I would say it is c, total = 12.

I did:

m = 1/3*T
m = 2/5*(T-2)

T/3 = 2/5*(T-2) => T = 12

Let $T$ be the total number of people originally in the group. Let $M$ be the initial number of men.
Let $W$ be the initial number of women.

Step 1: Initial Conditions
We are given:
1. The total number of people is $T=M+W$.
2. The number of men is $\(\frac{1}{3}\)$ of the total people:

$$
\(M=\frac{1}{3} T\)
$$

3. Therefore, the number of women is $\(\frac{2}{3}\)$ of the total people:

$$
\(W=T-M=T-\frac{1}{3} T=\frac{2}{3} T\)
$$

Step 2: After 2 Women Leave
- The number of men ( $M$ ) remains the same:

$$
\(M_{\mathrm{new}}=M\)
$$

- The number of women decreases by 2 :

$$
\(W_{\text {new }}=W-2\)
$$

- The new total number of people is $\(T_{\text {new }}\)$ :

$$
\(T_{\text {new }}=T-2\)
$$


Step 3: New Ratio
We are given that after 2 women leave, the number of men becomes $\(\frac{2}{5}\)$ of the remaining people ( $\(\left.T_{\text {new }}\right)\)$ :

$$
\(\begin{aligned}
& \text { New number of men }=\frac{2}{5} \times \text { Remaining people } \\
& \qquad M=\frac{2}{5}(T-2)
\end{aligned}\)
$$

Step 4: Solve for $T$
Now we have two expressions for $M$ (from Step 1 and Step 3). Set them equal to each other:

$$
\(\frac{1}{3} T=\frac{2}{5}(T-2)\)
$$


Multiply both sides by the least common multiple of 3 and 5 , which is 15 , to clear the fractions:

$$
\(\begin{aligned}
15 \times \frac{1}{3} T & =15 \times \frac{2}{5}(T-2) \\
5 T & =6(T-2) \\
5 T & =6 T-12
\end{aligned}\)
$$


Subtract $5 T$ from both sides and add 12 to both sides:

$$
\(\begin{gathered}
12=6 T-5 T \\
T=12
\end{gathered}\)
$$


The number of people present originally in the group was 12.

Thank you . I fixed the OA