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# Families in which some members argue with each other compete for a cha

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Intern
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Joined: 07 Feb 2015
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Families in which some members argue with each other compete for a cha  [#permalink]

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25 Jun 2015, 06:29
5
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65% (hard)

Question Stats:

55% (01:45) correct 45% (01:44) wrong based on 214 sessions

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Families in which some members argue with each other compete for a chance to appear on Barry Wringer’s TV show. The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

1. In each family, male members argue with each other, and female members argue with each other, but male members do not argue with female members and vice versa.

2. The Brown family has the same number of male members as female members. The Gonzales family has more male members than female members.
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Joined: 15 Aug 2017
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21 Sep 2017, 20:04
whoisthere wrote:
Families in which some members argue with each other compete for a chance to appear on Barry Wringer’s TV show. The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

1. In each family, male members argue with each other, and female members argue with each other, but male members do not argue with female members and vice versa.

2. The Brown family has the same number of male members as female members. The Gonzales family has more male members than female members.

Here is an explanation for the question since there hasnt been one posted yet. Typing this out made me understand why the answer is C .

The question stem is as follows: . The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

Rephrased as follows: Does the Brown family have a greater percentage of family members who argue with each other? (Sum Gonzales = Sum Brown).

(1) Gives the info: Males can only argue with Males while the inverse is true as well.
This alone is insufficient as it does not tell us what percentage of the family members argue with each other.

(2) Gives the info: Males = Females in Gonzales family and Males > Females in Brown family.
This alone is insufficient as it does not tell us anything about the percentage of family members that argue with each other.

However,

When taking (1)+(2) we have Males can only argue with males, and the inverse. In addition from 2 we have M=F in gonzales, and M>F in Brown.
Now we want to calculate if the Brown family has a greater Percentage of family members who argue with each other. Lets see if we can derive more than one case.

M=F in Gonzales, Therefore 50% of males argue amongst themselves, while 50% of females argue amongst themselves.
This results in 50% of the family argueing with each other.
Alternatively if M>F than we find that 51% (OR GREATER) of males argue amongst themselves, and resultantly, the Brown's family has a higher overall percentage of family members who argue amongst the family. This results in C.
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Re: Families in which some members argue with each other compete for a cha  [#permalink]

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22 Sep 2017, 00:24
Can you please explain how you are getting 50%?

if its equal number of males and females, lets say 5 and 5, all 5 males are arguing with one another ,i.e. 100 % of males. and all 5 females are arguing with one another, i..e 100% again.

So that family has 100% of its members arguing.
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Joined: 15 Aug 2017
Posts: 14
Re: Families in which some members argue with each other compete for a cha  [#permalink]

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22 Sep 2017, 13:01
rahulkashyap wrote:
Can you please explain how you are getting 50%?

if its equal number of males and females, lets say 5 and 5, all 5 males are arguing with one another ,i.e. 100 % of males. and all 5 females are arguing with one another, i..e 100% again.

So that family has 100% of its members arguing.

The question asks which family has a greater percentage of members that argue with each other. Therefore if males equals females and males argue only amongst each other then 50% of the family argues. Alternatively 50% of the females argue. In the case where we have more males than females the total percentage of family members who argue amongst each other is now greater than 50% they are for the chances the Brown family getting on the show is higher. Let me know if anything is still unclear. Remember the percentage of family members who argue amongst each other is percent arguing divided by total number.

Sent from my SM-G955F using GMAT Club Forum mobile app
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Re: Families in which some members argue with each other compete for a cha  [#permalink]

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22 Sep 2017, 20:54
Still unclear how you got only 50 percent of the family arguing. Would be helpful if you took numbers to show that. For example, 6 men and 6 women. 6 men argue with each other and 6 women argue with each other.
That's 12 people (100%) arguing. 8 men and 4 women would not change the percentage.

Posted from my mobile device
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25 Sep 2017, 11:53
1
rahulkashyap wrote:
Still unclear how you got only 50 percent of the family arguing. Would be helpful if you took numbers to show that. For example, 6 men and 6 women. 6 men argue with each other and 6 women argue with each other.
That's 12 people (100%) arguing. 8 men and 4 women would not change the percentage.

