Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Please help!?

A farm has pigs, cows and chickens. The number of cows is one more than the number of pigs. The number of chickens is three times the number of cows. If the farm has between 201 and 210 animals, what is the maximum number of chickens on the farm?

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  • 1 month ago

    cows+cows-1+cows*3=201>210

    excel

    cows=41, pigs =40, ckn=123= 204

    or

    cows=42, pigs=41, ckns=126 = 209

    max ckn =126

  • 1 month ago

    A farm has pigs, cows and chickens. 

    The number of cows is one more than the number of pigs. 

    The number of chickens is three times the number of cows. 

    If the farm has between 201 and 210 animals, what is the maximum number of chickens on the farm?

  • Ash
    Lv 7
    1 month ago

    Cows = Pigs + 1

    Pigs = Cows - 1 ......(1)

    Chicken = 3 Cows

    Cows = ⅓Chicken .....(2)

    plug (2) in (1)

    Pigs = ⅓Chicken - 1 ......(3)

    201≤ Total animals ≤ 210

    201≤ Cows + Pigs + Chicken ≤ 210

    201≤ ⅓Chicken + ⅓Chicken - 1 + Chicken ≤ 210

    201≤ (5/3)Chicken - 1 ≤ 210

    Add 1

    201+1≤ (5/3)Chicken - 1+1 ≤ 210+1

    202 ≤ (5/3)Chicken ≤ 211

    Multiply by ⅗

    202*⅗ ≤ (5/3)Chicken*⅗ ≤ 211*⅗

    121.2 ≤ Chicken*⅗ ≤ 126.6

    The max number of chicken are 126

    Check:

    Chicken = 126

    Cows =  ⅓Chicken = ⅓(126) = 42

    Pigs = Cows - 1 = 42 - 1 = 41

    Total 126+42+41 = 209 !!!

  • 1 month ago

    Let p = number of pigs

    Let w = number of cows

    Let n = number of chickens

    The number of cows is 1 more than the number of pigs:

    w = p + 1

    The number of chickens is 3 times the number of cows:

    n = 3w

    There are between 201 and 210 total animals:

    p + w + n = x where 201 ≤ x ≤ 210

    We know that the there variables must all be integers.  So let's do some substitutions to see what we end up with and compare the result to the possible values.

    Let's leave the equation in terms of x as-is for now.  I'll start with substituting the second equation into it and simplifying:

    p + w + n = x

    p + w + 3w = x

    p + 4w = x

    Now substitute the first equation into this and simplify:

    p + 4(p + 1) = x

    p + 4p + 4 = x

    5p + 4 = x

    Solve this for p:

    5p = x - 4

    p = (x - 4) / 5

    So in order for p to be an integer, x must be 4 more than a multiple of 5 (or one less than a multiple of 5).

    So that gives us possilbe values of x of:

    204 and 209

    Let's start with substituting those values into x and solving for p, n, and w and see if any of them don't make sense.

    p = (x - 4) / 5 and p = (x - 4) / 5

    p = (204 - 4) / 5 and p = (209 - 4) / 5

    p = 200 / 5 and p = 205 / 5

    p = 40 and p = 41

    w = p + 1

    w = 40 + 1 and w = 41 + 1

    w = 41 and w = 42

    n = 3w

    n = 3w and n = 3w

    n = 3(41) and n = 3(42)

    n = 123 and n = 126

    If we then add up the three values in each possible solution we get:

    40 + 41 + 123 and 41 + 42 + 126

    204 and 209

    Both of these sets are possible solutions.

    We are asked to find the maximum possible number of chickens so the answer is the larger of the two n's:

    The maximum number of chickens that satisfies this criteria is 126.

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  • fcas80
    Lv 7
    1 month ago

    Let K=cows, C=chickens, P=pigs.

    I used trial and error and got C=126 when K=42, P=41.

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