BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Verbal Review 2nd ed DS question 66

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 am
Thanked: 135 times
Followed by:7 members

by selango » Tue Oct 26, 2010 8:34 pm
...
Last edited by selango on Tue Oct 26, 2010 10:03 pm, edited 1 time in total.
--Anand--
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Tue Oct 26, 2010 8:51 pm
pharmxanthan wrote: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.
Say, the least of the n odd integers is a and the greatest is b.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
Given the average is 10 => (a + b) = 20

Statement 1: Range of n integers is 14 => Greatest - Least = 14 => b - a = 14 => b = a + 14
As, (a + b) = 20 => (a + a +14) = 20 => (2a + 14) = 20 => 2a = 6 => a = 3
Thus, least of the integers is 3.

Sufficient.

Statement 2: Greatest of the n integers is 17.
As, (a + b) = 20 => (a + 17) = 20 => a = 3
Thus, least of the integers is 3.

Sufficient.

The correct answer is D.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Oct 26, 2010 9:06 pm
pharmxanthan wrote: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 think answer should be B

1) Not sufficient.
But still i will not be surprised if somebody comes up with the solution
Because we know the difference in two terms that is 2,
We know the average of terms as 10.

But i am not able to figure out what how to use the information. That why for me its not sufficient


2) last term is 17 and it is the greatest

Sum of first 10 odd terms from [ 1 3 5 . ...19 ] = 100
average is 100/10 = 10

But the last term is 17 ( if we consider terms from ( 1 3 5 ..... 17 )
Sum will be equal to 81 { so the average will be 81 / 9 = 9 }

If we take 1 out then sum we will be 80 and average will be 10
80/ 8 = 10 First term is 3
Hence Sufficient.

What the OA?
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Oct 26, 2010 9:16 pm
Rahul@gurome wrote: Say, the least of the n odd integers is a and the greatest is b.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
Great Work,
I didn't knew that the sum of consecutive integers is = n/2 * { a + b }
I think we can use this formula then the different between two integers is constant { whether 2 or 3 or 4 or 5 ........ }


what is know is this is this formula n/2 * { n + 1 }
We use this formula when the difference between two integers is 1.

Please correct me Rahul Sir, if i am wrong.....
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Tue Oct 26, 2010 9:28 pm
goyalsau wrote:
Rahul@gurome wrote: Say, the least of the n odd integers is a and the greatest is b.
Therefore, sum of n consecutive odd integers = (n/2)*(a + b)
Average of these n consecutive integers = [(n/2)*(a + b)]/n = (a + b)/2
Great Work,
I didn't knew that the sum of consecutive integers is = n/2 * { a + b }
I think we can use this formula then the different between two integers is constant { whether 2 or 3 or 4 or 5 ........ }


what is know is this is this formula n/2 * { n + 1 }
We use this formula when the difference between two integers is 1.

Please correct me Rahul Sir, if i am wrong.....
We can use this formula whenever the common difference is constant, i.e. the numbers are in arithmetic progression. This formula is just an rearrangement of an well known formula.

We know, the sum, S of first n numbers of any arithmetic progression with common difference d and first term A1 is given by,

S = (n/2)*[2*A1 + (n - 1)*d]

Now, n-th term of an arithmetic progression, An = A1 + (n - 1)*d
Therefore, S = (n/2)*[A1 + A1 + (n - 1)*d] = (n/2)*[A1 + An]
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Senior | Next Rank: 100 Posts
Posts: 77
Joined: Fri Jan 08, 2010 7:55 am
Thanked: 2 times
GMAT Score:700

by pharmxanthan » Tue Oct 26, 2010 11:21 pm
OA is D.

Thanks Rahul!
Top
Post new topic Post Reply
• Page 1 of 1