2
$\begingroup$

A friend and I had two different answers to this seeming simple question. It goes as follows:

Jar A contains flour and sugar in the ratio 5 : 1. Jar B, which is three times larger than Jar A, contains flour and sugar in the ratio 8 : 1. When the contents of these jars are combined, the resulting mixture contains flour and sugar in the ratio x : 1. What is the value of x?

Thanks for any help. This is not homework of any sorts and we genuinely just want to figure out the proper way of doing it.

$\endgroup$

4 Answers 4

2
$\begingroup$

Jar A has 6 parts. 5 Flour and 1 sugar. Jar B has 18 parts (of the unit in A) - 8:1 -> 16 parts Flour and 2 parts sugar. Totally, 21 parts flour to 3 parts sugar. --> 7:1.

$\endgroup$
1
  • $\begingroup$ Thanks, that was my answer :P $\endgroup$
    – Jordan
    Commented Apr 6, 2018 at 22:58
2
$\begingroup$

Take the capacity of $Jar A$ as a unit, say $U$. Then $Jar A$ contains $(5/6)U$ of flour and $(1/6)U$ of sugar. Now $Jar B$ contains $(8/9)3U$ of flour and $(1/9)3U$ of sugar. If you combine them you get $$ (5/6)U+(8/9)3U=(5/6)U+(16/6)U=7/2U $$ of flour and $$ (1/6)U+(2/6)U=1/2U $$ of sugar then $x=7$.

Do not hesitate to interact.

$\endgroup$
2
  • $\begingroup$ Thanks, always neat to see a different approach than my own. "7/2" was one of the options which I didn't come across, but now realize why they chose it. :D $\endgroup$
    – Jordan
    Commented Apr 6, 2018 at 22:59
  • $\begingroup$ I had a lot of pleasure of writing this answer because, when I was a young boy, we had lots of problems like that (it is after all, the country of Lavoisier :), and I discovered that my mathematical vocation is based on such problems $\endgroup$ Commented Apr 6, 2018 at 23:06
2
$\begingroup$

Let

  • $F_A$ and $S_A$ the quantity of flour and sugar in the Jar A $\implies \frac{F_A}{S_A}=5\implies F_A=5S_A$
  • $F_B$ and $S_B$ the quantity of flour and sugar in the Jar B $\implies \frac{F_B}{S_B}=8\implies F_B=8S_B$

indicating with $V$ the volume of the jars, we also know that

  • $F_A+S_A=6S_A=V_A$
  • $F_B+S_B=9S_B=V_B=3V_A \implies 18S_A=9S_B\implies S_B=2S_A$

then we need to find

$$x=\frac{F_A+F_B}{S_A+S_B}=\frac{5S_A+8S_B}{S_A+S_B}=\frac{5S_A+16S_A}{S_A+2S_A}=\frac{21S_A}{3S_A}=7$$

$\endgroup$
2
  • $\begingroup$ Thanks a lot, I like the solution, but just chose the quickest one because all of these are practically analogous. $\endgroup$
    – Jordan
    Commented Apr 6, 2018 at 23:00
  • $\begingroup$ @Jordan You are welcome. Yes this one is more standard and with all the details, not the most efficient of course. Thanks for your appreciation, Bye $\endgroup$
    – user
    Commented Apr 6, 2018 at 23:03
2
$\begingroup$

I interpret "three times larger than" to mean "three times plus one the size of" instead of just "three times the size of". So, using a similar reasoning as Duchamp Gérard H. E.: $$\text{Flour:} \quad (5/6)U+(8/9)4U=(5/6)U+(32/9)U=(79/18)U\\ \text{Sugar:} \quad (1/6)U+(1/9)4U=(1/6)U+(4/9)U=(11/18)U.$$ The ratio $(79/18)$:$(11/18)$ is the same as $x$:$1$, so $x=79/11$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .