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

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

by VJesus12 » Fri May 11, 2018 7:22 am

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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.

The OA is the option D .

Could someone explain this question to me? Please. I'd really appreciate your help.

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by Vincen » Fri May 11, 2018 8:37 am

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

We know that the arithmetic mean of an evenly spaced set is $$\frac{x_1+x_n}{2}=10\ \ \ \Rightarrow\ \ \ x_1+x_n=20.$$

(1) The range of the n integers is 14

This implies the following: $$x_n-x_1=14$$ Now, we sum the two equations and we get: $$2x_n=34\ \Rightarrow\ x_n=17.\ This\ implies\ that\ x_1=3.$$ Hence, this statement is SUFFICIENT.

(2) The greatest of the n integers is 17.

This implies that $$x_1+17=20\ \ \ \ \Rightarrow\ \ \ x_1=3.$$ Again, SUFFICIENT..

Therefore, each statement alone is sufficient. Answer is the option D.

Hope it helps.

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by [email protected] » Fri May 11, 2018 3:50 pm

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Hi VJesus12,

We're told that the average of N CONSECUTIVE ODD integers is 10. We're asked for the least of the integers. This DS question hides an interesting Number Property that you'll find useful. If you're dealing with consecutive odd integers with an average of 10, then the possibilities will follow a pattern...

The numbers COULD be:

9, 11
7, 9, 11, 13
5, 7, 9, 11, 13, 15
3, 5, 7, 9, 11, 13, 15, 17
etc

The pattern is that there will be an EQUAL number of terms above 10 and below 10. You can use THAT pattern along with the other info to solve this question relatively quickly.

1) The range is 14.

Look at the examples above. There's only one group with a range of 14. The other groups would either have a smaller range or a larger range.
Fact 1 is SUFFICIENT.

2) The greatest value is 17

Again, look at the examples above. There's only one group with 17 as the largest number.
Fact 2 is SUFFICIENT.

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### Re: If the average (arithmetic mean) of n consecutive odd

by [email protected] » Fri Aug 06, 2021 4:48 am

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VJesus12 wrote:
Fri May 11, 2018 7:22 am
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.

The OA is the option D .

Could someone explain this question to me? Please. I'd really appreciate your help.
Solution:

We need to determine the least of the n consecutive odd integers given that the average of these n integers is 10. If we let x be the least integer, then the greatest integer is x + 2n - 2. Also, notice that the average of the n integers is also the average of the least and the greatest integers. Thus we have:

(x + x + 2n - 2)/2 = 10

x + n - 1 = 10

x + n = 11

Statement One Alone:

We can create the equation:

x + 2n - 2 - x = 14

2n = 16

n = 8

Since x + n = 11 (from stem analysis) and n = 8, then x = 3. Statement one alone is sufficient.

Statement Two Alone:

We can create the equation:

x + 2n - 2 = 17

x + 2n = 19

Since x + n = 11 (from stem analysis) and if we subtract this from x + 2n = 19, we have n = 8. Since n = 8, we have x = 3. Statement two alone is sufficient.