Question

Aman decides to buy only onion, in whatever maximum quantity possible (in positive integer kilogram), with the money he has come to the market with.
How much money will he be left with after the purchase?

• A. Rs. 12
• B. Rs. 9
• C. Rs. 7
• D. Rs. 5
• E. Rs. 1
• A. 3/10
• B. 5/6
• C. 2/9
• D. 7/20
• J. 45392

Answer 1

The Correct Option is A

Step 1: Write equations for the given conditions. 

Let the prices of cabbage, onion, and potato be c, o, and p (in Rs./kg), respectively. 

From the problem, we have:

5c + 4p + 20 = 100 = => 5c + 4p = 80, (1)

4c + 5o + 7 = 100 = => 4c + 5o = 93. (2)

Step 2: Solve the equations. From equation (1):

\(p = \frac{80 - 5c}{4}\).

For p to be an integer, 80 − 5c must be divisible by 4. Testing values of c:

• For c = 8:
• For c = 6:
• For c = 4:

Step 3: Find the money left after buying onions. With Rs. 100 and o = 15, Aman can buy:

$\frac{100}{15} = 6\frac{2}{3}$ kg of onions.

The cost is 6 × 15 = 90, leaving:

100 − 90 = 9 Rs.

Final Answer: Rs. 9.

Answer 2

Step 1: Write the total cost equation. The total cost is:

x ⋅ o + y ⋅ p = 100 − r,

where x and y are the kilograms of onion and potato, and r is the remaining money such that r \(<\) o and r \(<\) p. From Question 16, o = 15 and p = 15.

Step 2: Identify permissible combinations. Aman must spend at least 85 (to leave r \(<\) 15) but less than 100. Testing x and y such that x + y \(≤\) 6, the permissible combinations are:

(x, y) = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1).

Step 3: Count cases where x \(>\) y. For x \(>\) y:

(x, y) = (4, 2), (5, 1).

The total number of combinations is 5.

Probability:

\(P(x \(>\) y) = \frac{2}{5} = \frac{3}{10}\).

Final Answer: $\frac{3}{10}$