Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 07 Sep 2010
Posts: 293

In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
Updated on: 30 Sep 2013, 08:19
Question Stats:
66% (01:07) correct 34% (01:11) wrong based on 879 sessions
HideShow timer Statistics
In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from: A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by imhimanshu on 30 Sep 2013, 08:11.
Last edited by Bunuel on 30 Sep 2013, 08:19, edited 1 time in total.
Renamed the topic and edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 46271

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
30 Sep 2013, 08:38
imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 120 have TV and 90 have radio. Obviously the number of households that have both cannot be greater than 90. Eliminate B, D, and E. Next: {Total} = {TV} + {Radio}  {Both} + {Neither} 150 = 120 + 90  {Both} + {Neither} {Both} = 60 + {Neither} Thus {Both}>=60. Answer: C.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 07 Jan 2014
Posts: 4
Location: India

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
10 Mar 2015, 04:54
Bunuel wrote: imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 120 have TV and 90 have radio. Obviously the number of households that have both cannot be greater than 90. Eliminate B, D, and E. Next: {Total} = {TV} + {Radio}  {Both} + {Neither} 150 = 120 + 90  {Both} + {Neither} {Both} = 60 + {Neither} Thus {Both}>=60. Answer: C. Hi Bunuel, There are 2 options greater than 60 C & D. Can u pls explain how you narrowed down on C..?



Math Expert
Joined: 02 Sep 2009
Posts: 46271

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
10 Mar 2015, 04:57
TuringMachine wrote: Bunuel wrote: imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 120 have TV and 90 have radio. Obviously the number of households that have both cannot be greater than 90. Eliminate B, D, and E.Next: {Total} = {TV} + {Radio}  {Both} + {Neither} 150 = 120 + 90  {Both} + {Neither} {Both} = 60 + {Neither} Thus {Both}>=60. Answer: C. Hi Bunuel, There are 2 options greater than 60 C & D. Can u pls explain how you narrowed down on C..? Have you read the red part in my solution?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



eGMAT Representative
Joined: 04 Jan 2015
Posts: 1545

In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
Updated on: 07 Dec 2016, 04:06
Here's how you can solve this question using the Double Matrix Structure: We need to find the minimum and maximum possible values of X. From the matrix, it is clear that X cannot be greater than 90,000 (because x + y = 90,000. So, if x > 90,000 then y will have to take a negative value, and that is not allowed since y denotes the number of things. So it cannot be negative) Let's now try to find the lowest possible value of x. You can use either Equation 1: Using x + y = 90,000 => x = 90,000  y So, x will be minimum when y is maximum The maximum possible value of y = 30,000 So, minimum possible value of x from this equation = 90,000  30,000 = 60,000Or you can use Equation 2: Using x + z = 120,000 The maximum possible value of z = 60,000 So, minimum possible value of x from this equation is also 60,000anewbeginning, TuringMachine: Hope this helped! Please let me know if something is not clear Japinder
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



VP
Joined: 09 Jun 2010
Posts: 1188

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
12 May 2015, 02:14
I missed this question though it tests the same basic. practice is important.
_________________
visit my facebook to help me. on facebook, my name is: thang thang thang



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3511
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
22 May 2016, 02:55
imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive The highlighted part is the most important part of the question.... Posting the Venn Diagram approach hope it clears the concept further Attachment:
Venn Diagram.png [ 18.55 KiB  Viewed 13679 times ]
So, the possible range of values is between 40% to 60% 40% of 150,000 = 60,000 ( Minimum Value ) 60% of 150,000 = 90,000 ( Maximum Value ) Hence answer will be C. 60,000 to 90,000 inclusive
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club  Rules for Posting in QA forum  Writing Mathematical Formulas Rules for Posting in VA forum  Request Expert's Reply ( VA Forum Only )



Current Student
Joined: 28 Nov 2014
Posts: 895
Concentration: Strategy
GPA: 3.71

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
24 Sep 2016, 07:47
Bunuel  Thank you for the solution. Do you have any link to similar questions? I'm bad with these questions and I'd like to practise them. Please if you can help. Thank You!



