The ratio of balls in three boxes is $6:8:9$. In what ratio should the balls in the second and third boxes be increased, so that the final ratio becomes $1:3:4$?
To have a $1:3:4$ ratio, I think I need to change the given ratio to $6:18:24$. That means I need to add $10$ balls in the second box and $15$ balls in the third box. So the ratio in which second and third boxes to be increased should be $2:3$. Is this correct?
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I think you need of two distinct ratios: one for the second box and one for the third box. Since \begin{align} (1:3:4) &=(6:18:24)\\ &=\left(6:8\cdot\frac 94:9\cdot\frac 83\right)\\ &=\left(6:8\left(1+\frac 54\right):9\left(1+\frac 53\right)\right)\\ \end{align} hence the second box should be increased by $\frac 54$ while the third box by $\frac 53$.
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The best way to approach this problem is to work backwards. The final ratio is $1:3:4$. To make this answer more concrete, let's imagine that there are white balls, red balls, and black balls. This means that for $1$ white ball, there will be $3$ red balls, and $4$ black balls. One important thing to note about ratios is that they can be scaled up:
This means that $1:3:4=6:18:24$. Scaling this ratio is useful because it allows us to consider how we can get the final ratio to be $1:3:4$, without changing the number of white balls, as per the requirements of the question. This seems to be the approach that you have took.
We need to go from
$$ \text{6 white balls, 8 red balls, and 9 black balls} $$
to
$$ \text{6 white balls, 18 red balls, and 24 black balls} $$
The number of red balls we need is $10$, and the number of black balls we need is $15$. Can you take it from here?
You start with $6x,8x,9x$ and end with $6x,18x,24x$ since the first box doesn't change. You added $10x$ to the second and $15x$ to the third, so you're adding balls to the second and third in the ratio $2:3$ but the total number of balls you add is related to the (fixed) number in the first box by the ratio $25:6$, that's also important.