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Chicken Feed

This topic has 1 expert reply and 1 member reply
krishna kumar Junior | Next Rank: 30 Posts Default Avatar
Joined
11 Jun 2010
Posted:
15 messages

Chicken Feed

Post Sun Dec 12, 2010 2:53 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Can someone solve this please.


    If a farmer sells 75 of his chickens, his stock of feed will last 20 more days than planned,but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought the farmer will be exactly on schedule. How many chickens does the farmer have?

    A. 60

    B. 120

    C. 240

    D. 275

    E. 300


    OA E

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    limestone Master | Next Rank: 500 Posts
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    Post Sun Dec 12, 2010 3:22 am
    Let's say:

    The number of chicken : x
    The number of scheduled days that the farmer's stock will last: a
    And each chick eats 1 bag of food each day ( plug in 1 for the sake of simplicity. We can say "y" be the bags of food each chick eat per day, however, "y" will be simplify in 2 sides when we write down our equation. So why not take "1"?)

    Number of stock those chicken eat per day: x*1 = x
    The farmer's total stock: x*a

    In the first case:
    Number of chicken : x - 75
    Number of days: a + 20
    Then the equation is : (x-75)(a+20) = ax ( The farmer's stock is unchanged), or
    20x - 75a - 1500 = 0, or
    4x - 15a - 300 = 0 (I)

    In the second case:
    Number of chicken: x+100
    Number of days: a - 15
    Then the equation is: (x+100)(a-15) = ax, or
    -15x + 100a - 1500 = 0, or
    -3x + 20a - 300 = 0 (II)

    From I and II:
    4x -15a =-3x + 20a
    7x = 35a
    x = 5a
    Plug in x = 5a into (I):
    20a - 15a - 300 =0
    or 5a = 300 = x

    Thus Pick E.

    _________________
    "There is nothing either good or bad - but thinking makes it so" - Shakespeare.

    Thanked by: krishna kumar
    Post Sun Dec 12, 2010 4:53 am
    krishna kumar wrote:
    Can someone solve this please.


    If a farmer sells 75 of his chickens, his stock of feed will last 20 more days than planned,but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought the farmer will be exactly on schedule. How many chickens does the farmer have?

    A. 60

    B. 120

    C. 240

    D. 275

    E. 300


    OA E
    The number of chickens is inversely proportional to the number of days. As the number of chickens increases, the number of days must decrease, so that the same amount of feed is consumed. The product of the two values must remain constant. Thus, we can write the following equation:

    (number of chickens)*(number of days) = (number of chickens)*(number of days)

    Let d = number of days. Let's plug in the answer choices for the number of chickens.

    Answer choice C: 240 chickens
    Feed for 240 chickens lasts for d days, feed for 240-75=165 chickens lasts for d+20 days.
    240d = 165(d+20)
    d=44 days.
    Thus feed = 240*44 = 10,560.
    100 more chickens = 240+100=340 chickens. 10,560 is not divisible by 340.
    Eliminate C.

    Answer choice D: 275 chickens
    Feed for 275 chickens lasts for d days, feed for 275-75=200 chickens lasts for d+20 days.
    275d = 200(d+20)
    d=53.33 days. Doesn't work.
    Eliminate D.

    Answer choice E: 300 chickens
    Feed for 300 chickens lasts for d days, feed for 300-75=225 chickens lasts for d+20 days.
    300d = 225(d+20)
    d=60 days.
    Thus feed = 300*60 = 18000.
    100 more chickens = 300+100= 400 chickens. 18000/400 = 45 days.
    60-45 = 15 fewer days. Success!

    The correct answer is E.

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    Thanked by: krishna kumar
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