Let the total number of juniors = j, and seniors = s at the beginning of the year
Therefore, j : s = 3 : 4 j/s = 3/4
=> j = 3/4(s)...........eqn 1

During the year :
(j - 10) : (s - 20) = 4 : 5
$$\frac{\left(j-10\right)}{\left(s-20\right)}=\frac{4}{5}$$
5(j-10)= 4(s-20) where j = 3/4(s) from equation 1
5 [3/4(s) - 10 ] = 4 (s - 20)
$$\frac{15}{4}\left(s\right)-50=45-80$$
$$\frac{15}{4}\left(s\right)-\frac{4s}{1}=-80+50$$
$$\frac{15s-165}{4}=-30$$
$$15s-165=-30\cdot4$$
$$\frac{-1s}{-1}=\frac{-120}{-1}$$
$$s=120$$

Answer = E

Vincen wrote: ↑

Mon May 18, 2020 6:37 am

At the beginning of the year, the ratio of juniors to seniors in high school \(X\) was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school \(X.\) If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school \(X\) at the beginning of the year?

A. 80
B. 90
C. 100
D. 110
E. 120

[spoiler]OA=E[/spoiler]

Source: Manhattan GMAT

\begin{array}{|c|c|}
\hline
\text{Juniors} & \text{Seniors} \\ \hline
3x & 4x \text{(Initially)} \\ \hline
\end{array}
After transfers
\begin{array}{|c|c|}
\hline
\text{Juniors} & \text{Seniors} \\ \hline
3x-10 & 4x-20 \\ \hline
\end{array}

Given that new ratio \(=\dfrac{3x−10}{4x−20}=\dfrac{4}{5}\quad \Rightarrow\quad x=30\)

Seniors initially \(= 120\)

Therefore, E