You are here:

Advertisement

If A:B = 2:3 , B:C = 4:5 , C:D = 6:7. Find A:B:C:D.

Please answer it quickly sir

If we look at A:B and B:C, we see in the 1st, B is 3 and in the 2nd, B is 4.

These both divide evenly into 12, so change both to make B 12.

For A:B, since 3*4 = 12, we can take 2*4 and get 8, giving A:B as 8:12.

For B:C, since 4*3 = 12, we can take 5*3 and get 15, which gives B:C as 12:15.

Thus, A:B:C is 8:12:15.

Now if we look at the numbers used for C, they are 15 and 6.

The smallest number that is divisible by 15 and 6 is 30.

This means that 8:12:15 needs to be multiplied by 30/15 = 2, giving 16:24:30.

This means that since 30/6 = 5, multiply the C:D ratio by 5 and get 30:35.

We can now combine the final answer into the group we have so far since they both have a 30 for C. This gives A:B:C:D as 16:24:30:35.

- Add to this Answer
- Ask a Question

Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |

Comment | He should become Volunteer of the month. |

I can answer almost anything that is sent in. If I can't, I'll let you know, but I don't expect that to happen much.

I have known about number theory since the mid 80's. I have answered over 250 questions on Number Theory with this software. Altogether, I have answered over 8,500 questions in mathematics.**Publications**

You're looking at it ... I've answered over 8,500 quesitons in mathematics right here.
**Education/Credentials**

My credentials are an MS in Mathematics at Oregon State in 1986;
I received a BS in Mathematics at the same place in 1984.**Awards and Honors**

I graduated with honors in Mathematics when getting my BS degree and my MS degree.
**Past/Present Clients**

I have assisted many students in mathematics at OSU.
Perhaps I have assisted one of you're friends in math on a computer somewhere else,
but you don't even know... That would be late last night, perhaps with thousands of miles between us ...
Then again, if you're in Washington, so am I ...