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A group consisting of several families visited an amusement park where

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

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

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New post 04 Sep 2016, 10:48
When the original statement says that each apid an admission fee that was less than 10% of the original admissions fee, why are we assuming that the original admissions fee is not 5,500 but we're saying there is a different "standard" for adults and children?

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

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New post 24 Nov 2016, 22:43
Can any one please explain how 0.9 in equation ".9*5500 X" came into picture, please!!
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Re: A group consisting of several families visited an amusement park where [#permalink]

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anairamitch1804 wrote:
Can any one please explain how 0.9 in equation ".9*5500 X" came into picture, please!!

The question stem gives us the information that the regular admission fee for each adult is 5500 Yen; also note that each adult (and each child) receives a discount of 10% on his or her admission fee because the number of adults and children was at least 10. Therefore, the amount spent by the adults is x*0,9*5500.

In my opinion the discount-information is not really necessary to answer the question, maybe only introduced to confuse the reader...

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

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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.


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

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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.

Answer: C



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

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New post 12 Dec 2016, 08:27
ajdse22 wrote:
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.

Answer: C



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?



Anyone, please reply. I also struggle with the same problem.

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

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

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New post 12 Dec 2016, 23:09
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.




Thanks for your reply on the question. What I understood from your explanation is -- when there is addition in the equation then, in that case, the values would be limited, we have to figure out if those values are unique or not, but when there is subtraction (between the variable) in the equation then it will have infinite cases, so we should not waste the time in looking for possible values.

For instance--
For the question in the discussion, consider the number of children is c and the number of adults is a. If I just consider statement 2, it brings me to the below equation.

90/100(4800c-5500a)=4860
48c-55a=54
Now we have to see if we can find out the value of a and c by the above equation given that a and c are integers.

This will not give unique solution


Your first example
: Let's say that there are a apples and b oranges

0.15a +0.23b=2.9

This can give unique solution.

Please correct me if I am wrong.


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

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New post 30 Jan 2017, 11:56
Once again i did not read the question stem well enough. I read 10, not at least, and solved. I really need to slow down otherwise I will never break the 48/49 Q level. Good question
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Re: A group consisting of several families visited an amusement park where [#permalink]

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New post 30 Jan 2017, 20:21
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.


According to (2) - C * 4950 - A * 4320 = 4860. I could only come up with C = 6 and A = 8. There is no other way this statement holds true, atleast according to me. (C = no. of children and A = no. of adults)

Therefore why isn't the answer (B)?

Experts please advice.

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

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New post 31 Jan 2017, 10:45
manhasnoname 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.


According to (2) - C * 4950 - A * 4320 = 4860. I could only come up with C = 6 and A = 8. There is no other way this statement holds true, atleast according to me. (C = no. of children and A = no. of adults)

Therefore why isn't the answer (B)?

Experts please advice.


It has many more solutions. Check here: https://www.wolframalpha.com/input/?i=4 ... r+integers
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Re: A group consisting of several families visited an amusement park where [#permalink]

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New post 01 Feb 2017, 14:11
Bunuel wrote:
manhasnoname 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.


According to (2) - C * 4950 - A * 4320 = 4860. I could only come up with C = 6 and A = 8. There is no other way this statement holds true, atleast according to me. (C = no. of children and A = no. of adults)

Therefore why isn't the answer (B)?

Experts please advice.


It has many more solutions. Check here: https://www.wolframalpha.com/input/?i=4 ... r+integers


Hi Bunuel,

How can we be quite sure that there are more than one solution? Gmat makes some tricks on such possible solution cases?

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

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New post 02 Mar 2017, 20:51
Made a silly mistake. I assumed total number of people in the group is 10. bummer !!!

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

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New post 11 Jul 2017, 05:57
How can you divide 29700 by 4950 and get 6 in say 10 secs? Just by looking at the numbers it's obvious they don't want you to do calculations. It seems to me that most answers here are ppl using calculators for 10 mins and then post this stuff..

Is there the elegant solution?

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Re: A group consisting of several families visited an amusement park where   [#permalink] 11 Jul 2017, 05:57
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