Averages , Ratio and Proportion Question 1 - CAT 2020
A batsman played n+2 innings and got out on all occasions. His average score in these n+2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and the lowest score was x runs. The smallest possible value of x is:
Step-by-Step Explanation
- Find the number of initial innings (n):
The total runs scored across all n + 2 innings can be written by combining the sum of the first n innings and the absolute scores of the final two innings.
Total Runs = 30n + 38 + 15 = 30n + 53
We are given that the overall average for all n + 2 innings is 29 runs:
(30n + 53)/(n + 2) = 29
30n + 53 = 29n + 58
n = 5 innings
- Analyze the first 5 innings for optimization:
The total runs scored in these first 5 innings:
30 × 5 = 150 runs.
We want to find the smallest possible value of x (the lowest score). To minimize x, we must maximize the scores of the remaining 4 innings.
The problem states that the batsman scored less than 38 runs in each of these innings. Since cricket scores are integers, the maximum possible score for any single inning is 37 runs.
Maximum sum of the other 4 innings = 37 × 4 = 148 runs
x + 148 = 150
x = 2 runs
Correct Option: C
💡 FundaMakers Tip:
Use the Average Net Deviation Method to find n mentally. The first n innings have an average of 30, which is +1 above the final average of 29. This gives a total surplus of +n.
The last two innings contribute 38 and 15 runs. Their deviations from the target average of 29 are 38 − 29 = +9 and 15 − 29 = −14, creating a net deviation of +9 − 14 = −5.
For the overall average to balance out at 29, the net sum of all deviations must be zero:
+n − 5 = 0
n = 5
Once n = 5 is found, the optimization step simply requires pushing the other scores up to their upper ceiling boundary (≤ 37) to leave the minimum possible remainder for x.