Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is
Question:
Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is
Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is
Options
Answer: (3) 20
Explanation:
Step 1: Define variables Let the marks in increasing order be: Amit = a, and b < c < d < e. Total marks = 5 × 38 = 190 Exactly three students > 32 → c, d, e > 32; a, b ≤ 32 All marks distinct integers Step 2: Maximum Amit marks To maximize a, minimize b, c, d, e: Let b = a +1 Adjust c, d, e slightly above 32 while keeping total sum = 190 Optimizing gives maximum a = 31 Step 3: Minimum Amit marks To minimize a, maximize b, c, d, e: b ≤ 32 → b = 32 c, d, e = 50, 49, 48 → sum =147 a + b = 190 - 147 = 43 → a = 43 - 32 = 11 Step 4: Difference Difference = 31 - 11 = 20 Answer: 20
Explanation:
Step 1: Define variables Let the marks in increasing order be: Amit = a, and b < c < d < e. Total marks = 5 × 38 = 190 Exactly three students > 32 → c, d, e > 32; a, b ≤ 32 All marks distinct integers Step 2: Maximum Amit marks To maximize a, minimize b, c, d, e: Let b = a +1 Adjust c, d, e slightly above 32 while keeping total sum = 190 Optimizing gives maximum a = 31 Step 3: Minimum Amit marks To minimize a, maximize b, c, d, e: b ≤ 32 → b = 32 c, d, e = 50, 49, 48 → sum =147 a + b = 190 - 147 = 43 → a = 43 - 32 = 11 Step 4: Difference Difference = 31 - 11 = 20 Answer: 20
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