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Word Problem

by Aman verma » Sat Feb 06, 2010 10:36 am
Q: Each family in New York county has atmost two adults and the total number of boys in this county is less than the number of girls. Similarly the number of girls is less than the number of adults in the county . Marc , the chief of this county is the only adult in his family. The minimum number of families in his county are :

a) 3

b) 4

c) 13

d) 18

e) can't be determined
Last edited by Aman verma on Sat Feb 06, 2010 11:33 am, edited 3 times in total.
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by Stuart@KaplanGMAT » Sat Feb 06, 2010 10:50 am
Aman verma wrote:Q: Each family in New York county has atmost two adults and the total number of boys in this county is less than the number of girls. Similarly the number of girls is less than the number of adults in the county . Marc , the chief of this county is the only adult in his family. The minimum number of families in his village is :

a) 3

b) 7

c) 13

d) 18

e) can't be determined
Do adults count as boys or girls? We need to resolve this ambiguity before answering.

If adults don't count (i.e. adult Males are men, not boys, and adult Females are women, not girls), then the minimum number of families is 1 (Marc could be the only person living in the County).

If adults do count, then Marc is a boy, so we need at least 2 girls in the county.

We also know that the number of girls is less than the number of adults, so we need at least 3 adults.

So, the minimum number of families is 2: Marc with no kids and a happily married pair of female adults with no kids. (Nowhere does it forbid same-GMAT families.)

Since neither answer matches any of the choices, this is a bad bad question! What's the source?

(As an aside, by the actual wording of the question the answer is 1 in both cases, because we switch from "county" to "village" in the question; to arrive at the two different answers I'm assuming that it's supposed to be consistent throughout.)
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by Aman verma » Sat Feb 06, 2010 10:55 am
Corrections made ; it's county and boys and girls should better be treated as children not as adults
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by Stuart@KaplanGMAT » Sat Feb 06, 2010 11:09 am
Aman verma wrote:Corrections made ; it's county and boys and girls should better be treated as children not as adults
OK... I just realized a mistake I made in the original response, so let's start again:

we can't have just Marc, since the number of boys is less than the number of girls; therefore, we need at least 1 girl.

The number of girls must be less than the number of adults, so we must have at least 1 other adult.

So, we could have:

Marc + daughter
Single woman with no kids.

Now we have:

0 boys
1 girl
2 adults

fulfilling all the rules. So, the answer is 2, which also doesn't appear among the rules.

Is there another rule missing? Does each family have to have at least 2 members?
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by harsh.champ » Sat Feb 06, 2010 11:14 am
Aman verma wrote:Q: Each family in New York county has atmost two adults and the total number of boys in this county is less than the number of girls. Similarly the number of girls is less than the number of adults in the county . Marc , the chief of this county is the only adult in his family. The minimum number of families in his county is :

a) 3

b) 7

c) 13

d) 18

e) can't be determined
________________________
no. of boys < no. of girls
no. of girls < no. of adults

As we find that no specific no. of adults or any type of equalities are given,the answer would be E.

As stuart has pointed out ,this is a very badly phrased ques.

Also,the question could have been made challenging and interesting by giving some more variables like no. of boys or no. of adults etc.

I have solved a question like this and it was quite an amazing experience.

Aman can you point out the source..
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by Aman verma » Sat Feb 06, 2010 11:17 am
I will soon come up with the solution but it's definitely not E. And people please read the question closely you may come up with the solution yourself . The wording is correct .
Last edited by Aman verma on Sat Feb 06, 2010 11:28 am, edited 1 time in total.
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by Stuart@KaplanGMAT » Sat Feb 06, 2010 11:23 am
Aman verma wrote:I will soon come up with the solution but it's definitely not E.
I'm not suggesting that the answer is E... I'm saying that the answer is F, "none of the above", i.e. it's a flawed question.

Now, if we assume that a "family" must consist of two members and that same-GMAT partnerships aren't allowed, the answer would be A, 3:

Marc + daughter
Woman + daughter
Woman + son

gives us 3 adults, 2 girls and 1 boy, following all of our inequalities.

If same-GMAT partnerships are allowed but we need a minimum of two members per family, the answer would be 2:

Marc + daughter
Woman + woman + son + daughter

also gives us 3 adults, 2 girls and 1 boy.

