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
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
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
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.

130-point 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 of Five (from Manhattan GMAT)

Expert replies
by Sprite_TM » Wed May 20, 2009 6:18 pm
hi,

this is from the manhattan CAT test but i don't quite understand their explanation. could someone please explain the answer to me? Thanks!!


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?


a) 78
b) 77 1/5
c) 66 1/7
d) 55 1/7
e) 52

OA
a
Join the discussion
Source: — Problem Solving |

by cramya » Wed May 20, 2009 6:34 pm
x y 55 z 3x+20

Since we are asked for the maximum I started out with 78.


Let say 3x+20-x = 78
x=19

3x+20=77

77+19=96 (we are missing 14 from the mean if these were the numbers in the list(

Can I fit in 2 numbers to the right and left side of the median that would satisfy the mean and median being 55 plus compensate for the 14 and provide an ascending sequence.

YES

19 55 55 69 77

A

I am sure others may have a theoritical solution but I couldn't come up with any when I started the prob.

Regards,
CR
Join the discussion

by dumb.doofus » Thu May 21, 2009 12:41 am
I think I can try to give a more theoretical solution.. here it is..

Let's choose the numbers as x, y, 55, z, 3x+20

So range here is 3x + 20 - x = 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now when would x be maximum, only when y and z are minimum.

So
1. What is the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Well it has to be equal to x. It can't be less than x..

So with the above deductions, equation (2) becomes

4x + x + 55 = 200

or 5x = 145

or x = 29

Putting this value in (1)
we get 2*(29) + 20 = 78

So answer should be A i.e. 78

Hope this helps..
Last edited by dumb.doofus on Thu May 21, 2009 8:52 pm, edited 1 time in total.
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
Join the discussion

by Sprite_TM » Thu May 21, 2009 7:26 am
thanks a lot doofus! your explanation is very clear
Join the discussion

by Svedankae » Thu May 21, 2009 12:12 pm
dumb.doofus wrote:I think I can give a try to give a more theoretical solution.. here it is..

Let's choose the numbers as x, y, 55, z, 3x+20

So range here is 3x + 20 - x = 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now when would x be maximum, only when y and z are minimum.

So
1. What is the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Well it has to be equal to x. It can't be less than x..

So with the above deductions, equation (2) becomes

4x + x + 55 = 200

or 5x = 145

or x = 29

Putting this value in (1)
we get 2*(29) + 20 = 78

So answer should be A i.e. 78

Hope this helps..

wow youre insanely gifted. thanks a lot
Join the discussion