Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

GMATPrep - Average

Expert replies
by California4jx » Tue Nov 25, 2008 11:14 am
couldnt find the rationale

OA: D
You do not have the required permissions to view the files attached to this post.
Join the discussion
Source: — Data Sufficiency |

by muzali » Tue Nov 25, 2008 11:39 am
Let the eleven numbers be x, x+1, x+2,...,x+10

We need to find [x+(x+1)+ (x+2)+...+(x+10)]/11, i.e, if we know x, we can find the mean of these elven numbers.

STM1: Avg of first nine numbers is 7
=> [x+(x+1)+ (x+2)+...+(x+10)]/9 = 7 we can calculate x
(you don't need to do this, but (9x+36)/9=11, so x=3)

STM2: Avg of last nine numbers is 9
In a similar fashion as above, we can calculate x.

So both STMs are sufficient, therefore answer is D.
Join the discussion

by California4jx » Tue Nov 25, 2008 12:05 pm
Thanks !
Join the discussion

by Stuart@KaplanGMAT » Tue Nov 25, 2008 1:01 pm
The above solution is certainly correct, but seems like a lot of work. As experienced GMAT writers, we've all learned to hate extra work! 8)

Instead, let's use our knowledge of consecutive integers to answer the question:

Every set of n consecutive integers will have a unique average. So, if we know the average and the number of terms, we can figure out the exact set.

(1) n = 9, average is 7. Only one set of consecutive integers will fit this rule. We can just add on the next two integers to fill out our set of 11: sufficient.

(Note: if this were a problem solving question and we wanted to figure out the set, we'd see that we have an odd number of terms, which means that the average, 7, is the middle term, so our set must be {3, 4, 5, 6, 7, 8, 9, 10, 11}. Since it's data sufficiency, we couldn't care less what the members of the set are!)

(2) n = 9, average is 9. Only one set of consecutive integers will fit this rule. We can just add on the previous two integers to fill out our set of 11: sufficient.

Each of (1) and (2) are sufficient alone: choose (D).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Join the discussion

by California4jx » Tue Nov 25, 2008 2:46 pm
wonderful ! :)
Join the discussion

by vittalgmat » Tue Nov 25, 2008 3:11 pm
Yep got D here.
Here is my approach.

Stmt 1. Avg of first 9 ints is 7.
ie first int is x and the last int is 8.
We know avg of n consecutive ints = (first# + last#)/2
So here it is x + x+8 = 7 *2.
2x +8 = 14
x = 3
So we know the first element of the sequence. Now we derive the entire sequence and get anything else we want..
So sufficient.

Stmt 2. Similar to above. So sufficient.

So D

HTHelps
Join the discussion