MHB Ratio problem: The ratio of Liquid X to Liquid Y in a mixture is 3 to 7

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The ratio of Liquid X to Liquid Y in a mixture is established as 3 to 7. Given that the amount of Liquid Y is 161, the calculation for Liquid X can be derived using the ratio. By setting up the equation X = (3/7) * 161, Liquid X is determined to be approximately 69. The total amount in the mixture is not necessary for finding Liquid X. Thus, the solution focuses solely on the ratio and the known quantity of Liquid Y.
JohnDoe01
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The ratio of Liquid X to Liquid Y in a mixture is 3 to 7. If the amount of Liquid Y in a mixture is 161, then the amount of Liquid x in the mixture is equal to?
 
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Let $m$ be the total amount in the mixture. Then,

$$Y=\frac{7m}{10}=161$$

From that, can you find the amount for $X$, i.e. $\frac{3}{10}$ of the mixture?
 
Personally I wouldn't worry about the total amount at all. You are told that "ratio of Liquid X to Liquid Y in a mixture is 3 to 7" so [math]\frac{X}{Y}= \frac{3}{7}[/math]. You are told that "the amount of Liquid Y in a mixture is 161" so [math]\frac{X}{161}= \frac{3}{7}[/math]. [math]X= \frac{3}{7}(161)[/math].
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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