Averages , Ratio and Proportion Question 5 - CAT 2022
In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is:
Step-by-Step Explanation
Let the number of students in Section A be a. The number of students in Section B is a + 10.
Combined Average = (32a + 60(a + 10))/(a + a + 10)
= (92a + 600)/(2a + 10)
= (46a + 300)/(a + 5)
- Simplify the fraction algebraically:
Combined Average = (46 (a + 5) + -230+300)/(a + 5)
= 46 + 70/(a + 5)
- Apply the integer constraint:
For the combined average to be an integer, the term (a + 5) must be a clean factor of 70.
Since the student count a must be positive (a ≥ 1), we check the factors of 70 that are greater than or equal to 6:
Valid factors for (a + 5) ∈ {7, 10, 14, 35, 70}
Minimum size:
a + 5 = 7
a minimum = 2
Maximum size:
a + 5 = 70
a maximum = 65
Difference = a maximum − a minimum
= 65 − 2
= 63
Correct Answer: 63
💡 FundaMakers Tip:
Use polynomial long division to separate a complex fraction into a clean integer plus a remainder fraction (Integer + Remainder/Denominator). This approach turns a difficult substitution problem into a basic question about factors, allowing you to find the boundary limits for a easily.