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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

PS-2

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 129
Joined: Tue Jul 22, 2008 3:53 am
Thanked: 2 times
by sparsh.21 » Sat Dec 13, 2008 3:44 am
In a village of 100 households,75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one mp3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x-y is

a. 65
b. 55
c. 45
d. 35
e. 25
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members
Thanked: 146 times

Re: PS-2

by parallel_chase » Sat Dec 13, 2008 4:05 am
sparsh.21 wrote:In a village of 100 households,75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one mp3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x-y is

a. 65
b. 55
c. 45
d. 35
e. 25

getting the value of x is easy = 55

for the value of y i.e. lowest

lets say out of 55 nobody had cell phone

80-55 = 25

55-25 = 30

out of 30 nobody had DVD,

75-55 = 20

30-20 = 10

x - y = 55 - 10 = 45

I am not too sure about this. Kindly confirm the OA?
No rest for the Wicked....
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Tue Nov 18, 2008 2:23 pm

by mowie » Sat Dec 13, 2008 4:40 am
Yeah (C) should be right
a stands for all 3

i m+cm+dm+a=55
ii d+cd+dm+a=75
iii c+cd+cm+a=80

=> i+ii+iii
I c+d+m+(cd+cm+dm)2+3a=210
II c+d+m+cd+cm+dm+a=100

=> I-II
cd+cm+dm+2a=110

max 55 (I and II are ok!)
min 10 (c+d+m = 0)

IMO (C)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 129
Joined: Tue Jul 22, 2008 3:53 am
Thanked: 2 times

by sparsh.21 » Sat Dec 13, 2008 6:44 am
yes.... the OA is C
but i am still not able to understand how to get the value of y... :(
Can anybody please elaborate a little more ?

thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Tue Nov 18, 2008 2:23 pm

by mowie » Sat Dec 13, 2008 7:47 am
ok, lets have a look at these equotations:

I c+d+m+cd+cm+dm+a=100

II cd+cm+dm+2a=110

both need to be true.
We have no information about the value of cd,cm,dm (e.g. cd is cellphone AND dvd player; VENN DIAGRAMM!). So we get the max for a with cd=cm=dm=0 (sum of them is minimal).
=> 2a=110 <=> a=55

Let`s have a look for y now.
To find our min for a we know that the sum of cd,cm,dm should be maximal (a = (110-(cd+cm+dm))/2)
What is the max of this sum?
It is easy to find! Just look at I, the max should be reached when c,m,d all equal 0 ( cd+cm+dm=100-a-(c+m+d) ).


So: c=d=m=0
=>
(1) cd + cm + dm + a = 110
(2) cd + cm + dm + 2a = 100

(1)-(2) a=10
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1