Algebra

Linear Equation

Medium

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

Entered answer:

✅ Correct Answer: 12

Let us define:

x = number of Rs. 100 cheques

y = number of Rs. 250 cheques

z = number of Rs. 500 cheques

From the problem, we can write two equations:

Total number of cheques = 100

$x + y + z = 100$

Total value = Rs. 15250

$100x + 250y + 500z = 15250$

Since we want to maximise z, we need to eliminate one variable to work with a simpler equation.

From the first equation: $x = 100 - y - z$

Substituting into the second equation:

$100(100 - y - z) + 250y + 500z = 15250$

$10000 - 100y - 100z + 250y + 500z = 15250$

$10000 + 150y + 400z = 15250$

$150y + 400z = 5250$

Dividing everything by 50:

$3y + 8z = 105$

$y = \dfrac{105 - 8z}{3}$

For this to be a valid solution, $y$ must be a non-negative integer.

For $y \geq 0$ :

$\dfrac{105 - 8z}{3} \geq 0$

$105 - 8z \geq 0$

$z \leq \dfrac{105}{8} = 13.125$

Since z must be an integer: $z \leq 13$

For $y = \dfrac{105 - 8z}{3}$ to be an integer, $(105 - 8z)$ must be divisible by 3.

Since $105 = 3 \times 35$ is divisible by 3, we need $8z$ to also be divisible by 3. The largest multiple of $3$ that is $≤ 13$ is $z = 12$ .

The maximum possible number of Rs. 500 cheques is 12.

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