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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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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.

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

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30 Jan 2014, 23:35

4

This post received KUDOS

What we know: If n consecutive integers have arithmetic mean of 10; those numbers are distributed either side of 10 perfectly symmetrically.

Statement (1)

Range is the difference between the greatest and the least numbers in the series. If the range is known the least of these integers must be less than the mean, i.e. 10 - (14/2) = 3 (Note we don't really need to calculate this... to know we can is sufficient.)

Statement (2)

Similarly, the distance from the greatest number to mean equals to the distance from mean to the least number. So knowing the greatest number also sufficient to calculate the least number. (in this case 10 - (17 - 10) = 3)

The answer is D : Both Statement 1 and Statement 2 are sufficient ALONE .

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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.

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

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15 Aug 2014, 07:52

I did it this way.

Since consecutive odd integers are evenly spaced then avg = median = 10

Since median is even (10) for a set of consecutive odd integers, set must have even number of integers Ex: if set has 3,5,7 then median 5 (odd) if set has 3,5,7,9 then median = 5+7/2 = 6 (even)

so we can say (x+ (x+2))/2 = 10 => 2x+2=20 => x=9. So middle terms of the set are 9 & 11. If I know one more thing like range or first/last term I can create full list.

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

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13 Jun 2015, 04:47

(Min + Max)/2 = 10 --> Min + Max=20 1)Range means Max - Min=14, sow we have two equations and two unknowns --> can be solved Min = 3 2)Max=17, Plug it here Min+Max=20 --> Min=3
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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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

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20 Dec 2016, 16:05

Great Official Question. Here is what i did in this one =>

Given data => N consecutive odds Mean =Median (For any evenly spaced set)=10 AS medan 10 => N must be even Sum of deviations around the mean is always zero. E.g => 9,11 OR 7,9,11,13

Hence we are adding elements in pair around the mean 10. We need to get the least element now

Statement 1--> Range =14 Series => 3,5,7,9,11,13,15,17=> Least Integer will be 3 Hence Sufficient

Statement 2--> Greasers integer =17

Number of terms tot he right of the median = Number of terms to the left of median Median =10 11,13,15,17 => 4 terms to the right of it=> 4 terms must be there t the left of it. Series => 3,5,7,9.. Hence least term =3 Hence Sufficient

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