Posted from my mobile device

Hi Rahulkashyap, Here is how I got 50 percent arguing with each other.

The question states :The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show.
This translates to Family members who argue with each other =$$\frac{# arguing}{Whole family}$$
Thus, if we have 4 males and 4 females arguing we have Percent of family members who argue with each other =$$\frac{4}{8}$$ = 50% of members arguing with each other and the same for females.
Alternatively, if we have 5 females, and 4 males arguing we have percent of family members who argue = $$\frac{5}{9}$$ = 55% of family members arguing with each other.

While the total percentage of family members arguing is in fact 100%, the question does not ask for the total percent arguing, Rather it asks which family has a higher percentage of family members who argue with each other.
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Re: Families in which some members argue with each other compete for a cha  [#permalink]

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17 Oct 2017, 15:05
1
Arguing with each other is a combination solution. I think it's easiest to think of numbers like previously given.

Let's say there are 12 members in each family. For the Brown family, there are M=F, so in this example, 6 males and 6 females. The males all argue with each other and there are 6C2 ways for arguments which => 15. The women also have 6C2 ways, or 15. Thus 30 total arguments between individuals.

For the Gonzales family, they have more men than women M>F. Let's just say there are 8 males and 4 females. The arguments among men is 8C2 => 28 and arguments among women is 4C2 => 6. Thus the total arguments between individuals in this family is 34 which is higher than the Brown family. This discrepancy continues if you do 9 men and 3 women as well: 9C2 => 36 and 3C2 => 3 for a total of 39.

This is long winded, but it is possible to do these calculations quick in order to comes up with the answer or just use intuition since it's a DS question.
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Families in which some members argue with each other compete for a cha  [#permalink]

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23 Nov 2017, 12:21
rajarshee wrote:
RudeyboyZ wrote:
Families in which some members argue with each other compete for a chance to appear on Barry Wringer’s TV show. The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

1. In each family, male members argue with each other, and female members argue with each other, but male members do not argue with female members and vice versa.

2. The Brown family has the same number of male members as female members. The Gonzales family has more male members than female members.

Here is an explanation for the question since there hasnt been one posted yet. Typing this out made me understand why the answer is C .

The question stem is as follows: . The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

Rephrased as follows: Does the Brown family have a greater percentage of family members who argue with each other? (Sum Gonzales = Sum Brown).

(1) Gives the info: Males can only argue with Males while the inverse is true as well.
This alone is insufficient as it does not tell us what percentage of the family members argue with each other.

(2) Gives the info: Males = Females in Gonzales family and Males > Females in Brown family.
This alone is insufficient as it does not tell us anything about the percentage of family members that argue with each other.

However,

When taking (1)+(2) we have Males can only argue with males, and the inverse. In addition from 2 we have M=F in gonzales, and M>F in Brown.
Now we want to calculate if the Brown family has a greater Percentage of family members who argue with each other. Lets see if we can derive more than one case.

M=F in Gonzales, Therefore 50% of males argue amongst themselves, while 50% of females argue amongst themselves.
This results in 50% of the family argueing with each other.
Alternatively if M>F than we find that 51% (OR GREATER) of males argue amongst themselves, and resultantly, the Brown's family has a higher overall percentage of family members who argue amongst the family. This results in C.

Hi victor VeritasPrepKarishma,

I have one quesion.

Question premise says "Families in which some members" => there could be some members who don't get into argument at all.
As per statement1, can we assume that all male member will argue with each other or some may not argue and same for female members.?
Bringing in some CR techniques into DS

Can you please help?

Thanks
Intern
Joined: 15 Aug 2017
Posts: 14
Re: Families in which some members argue with each other compete for a cha  [#permalink]

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27 Nov 2017, 13:24
hellosanthosh2k2 wrote:
rajarshee wrote:
RudeyboyZ wrote:
Families in which some members argue with each other compete for a chance to appear on Barry Wringer’s TV show. The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

1. In each family, male members argue with each other, and female members argue with each other, but male members do not argue with female members and vice versa.