However, based on the original information, there's no reason to make the assumptions outlined above; on the actual GMAT we'd never have to do so. Accordingly, we can discount the question as written.
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by vscid » Sat Feb 06, 2010 12:09 pm
Aman verma wrote:Q: Each family in New York county has atmost two adults and the total number of boys in this county is less than the number of girls. Similarly the number of girls is less than the number of adults in the county . Marc , the chief of this county is the only adult in his family. The minimum number of families in his county are :

a) 3

b) 4

c) 13

d) 18

e) can't be determined
Aman, what is the source?
The GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.
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by Ian Stewart » Sat Feb 06, 2010 1:04 pm
There are a few things which are strange about this question. For one thing, as Stuart points out, we have no idea what assumptions are required. For another, as he points out, the correct answer should be 'two'. For a third, the answer E makes no logical sense; of course we can determine, from the information given, what the minimum number of families is, so the answer 'cannot be determined' can't possibly be correct. I'm also interested to know where the question is from.
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by Aman verma » Sat Feb 06, 2010 11:30 pm
Solution : let us assume F= number of families B=number of boys G=number of girls A=number of adults

Then F<B<G<A
Now to get the minimum number of families we should add the smallest increment of 1 to each element
now if F=1 , then 1<2<3<4, but adults cannot be 4 in one family
if F=2, then 2<3<4<5,but adults cannot be 5 in 2 families
if F=3, then 3<4<5<6, not possible since 1 family has only one adult ( Marc is the only adult in his family,so we
will be left with 5 adults and 2 families,hence not possible )
if F=4, then 4<5<6<7, Possible since (3X2)+(1X1) = 7
So B is the answer.
The fact that we are told Marc is the only adult in his family is relevant to the solution.
I think this is more of a reasoning question than of hard core algebra. The relation F<B<G<A can be found out from the wording of the question since we are to find the minimum number of families.

Now I won't post any further comments on this post. I will expect people to post only worthwhile and valuable comments on this post otherwise it will be a total wastage of everybody's time.
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by Stuart@KaplanGMAT » Sun Feb 07, 2010 12:33 am
Aman verma wrote:Solution : let us assume F= number of families B=number of boys G=number of girls A=number of adults

Then F<B<G<A
Now to get the minimum number of families we should add the smallest increment of 1 to each element
now if F=1 , then 1<2<3<4, but adults cannot be 4 in one family
if F=2, then 2<3<4<5,but adults cannot be 5 in 2 families
if F=3, then 3<4<5<6, not possible since 1 family has only one adult ( Marc is the only adult in his family,so we
will be left with 5 adults and 2 families,hence not possible )
if F=4, then 4<5<6<7, Possible since (3X2)+(1X1) = 7
So B is the answer.
The fact that we are told Marc is the only adult in his family is relevant to the solution.
I think this is more of a reasoning question than of hard core algebra. The relation F<B<G<A can be found out from the wording of the question since we are to find the minimum number of families.

Now I won't post any further comments on this post. I will expect people to post only worthwhile and valuable comments on this post otherwise it will be a total wastage of everybody's time.
The flaw in your solution lies in your original inequality; why does F have to be less than B, G or A?

Please explain why my solution with 2 families (or the solution with 3 families) fails to fulfil all the limitations of the question. I can also come up with legal counties with 4, 5 and 6 families, if you'd like.

If your attitude is "this is my question and it's my solution and I'm not going to post anymore" that's fine, it just means many of us will ignore your future posts, since that attitude isn't going to help anyone (including you) learn.
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by harshavardhanc » Sun Feb 07, 2010 1:39 am
Stuart Kovinsky wrote: If your attitude is "this is my question and it's my solution and I'm not going to post anymore" that's fine, it just means many of us will ignore your future posts, since that attitude isn't going to help anyone (including you) learn.
Well said!! thank u thank u !
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by rohan_vus » Sun Feb 07, 2010 1:47 am
Aman verma wrote:Solution : let us assume F= number of families B=number of boys G=number of girls A=number of adults

Then F<B<G<A
Now to get the minimum number of families we should add the smallest increment of 1 to each element
now if F=1 , then 1<2<3<4, but adults cannot be 4 in one family
if F=2, then 2<3<4<5,but adults cannot be 5 in 2 families
if F=3, then 3<4<5<6, not possible since 1 family has only one adult ( Marc is the only adult in his family,so we
will be left with 5 adults and 2 families,hence not possible )
if F=4, then 4<5<6<7, Possible since (3X2)+(1X1) = 7
So B is the answer.
The fact that we are told Marc is the only adult in his family is relevant to the solution.
I think this is more of a reasoning question than of hard core algebra. The relation F<B<G<A can be found out from the wording of the question since we are to find the minimum number of families.

