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In a certain city, 80 percent of the households have cable

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In a certain city, 80 percent of the households have cable  [#permalink]

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New post Updated on: 30 Sep 2013, 08:19
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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

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.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 30 Sep 2013, 08:38
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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.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 22 May 2016, 02:55
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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
Venn Diagram.png [ 18.55 KiB | Viewed 25141 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

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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 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..?
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 10 Mar 2015, 04:57
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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?
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In a certain city, 80 percent of the households have cable  [#permalink]

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New post Updated on: 07 Dec 2016, 04:06
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Here's how you can solve this question using the Double Matrix Structure:

Image

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 = 60,000

Or 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 60,000


anewbeginning, TuringMachine: Hope this helped! Please let me know if something is not clear :)

Japinder

Image
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Originally posted by EgmatQuantExpert on 12 May 2015, 01:13.
Last edited by EgmatQuantExpert on 07 Dec 2016, 04:06, edited 1 time in total.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 12 May 2015, 02:14
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I missed this question though it tests the same basic. practice is important.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 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.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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

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New post 25 Sep 2016, 06:49
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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
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 25 Sep 2016, 16:35
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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 non-tv holders we get maxima)

Hence range would be from 40% of 150,000 to 60% of 150,000.
So option C is correct.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 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
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 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.
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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New post 04 Apr 2017, 16:16
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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
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Re: In a certain city, 80 percent of the households have cable  [#permalink]

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In a certain city, 80 percent of the households have cable  [#permalink]

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New post 05 Mar 2019, 18:44
Line diagrams work well for these types of problems.

80
-------------------
60
-------------

Suppose those two lines represent the people who have cable (top line) and video cassette recorders.

The maximum overlap is 60.

How can we minimize the overlap?
We want to distribute as much of the 60 as we can to the 20 (don't have cable)

80...................20
--------------|-------
40...................... 20
--------------|-------

The minimum is 40% of 150,000 and the maximum is 60% of 150,000.

10% of 150,000 is 15,000. So 40% is 60,000. 60% is 1.5 times as much. 60,000+60,000/2=90,000
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In a certain city, 80 percent of the households have cable   [#permalink] 05 Mar 2019, 18:44
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