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Average of 10 observation in a series is...

Average of 10 observation in a series is 64. If average of first 5 observation is 45 and last 3 observation is 71. Find the average of remaining.

A

a. 91

B

b. 103

C

3,1

D

None of the above

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AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to find the average of the remaining observations after knowing the averages of the first five and the last three observations. ### Step-by-Step Solution: 1. **Calculate the Total of All Observations:** - Given that the average of 10 observations is 64, we can find the total sum of these observations. \[ \text{Total Sum} = \text{Average} \times \text{Number of Observations} = 64 \times 10 = 640 \] 2. **Calculate the Total of the First 5 Observations:** - The average of the first 5 observations is given as 45. \[ \text{Total of First 5 Observations} = 45 \times 5 = 225 \] 3. **Calculate the Total of the Last 3 Observations:** - The average of the last 3 observations is given as 71. \[ \text{Total of Last 3 Observations} = 71 \times 3 = 213 \] 4. **Calculate the Total of the Remaining Observations:** - We now need to find the total of the remaining observations. There are 10 observations in total, and we have accounted for 5 (first) + 3 (last) = 8 observations. - Therefore, the number of remaining observations is \(10 - 8 = 2\). - The total of the remaining observations can be found by subtracting the totals of the first 5 and last 3 observations from the total sum of all observations. \[ \text{Total of Remaining Observations} = \text{Total Sum} - (\text{Total of First 5} + \text{Total of Last 3}) \] \[ = 640 - (225 + 213) = 640 - 438 = 202 \] 5. **Calculate the Average of the Remaining Observations:** - Now, we can find the average of the remaining 2 observations. \[ \text{Average of Remaining Observations} = \frac{\text{Total of Remaining Observations}}{\text{Number of Remaining Observations}} = \frac{202}{2} = 101 \] ### Final Answer: The average of the remaining observations is **101**.
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