The average of n numbers is 42. If 60% of the numbers are increased by 5 each and the remaining numbers are decreased by 10 each, then what will be the average of the numbers so obtained?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 41.
Total sum = 42n
60% of the numbers (i.e., 0.6n) are increased by 5 , so total increase = 0.6n × 5 = 3n
40% of the numbers (i.e., 0.4) are decreased by 10 , so total decrease = 0.4n × 10 = 4n
So the net change in sum = +3n − 4n = −n
New total sum = 42n − n = 41n
New average = 41n/n = 41