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Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system

$A+4B=C, 4A+3B=D, 5A+6B=z$

to get

$z= \frac{9C+14D}{13} \rightarrow \frac{5A}{6B} = \frac{9C}{14D}?$ If not what am I doing wrong? What is the right method?

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5 Answers 5

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Each unit of $C$ contains $\frac15$ unit of $A$ and $\frac45$ unit of $B$. Each unit of $D$ contains $\frac47$ unit of $A$ and $\frac37$ unit of $B$. Thus, $c$ units of $C$ and $d$ units of $D$ contain $\frac{c}5+\frac{4d}7=\frac{7c+20d}{35}$ units of $A$ and $\frac{4c}5+\frac{3d}7=\frac{28c+15d}{35}$ units of $B$. (Sanity check: these two quantities do in fact sum to $c+d$.)

You want these two quantities to be in the proportion $5:6$, so you want

$$\frac{7c+20d}{35}=\frac56\cdot\frac{28c+15d}{35}\;,$$

or $6(7c+20d)=5(28c+15d)$. Simplifying yields $45d=98c$, or $\dfrac{c}d=\dfrac{45}{98}$: the correct proportion of $C$ to $D$ is $45:98$.


The calculations below resulted from a misreading of the problem; they answer a different question from the one actually asked, but I’ll leave them up, since someone may at some point find them useful.

From $1$ unit of $A$ and $4$ units of $B$ you get $5$ units of $C$. From $4$ units of $A$ and $3$ units of $B$ you get $7$ units of $D$. If you scale that second combination down by a factor of $\frac67$, you find that $\frac67\cdot4=\frac{24}7$ units of $A$ and $\frac67\cdot3=\frac{18}7$ units of $B$ will give you $\frac67\cdot7=6$ units of $D$. These data are summarized in first four rows of the table below.

$$\begin{array}{cccc|l} A&B&C&D\\ \hline 1&4&5&-&*\\ 4&3&-&7\\ 24/7&18/7&-&6&*\\ \hline 31/7&46/7&5&6&** \end{array}$$

Now add the starred rows to get the bottom row: $\frac{31}7$ units of $A$ and $\frac{46}7$ units of $B$ give you $5$ units of $C$ and $6$ of $D$. That’s $11$ units altogether, so for a single unit of a $5:6$ $CD$ mixture divide the quantities of $A$ and $B$ by $11$: you need $\frac{31}{11\cdot7}=\frac{31}{77}\approx0.4026$ units of $A$ and $\frac{46}{11\cdot7}=\frac{46}{77}\approx0.5974$ units of $B$.

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    $\begingroup$ I think that the OP is asking for what ratio of C to D will give a 5/6 ratio of A to B, not the other way around (from what I see, you worked from a 5/6 ratio of C to D to find the number of units of A and B needed? I like your approach, with the grid and such...but... $\endgroup$
    – amWhy
    Commented Nov 5, 2012 at 18:06
  • $\begingroup$ @amWhy: You’re right: I misread it. $\endgroup$ Commented Nov 5, 2012 at 18:08
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Let the amount of A and B in C be $m$ and $n$ respectively. Let the amount of A and B in D be $p$ and $q$ respectively. Then $\frac mn=\frac 14$ and $\frac pq=\frac 43$.

Let the amount of C and D in the mixture of C and D be $c$ and $d$ respectively. Let the ratio of C to D be $r$. Then $\frac{m+n}{p+q}=\frac cd=r$ and $\frac{m+p}{n+q}=\frac 56$.

Now $m=\frac n4$ and $p=\frac{4q}{3}$. Hence $\frac{5n}{4}=\frac{7qr}{3}$ and $\frac n4+\frac{4q}{3}=\frac{5n+5q}{6}$. Hence $r=\frac{15n}{28q}$ and $3n+16q=10n+10q$. Hence $7n=6q$ and $r=\frac{45}{98}$.

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A mixture of $c$ units of paint C and $d$ units of paint D will contain $$a={1\over 5}c+{4\over 7}d {\rm \ \ units\ of\ paint\ A}$$ and $$b={4\over 5}c+{3\over 7}d {\rm \ \ units\ of\ paint\ B}$$ You want to have $a : b = 5 : 6$, or $6a=5b$. Substituting for $a$ and $b$ and simplifying, you get that $98c=45d$, or that $c : d = 45 : 98$, which gives the right ratio of paint C and paint D.

