Last visit was: 20 Nov 2025, 03:22 It is currently 20 Nov 2025, 03:22
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
amanvermagmat
User avatar
Retired Moderator
Joined: 22 Aug 2013
Last visit: 28 Mar 2025
Posts: 1,148
Own Kudos:
2,855
 [14]
Given Kudos: 480
Location: India
Posts: 1,148
Kudos: 2,855
 [14]
2
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
u1983
User avatar
Current Student
Joined: 24 Aug 2016
Last visit: 06 Jun 2021
Posts: 710
Own Kudos:
852
 [6]
Given Kudos: 97
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
Products:
GMAT 2: 680 Q49 V33
Posts: 710
Kudos: 852
 [6]
3
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
avatar
ColumbiaEMBA
Joined: 29 Sep 2017
Last visit: 24 Mar 2019
Posts: 88
Own Kudos:
61
 [1]
Given Kudos: 10
Location: United States
Products:
Posts: 88
Kudos: 61
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,802
Own Kudos:
6,065
 [1]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,802
Kudos: 6,065
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey strivingFor800

The easiest way to find out the various combinations with sum = 90 is as follows:
Since \(90 = 2*3^2*5\), the various factors possible are 1,2,3,5,6,9,10,15,18,30,45,90

For the even number of consecutive integers, the mean/median is the average of the two
middle integers - which cannot be an integer.

For an odd number of consecutive integers, we will have a middle term that is an integer.
If N = 9, the middle term has a value of \(\frac{90}{9} = 10\)
The consecutive terms are from 6 to 14(9 in number with 10 as the middle term)

If N = 15, the middle term has a value of \(\frac{90}{15} = 6\)
The consecutive terms are from -1 to 13(15 in number with 6 as the middle term)

1. N could be 3 - 29,30,31(which adds up to give us 90)
N could be 9 - 6,7,8,9,10,11,12,13,14(again adds up to give us 90) (Insufficient)

2. 5 < N < 10. N could take values 6,7,8,or 9. (Insufficient)

Combining the information from both the statements and testing the other
values(N=6,7,8) we have a unique value of N. (Sufficient - Option C)
User avatar
akadiyan
User avatar
Retired Moderator
Joined: 31 May 2017
Last visit: 20 Jun 2025
Posts: 736
Own Kudos:
Given Kudos: 53
Concentration: Technology, Strategy
Products:
Posts: 736
Kudos: 698
Kudos
Add Kudos
Bookmarks
Bookmark this Post
What is the value of positive integer N?

(1) Sum of N consecutive positive integers is 90 - Insufficient - We could have more than one set of numbers that could yield 90.

(2) 5 < N < 10 = N = 6,7 or 8. More than 1 value - Insufficient.

Both (1) and (2) together - we have unique value of N.

Ans: C
User avatar
push12345
Joined: 02 Oct 2017
Last visit: 10 Feb 2019
Posts: 536
Own Kudos:
535
 [1]
Given Kudos: 14
Posts: 536
Kudos: 535
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Here is my approach

I) instead of counting or calculating let's do it other way

90=2*3*3*5

Now if we make pairs
Eg 15*6 means 6 numbers will give 90 i.e. N=6 whose average is 15 which will ultimately give 90.

So all pairs
15*6. N=6
18*5. N=5
30*3. N=3
45*2. N=2
90*1. N=1

We are not concerned the value of numbers as we are only concerned with value of N
So multiple values so Insufficient

II) n ranges from 6 to 9
Insufficient

Combining both
Only N=6 qualify

C is answer

Give kudos if it helps

Posted from my mobile device
avatar
@s
Joined: 12 Sep 2017
Last visit: 14 Aug 2018
Posts: 22
Own Kudos:
Given Kudos: 63
Posts: 22
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pushpitkc
Hey strivingFor800

The easiest way to find out the various combinations with sum = 90 is as follows:
Since \(90 = 2*3^2*5\), the various factors possible are 1,2,3,5,6,9,10,15,18,30,45,90

For the even number of consecutive integers, the mean/median is the average of the two
middle integers - which cannot be an integer.

For an odd number of consecutive integers, we will have a middle term that is an integer.
If N = 9, the middle term has a value of \(\frac{90}{9} = 10\)
The consecutive terms are from 6 to 14(9 in number with 10 as the middle term)

If N = 15, the middle term has a value of \(\frac{90}{15} = 6\)
The consecutive terms are from -1 to 13(15 in number with 6 as the middle term)

1. N could be 3 - 29,30,31(which adds up to give us 90)
N could be 9 - 6,7,8,9,10,11,12,13,14(again adds up to give us 90) (Insufficient)

2. 5 < N < 10. N could take values 6,7,8,or 9. (Insufficient)

Combining the information from both the statements and testing the other
values(N=6,7,8) we have a unique value of N. (Sufficient - Option C)


Hi Pushpitkc

i understood the solution provided by you. But i was thinking even sum of 90 can have 4 consecutive numbers (21, 22, 23, 24) which is even. So, in that case, how can we be sure that it is only an odd number.

Thank you.
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,802
Own Kudos:
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,802
Kudos: 6,065
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi @s

The answer to this problem is Option C meaning that both statements are needed to answer the question. Statement 2 says that N is a value between 5 and 10. That's the reason the case you have presented does not work.

Hope this helps you!
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,597
Own Kudos:
Posts: 38,597
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
105408 posts
496 posts