The average of five consecutive integers is 12. What is the smallest of these in

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Question: The average of five consecutive integers is 12. What is the smallest of these integers?

Options:

Correct Answer: 11

Solution:

Let the integers be x, x+1, x+2, x+3, x+4.\\n\\nMean = (5x + 10) / 5 = 12\\n\\n5x + 10 = 60\\n\\n5x = 50\\n\\nx = 10.

The average of five consecutive integers is 12. What is the smallest of these integers?

The average of five consecutive integers is 12. What is the smallest of these integers?

Step 1: Understand that we are looking for five consecutive integers. Let's call the smallest integer 'x'.
Step 2: The next four consecutive integers can be expressed as x+1, x+2, x+3, and x+4.
Step 3: To find the average (mean) of these five integers, we add them together: x + (x+1) + (x+2) + (x+3) + (x+4).
Step 4: Simplify the sum: x + x + 1 + x + 2 + x + 3 + x + 4 = 5x + 10.
Step 5: The average is calculated by dividing the total sum by the number of integers (5): (5x + 10) / 5.
Step 6: We know from the question that this average equals 12, so we set up the equation: (5x + 10) / 5 = 12.
Step 7: Multiply both sides of the equation by 5 to eliminate the fraction: 5x + 10 = 60.
Step 8: Subtract 10 from both sides to isolate the term with x: 5x = 50.
Step 9: Divide both sides by 5 to solve for x: x = 10.
Step 10: The smallest of the five consecutive integers is x, which is 10.

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