of the six possible newspaper pairs, an exactly-two-readers count of 20 per pair, so the total exactly-two-readers count is 6 × 20 = 120.
Substituting into the identity: (exactly-one) + 2(120) + 3(0) + 4(30) = 810, i.e. (exactly-one) + 240 + 120 = 810, so exactly-one readers = 450.
By identity (b), the number of students reading at least one newspaper = 450 + 120 + 0 + 30 = 600.
Since 20% of students read no newspaper, the 600 students reading at least one newspaper make up the remaining 80% of the college. So total students = 600 ÷ 0.8 = 750, and students reading no newspaper = 750 − 600 = 150.
Now apply the stated rule: every student adds exactly one newspaper (from the four) that they are not currently reading. A student's category can rise by at most one step — a student reading exactly k newspapers can reach at most exactly (k+1) newspapers, never more, since only one new newspaper is added.
Track where each pre-change group lands: the 150 no-newspaper readers rise to exactly-one; the 450 exactly-one readers rise to exactly-two; the 120 exactly-two readers rise to exactly-three; the exactly-three readers (0 of them) would rise to exactly-four; and the 30 students already reading all four remain at all four — there is no fifth newspaper left for them to add, so the instruction simply cannot apply to them and they are unaffected by the change.
So the only students reading all four newspapers after the change are the 30 who already did, plus however many rise from exactly-three to exactly-four — and that second group has zero members.
Cross-check: recombining the post-change groups — 150 (exactly-one) + 450 (exactly-two) + 120 (exactly-three) + 30 (exactly-four) = 750, matching the total student strength found above, so no student is left uncounted or double-counted by the shift. Notice also that this final count depends only on the two figures given directly in the problem — the already-all-four count (30) and the exactly-three count (0) — since only exactly-three-newspaper readers can rise into the all-four group when each student adds just one unread newspaper; the exactly-one, exactly-two breakdown and the total student strength (though needed to confirm the scenario is consistent) do not change this final count.
Hence, the least number of students reading all four newspapers after every student adds one unread newspaper is 30.