Combining ratios of quantities with different sizes

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

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

Answer 4

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 $x=79/11$.