Simplified Fruit

Let there be $4n$ apples and $5n$ oranges for some positive integer $n$ . Next, suppose she eats $a$ apples and $b$ oranges such that $a + b = 5$ . Then

$\dfrac{4n - a}{5n - b} = \dfrac{2}{3} \Longrightarrow 12n - 3a = 10n - 2b \Longrightarrow 2n = 3a - 2b = 3a - 2(5 - a) = 5a - 10 = 5(a - 2)$ .

Now as $n \gt 0$ we will require that $2 \lt a \le 5$ . Also, as $2$ does not divide $5$ we must then have that $2 | (a - 2)$ , and the only value in the given range for $a$ for which this occurs is $a = \boxed{4}$ .

This in turn means that $n = 5$ , so Peggy had $20$ apples and $25$ oranges to start with, and after her feast ended up with $16$ apples and $24$ oranges, which is indeed in a $2 : 3$ ratio.

We have that ratio of Apple $A$ to orange $O$ is $\ \dfrac{A}{O} = \dfrac{4x}{5x}$

If Peggy ate some fruits say she ate $A_n$ and $O_n$ apples and oranges respectively. Then $A_n +O_n =5$ and new ratio. $\dfrac{A_t}{O_t} = \dfrac{A-A_n}{O-O_n} = \dfrac{2}{3} \implies \dfrac{4x-A_n}{5x-O_n } = \dfrac{2y}{3y}$ Here, $\small{4x-A_n =2y\implies4x= {\color{#3D99F6}2y+A_n} \quad 5x-O_n=3y \implies 5x = {\color{#3D99F6}3y+O_n}}$ Since $A_n$ and $O_n \leq 5$ . Then $A_n$ cannot be an odd integer of eaten apples(blue colored part says) . So, Peggy must have eaten an even integer of apples either $2$ or $4$ . So it's clear that Peggy ate $\boxed{4}$ apples.

Note : If Peggy would have eaten $2$ apples & $3$ ornages.Then $x=\frac{y+1}{2}$ and $x = \frac{3(y+1)}{5}$ . These two fraction cannot to be true simultaneously for any value of $y$ since we don't find any integer that is both even and odd.