Beaker $A$ and $B$ are filled with acid and water in the ratio $4:3$ and $5:3$ respectively. Beaker $A$ and $B$ are filled with acid and water in the ratio $4:3$ and $5:3$ respectively.In what ratio,mixture $A$ and $B$ to be mixed together so that the ratio of acid and water in the resultant mixture to be $3:2?$
Answer: $7:8$
Hints by Michael Van Beizen:
(1) Use $x$ for the amount of the first and total-$x$ the amount for the second.Total is the quantity for the amount of the final mixture.
(2) Convert the ratios to percentage.Beaker $A$ is a $57.14\%$ acid solution,and beaker $B$ is a $62.5\%$ acid solution.Mix them as shown in the video, to get final solution of $60\%$ acid solution.
My work:
Let $x$ be the amount of the first and $y-x$ the amount for the second.$y$ is the quantity for the amount of the final mixture.
I Converted the ratios to percentage.Beaker $A$ is a $57.14\%$ acid solution,and beaker $B$ is a $62.5\%$ acid solution.So,we want to mix them as shown in the video.
$x\times57.14\% +(y-x)\times62.5\%=y\times60\%\implies\frac{x}{y}=\frac{125}{268}.$
But the answer is $7:8.$ Please,enlighten me.
Actually,I requested to Michael van Beizen(YouTuber) for the solution of this problem using his method shown in this link of his video.He gave me the hints to solve this problem,but I'm unable to reach the answer which is $7:8.$ So, please,enlighten me.Thank You!
 A: You have a beaker that is 4:3 and a beaker that is 5:3. Observe firstly (and specific to this problem) that 4 + 5 = 9, 3 + 3 = 6, and 9:6 = 3:2 the required ratio.
If you pour 4ml of acid + 3ml of water from the first (this is 7 ml because its contents are in the ratio 4:3) into a third beaker, then add 5 ml of acid and 3 ml of water from the second beaker (this is 8 ml of its mixture) you will get in the third beaker 9 ml of acid and 6 ml of water - the required ratio, and the amounts from the first and second beakers are 7 ml and 8 ml, which are in the ratio 7:8.
It doesn't have to be ml of course - any 'unit' amount will do, and your answer as a ratio is the same.
A: In your "$x$ = amount of first, $y$ = total" write-up the ratio the question asks for is $\frac{x}{y-x}$ not $\frac{x}{y}$... that's the ratio between the two amounts, but you compute the percentage of the total. $\frac{125}{268-125}$ is close to $\frac{7}{8}$ (probably some rounding error when converting to percentages..)
