Average, Mixture and Alligation Question 3 - CAT 2021
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is:
Step-by-Step Explanation
- Find the Total Marks:
Total marks = 25 × 50 = 1250
- Distribute to Maximize the Top Score:
Let the score of the 5 identical toppers be S. To maximize S, we must minimize the combined marks of the other 20 students.
We are given that the remaining 20 scores are distinct integers and the lowest score is 30. The absolute minimum values they can take are the consecutive integers from 30 up to 49:
Minimum sum of 20 scores= 30+31+32+.......+49
- Calculate the Arithmetic Progression Sum:
Sum = 20 / 2 × (30 + 49) = 10 × 79 = 790
- Solve for S:
5S + 790 = 1250
5S = 460
S = 92
The maximum possible score of the toppers is 92.
Correct Answer: 92
💡 FundaMakers Tip:
Maximum/Minimum optimization on datasets works like a see-saw. To push one side to its absolute highest limit, you must bind all other elements tightly down to their absolute minimal floor levels. Spotting that the non-toppers form a clean, consecutive Arithmetic Progression saves time during summation.