Maximum runs from mixed averages when two outliers are removed: A cricketer’s average over 42 innings is 30. The difference between his maximum and minimum scores is 100. If these two innings (max and min) are excluded, the average over the remaining 40 innings is 28. What is his maximum score in a single innings?

Introduction / Context:
We know the overall average, the average after removing two outliers, and the gap between the max and min scores. Using totals, we can form and solve a system to recover the maximum score.

Given Data / Assumptions:

• 42-innings average = 30 → total = 42 * 30 = 1260
• Average of 40 innings (without max and min) = 28 → total = 40 * 28 = 1120
• Let maximum = M, minimum = m, with M − m = 100

Concept / Approach:
Sum of removed innings = 1260 − 1120 = 140 → M + m = 140. Solve simultaneously with M − m = 100.

Step-by-Step Solution:
M + m = 140M − m = 100Adding: 2M = 240 → M = 120

Verification / Alternative check:
Then m = 20. Removing 120 and 20 reduces the total to 1120 and the average to 28 over 40 innings, satisfying all conditions.

Why Other Options Are Wrong:
125, 110, 100, and 130 do not pair with any minimum to produce both the 140 sum and a 100-point gap.

Common Pitfalls:
Using 42 as the divisor after removal, or forgetting that two innings are dropped (both count and total change).

Final Answer:
120