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.

word translations

Expert replies
by fruti_yum » Mon Sep 14, 2009 7:54 am
T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?

0
x
-x
(1/3)y
(2/7)y

I dont understand how 0 and -x are even possible choices of the median considering y <0 and average is only positive number.

Can anyone explain???

Source Mgmat

OA after some discussion
Join the discussion
Source: — Problem Solving |

IMO

by xcusemeplz2009 » Mon Sep 14, 2009 9:00 am
IMO E = (2/7)y

value of y can be 1,2,3,4,5,6

so in set T there can be 1,2,3,4,5 or 6 no.

avg of set T is x ( a +ve int)

suppose when
y=3 set T= -1,0,1(for eg); avg=0 and hence x=0 , now median is 0 and x
keeping y=3; t can be -2,-1and 6 ; avg=1;med=-1=-x
similarly d is posible bcoz (1/3 )y= whole no. when y is 3 or 6 so an integer
but for E (2/7)y can not be a whole no. bcoz y<7, so cannot be an int.
It does not matter how many times you get knocked down , but how many times you get up
Join the discussion

by vkb16 » Fri Oct 02, 2009 4:46 am
MGMAT SAYS

''As for answer choice E, there is no possible way to create Set T with a median of (2/7)y. Why? We know that y is either 1, 2, 3, 4, 5, or 6. Thus, (2/7)y will yield a value that is some fraction with denominator of 7.

The possible values of (2/7)y are as follows:
2/7, 4/7, 6/7, 1 1/7, 1 3/7, 1 5/7''

My question is, how did these values pop up?

thanks
Join the discussion