Question

A donation box can receive only cheques of ₹100,₹250 and ₹500. On one good day,the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained,is

Answer 1

Correct Answer: 12

Step 1: Define Variables

Let the count of ₹100, ₹250, and ₹500 cheques be \( x, y, z \) respectively.
The total number of cheques is \( x + y + z = 100 \).
The total value of the cheques is \( 100x + 250y + 500z = 15250 \).

Step 2: Substitute from the First Equation

From \( x + y + z = 100 \), we can express \( x \) as: \( x = 100 - y - z \).

Substituting this into the second equation yields: \[ 100(100 - y - z) + 250y + 500z = 15250 \] \[ 10000 - 100y - 100z + 250y + 500z = 15250 \] Combining like terms gives: \[ 150y + 400z = 5250 \]

Step 3: Simplify the Equation

Dividing the equation by 50 simplifies it to: \[ 3y + 8z = 105 \]. Our objective is to maximize \( z \), the number of ₹500 cheques.

Step 4: Test Integer Values

We test integer values for \( y \) to find an integer solution for \( z \):

• If \( y = 0 \), then \( 8z = 105 \Rightarrow z = 13.125 \) (Not an integer) ❌
• If \( y = 1 \), then \( 3(1) + 8z = 105 \Rightarrow 8z = 102 \Rightarrow z = 12.75 \) (Not an integer) ❌
• If \( y = 2 \), then \( 3(2) + 8z = 105 \Rightarrow 8z = 99 \Rightarrow z = 12.375 \) (Not an integer) ❌
• If \( y = 3 \), then \( 3(3) + 8z = 105 \Rightarrow 9 + 8z = 105 \Rightarrow 8z = 96 \Rightarrow z = 12 \) (Integer) ✅

The maximum integer value for \( z \) is 12, which occurs when \( y = 3 \).

Step 5: Determine Final Values

Using \( x + y + z = 100 \) with \( y = 3 \) and \( z = 12 \), we find \( x \):
\( x = 100 - 3 - 12 = 85 \).

Verification of total value: \[ 100 \times 85 + 250 \times 3 + 500 \times 12 = 8500 + 750 + 6000 = 15250 \] ✅ The total value matches the given amount.

Answer:

The maximum number of ₹500 cheques is: \[ \boxed{12} \]