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.
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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].
 
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