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The average of four positive integers is 73.5. The highest integer is 108 and the lowest integer is 29. The difference between the remaining two integers is 15. Which of the following is the smaller of the remaining two integers?

Question

The average of four positive integers is 73.5. The highest integer is 108 and the lowest integer is 29. The difference between the remaining two integers is 15. Which of the following is the smaller of the remaining two integers?

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Solution

To solve this problem, we first need to understand that the average of four numbers is the sum of those numbers divided by four.

Given that the average is 73.5, we can calculate the total sum of the four numbers by multiplying 73.5 by 4, which equals 294.

We know that the highest integer is 108 and the lowest is 29. So, we subtract these two numbers from the total sum of 294. This gives us 294 - 108 - 29 = 157.

The problem states that the difference between the remaining two integers is 15. If we let the smaller of the two integers be x, then the larger one would be x + 15.

Therefore, we can set up the equation x + (x + 15) = 157 to find the value of x. Solving this equation gives us 2x + 15 = 157, then 2x = 157 - 15 = 142, and finally x = 142 / 2 = 71.

So, the smaller of the remaining two integers is 71.

This problem has been solved

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