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
Start free 5-day trial
Read 563 reviewsTarget Test Prep

MEDIAN

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Mon Apr 18, 2011 4:34 pm
by SwatiDenre » Thu Jun 16, 2011 8:40 am
The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
Top
Join the discussion
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Thu Jun 16, 2011 8:44 am
Hi,
Where is the table, dude...
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 66
Joined: Wed Jul 07, 2010 11:06 am
Thanked: 1 times

by baladon99 » Thu Jun 16, 2011 10:06 am
SwatiDenre wrote:The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
Can you please post the table ?
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 461
Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members
Location: pune
Thanked: 36 times

by amit2k9 » Thu Jun 16, 2011 9:32 pm
since the number of years is 5, the median will be the number of accidents in the 3rd year.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Sat Nov 06, 2010 9:06 pmFollowed by: 1 member
Thanked: 10 times

by edvhou812 » Thu Jun 16, 2011 10:03 pm
amit2k9 wrote:since the number of years is 5, the median will be the number of accidents in the 3rd year.
No. You will need to reorder the amounts from least to greatest, and then take the third largest number.
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Mon Apr 18, 2011 4:34 pm

by SwatiDenre » Sun Jun 19, 2011 3:51 pm
Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 - 17
1 - 13
2 - 21
3 - 4
4 - 2
5 - 2
6 - 1
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 401
Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members
Thanked: 37 times

by MBA.Aspirant » Sun Jun 19, 2011 4:10 pm
The middle of this set will = the number of drivers/2 = 43/2 = 21.5

This number is between two accidents values 1 and 2

so we take the average of them = 1+2/2 = 1.5
Top
Join the discussion
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sun Jun 19, 2011 8:48 pm
SwatiDenre wrote:Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 (1-17) - 17
1 (18-30) - 13
2 (31-51) - 21
3 (52-55) - 4
4 (56-57) - 2
5 (58-59) - 2
6 (60) - 1
Hi,
Total number of drivers is 17+13+21+... = 60
Median is the average of 30th and 31st elements
30th element is 1 and 31st element is 2
So, median is (1+2)/2 = 1.5

Hence, C
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion