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OG DS #66

by tonebeeze » Wed Jan 12, 2011 8:15 am
What is the best approach to solve these types of problems? The answer in the back of the OG seem a bit time intensive. Thanks!

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

1. The range of the n integers is 14

2. The greatest of the n integers is 17
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by jaymw » Wed Jan 12, 2011 8:34 am
First of all, it might be helpful to examine the question text a little bit.

n consecutive integers with a mean of 10 means that n must be even. If n were odd, the mean could never be even because one would have to divide an odd number by another odd number and that can never yield an even number.

Next, we know that the set of integers is evenly spaced (consecutive odd integers).

The middle pair of the set will therefore be 9 and 11, since both of these numbers have the same distance from the mean. For each number added to one side of these two numbers, a counterpart has to be added on the opposite side to 'keep the balance'.

Statement I:

This tells us that one integer must be 7 bigger than the mean while the other integer must be 7 smaller than the mean. Of course the consecutive odd integers from 9 to 23 would also have a range of 14, but then the mean would not be 10 anymore. To keep the balance around the two middle number 9 and 11, just add the same amount of odd integers to both sides until you get a range of 14.

9,11 range is 2
7,9,11,13 range is 6
5,7,9,11,13,15 range is 10
3,5,7,9,11,13,15,17 range is 14. Bingo!

So, 3 is the smallest integer.

SUFFICIENT



Statement II:

Basically the same as statement I. When the greatest integer is 17, when the integers are consecutive odd integers and when the mean is 10, the only possible set of numbers is the one above with a range of 14.

In that set, 17 is the biggest number, and 3 is the least integer.

SUFFICIENT

Hence, D is correct.
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by Anurag@Gurome » Wed Jan 12, 2011 8:35 am
tonebeeze wrote:If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

1. The range of the n integers is 14
2. The greatest of the n integers is 17
Note that n consecutive odd integers will form an arithmetic progression with 2 as common difference. Say the first term of the progression, i.e. the least of the integers is a. Hence n-th term of the progression, i.e. the largest of the integers will be [a + 2(n - 1)].

Therefore, Range = Max - Min = [a + 2(n - 1)] - a = 2(n - 1)
and, Arithmetic Mean = (Max + Min)/2 = [a + 2(n - 1) + a]/2 = [a + (n - 1)]

Now, [a + (n - 1)] = 10 => (a + n) = 11

Thus, we can determine the value of a, once we know the value of n.

Statement 1: Range = 14
Hence, 2(n - 1) = 14
=> n = 8 => We can determine the value of a.

Sufficient

Statement 2: [a + 2(n - 1)] = 17
We have two equations in two unknowns. Hence, we can determine the value of a.

Sufficient

The correct answer is D.
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by anshumishra » Wed Jan 12, 2011 8:39 am
tonebeeze wrote:What is the best approach to solve these types of problems? The answer in the back of the OG seem a bit time intensive. Thanks!

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

1. The range of the n integers is 14

2. The greatest of the n integers is 17
Odd consecutive integers is an evenly spaced set (with common difference 2). For any evenly spaced set, the average is equal to the average of the first and last term.

So, mean (10) = (x1+xn)/2 => x1+xn = 20 ----- (1)
x1 = ?

Statement 1:
xn - x1 = 14
Sufficient, as we have two linear equations in x1 and xn.

Statement 2:
Xn = 17, we can find x1 using equation (1)
Sufficient.

Hence, D
Thanks
Anshu

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by GMATGuruNY » Wed Jan 12, 2011 2:08 pm
tonebeeze wrote:What is the best approach to solve these types of problems? The answer in the back of the OG seem a bit time intensive. Thanks!

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

1. The range of the n integers is 14

2. The greatest of the n integers is 17
When a set of numbers is evenly spaced, the average = the median, and the set is symmetrical about the median.
In other words, for every number that is x places below the median, there must be a corresponding number that is x places above the median.
Thus, the n consecutive odd integers will be symmetrical about the median of 10: for every odd integer that is x places below 10, we'll need a corresponding odd integer that is x places above 10.

Statement 1: Range = 14.
If the range is 14 and 10 is in the middle, the smallest integer must be 10-7 = 3 and the largest integer must be 10+7 = 17, giving us a range of 17-3 = 14.
Sufficient.

Statement 2: Largest integer = 17.
If the greatest integer is 17, then there are 4 consecutive odd integers greater than 10 (11, 13, 15 and 17) and 4 consecutive odd integers less than 10 (3, 5, 7, 9). Thus, the smallest integer is 3.
Sufficient.

The correct answer is D.
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by frankodern » Tue Jan 21, 2014 5:47 am
Dr. Anurag,

Can you please explain the use of 2(n-1); I have seen this used for range questions but do not understand it. I would really appreciate your assistance.

Anurag@Gurome wrote:
tonebeeze wrote:If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

1. The range of the n integers is 14
2. The greatest of the n integers is 17
Note that n consecutive odd integers will form an arithmetic progression with 2 as common difference. Say the first term of the progression, i.e. the least of the integers is a. Hence n-th term of the progression, i.e. the largest of the integers will be [a + 2(n - 1)].

Therefore, Range = Max - Min = [a + 2(n - 1)] - a = 2(n - 1)
and, Arithmetic Mean = (Max + Min)/2 = [a + 2(n - 1) + a]/2 = [a + (n - 1)]

Now, [a + (n - 1)] = 10 => (a + n) = 11

Thus, we can determine the value of a, once we know the value of n.

Statement 1: Range = 14
Hence, 2(n - 1) = 14
=> n = 8 => We can determine the value of a.

Sufficient

Statement 2: [a + 2(n - 1)] = 17
We have two equations in two unknowns. Hence, we can determine the value of a.

Sufficient

The correct answer is D.
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by dimochka » Tue Jan 21, 2014 7:34 am
The more generic formula would be x*(n-1), where x is the difference between each chronological pair of numbers in the set.

The 1st number is A
The 2nd number is A+x
The 3rd number is A+x*2
The 4th number is A+x*3
The nth number is A+x*(n-1)

In this question we are dealing consecutive odd integers, in which case x = 2.
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