Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

5 Answers

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 $B$.

share|improve this answer
add comment

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 $r=\frac{45}{98}$.

share|improve this answer
add comment

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 $$\frac{45}{143} C + \frac{98}{143}D = \frac{9(A+4B)}{143} C + \frac{14(4A+3B)}{143}D = \frac{5}{11} A + \frac{6}{11}B$$

share|improve this answer
    
@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. –  amWhy Nov 5 '12 at 17:51
    
Personally, I find the translation of problems into symbols to be rather error prone. –  copper.hat Nov 5 '12 at 18:13
    
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. –  copper.hat Nov 5 '12 at 18:28
2  
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. –  amWhy Nov 5 '12 at 18:36
    
Oops, thanks for catching this, I think I have fixed it. –  copper.hat Nov 5 '12 at 18:40
add comment

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 $B$.

share|improve this answer
1  
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... –  amWhy Nov 5 '12 at 18:06
    
@amWhy: You’re right: I misread it. –  Brian M. Scott Nov 5 '12 at 18:08
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.