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Substance A has butter to sugar ratio of 8:1 and substance B has butter to sugar ratio of 175:1. The whole of A is combined with part of B to get substance C of Butter to sugar ratio of 30:1. What ratio of substance B was used?

Kwame kwofie
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  • Welcome to Math SE! Please add some details to your question about what you've tried and where you're stuck. – DMcMor Apr 22 '19 at 12:46
  • Where's the hard part ? I tried to become a pharmacy tech and had to figure these things out before. –  Apr 22 '19 at 19:13
  • If the ratios are by weight, we need to know the weights of whole A and whole B in order to get an answer. – KY Tang Apr 23 '19 at 02:29
  • did the math wrong, but you actually don't have to know how much of A and B, just their relative ratio. alligation is a method useful for this. –  Apr 23 '19 at 13:24

2 Answers2

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I'll look at doing alligation:

$$\begin{array}{c|c|c}\hspace{-1px}11\;\frac{1}{9}&&2\frac{897}{1364}\\\hline&3\frac{7}{31}&\\\hline \frac{25}{44}&&7\frac{247}{279}\end{array}$$

all numbers in the left columns are percentages for the mixtures (8:1 has 9 parts total, 175:1 has 176 parts total). middle column has the wanted ratio ( 30:1 has 31 parts). the last column has parts in the final mixture, turning them to full integers we get them equivalent to 32625 and 96800 reducing to 1305 and 3872 which then works to produce 5177 parts of $\frac{1}{31}$ which is the ratio $30:1$ for the two ingredients.

Probably not the way your teacher wants it done, but it works via subtraction along diagonals, and converting to same denominator fractions on the right then reduction of numerators. Thankfully, I had a calculator to check my work.

technical notes: Assumes percentages are in same units or equivalent units. Doesn't take into account contraction of by-product mixture.

Community
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A has $8x$ butter to $1x$ sugar.

B has $175y$ butter to $1y$ sugar.

C has $30z$ butter to $1z$ sugar.

So

\begin{align} 8x + 175y &= 30z \\ 1x + 1y &= 1z \\ \hline 8x + 175y &= 30z \\ 8x + 8y &= 8z \\ \hline 167y &= 22z \\ y &= \dfrac{2}{17}z \\ x &= \dfrac{15}{17}z \end{align}

We could say

$$\dfrac{15}{17}A + \dfrac{2}{17}B = C$$

To use the whole of $A$, we need to multiply through by $\dfrac{17}{15}$.

$$1A + \dfrac{2}{15}B = \dfrac{17}{15}C$$

So $\dfrac{2}{15}$ of substance $B$ was used.

Steven Alexis Gregory
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