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📌 Section 1 is Logical reasoning specially designed for CAT
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SET 1 — 🔍 CAT Level
Five friends, A through E, has a certain number of pens with them. The five persons have pens of two different colours – Black and Red. It is known that the number of Red pens with C is equal to the average of the number of Black pens with the five persons, while the number of Black pens with A is equal to the average number of Red pens with the five persons. The total number of pens with the five persons was 60 and each person has at least two black pens. The difference in the number of Red pens between any two persons was at least 3 and the difference in the number of Black pens between any two persons was at least 2.
Further, D has more number of Red pens than that with C and also has more number of Black pens than Red pens. Each of A, B and C has more number of Red pens than Black pens.
Question 1
What is the difference between the number of Red pens with D and the number of Black pens with B?
Question 2
The difference between the total number of pens with any pair of persons is at most ______.
Question 3
One of the five persons gave a certain number of pens of exactly one colour to another of the five persons. After this, the person who gave the pens had an equal number of Red and Black pens with him; and the same was the case with the person who received the pens.
What is the total number of pens with these two persons?
Question 4
One of the five persons gave a certain number of pens of exactly one colour to one other person. After this, the person who gave the pens had an equal number of Red and Black pens with him; and the same was the case with the person who received the pens.
What is the maximum possible number of pens with any of these two persons after this?
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✅ Answers & Explanations
We can determine the entire distribution uniquely.
Since the total number of pens is 60, let total Black pens = B and total Red pens = R.
Given:
Red pens with C = average Black pens = B/5
Black pens with A = average Red pens = R/5
B+R=60
Because every person has at least 2 Black pens and the Black counts differ by at least 2, the minimum possible five Black counts are:
2, 4, 6, 8, 10 → Total = 30
Similarly, the five Red counts differ by at least 3. Their minimum possible total is:
0, 3, 6, 9, 12 → Total = 30
Hence:
Total Black = 30
Total Red = 30
Therefore:
C has 30/5=6 Red pens.
A has 30/5=6 Black pens.
Now D has more Red than C, so D has either 9 or 12 Red pens. But D also has more Black than Red, and the maximum Black count is 10. Hence:
D → Red = 9, Black = 10
A, B and C have more Red than Black.
A already has 6 Black. Since 9 Red belongs to D, A must have:
A → Red = 12, Black = 6
C has 6 Red, so its Black count must be below 6.
B must also have Red > Black. The remaining Red values are 0 and 3 for B and E. Hence B must have 3 Red and therefore 2 Black.
This gives the complete table:
Person | Red | Black | Total |
|---|---|---|---|
A | 12 | 6 | 18 |
B | 3 | 2 | 5 |
C | 6 | 4 | 10 |
D | 9 | 10 | 19 |
E | 0 | 8 | 8 |
1️⃣ Difference between Red pens with D and Black pens with B
9−2=7
2️⃣ Maximum difference between total pens of any two persons
Maximum = D = 19
Minimum = B = 5
19−5=14
3️⃣ Total number of pens with the two persons involved in the transfer
Initially:
B → 3 Red, 2 Black
D → 9 Red, 10 Black
B can give 1 Red pen to D:
B → 2 Red, 2 Black
D → 10 Red, 10 Black
Both then have equal numbers of Red and Black pens.
Combined total:
4+20=24
4️⃣ Maximum possible number of pens with either of these two persons after the transfer
After B gives 1 Red pen to D:
D has:
10+10=20
✅ Final Answers
1️⃣ 7
2️⃣ 14
3️⃣ 24
4️⃣ 20
⚡ Brain teasers are not just about getting the right answer — they train you to think differently under pressure.
Stay consistent with your daily practice, challenge your assumptions, and keep sharpening your reasoning skills.
See you in tomorrow’s challenge!