A3. Yesterday afternoon, the ratio of the number of boys to the number of girls on the playground was $3: 2$. Today, the number of boys on the playground is the square of the number of girls, and there are 6 fewer boys and 7 fewer girls on the playground than yesterday afternoon. How many children were on the playground yesterday afternoon?
(A) 12
(B) 13
(C) 15
(D) 25
(E) 30
A3. Yesterday, there were $3 t$ boys and $2 t$ girls on the playground. Today, there are $3 t-6$ boys and $2 t-7$ girls on the playground. From the equation $3 t-6=(2 t-7)^{2}$, it follows that $4 t^{2}-31 t+55=0$ or $t_{1,2}=\frac{31 \pm \sqrt{81}}{8}$. The only possibility is $t=5$, since $t$ is an integer. Yesterday, there were $3 t=15$ boys and $2 t=10$ girls on the playground, making a total of 25 children.

2nd method Let $f$ and $d$ denote the number of boys and girls on the playground yesterday afternoon. It holds that $\frac{f}{d}=\frac{3}{2}$ or $f=\frac{3 d}{2}$. Today, there are $f-6$ boys and $d-7$ girls on the playground, and it holds that $f-6=(d-7)^{2}$. If we substitute $f=\frac{3 d}{2}$, we get $2 d^{2}-31 d+110=0$ or

$$
d_{1,2}=\frac{31 \pm \sqrt{31^{2}-880}}{4}=\frac{31 \pm 9}{4}
$$

The only possibility is $d=10$, since $d$ is an integer. Then $f=15$. Yesterday afternoon, there were 25 children on the playground.