In a town of 10 million households : GMAT Problem Solving (PS)
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# In a town of 10 million households

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In a town of 10 million households [#permalink]

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11 Oct 2011, 23:29
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In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type.

a. At least 3 million

b. Exactly 3 million

c. Less than 3 million

d. At least 2 million

e. At least 1.5 million.
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Re: In a town of 10 million households [#permalink]

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12 Oct 2011, 00:36
I think its A .. 12m subs - 9m households so atleast 3m households should subscribe for more that one type
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12 Oct 2011, 01:25
In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type.

a. At least 3 million

b. Exactly 3 million

c. Less than 3 million

d. At least 2 million

e. At least 1.5 million.

The difference between Total news paper sold and total households subscribed for newspapers is 3M

so the maximum number of households that subscribing more than 1 Newspaper is 3M.--(1)

If all the households which are subscribing more than 1 news paper are subscribing to three news papers then we can find the minimum number of households, which is 1.5 M---(2)

From (1) and (2) we can ignore options a ,b, c,d

So Ans E
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Re: In a town of 10 million households [#permalink]

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12 Oct 2011, 01:27
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Priyanka2011 wrote:
I think its A .. 12m subs - 9m households so atleast 3m households should subscribe for more that one type

A is not the right answer as the difference is 3M and if any household subscribes three papers then the number will be less than 3M

3M is the maximum limit for the households which are subscribing more than one paper.
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Re: In a town of 10 million households [#permalink]

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12 Oct 2011, 01:42
gopu106 wrote:
In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type.

a. At least 3 million

b. Exactly 3 million

c. Less than 3 million

d. At least 2 million

e. At least 1.5 million.

The difference between Total news paper sold and total households subscribed for newspapers is 3M

so the maximum number of households that subscribing more than 1 Newspaper is 3M.--(1)

If all the households which are subscribing more than 1 news paper are subscribing to three news papers then we can find the minimum number of households, which is 1.5 M---(2)

From (1) and (2) we can ignore options a ,b, c,d

So Ans E

i Think below method is easy to understand.

The total number of house holds buying newspaper = X+Y+Z+XY+XZ+YZ+XYZ = 9---(1)
Total number of newspapers sold = X+Y+Z+2XY+2XZ+2YZ+3XYZ = 12---(2)
If we need maximum number of households then no house hold should buy all three.
applying this in 1 and 2

X+Y+Z+XY+XZ+YZ = 9
X+Y+Z+2XY+2XZ+2YZ=12
So XY+XZ+YZ = 3(Maximum)

If we need the minimum number of households then no one should buy 2 newspapers.
1,2 will become
X+Y+Z+XYZ = 9
X+Y+Z+3XYZ=12
2XYZ=3
XYZ=1.5(minimum)

Applying this on options we can conclude that E is the answer
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Last edited by g106 on 12 Oct 2011, 02:08, edited 1 time in total.
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Re: In a town of 10 million households [#permalink]

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12 Oct 2011, 02:01
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Expert's post
In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type.

a. At least 3 million

b. Exactly 3 million

c. Less than 3 million

d. At least 2 million

e. At least 1.5 million.

12 mil is the number of newspapers distributed.
9 mil is the number of households in which they are distributed.
Since each household gets at least one paper, say 9 mil are distributed, one each, in the 9 mil households. We are left with 3 mil newspapers. Each one of these 3 mil papers is distributed as second/third paper in some household.

Let's focus on distributing just these 3 mil papers now.

The maximum number of households in which these 3 mil papers can be distributed is 3 mil. Each paper goes to a different household. So number of households subscribing to more than 1 newspaper is $$\leq 3$$mil. But there is no such option.

