Question

• If the average of 12 consecutive odd Integers is 328, what is the least of these integers
(a)317 (b) 319 (c) 321 (d)323 (e)325
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Answer 1

Answer:here is your answer

Explanation:

(x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) + (x+12) + (x+14) + (x+16) + (x+18) + (x+20) + (x+22)) / 12 = 328

We divide by 12 because that is the number of given terms; that’s how you find the average.

Add like terms in the equation, and then divide by 12:

(12x + 132) / 12 = 328

Multiply both sides by 12 to get rid of the fraction:

12x + 132 = 3,936

Subtract 132 from both sides:

12x = 3,804

Divide by 12:

x = 317

So the smallest number is 317. The remaining numbers are:

319, 321, 323, 325, 327, 329, 331, 333, 335, 337, 339

To prove it, add all of these together:

3,936

divide by 12:

328

So, again, the smallest number is 317.