Four students A, B, C and D distributed 30 marbles among themselves. No two students got..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> Four students A, B, C and D distributed 30 marbles among themselves. No two students got..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> Four students A, B, C and D distributed 30 marbles among themselves. No two students got..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 62

Logical Reasoning Puzzles Easy

<p>Four students A, B, C and D distributed 30 marbles among themselves. No two students got equal number of marbles. No student got more than 10 marbles. No student got less than 5 marbles. A and C got odd number of marbles. B and D got even number of marbles. A got more marbles than B, C got more marbles than D, B got more marbles than D.&nbsp;</p> <p>What is the number of marbles with A?&nbsp;</p>

(A) 6
(B) 7
(C) 8
(D) 9
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Step-by-Step Logical Deduction

To find the number of marbles with A, we must find the unique distribution of 30 marbles among the four students that satisfies all the given conditions.

1. List All Constraints

  • Total Marbles: A + B + C + D = 30
  • Unique Amounts: All values for A, B, C, D are different.
  • Range: Each person has between 5 and 10 marbles, inclusive. {5, 6, 7, 8, 9, 10}
  • Parity:
    • A and C have odd numbers of marbles: {5, 7, 9}
    • B and D have even numbers of marbles: {6, 8, 10}
  • Inequalities:
    • A > B
    • C > D
    • B > D (Note: This is the interpretation that leads to the correct answer, as the text in some sources is ambiguous)

2. Deduce the Values for B and D

From the inequalities, we have a chain: B > D. Since B and D must be two different even numbers from {6, 8, 10}, the possible pairs for (B, D) are:

  • (8, 6)
  • (10, 6)
  • (10, 8)

3. Test Each Case

We'll now test each possible pair for (B, D) to see if a valid solution can be found.

Case 1: B = 8 and D = 6

  • From the inequality C > D, we know C > 6. The possible odd values for C are 7 or 9.
    • If C = 7: The sum is A + 8 + 7 + 6 = 30 → A + 21 = 30 → A = 9.
      Let's check the final condition A > B: Is 9 > 8? Yes. This is a valid solution: {A=9, B=8, C=7, D=6}.
    • If C = 9: The sum is A + 8 + 9 + 6 = 30 → A + 23 = 30 → A = 7.
      Let's check the final condition A > B: Is 7 > 8? No. This is not a valid solution.

Case 2: B = 10 and D = 6

  • From C > D, we know C > 6. Possible C values are 7 or 9.
    • If C = 7: A + 10 + 7 + 6 = 30 → A + 23 = 30 → A = 7. C and A cannot have the same value. Invalid.
    • If C = 9: A + 10 + 9 + 6 = 30 → A + 25 = 30 → A = 5. Check A > B: Is 5 > 10? No. Invalid.

Case 3: B = 10 and D = 8

  • From C > D, we know C > 8. The only possible value for C is 9.
  • The sum is A + 10 + 9 + 8 = 30 → A + 27 = 30 → A = 3. This is outside the valid range of [5, 10]. Invalid.

Conclusion

The only combination of marbles that satisfies all the given conditions is A=9, B=8, C=7, and D=6.

Therefore, the number of marbles with A is 9.

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