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Algebra Statistics

This topic has 5 expert replies and 1 member reply
datonman Senior | Next Rank: 100 Posts Default Avatar
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Algebra Statistics

Post Wed Nov 11, 2015 11:31 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5

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    Post Wed Nov 11, 2015 11:46 am
    If you start at 56 cents, the combination that you must have is 8 oranges and 2 apples.

    An equation for this is [40(a) + 60(10 - a)]/10 = 56

    a is the number of apples.

    The next step is to solve the following:

    [40(2) + 60(o)]/(o+2) = 52

    Solving for o gives you 3, meaning she has to put back 5.

    There are 'easier' ways to get there, but understand this method first.

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    Post Thu Nov 12, 2015 3:09 am
    Quote:
    At a certain food stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the food stand, and the average (mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
    1
    2
    3
    4
    5
    This is a mixture problem.
    An average of 40 and an average of 60 are being mixed to yield an average of 56 in the first case and of 52 in the second case.
    To determine the ratio of apples to oranges in each case, use ALLIGATION.

    Case 1: average cost = 56.
    Step 1: Plot the 3 averages on a number line, with the average apple cost (40) and the average orange cost (60) on the ends and the average cost of all the fruit (56) in the middle.
    A(40)--------------------56-------O(60)

    Step 2: Calculate the distances between the averages.
    A(40)----------16----------56---4----O(60)

    Step 3: Determine the ratio of apple to oranges.
    The ratio of A to O is the RECIPROCAL of the distances in red.
    A : O = 4:16 = 1:4.

    Since A : O = 1:4 = 2:8, we know that A=2 and O=8, for a total of 10 pieces of fruit.

    Case 2: average cost = 52.
    A(40)----------12----------52---8----O(60)
    A : O = 8:12 = 2:3.
    Since the number of apples isn't changing, A=2 (same as above) and new O=3.

    Thus:
    Old O - new O = 8-3 = 5.

    The correct answer is E.

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    Post Thu Nov 12, 2015 3:15 am
    Quote:
    At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
    Alternate approach:

    The original total cost of the 10 pieces of fruit = 10*56 = 560.
    According to the answers, after 1, 2, 3, 4, or 5 pieces are removed -- so that 9, 8, 7, 6, or 5 pieces remain -- the average cost decreases to 52.
    Since the prices are each a multiple of 10, the new total cost after the oranges are removed must also be a multiple of 10.
    Only the option in red -- implying 5 remaining pieces of fruit -- will yield a new total cost that is a multiple of 10:
    5*52 = 260.

    The correct answer is E.

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    Post Thu Nov 12, 2015 3:29 am
    datonman wrote:
    At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
    An alternate approach is to PLUG IN THE ANSWERS, which represent the number of oranges that must be put back.
    Before any oranges are put back, the total cost of the 10 pieces of fruit = 10*56 = 560.
    When the correct answer choice is plugged in, the average cost will decrease to 52 cents.

    Answer choice D: 4 oranges put back
    Resulting total cost after 4 60-cent oranges are put back = 560 - (4*60) = 320.
    Average price of the remaining 6 pieces of fruit = 320/6 = 53+.
    The average price is too high.
    Eliminate D.

    Since oranges are more expensive than apples, the average price will decrease to 52 cents only if MORE oranges are put back.

    The correct answer is E.

    Answer choice E: 5 oranges put back
    Resulting total cost after 5 60-cent oranges are put back = 560 - (5*60) = 260.
    Average price of the remaining 5 pieces of fruit = 260/5 = 52.
    Success!

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    manik11 Master | Next Rank: 500 Posts Default Avatar
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    Post Thu Nov 12, 2015 7:46 am
    GMATGuruNY wrote:
    This is a mixture problem.
    An average of 40 and an average of 60 are being mixed to yield an average of 56 in the first case and of 52 in the second case.
    To determine the ratio of apples to oranges in each case, use ALLIGATION.
    Mitch,
    Could you please post links to some problems, which can be solved using ALLIGATION method.
    I want to familiarize myself with different situations in which I can use this method.

    Thanks!
    Manik

    Post Thu Nov 12, 2015 4:22 pm
    manik11 wrote:
    GMATGuruNY wrote:
    This is a mixture problem.
    An average of 40 and an average of 60 are being mixed to yield an average of 56 in the first case and of 52 in the second case.
    To determine the ratio of apples to oranges in each case, use ALLIGATION.
    Mitch,
    Could you please post links to some problems, which can be solved using ALLIGATION method.
    I want to familiarize myself with different situations in which I can use this method.

    Thanks!
    Manik
    See here: http://www.beatthegmat.com/seed-mixture-t21603.html
    And here: http://www.beatthegmat.com/managers-and-directors-t38687.html
    And here: http://gmatclub.com/forum/during-a-certain-season-a-team-won-80-percent-of-its-first-144454.html

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