To check the answer, you can mix 45 units of paint C and $98$ units of paint D. You will then have a mixture which contains $${1\over 5}45+{4\over 7}98=9+56=65 {\rm\ \ units\ of\ paint\ A}$$ and $${4\over 5}45+{3\over 7}98=36+42=78 {\rm\ \ units\ of\ paint\ B}$$ Since $65 : 78 = (5\cdot 13) : (6 \cdot 13) = 5 : 6$, the ratio between paint A and paint B in the mixture it the desired one.

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When you mix up a m/n ratio of A/B to get C, then $m+n$ units of C will contain m units of A and n units of B, or $(m+n)C = mA+nB$. I prefer to express this as $C = \frac{m}{m+n}A + \frac{n}{m+n} B$. (Note that the sum of the fractions is exactly 1, so I can work with just the fraction of A, knowing that the fraction of B is just one minus A's fraction.)

This gives the two equations $C = \frac{1}{5}A + \frac{4}{5} B$, $D = \frac{4}{7}A + \frac{3}{7} B$

Now I mix up paint E with a fraction $x$ of C and fraction $1-x$ of D (that would be a ratio $\frac{x}{1-x}$ of C/D). This gives me $E = x C + (1-x)D$, and we also want $E=\frac{5}{11}A + \frac{6}{11} B$. Equating the two formula and and comparing the quantities of A, B will give me a formula for $x$. In fact, we need only look at the fraction of A in E: This gives the formula $$x \frac{1}{5} + (1-x)\frac{4}{7} = \frac{5}{11}$$

Working through the tedious details gives $x = \frac{45}{143}$.

To validate the answer, we take $$\frac{45}{143} C + \frac{98}{143}D = \frac{9(A+4B)}{143} C + \frac{14(4A+3B)}{143}D = \frac{5}{11} A + \frac{6}{11}B$$

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  • $\begingroup$ @Danielle - if you substitute the first equation for E into "E" of the second equation E = ..., you'll have an equation $xC + (1-x)D = \frac {5}{11} A + \frac {6}{11}B$. Then substitute the exressions C = .. and D = .. into the left side of that equation, and the A's and B's will cancel out, leaving only an equation in x, which will be your ratio of C to D. $\endgroup$
    – amWhy
    Commented Nov 5, 2012 at 17:51
  • $\begingroup$ Personally, I find the translation of problems into symbols to be rather error prone. $\endgroup$
    – copper.hat
    Commented Nov 5, 2012 at 18:13
  • $\begingroup$ No criticism taken at all. It was more a comment for the OP to realize that even with lots of experience, we can still can trip on simple problems. $\endgroup$
    – copper.hat
    Commented Nov 5, 2012 at 18:28
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    $\begingroup$ You might want to clarify in your answer that x represents the percentage of C needed, and 1 - x the percentage of D needed, not the ratio of C to D which is 45/98. I was looking at your x originally as representing the ratio C/D. $\endgroup$
    – amWhy
    Commented Nov 5, 2012 at 18:36
  • $\begingroup$ Oops, thanks for catching this, I think I have fixed it. $\endgroup$
    – copper.hat
    Commented Nov 5, 2012 at 18:40
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The mixing ratios give $$ \begin{bmatrix} \frac15&\frac45\\ \frac47&\frac37 \end{bmatrix} \begin{bmatrix}A\\B\end{bmatrix}=\begin{bmatrix}C\\D\end{bmatrix} $$ That is, $1$ can of $C$ is $\frac15$ can of $A$ and $\frac45$ can of $B$, $1$ can of $D$ is $\frac47$ can of $A$ and $\frac37$ can of $B$.

Because $$ \begin{align} \begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix} \begin{bmatrix} \frac15&\frac45\\ \frac47&\frac37 \end{bmatrix}^{-1} &=\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix} \begin{bmatrix} -\frac{15}{13}&\frac{28}{13}\\ \frac{20}{13}&-\frac7{13} \end{bmatrix}\\[6pt] &=\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix} \end{align} $$ we get $$ \begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix} \begin{bmatrix}A\\B\end{bmatrix} =\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix} \begin{bmatrix}C\\D\end{bmatrix} $$ Therefore, a $45:98$ mixture of $C$ and $D$ requires a $5:6$ ratio of $A$ and $B$.

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