The average of 18 numbers is 65. The average of the first 10 numbers is 72 and that of the last 9 numbers is 58. If the `10^(th)` number is excluded, then what is the average of the remaining numbers (correct to one decimal place) ?

Options

A. `64.6`
B. `66.5`
C. `65.2`
D. `65.4`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the total sum of the 18 numbers.
Given that the average of 18 numbers is 65, we can find the total sum (S) of these numbers using the formula:
\[ S = \text{Average} \times \text{Number of items} \]
\[ S = 65 \times 18 = 1170 \]

### Step 2: Calculate the total sum of the first 10 numbers.
The average of the first 10 numbers is given as 72. Therefore, the total sum (S1) of the first 10 numbers is:
\[ S1 = 72 \times 10 = 720 \]

### Step 3: Calculate the total sum of the last 9 numbers.
The average of the last 9 numbers is given as 58. Therefore, the total sum (S2) of the last 9 numbers is:
\[ S2 = 58 \times 9 = 522 \]

### Step 4: Determine the 10th number.
Since the 10th number is included in both the first 10 numbers and the last 9 numbers, we can find the 10th number (N) using the equation:
\[ S1 + S2 - N = S \]
Substituting the values we have:
\[ 720 + 522 - N = 1170 \]
\[ 1242 - N = 1170 \]
\[ N = 1242 - 1170 = 72 \]

### Step 5: Calculate the sum of the remaining 17 numbers.
Now that we have the 10th number, we can calculate the sum of the remaining 17 numbers:
\[ \text{Sum of remaining 17 numbers} = S - N = 1170 - 72 = 1098 \]

### Step 6: Calculate the average of the remaining 17 numbers.
Finally, we can find the average of the remaining 17 numbers:
\[ \text{Average} = \frac{\text{Sum of remaining 17 numbers}}{\text{Number of remaining numbers}} = \frac{1098}{17} \]
Calculating this gives:
\[ \text{Average} = 64.5882 \approx 64.6 \text{ (to one decimal place)} \]

### Final Answer:
The average of the remaining numbers, correct to one decimal place, is **64.6**.

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