Aman's Vegetable Purchases

Let $C, P, O$ be the prices per kg of cabbage, potato, and onion. $5C + 4P = 80$ ... (i) $4C + 5O = 93$ ... (ii)

From (i), $5C$ ends in 0 or 5, so $4P$ must end in 0 or 5. Since $P$ is integer, $4P$ ends in 0, so $P$ is a multiple of 5. Possible values for $P$: $5, 10, 15$.

• If $P = 5$: $5C = 60 \implies C = 12$. Then $4(12) + 5O = 93 \implies 48 + 5O = 93 \implies 5O = 45 \implies O = 9$.
• If $P = 10$: $5C = 40 \implies C = 8$. Then $4(8) + 5O = 93 \implies 32 + 5O = 93 \implies 5O = 61$ (Not integer).
• If $P = 15$: $5C = 20 \implies C = 4$. Then $4(4) + 5O = 93 \implies 16 + 5O = 93 \implies 5O = 77$ (Not integer).

So $P = 5, C = 12, O = 9$.

Buying maximum onions with Rs. 100: $\lfloor 100 / 9 \rfloor = 11$ kg of onions costing $11 \times 9 = 99$. Remaining money $= 100 - 99 = \text{Rs. } 1$.

Prices: Potato $P = 5$, Onion $O = 9$. Leftover money must be $< \min(5, 9) = 5$, i.e., $100 - (5p + 9o) \in \{0, 1, 2, 3, 4\}$. So $5p + 9o \in \{96, 97, 98, 99, 100\}$.

For each $o \in \{1, 2, \dots, 10\}$ (positive integers):

• $o = 10 \implies 9o = 90 \implies 5p \in \{6, 7, 8, 9, 10\} \implies p = 2$.
• $o = 9 \implies 9o = 81 \implies 5p \in \{15, 16, 17, 18, 19\} \implies p = 3$.
• $o = 8 \implies 9o = 72 \implies 5p \in \{24, \dots, 28\} \implies p = 5$.
• $o = 7 \implies 9o = 63 \implies 5p \in \{33, \dots, 37\} \implies p = 7$.
• $o = 6 \implies 9o = 54 \implies 5p \in \{42, \dots, 46\} \implies p = 9$.
• $o = 5 \implies 9o = 45 \implies 5p \in \{51, \dots, 55\} \implies p = 11$.
• $o = 4 \implies 9o = 36 \implies 5p \in \{60, \dots, 64\} \implies p = 13$.
• $o = 3 \implies 9o = 27 \implies 5p \in \{69, \dots, 73\} \implies p = 15$.
• $o = 2 \implies 9o = 18 \implies 5p \in \{78, \dots, 82\} \implies p = 17$.
• $o = 1 \implies 9o = 9 \implies 5p \in \{87, \dots, 91\} \implies p = 18$.

Total permissible combinations $= 10$. Cases where $o > p$:

• $(o=10, p=2)$
• $(o=9, p=3)$
• $(o=8, p=5)$ Total favorable cases $= 3$. Probability $= \frac{3}{10}$.