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: Formulate the equations based on the given conditions.

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

The problem yields the following equations:

5c + 4p + 20 = 100, which simplifies to 5c + 4p = 80 (1).

4c + 5o + 7 = 100, which simplifies to 4c + 5o = 93 (2).

Step 2: Solve the system of equations. From equation (1), we get:

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

For p to be an integer, (80 - 5c) must be divisible by 4. We test values for c:

• When c = 8:
• When c = 6:
• When c = 4:

Step 3: Calculate the remaining money after purchasing onions. With Rs. 100 and an onion price of o = 15, Aman can purchase:

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

The cost of these onions is 6 × 15 = 90 Rs. The remaining amount is:

100 − 90 = 9 Rs.

Final Answer: Rs. 9.

Answer 2

Step 1: Formulate the total cost equation. The total cost is expressed as:

x ⋅ o + y ⋅ p = 100 − r,

where x and y represent the kilograms of onion and potato, respectively, and r is the remaining money, with the conditions r \(<\) o and r \(<\) p. Based on Question 16, o = 15 and p = 15.

Step 2: Determine valid purchase combinations. Aman must spend a minimum of 85 (to ensure r \(<\) 15) and a maximum of 100. By evaluating pairs of x and y such that x + y \(≤\) 6, the valid combinations are:

(<i>x</i>, <i>y</i>) = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1).

Step 3: Count combinations where x \(>\) y. The combinations satisfying x \(>\) y are:

(<i>x</i>, <i>y</i>) = (4, 2), (5, 1).

The total number of possible combinations is 5.

Probability calculation:

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

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