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# A group consisting of several families visited an amusement park where

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Re: A group consisting of several families visited an amusement park where [#permalink]
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IMO C
There were more than or equal to 10 ppl in group
Child:4800-10%=4320
but number of children not known so insufficient
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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Video solution from Quant Reasoning starts at 14:15 here:
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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
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JeffTargetTestPrep wrote:
Bunuel wrote:
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.

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.

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?
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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:-

Attachment:

DS Soln.jpg [ 126.94 KiB | Viewed 25124 times ]
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Re: A group consisting of several families visited an amusement park where [#permalink]
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chetan2u

Hey Chetan

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
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goaltop30mba wrote:
chetan2u

Hey Chetan

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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Bunuel wrote:
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.

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Re: A group consisting of several families visited an amusement park where [#permalink]
goaltop30mba wrote:
chetan2u

Hey Chetan

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

True -

We can solve the equation and find Head count - Time and time again B is correct answer because it follows " Only integer Solution".
If you were to try finding next integer solutions, You'll have to go all the way over 100 .

Hence , How can we chose ( c ) as an answer in such Questions else where.

Or how do we know that the Soln we have got here is not UNIQUE

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Bunuel wrote:
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.

goaltop30mba, it seems I missed out on your question earlier.
hD13

I. Type of equation 5x=7y+23, where x and y are integers.
The reason I will not try to look for additional values is that there is no upper limit to the value of variables, and in all probabilities you will have another set of variables fitting in.

II. Type of equation 5x+7y=34, where x and y are integers.
The reason I will look for additional values is that there is upper limit to the value of variables, and you cannot have 5x or 7y going above 34.

But if we still want to look for a solution with other possible values.

Solution

Let there be a adults and c children.
We have a 10% reduction in cost of admission fee.
Admission fee for a child = 0.9*4800=4320

(1) The total of the admission fees paid for the adults in the group was ¥29,700
So, 4950a=29700…..a=6.
Insufficient

(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.
$$4950a+4860=4320c$$
Simplify to get $$55a+54=48c$$
There are no restrictions on the upper limit, so in GMAT, it is highly unlikely that there is a unique solution.
But let us still work on the values and why we can have another solution.
Let$$a_1$$ be the least value that fits in for a.
$$55a_1+54=48c_1$$,
Let us add 48 to$$a_1$$, that is $$a_2=a_1+48$$
Why 48? : Because 48 is coefficient of the other variable.

$$55(a_1+48)+54=48c_2$$
$$55a_1+55*48+54=48c_2$$
$$(55a_1+54)+55*48=48c_2$$
$$(48c_1)+55*48=48c_2$$
$$48(c_1+55)=48c_2$$
$$c_1+55=c_2$$

So a increases by 48, and c increases by 55.
If a=6 and c=8 is the first solution, then next will be
a=6+48=54 and c=8+55=63.
Next values: a=54+48=102, and c=63+55=118
Insufficient

Combined
a=6, so c=8.
Sufficient

C

I hope it clears a bit of confusion on statement II.
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hD13 wrote:
We can solve the equation and find Head count - Time and time again B is correct answer because it follows " Only integer Solution".
If you were to try finding next integer solutions, You'll have to go all the way over 100 .

Hence , How can we chose ( c ) as an answer in such Questions else where.

Or how do we know that the Soln we have got here is not UNIQUE

I recall one of my math teachers saying, after the class gave the wrong answer to a tricky question, that "math is not a democracy". And sometimes that's an issue with gmatclub's system, where the forum elevates posts with the most 'votes' (or kudos). Atop this thread, none of the posts is correct - one just ignores the issue of unique integer solutions altogether, and that's the only interesting feature of this question. Another says you need number theory "way beyond the scope of the GMAT" to see if you have unique solutions, and you never need math beyond the scope of the test to definitively prove an answer is correct on the GMAT. And another post claims we need to go past adults = 100 to find the second smallest solution here, which also is not the case.

