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A group consisting of several families visited an amusement

This topic has 1 expert reply and 0 member replies
jjjinapinch Senior | Next Rank: 100 Posts Default Avatar
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24 Jul 2017
Posted:
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A group consisting of several families visited an amusement

Post Tue Aug 08, 2017 7:41 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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.

    Official Guide question
    Answer: C

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    Post Thu Aug 10, 2017 11:09 am
    Hi jjjinapinch,

    We're told that admission fees at a park were ¥5,500 for each adult and ¥4,800 for each child and that because there were at least 10 people in the group, each paid an admission fee that was 10% less that the regular admission fee (meaning that each adult fee was 4950 yen and each child fee was 4320 yen). We're asked for the number of children in the group. While this question is wordy, it's built around a series of standard Algebra concepts.

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

    With the information in Fact 1, we can determine the number of adults in the group (using the following equation):
    4950(A) = 29,700
    A = 29,700/4950 = 6 adults
    However, we do not know the number of children in the group.
    Fact 1 is 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.

    With the information in Fact 2, we can create the following equation):
    4320(C) = 4950(A) + 4860
    While this equation is 'thick', it's still two variables and just one equation; since the numbers involved don't include anything too strange (weird decimals or primes, for example) there are likely to be multiple solutions.
    Fact 2 is INSUFFICIENT

    Combined, we know:
    A = 6
    4320(C) = 4950(A) + 4860
    With the value of A, we can 'plug in' and solve for C.
    Combined, SUFFICIENT

    Final Answer: C

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