The average of 19 numbers is 42. The average of the first 6 numbers is 38.5 and that of the last 14 numbers is 45.5. If the sixth number is excluded then what’s the average of the remaining numbers?(Correct to one decimal Place)41.640.440.841.2
Question
The average of 19 numbers is 42. The average of the first 6 numbers is 38.5 and that of the last 14 numbers is 45.5. If the sixth number is excluded then what’s the average of the remaining numbers?(Correct to one decimal Place)41.640.440.841.2
Solution
The problem can be solved in the following steps:
- The total sum of the 19 numbers is 42 * 19 = 798.
- The total sum of the first 6 numbers is 38.5 * 6 = 231.
- The total sum of the last 14 numbers is 45.5 * 14 = 637.
- The sixth number is included in both the first 6 numbers and the last 14 numbers. So, it has been counted twice in the total sum. To find the value of the sixth number, subtract the total sum of the 19 numbers from the sum of the first 6 numbers and the last 14 numbers: (231 + 637) - 798 = 70.
- Now, to find the average of the remaining 18 numbers, subtract the sixth number from the total sum of the 19 numbers and divide by 18: (798 - 70) / 18 = 40.4.
So, the average of the remaining numbers, correct to one decimal place, is 40.4.
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