Average, Mixture and Alligation Question 9 - CAT 2022
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:
Step-by-Step Explanation
- Find the Total Sum of Marks:
Five distinct students have an overall average score of 38, which means the total sum of their marks is:
Total Sum = 38 Γ 5 = 190 marks
- Analyze the Score Distribution Constraints:
The marks are distinct integers in the range 0, 50. Exactly three students scored strictly above 32, which implies the remaining two students scored 32 or less. Amit scored the fewest marks overall, meaning he occupies the absolute lowest position in this two-student lower pool.
A < X β€ 32 < Sβ < Sβ < Sβ β€ 50
- Find the Lowest Possible Marks for Amit (Aβα΅’β):
To make Amit's score as small as possible, push the scores of the other four students to their highest possible limits.
- Maximize the top three scores: Sβ = 50, Sβ = 49, Sβ = 48.
- Maximize the intermediate score below the threshold: X = 32.
Sum of the other four scores = 50 + 49 + 48 + 32 = 179
Aβα΅’β = 190 β 179 = 11
- Find the Highest Possible Marks for Amit (Aβββ):
To make Amit's score as large as possible, push it up against its structural ceiling. Amit belongs to the lower group where A < X β€ 32. The largest integer score Amit can obtain while remaining strictly lower than a distinct integer X (where X β€ 32) is when X takes its maximum value of 32 and Amit takes the next lower integer value:
Aβββ = 31
(A valid choice for the top three scorers exists here because their required sum would be 190 β 32 β 31 = 127, which is well within the possible range for three distinct integers chosen from 33, 50.)
- Calculate the Difference:
Difference = Aβββ β Aβα΅’β = 31 β 11 = 20
Correct Option: D
π‘ FundaMakers Tip:
Extreme-value optimization relies on a simple balance: to minimize one component, maximize everything else to its absolute ceiling. Conversely, to find a maximum value within a constrained sequence, look directly at the immediate structural boundary wall. Recognizing that Amit is locked below the threshold limit allows you to find Aβββ = 31 instantly, without needing to calculate the specific values for the higher scores.