4. In the box, there are red and blue balls. The number of red balls is to the number of blue balls as $7: 3$. By what percentage does the number of red balls need to be increased relative to the red balls already in the box so that the new number of red balls is to the number of blue balls as $14: 5$?
First method:

Let $p$ be the total number of blue balls in the box, $c$ be the total number of red balls before adding, and $C$ be the total number of red balls after adding the necessary number of red balls.

Due to the ratio of the number of blue and red balls in the box before adding red balls, it holds that $p: c=3: 7$ or $p=\frac{3}{7} c$.

1 POINT

Due to the ratio of the number of blue and red balls after adding a certain number of red balls,

it holds that $p: C=5: 14$ or $p=\frac{5}{14} C$.

It follows that $\frac{3}{7} c=\frac{5}{14} C \longrightarrow C=\frac{6}{5} c$.

Since $C=\frac{6}{5} c=\frac{120}{100} c=120 \% c$, the new number of red balls is $120 \%$ of the initial number of red balls.

1 POINT

The number of red balls needs to be increased by $20 \%$.

1 POINT

TOTAL 6 POINTS