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

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

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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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.

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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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 = 60,000

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

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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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]
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!
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Keats wrote:
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!

Data Sufficiency Questions on Overlapping Sets
Problem Solving Questions on Overlapping Sets
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In a certain city, 80 percent of the households have cable [#permalink]
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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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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]
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.

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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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.

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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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.

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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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Re: In a certain city, 80 percent of the households have cable [#permalink]
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% = 120,000
60% = 90,000

Therefore the maximum is 90,000. We can eliminate B, D, and E. We have a 50/50 chance if we guess between A and C.

To find the minimum: We know we have 60% of the households with videocassette recorders. Since 80% of the households have cable television, the MINIMUM overlap will be 40%.

40% = 60,000

The range will be 60,000 to 90,000, inclusive. Answer is C.
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Re: In a certain city, 80 percent of the households have cable [#permalink]
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Re: In a certain city, 80 percent of the households have cable [#permalink]
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