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let us consider following problem :

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are $5 : 3$ and $1 : 2$, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio $4 : 3$. What is the ratio of the composition of the two basic elements in alloy X ?

i would like to know ways of solution of this problem,i can take any numbers ,for example suppose in alloy $x$,elements are $15$ and $9$,and in $B$ $3$ and $6$,if portions are taken from $A$ and $B$ as $4:3$;how can i calculate elements ?clearly in portion $4$,ratio also would be $5$ to $3$,so would it be $5/4$ and $3/4$? or?if it would be same we can take the same procedure for second one,or for second it would be $3/3$ and $6/3$,if we add to each other,we get $27/44$,but answers are not like this,so what should be another way,i am looking for general solution,suppose that in one container ration of basic elements are $a:b$,in another container $c:d$,if we take amount of water from both container with ration $E:F$,what would be ratios of two elements?

i have solved it like this: suppose we have $16$ first substance and $12$ second,because $16/12=4/3$

in each category we would have

$5*x+3*x=16$

$x+2*x=12$

i have solved, and got

$10$ and $6$ in first and $4$ and $8$ in second,so i have added and got ratio $14/14=1$,is it correct?

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2 Answers 2

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If we take $40x$ unit of $A,$ the quantity of the first basic element will be $40x\cdot\frac5{(5+3)}=25x$ units

So, we need to take $30x$ unit of $A,$ the quantity of the first basic element will be $30x\cdot\frac1{(1+2)}=10x$ units

So, in total of $40x+30x=70x$ units of $X$,

the quantity of the first basic element will be $25x+10x=35x$ units

and consequently the quantity of the second basic element will be $70x-35x=35x$ units

So, the ratio of the first & the second basic elements in $X$ will be $35x:35x=1:1$

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  • $\begingroup$ it is also easy by substitution right? $\endgroup$
    – dato datuashvili
    Commented Aug 10, 2013 at 16:11
  • $\begingroup$ @dato, not sure what is meant by substitution? $\endgroup$
    – lab bhattacharjee
    Commented Aug 10, 2013 at 16:13
  • $\begingroup$ i mean by values,like i took $16$ and $12$ $\endgroup$
    – dato datuashvili
    Commented Aug 10, 2013 at 16:14
  • $\begingroup$ @dato, yes no problem, you can just put $x=\frac25$ in my answer. But, some suitable value of $x$ can help us to avoid fractions $\endgroup$
    – lab bhattacharjee
    Commented Aug 10, 2013 at 16:17
  • $\begingroup$ yes thanks in advance $\endgroup$
    – dato datuashvili
    Commented Aug 10, 2013 at 16:18
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To save paper:

$\frac{5}{8}$ of $A$ is Element $\#1$, so when you take 4 units of $A$ for your mix, you have $4\times \frac{5}{8}$ or $2.5$ units of Element $\#1$

$\frac{1}{3}$ of $B$ is Element $\#1$, so when you take 3 units of $B$ for your mix, you have $3\times \frac{1}{3}$ or $1$ units of Element $\#1$

Total of $3.5$ units of Element $\#1$ in the $7$ units of mix implies $1:1$

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