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?
- A41
- B45
- C43
- D42
Solution & Step-by-step Explanation
Let's assume there are 100 numbers (n=100).
The initial total sum of the numbers is:
Initial Sum=100×42=4200
Now, let's look at the changes:
60% of the numbers (60 numbers) are increased by 5 each:
Total increase=60×5=+300
The remaining numbers (40% or 40 numbers) are decreased by 10 each:
Total decrease=40×10=−400
The net change in the sum is:
Net Change=+300−400=−100
The new total sum is:
New Sum=4200−100=4100
The new average across the 100 numbers is:
New Average=
100
4100
=41
The initial total sum of the numbers is:
Initial Sum=100×42=4200
Now, let's look at the changes:
60% of the numbers (60 numbers) are increased by 5 each:
Total increase=60×5=+300
The remaining numbers (40% or 40 numbers) are decreased by 10 each:
Total decrease=40×10=−400
The net change in the sum is:
Net Change=+300−400=−100
The new total sum is:
New Sum=4200−100=4100
The new average across the 100 numbers is:
New Average=
100
4100
=41