That's a good question. It's the first thing I thought of while solving this problem!

On most problems where integer constraints matter, you're given an upper limit. This limit can either be stated in the problem, or implied by the equations. For instance, if the problem says this:

Each apple costs 15 cents, and each orange costs 23 cents. If a certain purchase of apples and oranges costs $2.90, how many apples and oranges are there in total?

The number of apples and oranges has an upper limit, implied by the equations. You can't logically have more than 19 apples, since then the apples alone would cost more than $2.90.

Or, the problem might say something like this:

Each apple costs 15 cents, and each orange costs 23 cents. If John purchases no more than 10 apples and oranges, and the apples he purchases cost exactly 14 cents more than the oranges he purchases, how many apples and oranges did he purchase?

In that one, there's an upper limit stated in the problem (no more than 10).

In both of those problems, there should be a single good solution. You'd only have to test cases up to a certain number, and then all other cases are 'invalid'.

In this problem, there's no upper limit. Theoretically, there's no reason you can't have an infinite number of adults and children! So, you'd have to keep testing cases on up to infinity. It's very, very likely (although I'll stop short of saying 'inevitable', since that would take some number theory and is way beyond the scope of the GMAT) that there will be two different cases that would work.
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Video solution from Quant Reasoning starts at 14:15 here:

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IMO C
There were more than or equal to 10 ppl in group
so adult:5500-10%=4950
Child:4800-10%=4320
A)total adult fees=29700
so number of adults=29700/4950=6
but number of children not known so insufficient
B)no of adults=x
no of children=y
4320y-4950x=4860
but x+y=?
not sufficient
A+B coimbiled
4320y-4950(6)=4860
y=6
so A+B is sufficient
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A group consisting of several families visited an amusement park where the regular admission fees were ¥5,500 for each adult and ¥4,800 for each child. Because there were at least 10 people in the group, each paid an admission fee that was 10% less that the regular admission fee. How many children were in the group?

(1) The total of the admission fees paid for the adults in the group was ¥29,700
(2) The total of the admission fees paid for the children in the group was ¥4,860 more than the total of the admission fees paid for the adults in the group

let X and Y be for adults and children respectively.
Stmt 1 .9*5500 X = 29700
X = 6 Not suff

Stmt 2 says,
0.9*4800 Y = 4860 + .9*5500 X Not suff
Together suff. C is the answer

We are given that at an amusement park, regular admission fees were ¥5,500 for each adult and ¥4,800 for each child. However, a particular group paid 10% less than the regular admission fee. Thus, the admission fee per adult was 5,500 x 0.9 = 4,950 yen and per child was 4,800 x 0.9 = 4,320 yen. We need to determine the number of children in the group.

Statement One Alone:

The total of the admission fees paid for the adults in the group was ¥29,700.

If we let the number of adults = A, we can create the following equation:

4,950A = 29,700

A = 6

However, since we do not know the number of children, statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The total of the admission fees paid for the children in the group was ¥4,860 more than the total of the admission fees paid for the adults in the group.

If we let C = the number of children, we can create the following equation:

4,320C = 4,860 + 4,950A

We see that we do not have enough information to determine C. We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two, we know that A = 6 and 4,320C = 4,860 + 4,950A. We can see that if we substitute 6 for A in 4,320C = 4,860 + 4,950A, we can determine the value of C. The two statements together are sufficient.

Answer: C
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I have seen situations where statements such as 2 are sufficient because of integer constraints. Here too there are integer constraints but as many have pointed one equations in two unknowns have many solutions. How do we differentiate those two circumstances in the test when we have limited time?

There is a doubt raised by many that maybe statement 2 is sufficient. This question uses Diophantine equation, i.e. only integer values are acceptable. So many here are having doubts if only one set of numbers work?

Kindly make a note of other solutions and my insight on why you should not bother about this dilemna:-


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chetan2u

Hey Chetan

Seriously Need your help..

As per statement two, considering that there are x children and y adults, we have the following equation—

4320x - 4950y = 4860

As per the information present in the question itself, we know that x+y>= 10

My doubt is — how do I check the uniqueness of the above equation ie 4320x - 4950y = 4860 ? Just because they are so many values doesn’t mean that we cannot have a unique solution right ? The reason I’m asking this is that I have seen a few similar equations having unique solutions.

Looking forward to hearing from you

Regards,

Posted from my mobile device

I explain that in my video solution
https://gmatclub.com/forum/a-group-cons ... 2#p2653474
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Answer: Option C

Video solution by GMATinsight



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