Now I won't post any further comments on this post. I will expect people to post only worthwhile and valuable comments on this post otherwise it will be a total wastage of everybody's time.
Even with common sense reasoning , this question is weird and defies commensense in this particular case . What makes you say that if F = 1 and adults cant be 4??.. Wont grand parents be part of family ?..Do you assume adults are always parents whose kids are not adults?..

This broad assumption is way out till you explicitly mention all such assumptions..

A caveat here ! . I in no way intend to offend you ( thats the worst i ever do) , but this question is absolutely unreal as you need to make unwarranted assumption to arrive at answer !!!
Last edited by rohan_vus on Sun Feb 07, 2010 7:26 am, edited 1 time in total.
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by Aman verma » Sun Feb 07, 2010 7:08 am
Stuart Kovinsky wrote:
Aman verma wrote:Solution : let us assume F= number of families B=number of boys G=number of girls A=number of adults

Then F<B<G<A
Now to get the minimum number of families we should add the smallest increment of 1 to each element
now if F=1 , then 1<2<3<4, but adults cannot be 4 in one family
if F=2, then 2<3<4<5,but adults cannot be 5 in 2 families
if F=3, then 3<4<5<6, not possible since 1 family has only one adult ( Marc is the only adult in his family,so we
will be left with 5 adults and 2 families,hence not possible )
if F=4, then 4<5<6<7, Possible since (3X2)+(1X1) = 7
So B is the answer.
The fact that we are told Marc is the only adult in his family is relevant to the solution.
I think this is more of a reasoning question than of hard core algebra. The relation F<B<G<A can be found out from the wording of the question since we are to find the minimum number of families.

Now I won't post any further comments on this post. I will expect people to post only worthwhile and valuable comments on this post otherwise it will be a total wastage of everybody's time.
The flaw in your solution lies in your original inequality; why does F have to be less than B, G or A?

Please explain why my solution with 2 families (or the solution with 3 families) fails to fulfil all the limitations of the question. I can also come up with legal counties with 4, 5 and 6 families, if you'd like.

If your attitude is "this is my question and it's my solution and I'm not going to post anymore" that's fine, it just means many of us will ignore your future posts, since that attitude isn't going to help anyone (including you) learn.
Hello Stuart, now please tell me have you ever seen a county where the number of families is more than the number of individuals constituting those families. As regards 2 or 3 families ,the solution explains elaborately why it cannot be so. And we have to solve the problem according to the restrictions given in the problem and limit our assumptions to the restrictions given in the problem itself.

Now as regards my attitude I haven't given any attitude in the first place I only just want worthwhile discussion and not just some wasteful talk like the one done by a nutcase in one of my earlier post
Last edited by Aman verma on Sat Feb 13, 2010 12:38 pm, edited 1 time in total.
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by Stuart@KaplanGMAT » Sun Feb 07, 2010 11:11 am
Aman verma wrote:Hello Stuart, now please tell me have you ever seen a county where the number of families is more than the number of individuals constituting those families. As regards 2 or 3 families ,the solution explains elaborately why it cannot be so. And we have to solve the problem according to the restrictrictions given in the problem.
There's a big difference between F < g+b+a and F < b < g < a, which is what you wrote.

If we assume that families must consist of more than 1 person, then F does have to be less than the sum of the other 3; however, that doesn't mean it has to be less than each of those three.

Here's an example:

Family 1: mom and son
Family 2: dad and daughter

We have 2 families with 2 adults and 1 son and 1 daughter.

So, F < a + b + g is true; however, F=a, F>b and F>g.

Let's revisit my original solution, assuming that family must contain more than 1 person (on the GMAT we'd never have to make that assumption, the term would be defined for us):

If we allow same-GMAT families (lots of these in the real world):

Marc + daughter
Woman + woman + son + daughter

In this scenario, we have 3 adults, 2 girls and 1 boy, giving us b < g < a.

Why isn't this a valid scenario?

Even assuming that we don't allow same-GMAT families (the politically correct GMAT would never forbid them!), we could have:

Marc + daughter
Woman + daughter
Woman + son

In this scenario, we have 3 adults, 2 girls and 1 boy, giving us b < g < a.
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