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If x is the average (arithmetic mean) of 5 consecutive even integers,
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10 Nov 2021, 03:42
10 Nov 2021, 03:42
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If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?
I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.
(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III
I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.
(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III
If x is the average (arithmetic mean) of 5 consecutive even integers,
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11 Nov 2021, 03:50
11 Nov 2021, 03:50
1
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
\(x = \frac{(a-4) + (a-2) + a + (a+2) + (a+4)}{5} = \frac{5a}{5} = a\)
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
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Re: If x is the average (arithmetic mean) of 5 consecutive even integers,
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16 Sep 2022, 22:17
16 Sep 2022, 22:17
Given that x is the average (arithmetic mean) of 5 consecutive even integers and we need to find which of the following must be true
I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.
============================================================
Theory
============================================================
Let the 5 consecutive even numbers be y - 4, y - 2 , y , y + 2, y + 4
=> Mean = \(\frac{Sum}{5}\) = \(\frac{y - 4 + y - 2 + y + y + 2 + y + 4 }{ 5}\) = \(\frac{5y}{5}\) = y
=> Mean = y = x
So, the numbers are x - 4, x - 2 , x , x + 2 , x + 4
I. x is an even integer.
This is true as x is the middle term and we know that all numbers are even
II. x is a nonzero integer.
This doesn't have to be true as x can very well be 0 and the set will become
0 -4, 0 -2 , 0 , 0 +2 , 0 + 4 or -4, -2, 0, 2, 4
III. x is a multiple of 5.
This doesn't have to be true. Same example as II proves this one wrong too
-4, -2, 0, 2, 4
So, Answer will be A.
Hope it helps!
Watch the following video to Learn the Basics of Statistics
I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.
============================================================
Theory
- ‣‣‣ Mean = (Sum Of All The Numbers) / (Total Number Of Numbers).
============================================================
Let the 5 consecutive even numbers be y - 4, y - 2 , y , y + 2, y + 4
=> Mean = \(\frac{Sum}{5}\) = \(\frac{y - 4 + y - 2 + y + y + 2 + y + 4 }{ 5}\) = \(\frac{5y}{5}\) = y
=> Mean = y = x
So, the numbers are x - 4, x - 2 , x , x + 2 , x + 4
I. x is an even integer.
This is true as x is the middle term and we know that all numbers are even
II. x is a nonzero integer.
This doesn't have to be true as x can very well be 0 and the set will become
0 -4, 0 -2 , 0 , 0 +2 , 0 + 4 or -4, -2, 0, 2, 4
III. x is a multiple of 5.
This doesn't have to be true. Same example as II proves this one wrong too
-4, -2, 0, 2, 4
So, Answer will be A.
Hope it helps!
Watch the following video to Learn the Basics of Statistics
