Families in which some members argue with each other compete for

Question

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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VibhuAnurag · Answer

The answer should be (C).

(1) This statement clarifies that within each family, arguments occur among male members separately and among female members separately. However, it doesn’t provide information about the actual percentage or proportion of family members who argue --> Insufficient.

(2) This statement indicates that the Brown family has an equal number of male and female members, while the Gonzales family has more male members than female members. No information is provided about who can argue with whom --> Insufficient.

Combined:
Let the total members per family = 6
--> Brown family = 3 men (A, B & C) and 3 women (D, E & F)
 Possibilities of arguments = AB, AC, BC, DE, DF, EF = 6
whereas, for Gonzales family let's take 4 men (A, B, C & D) and 2 women (E & F)
--> Possibilities of arguments = AB, AC, AD, BC, BD, CD, EF = 7
or let's take 5 men (A, B, C, D & E) and 1 woman (F)
--> Possibilities of arguments = AB, AC, AD, AE, BC, BD, BE, CD, CE, DE = 10

Hence, taken together the statements indicate that the Gonzales family has a higher chance of appearing on the show.

VivekSri · Answer

OA is C.

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djangobackend · Answer

say each family has 10 members, combining both statements say - Brown has 5M, 5F and GZ has 6M, 4F so argument pairs in brown = 5C2+ 5C2 = 20 and in GZ it will be 6C2 + 4C2 = 21 argument pairs (members arguing with each other) so definitely 21>20 so higher chances for GZ. 
Note - other combinations of GZ (7M,3F), (8M,2F), (9M,1F), (10M,0F) all will result in combinations > 21 - for (9M,1F), (10M,0F) only Males pairs can be formed, no females. hence C sufficient

HassanZafar · Answer

For question like this is it better to attempt based on intuition? Since, clearly answer was C or E. Takes less time than solving the question?

arushi118 · Answer

In this question won't we have to assume that it is 2 people fighting each other?

KRNSW · Answer

Where is it written that only 2 members can argue amongst them?

I think B is the correct answer, if we consider that an argument can take palce between more than 2 members.For example, Brown family has 3 Men and 1 Woman, so the lone woman cannot enter an arguemnet as per question. While the other family has 2 Men and 2 Women, making all of them participate in an arguement. Hence, the ratio of particiaption in an arguement is higher in Gonzales family..

Is my interpretation valid?

Flyingdutch1307 · Answer

No, we dont know how many members in Family B and Family G! the only right way is to combine both

1. Family B - each person argues with 50% of members
2. Family G - M ~ 60%, F ~ 40% => avg. member argues >50% of family members

can also be thought with # instead of % values

Families in which some members argue with each other compete for

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Families in which some members argue with each other compete for

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