Question : The average of 10 consecutive integers is $\frac{11}{2}$ . What is the average of the "smallest four" of these integers?
Option 1: 2.5
Option 2: 3.5
Option 3: 3
Option 4: 2

Question

Question : The average of 8 consecutive integers is $\frac{27}{2}$ . What is the average of the largest four of these integers?
Option 1: 16.5
Option 2: 15
Option 3: 14.5
Option 4: 15.5

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Answer 1

Correct Answer: 2.5

Solution : Let the 10 consecutive integers be $x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), (x + 6), (x + 7), (x + 8), (x + 9)$
Average $= \frac{x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) (x + 8) (x + 9)}{10} = \frac{11}{2}$
⇒ $10 x + 45 = \frac{11}{2} \times 10$
⇒ $10 x = 55 - 45$
⇒ $x = \frac{10}{10} = 1$
The Smallest four integers = $x + (x + 1) + (x + 2) + (x + 3)$
= 1 + 2 + 3 + 4
= 10
Average of the smallest four integers = $\frac{10}{4}$ = 2.5
Hence, the correct answer is 2.5.

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Question : The average of 10 consecutive integers is 112. What is the average of the "smallest four" of these integers?Option 1: 2.5Option 2: 3.5Option 3: 3Option 4: 2

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Question : The average of 10 consecutive integers is $\frac{11}{2}$ . What is the average of the "smallest four" of these integers?
Option 1: 2.5
Option 2: 3.5
Option 3: 3
Option 4: 2