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
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.

greatest And Lowest Problem !! tough one :(

Expert replies
Source: — Problem Solving |

by Morgoth » Wed Oct 01, 2008 10:15 am
greatest number of household who have all three = x

80 - 75 = 15
75 - 55 = 20
80 - 55 = 25

x = 15+20+25 = 60

Lowest number = y = 15 [compare which is the lowest among the above cases]


x-y = 60 - 15 = 45.

OA?
Join the discussion

by vishubn » Wed Oct 01, 2008 5:38 pm
Ya thats the OA :)
If u could please elaborate a bit more on this part
80 - 75 = 15
75 - 55 = 20
80 - 55 = 25
Vishu
Join the discussion

by gmat009 » Wed Oct 01, 2008 6:30 pm
Morgoth wrote:greatest number of household who have all three = x

80 - 75 = 15
75 - 55 = 20
80 - 55 = 25

x = 15+20+25 = 60

Lowest number = y = 15 [compare which is the lowest among the above cases]


x-y = 60 - 15 = 45.

OA?
CAn you plz. explain this.
Join the discussion

by Morgoth » Wed Oct 01, 2008 7:36 pm
I apologize for the strange method used. Even though I got the correct answer method used is incorrect. Dont know what I was thinking.

Here is why

Since 55 MP3 at least, the maximum number with all three cannot be 60.

Here is the actual method rather correct method :D

x = maximum = 55

Maximum people with at least 3 - maximum number with no three = minimum = y

maximum without all three
75-55 = 20 no MP3
80-55 = 25 no MP3

55 - 45 = 10

x-y = 55-10 = 45.


Hope its clear.
Join the discussion

by adxb » Sat Oct 04, 2008 3:13 am
I understand how you got x, but I am still confused about how you determined y?
Join the discussion

by Ian Stewart » Tue Oct 07, 2008 8:40 am
Clearly the maximum number that could have all three devices is 55; that is, the 55 people who own an MP3 player could also each own a DVD player and a cell phone. So x = 55.

To find y, we need to minimize the number of people who own all three devices. If 80 people own a cell phone, and 75 own a DVD player, let's find the smallest possible number that might own both. We'd want to be sure that all 20 people who do *not* own a cell phone *do* own a DVD player. We'd then have:

own cell, do not own DVD: 25
own DVD, do not own cell: 20
own DVD and cell: 55

Now we have 55 who own both a DVD and a cell. We also know that 55 own an MP3 player. Again, to find the smallest number who own an MP3, a cell and DVD, we want to be sure that all 45 people who do *not* own an MP3 *do* own a cell+DVD. So we have:

owns cell+DVD, but no MP3: 45
owns MP3, but no cell+DVD: 45
owns all three: 10

So at least 10 people must own all three devices, and y = 10.

Finally, x - y = 45.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Join the discussion