2. The Brown family has the same number of male members as female members. The Gonzales family has more male members than female members.

Here is an explanation for the question since there hasnt been one posted yet. Typing this out made me understand why the answer is C .

The question stem is as follows: . The greater the percentage of family members who argue with each other, the greater the family’s chances of appearing on Wringer’s show. If the Brown family and the Gonzales family both have the same number of members, does the Brown family have a better chance of appearing on Wringer’s show than does the Gonzales family?

Rephrased as follows: Does the Brown family have a greater percentage of family members who argue with each other? (Sum Gonzales = Sum Brown).

(1) Gives the info: Males can only argue with Males while the inverse is true as well.
This alone is insufficient as it does not tell us what percentage of the family members argue with each other.

(2) Gives the info: Males = Females in Gonzales family and Males > Females in Brown family.
This alone is insufficient as it does not tell us anything about the percentage of family members that argue with each other.

However,

When taking (1)+(2) we have Males can only argue with males, and the inverse. In addition from 2 we have M=F in gonzales, and M>F in Brown.
Now we want to calculate if the Brown family has a greater Percentage of family members who argue with each other. Lets see if we can derive more than one case.

M=F in Gonzales, Therefore 50% of males argue amongst themselves, while 50% of females argue amongst themselves.
This results in 50% of the family argueing with each other.
Alternatively if M>F than we find that 51% (OR GREATER) of males argue amongst themselves, and resultantly, the Brown's family has a higher overall percentage of family members who argue amongst the family. This results in C.

Hi victor VeritasPrepKarishma,

I have one quesion.

Question premise says "Families in which some members" => there could be some members who don't get into argument at all.
As per statement1, can we assume that all male member will argue with each other or some may not argue and same for female members.?
Bringing in some CR techniques into DS

Can you please help?

Thanks

While it is stated that some members argue with each other, as per question one we are safe to assume all male members argue with one another and vice versa. If statement 1 had said some males argue amongst each other than we would have to consider the case in which members do not argue at all.
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Joined: 23 Mar 2017
Posts: 2
Re: Families in which some members argue with each other compete for a cha  [#permalink]

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04 Jan 2019, 23:49
The statements 1 and 2 are not sufficient. But on combining the two, they again become insufficient.

It is written in the question stem that the male/female members argue with EACH other and not ONE ANOTHER, in which it is safe to assume that the argument at any point of time is between 2 members of the same sex.
Then going ahead and taking an example of 6 members in both the families.

Brown family: 3M and 3W => 3C2 x 3C2 = 9
Gonzales family : 4M and 2W => 4C2 x 2C2 = 6
5M and 1W => 5C2 = 10
Therefore, either of the families has the chance to appear on the show. My assumption of the argument is based on the 'EACH' keyword in the question.
Hence, I think the answer should be E.
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Joined: 06 May 2019
Posts: 7
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21 May 2019, 01:01
aman23 wrote:
The statements 1 and 2 are not sufficient. But on combining the two, they again become insufficient.

It is written in the question stem that the male/female members argue with EACH other and not ONE ANOTHER, in which it is safe to assume that the argument at any point of time is between 2 members of the same sex.
Then going ahead and taking an example of 6 members in both the families.

Brown family: 3M and 3W => 3C2 x 3C2 = 9
Gonzales family : 4M and 2W => 4C2 x 2C2 = 6
5M and 1W => 5C2 = 10
Therefore, either of the families has the chance to appear on the show. My assumption of the argument is based on the 'EACH' keyword in the question.
Hence, I think the answer should be E.

I think it should be:
Brown family: 3M and 3W => 3C2 + 3C2 = 6
Gonzales family : 4M and 2W => 4C2 + 2C2 = 7
5M and 1W => 5C2 = 10
Re: Families in which some members argue with each other compete for a cha   [#permalink] 21 May 2019, 01:01
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