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 number of cheques of ₹100, ₹250, and ₹500 be \( x, y, z \) respectively.
Total number of cheques: \( x + y + z = 100 \) 
Total value of cheques: \( 100x + 250y + 500z = 15250 \)

Step 2: Substitute from the First Equation

From \( x + y + z = 100 \), we get: 
\( x = 100 - y - z \)

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

Step 3: Simplify the Equation

Divide the equation by 50: \[ 3y + 8z = 105 \] Now, we want to maximize the value of \( z \) (₹500 cheques).

Step 4: Try Integer Values

We test integer values of \( y \) that make \( z \) an integer:

• \( y = 0 \Rightarrow z = 105 / 8 = 13.125 \) ❌
• \( y = 1 \Rightarrow z = 102 / 8 = 12.75 \) ❌
• \( y = 2 \Rightarrow z = 99 / 8 = 12.375 \) ❌
• \( y = 3 \Rightarrow z = 96 / 8 = 12 \) ✅

So the maximum integer value of \( z \) is 12 when \( y = 3 \).

Step 5: Final Values

Given \( x + y + z = 100 \), and \( y = 3, z = 12 \), we get: 
\( x = 100 - 3 - 12 = 85 \)

Total value check: 
\[ 100 \times 85 + 250 \times 3 + 500 \times 12 = 8500 + 750 + 6000 = 15250 \] ✅ Total matches correctly.

Answer:

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