If x is the average (arithmetic mean) of 5 consecutive even integers,
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11 Nov 2021, 03:50
Let the 5 consecutive even integers be \(a-4, a-2, a, a+2, a+4\)
\(x = \frac{(a-4) + (a-2) + a + (a+2) + (a+4)}{5} = \frac{5a}{5} = a\)
Now, \(a\) is even integer
Which means \(a\) can be 0 or any multiple of 2 positive and negative both included
Since, \(x = a\) and \(a\) is an even integer, which means \(x\) is also an even integer. I is true
\(x\) can be 0 since 0 is also considered an even integer. II is false
Any number is multiple of 5 has its one's digit is either 0 or 5.
Since, \(x\) is even, its one's digit cannot be 5. So, \(x\) is multiple of 5 only in certain cases not all. Hence, III is false
Hence, Answer is A