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

Redeem

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.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
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
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

set/median

Expert replies
by advita » Tue Jan 11, 2011 2:09 am
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
Join the discussion
Source: — Problem Solving |

by jaymw » Tue Jan 11, 2011 2:35 am
What's the source of this question?

I backsolved this problem, meaning I plugged in the answer choices.

Before I did so I wrote down the set as: x,>=x,55,>=55,3x+20

Now the range is always MAXvalue - MINvalue, so here we have: 3x+20-x=78(answer choice A)
Solving for X yields 29. That means, for answer choice A, the smallest number of the set must be 29. The biggest number must be 3(29)+20=107.

Now that we have three values, we can calculate the sum of those.

29+55+107=191

In total our set should have a sum of 5*55=275.

That leaves a difference of 84 to be distributed between the two missing numbers in the set.

The second number in the ascending ordered set can now be 29 and the fourth number can be 55. Bingo! A must be correct and there's no need to plug in other answer choices.

Hope this was helpful.
Join the discussion

by Anurag@Gurome » Tue Jan 11, 2011 3:38 am
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?
Say the numbers are a, b, c, d, and e; where a ≤ b ≤ c ≤ d ≤ e
Now, median = c = 55
and, e = (3a + 20)

Therefore, range = (max - min) = (e - a) = (2a + 20)
Therefore to maximize the range we have to maximize a.

Also, (a + b + c + d + e) = 5*55 = 275
=> (a + b + 55 + d + 3a + 20) = 275
=> (4a + b + d) = 200
=> 4a = (200 - b - d)

Now, a will be maximum when both b and d will be minimum.
Minimum value of b is a and that of d is c, i.e. 55.

Hence, when a is minimum,
  • 4a = (200 - a - 55) => 5a = 145 => a = 29
Hence maximum value of the range is (2*29 + 20) = 78

The correct answer is A.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by ankur.agrawal » Tue Jan 11, 2011 3:47 am
I have read sumehere that for evenly spaced integers in a set mean & median are same.

Also we can find the mean of an evenly spaced integers in a set by just averaging the first & last terms.

Can we make use of the above two concepts to solve this problem. I tried but failed.

Help!!!!
Join the discussion

by jaymw » Tue Jan 11, 2011 4:15 am
@ankur.agrawal:

Unfortunately, you can't use these concepts for the above problem because it doesn't revolve around a set of evenly spaced integers.

Also, you aren't asked to find the mean or some other average but you're asked to find the range.

Therefore, you'd have to solve it my way or Anrug's, which, admittedly, is more scientific than mine.
Join the discussion

by GMATGuruNY » Tue Jan 11, 2011 4:56 am
advita wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
To maximize the range, we want to make the smallest number as small as possible and the largest number as large as possible. So the five numbers would be:

x, x, 55, 55, y

The set above will minimize the smallest number (x) because the two smallest values are the same. It will maximize the largest number (y) because the next largest number (55) is the same as the median.

Since the largest number is 20 more than 3 times the smallest number, y = 20+3x. So our list of numbers can be rewritten as:

x, x, 55, 55, 20+3x

Since the average is 55, the sum of the five numbers = 5*55 = 275.
Thus, x + x + 55 + 55 + 20+3x = 275.
5x = 145
x= 29.

Thus, y = 20+3x = 20 + 3*29 = 107.

Thus, the largest possible range is y-x = 107-29 = 78.

The correct answer is A.
Last edited by GMATGuruNY on Fri Oct 21, 2011 10:39 am, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by ankurmit » Sun Jan 16, 2011 10:11 pm
ankur.agrawal wrote:I have read sumehere that for evenly spaced integers in a set mean & median are same.

Also we can find the mean of an evenly spaced integers in a set by just averaging the first & last terms.

Can we make use of the above two concepts to solve this problem. I tried but failed.

Help!!!!
I tried the same concept and got 65 as answer :(

which is not the provided answer.
--------
Ankur mittal
Join the discussion

by pesfunk » Thu Jan 27, 2011 6:23 am
Excellent idea.

Basically the highest range can only be achieved when the smallest number is really small and biggest number is really big. In that case....Y need to take the load of right side and both numbers on the left needs to be as small as possible.


GMATGuruNY wrote:
advita wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
55 1/7
66 1/7
77 1/5
52

oa-78
To maximize the range, we want to make the smallest number as small as possible and the largest number as large as possible. So the five numbers would be:

x, x, 55, 55, y

The set above will minimize the smallest number (x) because the two smallest values are the same. It will maximize the largest number (y) because the next largest number (55) is the same as the median.

Since the largest number is 20 more than 3 times the smallest number, y = 20+3x. So our list of numbers can be rewritten as:

x, x, 55, 55, 2x+30

Since the average is 55, the sum of the five numbers is 5*55 = 275.
Thus, x + x + 55 + 55 + 20+3x = 275.
5x = 145
x= 29.

Thus, y = 20+3x = 20 + 3*29 = 107.

Thus, the largest possible range is y-x = 107-29 = 78.

The correct answer is A.
Join the discussion