At the beginning of the winter, there were at least 66 students registered in a ski class. After the class started, eleven boys transferred into this class and thirteen girls transferred out. The ratio of boys to girls in the class was then $1: 1$. Which of the following is not a possible ratio of boys to girls before the transfers?
(A) $4: 7$
(B) $1: 2$
(C) $9: 13$
(D) $5: 11$
(E) $3: 5$
Solution 1

Let $b$ represent the number of boys initially registered in the class.

Let $g$ represent the number of girls initially registered in the class.

When 11 boys transferred into the class, the number of boys in the class was $b+11$.

When 13 girls transferred out of the class, the number of girls in the class was $g-13$.

The ratio of boys to girls in the class at this point was $1: 1$.

That is, the number of boys in the class was equal to the number of girls in the class, or $b+11=g-13$ and so $g=b+24$.

Since there were at least 66 students initially registered in the class, then $b+g \geq 66$.

Substituting $g=b+24, b+g$ becomes $b+(b+24)=2 b+24$, and so $2 b+24 \geq 66$ or $2 b \geq 42$, so $b \geq 21$.

At this point, we may use the conditions that $g=b+24$ and $b \geq 21$ to determine which of the 5 answers given is not possible.

Each of the 5 answers represents a possible ratio of $b: g$ or $b:(b+24)$ (since $g=b+24)$.

We check whether $b:(b+24)$ can equal each of the given ratios while satisfying the condition that $b \geq 21$.

In (A), we have $b:(b+24)=4: 7$ or $\frac{b}{b+24}=\frac{4}{7}$ or $7 b=4 b+96$ and then $3 b=96$, so $b=32$.

In (B), we have $b:(b+24)=1: 2$ or $\frac{b}{b+24}=\frac{1}{2}$ or $2 b=b+24$, so $b=24$.

In (C), we have $b:(b+24)=9: 13$ or $\frac{b}{b+24}=\frac{9}{13}$ or $13 b=9 b+216$ and then $4 b=216$, so $b=54$.

In (D), we have $b:(b+24)=5: 11$ or $\frac{b}{b+24}=\frac{5}{11}$ or $11 b=5 b+120$ and then $6 b=120$, so $b=20$.

In (E), we have $b:(b+24)=3: 5$ or $\frac{b}{b+24}=\frac{3}{5}$ or $5 b=3 b+72$ and then $2 b=72$, so $b=36$.

Thus, the only ratio that does not satisfy the condition that $b \geq 21$ is $5: 11$ (since $b=20$ ).

## Solution 2

As in Solution 1, we establish the conditions that $g=b+24$ and $b \geq 21$.

Again, the ratio $b: g$ or $\frac{b}{g}$ then becomes $\frac{b}{b+24}$ (since $g=b+24$ ).

Since $b \geq 21$, then $b$ can equal $21,22,23,24, \ldots$, but the smallest possible value for $b$ is 21 .

When $b=21, \frac{b}{b+24}$ becomes $\frac{21}{21+24}=\frac{21}{45}=0.4 \overline{6}$.

When $b=22, \frac{b}{b+24}$ becomes $\frac{22}{22+24}=\frac{22}{46} \approx 0.478$.

When $b=23, \frac{b}{b+24}$ becomes $\frac{23}{23+24}=\frac{23}{47} \approx 0.489$.

When $b=24, \frac{b}{b+24}$ becomes $\frac{24}{24+24}=\frac{24}{48}=0.5$.

As $b$ continues to increase, the value of the ratio $\frac{b}{b+24}$ or $\frac{b}{g}$ continues to increase.

Can you verify this for yourself?

Thus, the smallest possible value of the ratio $\frac{b}{g}$ is $\frac{21}{45}$ or $0.4 \overline{6}$.

Comparing this ratio to the 5 answers given, we determine that $\frac{5}{11}=0 . \overline{45}<0.4 \overline{6}=\frac{21}{45}$, which is not possible.

(You may also confirm that each of the other 4 answers given are all greater than $\frac{21}{45}$ and are obtainable, as in Solution 1.)

Therefore, the only ratio of boys to girls which is not possible is (D).

ANSWER: (D)