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Difficulty:
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Question Stats:
42% (02:18) correct
58%
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Is the average of n consecutive integers equal to 1 ?
(1) n is even
(2) if S is the sum of the n consecutive integers, then 0 < S < n
(1) n is even
(2) if S is the sum of the n consecutive integers, then 0 < S < n
Show SpoilerMy approach
Correct answer D
Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?
My approach to solve this:
Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.
Statement 2: I understand what the statement means, but I am struggling to understand the concept.
My understanding is
Average = Sum/Number of terms
Therefore, Sum = Average * Number of Terms
Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.
So the answer is D. But I don't understand the Statement 2.
Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?
My approach to solve this:
Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.
Statement 2: I understand what the statement means, but I am struggling to understand the concept.
My understanding is
Average = Sum/Number of terms
Therefore, Sum = Average * Number of Terms
Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.
So the answer is D. But I don't understand the Statement 2.
Great Question.
Here is what i did in this one ->
We are asked if the mean of n consecutive is an integer or not.
N consecutive integer => Evenly spaced set =>Mean = Median =Average of the first and the last term
Mean for N consecutive integers can take only 2 forms =>
x => For N = odd
x.5=> for N=even
for some integer x
Statement 1-->
N is even
So mean will never be an integer(Because median will never be an integer)
Hence mean can never be "one".
Hence Sufficient
Statement 2-->
Sum=> S
And 0<S<N
\(Mean = \frac{Sum}{#}\)
Mean = \(\frac{S}{N}\)
As \(0<S<N\)=>
\(\frac{S}{N}\) will be a decimal between (0,1)
Hence mean will never be one.
Hence Sufficient
Hence D
Here is what i did in this one ->
We are asked if the mean of n consecutive is an integer or not.
N consecutive integer => Evenly spaced set =>Mean = Median =Average of the first and the last term
Mean for N consecutive integers can take only 2 forms =>
x => For N = odd
x.5=> for N=even
for some integer x
Statement 1-->
N is even
So mean will never be an integer(Because median will never be an integer)
Hence mean can never be "one".
Hence Sufficient
Statement 2-->
Sum=> S
And 0<S<N
\(Mean = \frac{Sum}{#}\)
Mean = \(\frac{S}{N}\)
As \(0<S<N\)=>
\(\frac{S}{N}\) will be a decimal between (0,1)
Hence mean will never be one.
Hence Sufficient
Hence D
General Discussion
enigma123
2) To get an average=1; the Sum of elements must be equal to number of elements.
\(If S=n; \frac{S}{n}=1\);
If S is NOT equal to n, the average can not be 1.
Sum=100; n=100; S/n=100/100=1
Sum=1; n=1; S/n=1/1=1
Sum=10; n=9; S/n=10/9=1.##(NOT EQUAL TO 1)
Sufficient.
Is the average of n consecutive integers equal to 1 ?
Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.
(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.
(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.
Answer: D.
Hope it's clear.
_________________
Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.
(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.
(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.
Answer: D.
Hope it's clear.
_________________
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Bunuel
KarishmaB
I hope that you are having a wonderful week
I was confused as to how you can just divide the whole inequality by n for statement 2. I have only seen manipulation of inequalities when you divide the middle term in the inequality (e.g., in which the inequality would be divided by s in this case --> 0/s < 1 < n/s). To clarify, you can divide by any term in the inequality as long as you consistently divide each variable or expression in the inequality by that term (except I guess in this example you cannot divide the inequality by the first term because it is a 0).
Many thanks
woohoo921
You can divide an inequality by any positive number without impacting the inequality.
You might have seen division by middle term (provided we know the term is positive) because in that question it would have helped simplify. Otherwise, you can divide by any positive number.
a < b
a/2 < b/2 (valid)
a < b < c
a/10 < b/10 < c/10 (valid)
a < b < c
1 < b/a < c/a (valid if a is known to be a positive number)
etc.
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Bunuel
Bunuel
I thought that we can only work with manipulating the inner term (e.g., s) in this case per my question above for Karishmab
In other words, if I have: 5 < x+2 < 9
Can you do the following for division?
Dividing all terms by 5 --> 1 < (x+2)/5 < 9/5
Dividing all terms by x+2 --> x/(x+2) < 1 < 9/(x+2)
Dividing all terms by 9 --> 5/9 < (x+2)/9 < 1
Can you do the following for subtraction?
Again, if I have: 5 < x+2 < 9
Subtracting 9 from all terms --> -4 < x-7 < 0
Subtracting x+2 from all terms --> -x+3 < 0 < -x-2
and vice versa for multiplication?
Many thanks
KarishmaB 
MacroMX
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I think Statement 1 is not sufficient: assume -1;0;1;2 here average is 1, 1;2;3;4 here average is 5. Isn't it insufficient
woohoo921
We can add or subtract the same term on both sides of the inequality without changing the sign of the inequality. We can multiply/divide the same term on both sides without impacting the inequality sign provided the term is positive.
Here, since (x + 2) lies between 5 and 9, we know it must be positive. So we can multiply or divide the inequality by (x + 2) and the inequality sign remains the same.
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MacroMX
When we have even number of consecutive integers, average will be the average of the two middle integers so it will not be an integer.
Avg of -1, 0, 1, 2 is the average of 0 and 1 which is 0.5.
Avg of 1, 2, 3, 4 is the average of 2 and 3 which is 2.5.
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KarishmaB
I thought that you can't multiply nor divide inequalities together, or does that rule only apply to when you have two separate inequalities?
I thought that you can't multiply nor divide inequalities together, or does that rule only apply to when you have two separate inequalities?
woohoo921
We can multiply and divide inequalities as long as both sides are non negative in both inequalities (and there is no 0 in the denominator in case of division).
We can multiply when both inequalities have the same inequality sign and we can divide when they have opposite inequality signs.
A < B
C < D
-----
AC < BD - valid if A, B, C and D are all non negative
A < B
C > D
-----
A/C < B/D - valid if A and B are non negative and C and D are positive.
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