Results 1 to 4 of 4

Math Help - Proportion Problem Help

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    13

    Proportion Problem Help

    I want to know how to solve this, not just the answer. Thanks.

    The ratio of rubies to emeralds was 3 to 1, and the ratio of emeralds to diamonds was 2 to 1. If there were 18 rubies, emeralds, and diamonds in all, how many of each were there?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Ruby:Emerald are in the following ratio:
    3:1
    Can you appreciate that this is equivalent to:
    6:2?

    So If I have R:E:D, I could write it all in one ratio as:
    6:2:1
    Working backwards, for every one diamond, there are two emeralds. And for every emerald, there are three times as many rubies. This is the core statement.

    Now, in the ratio above there are 9 gems. We could say that for every 9 gems, \frac{6}{9} are rubies. So, if that is true, how many rubies would there be in a collection of 18 gems? That is to say, for every 18 gems, \frac{?}{18} would be rubies? Use equivalent fractions to work it out.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member lanierms's Avatar
    Joined
    May 2010
    From
    Seoul, South Korea
    Posts
    50
    You can also use a constant k to figure it out quite easily.

    Solution:

    Ruby:Emerald = 3:1.

    So, Ruby = 3k, Emerald = k.

    Emerald: Diamond = 2:1 = k: Diamond.

    \therefore Diamond=\frac{1}{2}k

    So, there are 3k rubies, k emeralds and half k diamonds.

    Now, if we add all of them,

    3k+k+\frac{1}{2}k=18

    \therefore k=4.

    So, there are 12 rubies, 4 emeralds and 2 diamonds.

    The solution posted by Quacky is great, but I don't think problems will always be simple like this one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2011
    Posts
    13
    Quote Originally Posted by lanierms View Post
    You can also use a constant k to figure it out quite easily.

    Solution:

    Ruby:Emerald = 3:1.

    So, Ruby = 3k, Emerald = k.

    Emerald: Diamond = 2:1 = k: Diamond.

    \therefore Diamond=\frac{1}{2}k

    So, there are 3k rubies, k emeralds and half k diamonds.

    Now, if we add all of them,

    3k+k+\frac{1}{2}k=18

    \therefore k=4.

    So, there are 12 rubies, 4 emeralds and 2 diamonds.

    The solution posted by Quacky is great, but I don't think problems will always be simple like this one.
    Ah, very simple. Thank you both.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Inverse proportion/Direct proportion question
    Posted in the Statistics Forum
    Replies: 3
    Last Post: September 26th 2011, 07:22 AM
  2. Proportion problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 15th 2011, 06:40 PM
  3. proportion problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 15th 2010, 11:50 PM
  4. Replies: 3
    Last Post: July 6th 2009, 07:25 PM
  5. proportion problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 15th 2008, 06:55 PM

Search Tags


/mathhelpforum @mathhelpforum