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The average of 8 consecutive integers is 23/2. What is the average of first three integers?
- a)9
- b)19/2
- c)8
- d)10
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The average of 8 consecutive integers is 23/2. What is the average of ...
Let the 8 consecutive integers be x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), (x + 6), (x + 7)
Average = [x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7)]/8 = 23/2
⇒ (8x + 28) = (23/2) × 8 = 92
⇒ 8x = 92 - 28 = 64
⇒ x = 64/8 = 8
So, the first three numbers are 8, (8 + 1), (8 + 2)
Average of the first three integers = (8 + 9 + 10)/3 = 27/3 = 9
∴ Required Average = 9
Average = [x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7)]/8 = 23/2
⇒ (8x + 28) = (23/2) × 8 = 92
⇒ 8x = 92 - 28 = 64
⇒ x = 64/8 = 8
So, the first three numbers are 8, (8 + 1), (8 + 2)
Average of the first three integers = (8 + 9 + 10)/3 = 27/3 = 9
∴ Required Average = 9
Most Upvoted Answer
The average of 8 consecutive integers is 23/2. What is the average of ...
Given:
- The average of 8 consecutive integers is 23/2.
To find:
- The average of the first three integers.
Solution:
Let's assume the first integer as 'x'.
Since we are given that the integers are consecutive, we can write the other consecutive integers as x+1, x+2, x+3, x+4, x+5, x+6, and x+7.
Finding the average of the 8 consecutive integers:
The average of these 8 consecutive integers can be calculated by adding them all and dividing by 8.
(x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7))/8 = 23/2
Simplifying the equation:
8x + 28 = 92
8x = 92 - 28
8x = 64
x = 64/8
x = 8
Calculating the average of the first three integers:
The first three integers are x, x+1, and x+2.
Average = (x + (x+1) + (x+2))/3
Average = (8 + (8+1) + (8+2))/3
Average = (8 + 9 + 10)/3
Average = 27/3
Average = 9
Therefore, the average of the first three integers is 9.
Hence, the correct answer is option A (9).
- The average of 8 consecutive integers is 23/2.
To find:
- The average of the first three integers.
Solution:
Let's assume the first integer as 'x'.
Since we are given that the integers are consecutive, we can write the other consecutive integers as x+1, x+2, x+3, x+4, x+5, x+6, and x+7.
Finding the average of the 8 consecutive integers:
The average of these 8 consecutive integers can be calculated by adding them all and dividing by 8.
(x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7))/8 = 23/2
Simplifying the equation:
8x + 28 = 92
8x = 92 - 28
8x = 64
x = 64/8
x = 8
Calculating the average of the first three integers:
The first three integers are x, x+1, and x+2.
Average = (x + (x+1) + (x+2))/3
Average = (8 + (8+1) + (8+2))/3
Average = (8 + 9 + 10)/3
Average = 27/3
Average = 9
Therefore, the average of the first three integers is 9.
Hence, the correct answer is option A (9).
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The average of 8 consecutive integers is 23/2. What is the average of first three integers?a)9b)19/2c)8d)10Correct answer is option 'A'. Can you explain this answer?
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