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Thread: serious problem help needed: there are 400 llamas and alpacas in total....

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    serious problem help needed: there are 400 llamas and alpacas in total....

    there are 400 llamas and alpacas in total.
    3/4 of llamas were sold at auction.
    1/3 of alpacas were sod at auction.
    there are 125 llamas and alpacas left in total.
    how many llamas were sold at auction.

    please post the answer and how you worked it out.

  2. #2
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    How many llamas were there initially? Call it something. L?
    How many alpacas were there initially? Call it something. A?

    That was about half of the problem. A very important place to start.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    I don't know how many of each there was initially this Is the problem my daughter brought home from school and I've got no idea how to help her out. Please help

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    Quote Originally Posted by lajohno View Post
    there are 400 llamas and alpacas in total.
    3/4 of llamas were sold at auction.
    1/3 of alpacas were sold at auction.
    there are 125 llamas and alpacas left in total.
    how many llamas were sold at auction.

    please post the answer and how you worked it out.
    You posted this under arithmetic, suggesting that algebra may not be an appropriate method. Can you confirm that?

    If you read the "Read before posting" note at the top of the forum, you know that we don't just give answers; we want to help you with whatever work you have done. We need to see work in order to see what method you are using, and where you need help in it.

    If you can't use algebra, and it is for a class, then at least give us a description of what methods you have seen. There are some methods I can imagine, but anything other than trial and error will ultimately be algebra in disguise.

  5. #5
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by lajohno View Post
    I don't know how many of each there was initially this Is the problem my daughter brought home from school and I've got no idea how to help her out. Please help
    Unfortunately, we have learned from hard experience that attempting to converse through a "translator" who "doesn't speak the language" is generally doomed. So please have your daughter reply (maybe with you doing the typing), explaining how she thinks her teacher expects her to do this, what your daughters thoughts are, what she has tried, and how far she has gotten.

    Quote Originally Posted by lajohno View Post
    please post the answer and how you worked it out.
    Um... that's kinda cheating, and is generally against the rules. Sorry, but this isn't a "cheetz" site.

    Quote Originally Posted by lajohno View Post
    Originally, there were 400 llamas and alpacas in total.
    3/4 of llamas were sold at auction.
    1/3 of alpacas were sold at auction.
    Then there were 125 llamas and alpacas left in total.
    How many llamas were sold at auction?
    a. You know that the original herd contained four hundred animals.

    b. At auction, 3/4 of the llamas were sold, as were 1/3 of the alpacas.

    c. At action, 125 animals were sold.

    d. Make the fractions comparable: nine-twelfths of the llamas sold, and four-twelfths of the alpacas sold, for a total of 275 animals.

    e. This means that three-twelfths of the llamas remained, as did eight-twelfths of the alpacas, for a total of 125 animals.

    f. Let's get rid of denominators by multiplying everything by 12: three llama-units remain, and eight alpaca-units remain, for a total of 1500 animals.

    g. Let's make the original situation comparable by multiplying everything by 3: three llama-units and three alpaca-units (for the original herd) totalled 1200 animals.

    h. What is the difference between the herds in (f) and (g)? Can you use this to figure out how many animals are in five alpaca units? (Hint: Subtract.)

    i. Then how many animals are in one alpaca-unit?

    j. Then how many animals are in three alpaca-units?

    k. Then how many animals are in three llama units?

    l. Dividing by three to get back to the original herd, how many llamas were there originally? And how many sold?

    If your daughter gets stuck in following these steps, please have her reply with a clear listing of her efforts, specifying where she is getting stuck. Thank you!

  6. #6
    Elite Member mmm4444bot's Avatar
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    Here's an arithmetic approach.

    3/4ths of the llamas were sold. This means (at least) two things:

    1/4th of the llamas remain, and the original number of llamas must be a multiple of 4:

    4, 8, 12, 16, 20, 24,


    1/3rd of the alpacas were sold. This means (at least) two things:

    2/3rds of the alpacas remain, and the original number of alpacas must be a multiple of 3:

    3, 6, 9, 12, 15, 18,


    We can discover patterns, by writing out specific cases and comparing the resulting numbers. For example, we know that the original number of alpacas is a multiple of 3, so let's see what happens if there were originally three alpacas (i.e., the first multiple of 3).

    3 alpacas means 397 llamas because there were 400 animals originally. But 397 is not a multiple of 4 because all multiples of 4 are even numbers. (Dividing by 4 is the same as dividing by 2 twice.)

    Okay, next case: 6 alpacas means 394 llamas. But 394 is not a multiple of 4 because 3944 has a remainder of 2.

    Okay, next case: 9 alpacas means 391 llamas. 391 is odd, so it's not a multiple of 4.

    Okay, next case: 12 alpacas means 388 llamas. Success! 388 is the 97th multiple of 4 because 3884 is 97 with no remainder.

    Therefore, after the sale: 1/4th of 388 llamas (which is 97 animals) and 2/3rds of 12 alpacas (which is 8 animals) remain.

    8 + 97 = 105

    We need the remaining animals to total 125, so we keep going.

    Next case: 15 alpacas means 385 llamas. 385 is odd, so it's not a multiple of 4.

    Next case: 18 alpacas means 382 llamas. But 382 is not a multiple of 4 because 3824 has a remainder of 2.

    If you have been organized, you might notice a pattern in the results, so far.

    Code:
    Alpacas  Llamas  After Sale
    -------  ------  ----------
       3       397
       6       394
       9       391
      12       388   8+97 = 105
      15       385
      18       382
    
    We need the original number of llamas to be even. That only happens when the number of alpacas is even. Therefore, we can skip all the cases where the multiple of 3 (i.e., original alpacas) is odd. Keep going. Be organized.

    Code:
    Alpacas  Llamas  After Sale
    -------  ------  ----------
       6       394
      12       388    8+97 = 105
      18       382
      24       376   16+94 = 110
      30       370
      36       364   24+91 = 115
    
    Two more patterns have emerged.

    1) The number of llamas is a multiple of 4 only when the number of alpacas is a multiple of 12.

    2) Each new multiple of 12 alpacas increases the number of remaining animals by 5.

    I hope that the patterns I have shown help you to answer the exercise.
    Last edited by mmm4444bot; 12-01-2017 at 02:43 AM. Reason: Typo in table 2
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    testing...
    (my post here about 45 min ago disappeared!)
    I'm just an imagination of your figment !

  8. #8
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    Quote Originally Posted by lajohno View Post
    there are 400 llamas and alpacas in total.
    3/4 of llamas were sold at auction.
    1/3 of alpacas were sod at auction.
    there are 125 llamas and alpacas left in total.
    how many llamas were sold at auction.
    x = llamas; so 400-x = alpacas

    total sold = 400 - 125 = 275

    (3/4)x + (1/3)(400-x) = 275

    Solve for x.
    I'm just an imagination of your figment !

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