Say, if I give 2 papers (a total of 3 papers) in the same household, and the rest of them in different households, how many households will get covered? (3 mil - 1)
How can I minimize the total number of households? By giving 2 papers in as many households (a total of 3 papers) as possible. If I give 2 papers in each household, I will need to cover 1.5 mil households. So number of households receiving more than 1 paper should be at least 1.5 mil.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 23 Oct 2011 Posts: 84 Followers: 0 Kudos [?]: 73 [0], given: 34 Re: In a town of 10 million households [#permalink] ### Show Tags 27 Nov 2011, 04:58 VeritasPrepKarishma wrote: anjanapadmaraj wrote: In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type. a. At least 3 million b. Exactly 3 million c. Less than 3 million d. At least 2 million e. At least 1.5 million. 12 mil is the number of newspapers distributed. 9 mil is the number of households in which they are distributed. Since each household gets at least one paper, say 9 mil are distributed, one each, in the 9 mil households. We are left with 3 mil newspapers. Each one of these 3 mil papers is distributed as second/third paper in some household. Let's focus on distributing just these 3 mil papers now. The maximum number of households in which these 3 mil papers can be distributed is 3 mil. Each paper goes to a different household. So number of households subscribing to more than 1 newspaper is $$\leq 3$$mil. But there is no such option. Say, if I give 2 papers (a total of 3 papers) in the same household, and the rest of them in different households, how many households will get covered? (3 mil - 1) How can I minimize the total number of households? By giving 2 papers in as many households (a total of 3 papers) as possible. If I give 2 papers in each household, I will need to cover 1.5 mil households. So number of households receiving more than 1 paper should be at least 1.5 mil. Answer (E) Karishma if we tried to solve this algebraically: (numbers in million) AuBuC=A+b+C-{exactly 2}-2*{exactly 3} --->9=12-{exactly 2}-2*{exactly 3} ---> {exactly 2}+2*{exactly3}=3 (1) Is this the 3 that you mention in your explanation? (red phrase). Then what do we say? we want to find {exactly 2}+{exactly 3}... Based on (1) i guess I have to find the value of {exactly 3}.. then what? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7119 Location: Pune, India Followers: 2130 Kudos [?]: 13630 [1] , given: 222 Re: In a town of 10 million households [#permalink] ### Show Tags 28 Nov 2011, 01:34 1 This post received KUDOS Expert's post SonyGmat wrote: VeritasPrepKarishma wrote: anjanapadmaraj wrote: In a town of 10 million households there are 3 newspapers in circulation - X, Y and Z. There are 9 million households that subscribe to one or more newspapers.The number of total circulation of all the newspapers is 12 million. The number of households subscribing to X,Y are 2 million and 5 million respectively. What is the number of households subscribing to more than 1 newspaper.Assume that any household subscribes only one newspaper of a particular type. a. At least 3 million b. Exactly 3 million c. Less than 3 million d. At least 2 million e. At least 1.5 million. 12 mil is the number of newspapers distributed. 9 mil is the number of households in which they are distributed. Since each household gets at least one paper, say 9 mil are distributed, one each, in the 9 mil households. We are left with 3 mil newspapers. Each one of these 3 mil papers is distributed as second/third paper in some household. Let's focus on distributing just these 3 mil papers now. The maximum number of households in which these 3 mil papers can be distributed is 3 mil. Each paper goes to a different household. So number of households subscribing to more than 1 newspaper is $$\leq 3$$mil. But there is no such option. Say, if I give 2 papers (a total of 3 papers) in the same household, and the rest of them in different households, how many households will get covered? (3 mil - 1) How can I minimize the total number of households? By giving 2 papers in as many households (a total of 3 papers) as possible. If I give 2 papers in each household, I will need to cover 1.5 mil households. So number of households receiving more than 1 paper should be at least 1.5 mil. Answer (E) Karishma if we tried to solve this algebraically: (numbers in million) AuBuC=A+b+C-{exactly 2}-2*{exactly 3} --->9=12-{exactly 2}-2*{exactly 3} ---> {exactly 2}+2*{exactly3}=3 (1) Is this the 3 that you mention in your explanation? (red phrase). Then what do we say? we want to find {exactly 2}+{exactly 3}... Based on (1) i guess I have to find the value of {exactly 3}.. then what? Yes it is. 3 mil newspapers are second or third papers. {exactly 2}+2*{exactly3}=3 Maximize households -> Distribute each of the 3 mil papers in two paper homes. {exactly3} = 0 {exactly2} = 3 Number of households subscribing to more than 1 paper = 3 mil Minimize households -> Distribute each of the 3 mil papers in 3 paper homes. {exactly 2} = 0 2*{exactly 3} = 3 {exactly 3} = 1.5 mil Number of households subscribing to more than 1 paper = 1.5 mil _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In a town of 10 million households [#permalink]

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28 Nov 2011, 02:35
VeritasPrepKarishma wrote:
Yes it is. 3 mil newspapers are second or third papers.

{exactly 2}+2*{exactly3}=3
Maximize households -> Distribute each of the 3 mil papers in two paper homes. {exactly3} = 0
{exactly2} = 3
Number of households subscribing to more than 1 paper = 3 mil

Minimize households -> Distribute each of the 3 mil papers in 3 paper homes. {exactly 2} = 0
2*{exactly 3} = 3
{exactly 3} = 1.5 mil
Number of households subscribing to more than 1 paper = 1.5 mil

Perfect explanation as always. Thanks.
Re: In a town of 10 million households   [#permalink] 28 Nov 2011, 02:35
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