The average of 12 numbers is 48. The average of the first 5 numbers is 45 and the average of next 4 numbers is 52. If the 10th number is 10 less than the 11th number and is 5 more than the 12th number, then what is the average of the 11th and 12th numbers?

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Accepted Answer

Let the 12 numbers be: a1, a2, ..., a12.

Step 1: Total sum of all 12 numbers = 12 × 48 = 576.

Step 2: Sum of first 5 numbers = 5 × 45 = 225.

Step 3: Sum of next 4 numbers (6th to 9th) = 4 × 52 = 208.

Step 4: Let 10th number = x, 11th number = y, 12th number = z.

So, sum of all numbers:
225 (first 5) + 208 (next 4) + x + y + z = 576
=> 433 + x + y + z = 576
=> x + y + z = 576 - 433 = 143

Step 5: Given:
- 10th number is 10 less than 11th: x = y - 10
- 10th number is 5 more than 12th: x = z + 5

So, y = x + 10, z = x - 5

Step 6: Substitute in sum:
x + (x + 10) + (x - 5) = 143
=> x + x + 10 + x - 5 = 143
=> 3x + 5 = 143
=> 3x = 138
=> x = 46

So,
y = x + 10 = 56
z = x - 5 = 41

Step 7: Average of 11th and 12th numbers = (y + z)/2 = (56 + 41)/2 = 97/2 = 48.5

But this matches option 4, not option 3. Let's double-check:
- x = 46, y = 56, z = 41
- x + y + z = 46 + 56 + 41 = 143 (matches)
- x = y - 10 = 56 - 10 = 46 (matches)
- x = z + 5 = 41 + 5 = 46 (matches)

So, the correct answer is 48.5, which is option 4.

Answer

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