Math Expert
Joined: 02 Sep 2009
Posts: 46271

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
25 Sep 2016, 03:17



Intern
Joined: 20 Sep 2016
Posts: 6

In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
25 Sep 2016, 06:49
General Formula: Sum = Gr A + Gr B  (A and B) + (neither A nor B) With that we have: (Z = 150,000) Z = 0.8Z + 0.6Z  (A and B) + (none) (A and B) = 0.4Z + none To get min(A and B), (none) will be 0, hence min(A and B) = 0.4Z = 60,000 To get max(A and B), we know that (A and B: the number of pp who own both items) will be maximum only when all of those who own recorders also have cable TV, therefore the largest value of (A and B) is the number of people who own recorders, which is equal to 0.6Z=90,000



Manager
Joined: 18 Jun 2016
Posts: 97
Location: India
Concentration: Technology, Entrepreneurship
WE: Business Development (Computer Software)

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
25 Sep 2016, 16:35
imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 80 has tv 60 has videocassettes There are 100 people in total. Manima would be (80+60)  100 = 40(consider some videocassete holders do not own tv, maximizing those people we get minima) Mixima would be 60(consider all the videocassete holder has tv, minimizing nontv holders we get maxima) Hence range would be from 40% of 150,000 to 60% of 150,000. So option C is correct.
_________________
If my post was helpful, feel free to give kudos!



Manager
Joined: 13 Dec 2013
Posts: 162
Location: United States (NY)
Concentration: Nonprofit, International Business
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
30 Mar 2017, 21:06
Bunuel wrote: imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 120 have TV and 90 have radio. Obviously the number of households that have both cannot be greater than 90. Eliminate B, D, and E. Next: {Total} = {TV} + {Radio}  {Both} + {Neither} 150 = 120 + 90  {Both} + {Neither} {Both} = 60 + {Neither} Thus {Both}>=60. Answer: C. Hi Bunuel, the formula you use, {Total} = {TV} + {Radio}  {Both} + {Neither}, is a modified version of the following formula for three overlaps, correct? Total = (members of one group)  (members of two groups) + (members of all three groups) + neither



Math Expert
Joined: 02 Sep 2009
Posts: 46271

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
30 Mar 2017, 22:10
Cez005 wrote: Bunuel wrote: imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive 120 have TV and 90 have radio. Obviously the number of households that have both cannot be greater than 90. Eliminate B, D, and E. Next: {Total} = {TV} + {Radio}  {Both} + {Neither} 150 = 120 + 90  {Both} + {Neither} {Both} = 60 + {Neither} Thus {Both}>=60. Answer: C. Hi Bunuel, the formula you use, {Total} = {TV} + {Radio}  {Both} + {Neither}, is a modified version of the following formula for three overlaps, correct? Total = (members of one group)  (members of two groups) + (members of all three groups) + neither This is a formula for 2 overlapping sets. Don't know what formula you are referring though. Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2771
Location: United States (CA)

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
04 Apr 2017, 16:16
imhimanshu wrote: In a certain city, 80 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
A. 30,000 to 90,000 inclusive B. 30,000 to 120,000 inclusive C. 60,000 to 90,000 inclusive D. 60,000 to 120,000 inclusive E. 90,000 to 120,000 inclusive We are given that there are 150,000 households in the city, that 60%, or 0.6 x 150,000 = 90,000, have videocassette recorders, and that 80%, or 0.8 x 150,0000 = 120,000, have cable television. We can create the following equation: 150,000 = 90,000 + 120,000  number who have both (b) + number who have neither (n) 150,000 = 210,000  b + n b = 60,000 + n To minimize b, we can let n = 0, and thus b = 60,000. To maximize b, we need to use the fact that b is the number of households that have both cable television and videocassette recorders; thus, it can’t be more than the smaller of the number of households that have cable television and the number of households with videocassette recorders. Since 90,000 households have videocassette recorders, which is less than the 150,000 households that have cable television, the maximum value of b is 90,000. Answer: C
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



NonHuman User
Joined: 09 Sep 2013
Posts: 7021

Re: In a certain city, 80 percent of the households have cable [#permalink]
Show Tags
09 May 2018, 13:19
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: In a certain city, 80 percent of the households have cable
[#permalink]
09 May 2018, 13:19






