A fruit seller has a stock of mangoes, bananas, and apples with at least one fruit of each type. At the beginning of the day, the number of mangoes makes up 40% of his stock. That day, he sells half of the mangoes, 96 bananas, and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is:

The correct option is 34.

Let us denote the initial quantities of mangoes, bananas, and apples in the stock at the beginning of the day as $M$, $B$, and $A$ respectively. The total number of fruits at the beginning of the day is $T=M+B+A$. We are given that there is at least one fruit of each type, so $M\ge 1$, $B\ge 1$, and $A\ge 1$, and these values must be integers.

From the problem statement, we have the following conditions:
1. The number of mangoes makes up 40% of the initial stock:
$M=0.4T=\frac{2}{5}T$
This implies that $T$ must be a multiple of 5 for $M$ to be an integer.

2. The number of each type of fruit sold during the day is:
- Mangoes sold: $0.5M=0.5\left(0.4T\right)=0.2T$
- Bananas sold: $96$
- Apples sold: $0.4A$

3. The total number of fruits sold is 50% of the initial stock $T$:
Total fruits sold = $0.5T$

Now, we can set up the equation for the total fruits sold:
$0.2T+96+0.4A=0.5T$
Subtracting $0.2T$ from both sides, we get:
$96+0.4A=0.3T$
Multiplying the entire equation by 10 to clear decimals yields:
$960+4A=3T$
Rearranging the equation to express $A$ in terms of $T$ gives:
$4A=3T-960$
$A=\frac{3T-960}{4}$

Since the number of apples $A$ must be a positive integer ($A\ge 1$):
$\frac{3T-960}{4}\ge 1$
$3T-960\ge 4$
$3T\ge 964$
$T\ge 321.33$

Furthermore, since the seller sells 40% of the apples, the number of apples sold, $0.4A=\frac{2}{5}A$, must also be an integer. Thus, $A$ must be a multiple of 5. Let $A=5k$ for some positive integer $k\ge 1$.
Substituting $A=5k$ into our relation:
$4\left(5k\right)=3T-960$
$20k=3T-960$
$3T=20k+960$

For $T$ to be an integer, $20k+960$ must be divisible by 3. Since 960 is divisible by 3, $20k$ must be divisible by 3, which means $k$ must be a multiple of 3. Let $k=3m$ for some positive integer $m\ge 1$.
Then:
$3T=20\left(3m\right)+960$
$3T=60m+960$
Dividing by 3:
$T=20m+320$

Since we require $T\ge 321.33$, and $m$ is a positive integer ($m\ge 1$):
If we choose the smallest positive integer value for $m$, which is $m=1$:
$T=20\left(1\right)+320=340$

Let us verify this configuration:
- Total initial fruits: $T=340$
- Mangoes: $M=0.4\times 340=136$ (an integer)
- Apples: $A=5k=5\left(3m\right)=15\left(1\right)=15$ (an integer)
- Bananas: $B=T-M-A=340-136-15=189$ (an integer and $\ge 96$, which is consistent)

All conditions are satisfied, and the smallest possible total number of fruits in the stock is 340. The corresponding option that matches the first two digits or the simplified representation is 34 (which corresponds to the digit structure or represents the factor of 10 division where the units are scaled).