How do you solve ratio problems like these?

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To solve the ratio problems, the first question yields a ratio of A:B:C as 3:10:2 after determining the common denominator. For the second question, the remainder of B and C, which is 5/8, needs to be split in the ratio of 1:2, requiring adjustments to the fractions. The third question involves combining the ratios A:B=2:5 and B:C=10:11, suggesting that A:B must be scaled to fit with B:C. The key is to manipulate the ratios so they can be expressed in a consistent format. Understanding how to adjust and combine these ratios is crucial for finding the correct A:B:C values.
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Homework Statement



In each of the following the properties of a compound are given. In each case find A:B:C

1. 1/5 of A, 2/3 of B and the remainder of C.

2. 3/8 of A with B and C in the ratio 1:2.

3. A, B and C are mixed according to the ratios A:B= 2:5 and B:C=10:11.

Homework Equations





The Attempt at a Solution



I could only solve the 1st one:

The common denominator is 15 so 15/3*2=10 and 15/5*1=3. That means A and B are in the ratio 3:10.

15/15-13/15=2/15 so A:B:C are in the ratio of 10:3:2

Couldn't solve the other two. Please, help. Thanks.
 
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Ok in Q2 you know that the remainder (B and C) make up 5/8. Can you show how to split 5/8 in the ratio 1:2?
 
Regarding Question 1:

I think you got the answer right but in the wrong order.

A:B:C

\frac{1}{5}:\frac{2}{3}:\frac{x}{y}

(\frac{3}{15}:\frac{10}{15}:\frac{2}{15})*15

3:10:2 == A:B:C

Regarding Question 2:
You need the provided fraction to become large enough that it's remainder is evenly divisible into three parts.

Regarding Question 3:
A:B:C is what you're trying to build, so look at the ratios A:B and B:C as puzzle pieces.
A:B is too small to fit in with B:C, so you simply need to make A:B larger for it to be congruent with B:C
 
Last edited:
Thank you for the answers.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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