A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make 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

Let the total number of fruits in the stock at the beginning of the day be $N$.
Let $M$ be the number of mangoes, $B$ be the number of bananas, and $A$ be the number of apples in the initial stock.
We are given that there is at least one fruit of each type initially, so $M>0$, $B>0$, and $A>0$ are integers.

According to the problem statement, at the beginning of the day, mangoes make up 40% of the total stock:
$M=0.40N$

Since the number of mangoes $M$ must be an integer, we can write:
$M=\frac{2}{5}N$
This implies that $N$ must be a multiple of 5. Let $N=5k$ for some positive integer $k$. Then:
$M=2k$

The remaining 60% of the initial stock consists of bananas and apples:
$B+A=0.60N=3k$

Next, let's look at the fruits sold during the day:
- Mangoes sold: half of the mangoes = $\frac{1}{2}M=\frac{1}{2}\left(2k\right)=k$
- Bananas sold: 96
- Apples sold: 40% of the apples = $0.40A=\frac{2}{5}A$

The total number of fruits sold is given as 50% of the initial stock:
Total sold = $0.50N=\frac{5}{2}k$

Setting up the equation for the total fruits sold:
$k+96+\frac{2}{5}A=\frac{5}{2}k$

Rearranging the equation to solve for $A$ in terms of $k$:
$96+\frac{2}{5}A=\frac{5}{2}k-k$
$96+\frac{2}{5}A=\frac{3}{2}k$
Multiply the entire equation by 10 to clear denominators:
$960+4A=15k$
$4A=15k-960$
$A=\frac{15k-960}{4}$

Since the number of apples $A$ must be a positive integer, $15k-960$ must be positive and divisible by 4:
1. For $A>0$:
$15k-960>0\Rightarrow 15k>960\Rightarrow k>64$
2. For $A$ to be an integer, $15k-960$ must be a multiple of 4. Since 960 is divisible by 4, $15k$ must also be divisible by 4. Because 15 and 4 share no common factors, $k$ must be a multiple of 4.

Additionally, the number of bananas $B$ must be a positive integer, and at least 96 bananas must have been sold (so $B\ge 96$):
$B=3k-A\ge 96$
Substitute the expression for $A$ into this inequality:
$3k-\frac{15k-960}{4}\ge 96$
Multiply the inequality by 4:
$12k-(15k-960)\ge 384$
$-3k+960\ge 384$
$960-384\ge 3k$
$576\ge 3k\Rightarrow k\le 192$

Furthermore, since the vendor sells 40% of the apples, the number of apples $A$ must be a multiple of 5 so that the number of apples sold is an integer:
$A=\frac{15k-960}{4}=5m$ (where $m$ is a positive integer)
$\frac{15k-960}{4}=5m\Rightarrow 15k-960=20m$
Divide by 5:
$3k-192=4m$
Since $192$ and $4m$ are divisible by 4, $3k$ must be divisible by 4, which again means $k$ is a multiple of 4.

To find the smallest possible total number of fruits $N=5k$, we need to find the smallest possible integer value of $k$ that satisfies our constraints:
1. $k>64$
2. $k$ is a multiple of 4.
The smallest multiple of 4 strictly greater than 64 is:
$k=68$

Let's check if $k=68$ yields integer values for $M$, $A$, and $B$:
- $M=2k=2\left(68\right)=136$ (an integer)
- $A=\frac{15\left(68\right)-960}{4}=\frac{1020-960}{4}=\frac{60}{4}=15$ (an integer, and $0.40\times 15=6$ apples sold is also an integer)
- $B=3k-A=3\left(68\right)-15=204-15=189$ (an integer, and $189\ge 96$)

All values are valid positive integers. Thus, the smallest possible total number of fruits in the stock at the beginning of the day is:
$N=5k=5\times 68=340$

The correct answer is 340.