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If the average (arithmetic mean) of n consecutive odd intege

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If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Dec 2012, 08:16, edited 1 time in total.
Edited the question and added the OA.

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Re: Quant Guide DS 66 [#permalink]

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New post 26 Nov 2009, 14:13
JimmyWorld wrote:
Hi,

I am unsure about the explanation on DS #66 of Quant Guide.

Can someone explain the answer to 2) for me?

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

1) The range of n integers is 14.
2) The greatest of the n integers is 17.

I understood 1) but didnt know how to get 2).


n is a set of consecutive odd integers and the last one being 17

17 + 15 = 22/2 = 11
17 + 15 + 13 = 45/3 = 15
working backwards you'll find 3 is the smallest number, 8 numbers total 80 and the average 10

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Re: Quant Guide DS 66 [#permalink]

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New post 26 Nov 2009, 15:00
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JimmyWorld wrote:
Thanks for the response. I was just wondering what would be a faster way to do this problem. If I were to work backwards to find when the average is 10, it would take me longer then 2 minutes to do that. Is there a quick way to figure out that the least number would be 3?




in a group of consecutive integers the average is also the median
10 would be the median
17 the largest

that would have to make 3 the smallest

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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]

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New post 16 Dec 2012, 00:05
Bunuel,
how to solve this using algebraic approach?

Regards,
Sach
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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]

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Sachin9 wrote:
Bunuel,
how to solve this using algebraic approach?

Regards,
Sach



Let me give it a try.

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.
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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]

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Sachin9 wrote:
Bunuel,
how to solve this using algebraic approach?

Regards,
Sach


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

Odd consecutive integers is an evenly spaced set. For any evenly spaced set the mean equals to the average of the first and the last terms, so in our case \(mean=10=\frac{x_1+x_{n}}{2}\) --> \(x_1+x_{n}=20\). Question: \(x_1=?\)

(1) The range of the n integers is 14 --> the range of a set is the difference between the largest and smallest elements of a set, so \(x_{n}-x_1=14\) --> solving for \(x_1\) --> \(x_1=3\). Sufficient.

(2) The greatest of the n integers is 17 --> \(x_n=17\) --> \(x_1+17=20\) --> \(x_1=3\). Sufficient.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-average-arithmetic-mean-of-n-consecutive-odd-106036.html
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Re: If the average(arithmetic mean) of n consecutive odd integer   [#permalink] 16 Dec 2012, 08:18
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