Statement 1 is clearly insufficient alone. For Statement 2, we can write an equation -- notice that the numbers in the question are nice to work with, but if we reduce them by 10%, they become annoying. So we should leave them as is. If we have c children and a adults, then, multiplying by 0.9 to apply the 10% discount, Statement 2 tells us:

(0.9)(4800)*c - (0.9)(5500)*a = 4860

4800c - 5500a = 4860*10/9

and since 4500 + 360 = 4860, we can see that 4860/9 = 500 + 40 = 540, so multiplying that by the extra 10, we get

4800c - 5500a = 5400

48c - 55a = 54

48c = 55a + 54

Notice what this equation says, if a and c are integers: it just says "55a + 54 is a multiple of 48". Since this is DS, we absolutely know there is at least one solution to this equation (it is never true in DS that there are zero solutions). So we know there is at least one integer a for which 55a + 54 is a multiple of 48. But if we take that integer a, and add 48 to it, this is what we get:

55(a + 48) + 54 = 55a + 54 + 55*48

and since 55a + 54 is a multiple of 48, and 55*48 is also a multiple of 48, this sum must be a multiple of 48 too. So if a gives one integer solution to the equation, so will a + 48, and a + 96, and so on. So we'll have an infinite number of solutions. There's no reason to actually find any of those solutions, since it's a DS problem (as soon as we use Statement 1+2 together, we'll clearly get one solution and we don't care what it is), but if you did solve here, you'd find that the smallest solution is a = 6 and c = 8, but we'll also have a solution, adding 48, when a = 54 and c = 63, and another solution when a = 102, and so on.

In general, if you want to know when these types of equations have unique integer solutions, there are a few different techniques that you might use, depending on the numbers involved. So it's too big of a topic to get into in one post here, but my Word Problems book discusses issues like this in detail.
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Re: A group consisting of several families visited an amusement park where [#permalink]
mehtakaustubh wrote:
no of children=y
4320y-4950x=4860

Hi VeritasKarishma, can you explain why the above equation (4320y-4950x=4860) should not be sufficient for us to find out the value of x and y, since we also know that x and y can only be integers. Wouldn't there be a unique combination of integer values of x and y that will satisfy this equation?
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sahilvermani wrote:
mehtakaustubh wrote:
no of children=y
4320y-4950x=4860

Hi VeritasKarishma, can you explain why the above equation (4320y-4950x=4860) should not be sufficient for us to find out the value of x and y, since we also know that x and y can only be integers. Wouldn't there be a unique combination of integer values of x and y that will satisfy this equation?

Yes, I would look at this equation and know that it will not have a unique solution. No calculations needed.
Here is why:

An equation with two variables will give an infinite number of solutions.
e.g. x + y = 10
x = 1, y = 9
x = 2, y = 8
x = 0.5, y = 9.5
x = -1, y = 11
and so on..
We keep changing the value of one variable and getting a corresponding value for the other.

Now think why sometimes we say that we got a unique solution. It is not because the equation didn't give infinite values. It is because only certain solutions are acceptable to us.

1. Only integers are acceptable to us. In some cases, we may get no integer solution such as 3x + 6y = 40 (think why).
If we get one set of integer values, we see that we get infinite integer solutions. For this, check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -of-thumb/
Focus on point 2.

2. Negative values are not acceptable so we get a limited number of solutions because one or the other variable becomes negative. Then this limits the number of solutions we get and sometimes the numbers are so used that we get only a handful of solutions or only 1 acceptable solution.

Now consider what happens when the two terms are subtracted e.g.
3x – 5y = 42
x = 14, y = 0
x = 19, y = 3
x = 24, y = 6
...
Since the terms are subtracted, when one term increases, the other will also increase to keep the difference between them the same.

Now coming to the original question. This is what comes to my mind when I see:
4320y-4950x=4860
It must have one integer positive solution since the adults and children did go and buy tickets. Now, the moment we get one set of solution, we will get infinite more by increasing values of y and x appropriately. Since there is no upper limit on how many people visited (there is a lower limit of 10 but no upper limit), there will be infinite acceptable solutions using this statement alone.

The moment we set one value for x, we will get a unique value for y and hence both statements together will be sufficient to give a unique value.

sahilvermani, hD13
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Re: A group consisting of several families visited an amusement park where [#permalink]
ccooley wrote:
ajdse22 wrote:
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?

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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Posted from my mobile device
Let me try and summarise, if no upper /lower limit on the integers is give and two integers are related in a manner that causes one integer to increase/decrease hand in hand with the other integer increasing/decreasing then you don't have a conclusion. If one increases and other decreases or vice-versa, in such cases we can conclude.
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Re: A group consisting of several families visited an amusement park where [#permalink]
Bunuel wrote:
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.

Hi experts,
I have in my notes on this problem that you can use the fundamental algebra principle to disprove statement B. In that you can add the same number to both sides of the equation and the equation will still be true.

(4800*.9*c)-(5500*.9*a)=4860
a=6 and c=8

(4800*.9*c)+(4800*55)=4860+(5500*.9*a)+(5500*48)
c=63 and a=54

I am a bit confused by this above.

I would be so appreciative if an expert can confirm if this is correct/further explain this. I am not sure where I wrote this down from.

Thank you for your time and help.
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Re: A group consisting of several families visited an amusement park where [#permalink]
woohoo921 wrote:
Bunuel wrote:
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.

Hi experts,
I have in my notes on this problem that you can use the fundamental algebra principle to disprove statement B. In that you can add the same number to both sides of the equation and the equation will still be true.

(4800*.9*c)-(5500*.9*a)=4860
a=6 and c=8

(4800*.9*c)+(4800*55)=4860+(5500*.9*a)+(5500*48)
c=63 and a=54

I am a bit confused by this above.

I would be so appreciative if an expert can confirm if this is correct/further explain this. I am not sure where I wrote this down from.

Thank you for your time and help.

Hello woohoo921,

You make a great point which may be helpful to others as well. So, let me explain this by analyzing what you did in this solution.

EQUATION AND FIRST SOLUTION (a = 6, c = 8)

From statement 2, we get the equation: (4800 × 0.9 × c) – (5500 x 0.9 × a) = 4860--- (1)

Before we start trying to solve this, we will simplify the equation. Dividing by 0.9 throughout, we get (4800 × c) – (5500 x a) = 5400 --- (2)

(Though we can simplify further by dividing by 100 throughout, I will not do that here since your solution beyond this point uses (2) only.)

After some trial & error, we will get a = 6 and c = 8 just as you did. But now the question is how do we get all other possible values of a and c? This is precisely where you got confused. Let me explain the next steps:

OTHER SOLUTIONS (a = 54, c = 63)

The term you added on both sides of equation (2) is nothing but the LCM of the coefficients of a and c. Since adding the same number on both sides of the equation does not change it, we were free to do this manipulation to simplify calculations.

In our case, coefficient of a = 5500 and coefficient of c = 4800
So, their LCM = 4800 x 55 (can also be written as 5500 × 48). Adding this on both sides, we get:

• (4800 × c) – (5500 x a) + 4800 x 55 = 5400 + 4800 x 55
• Rewrite 4800 x 55 on the LHS as 5500 x 48 to get:
• (4800 × c) – (5500 x a) + 5500 x 48 = 5400 + 4800 x 55
• Next, just take the 4800 x 55 from the RHS to the LHS to get:
• (4800 × c) - 4800 x 55 – (5500 x a) + 5500 x 48 = 5400.
• Finally, use the distributive law [a x (b-c) = ab - ac] to get:
• So, 4800(c - 55) – 5500(a - 48) = 5400
• This can be represented as 4800x – 5500y = 5400 ----(3)
• Here, x = c – 55 and y = a – 48.

NOTE: Equation (3): 4800x – 5500y = 5400 is identical to equation (2): 4800c – 5500a = 5400!
So, the solutions for equation (3) would be same as the solutions for equation (2).

• That is, x = 8 and y = 6.
• So, x = c – 55 = 8 and hence, c = 8 + 55 = 63
• Similarly, y = a – 48 = 6 and hence, a = 6 + 48 = 54

And that’s it! You saw how we used fundamental algebraic principles to get to the next possible solutions from an apparently unsolvable equation!

Hope this helps!

Best Regards,
Ashish Arora
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