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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

Let no. of ₹100, ₹250 and ₹500 cheques are x, y and z respectively. And the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250.

So, x+y+z = 100  &  100x+250y+500z = 15250  => 2x+5y+10z = 305

By eliminating x from 2nd equation we get, 2(100-y-z)+5y+10z=305  => 3y+8z = 105

The range of values of x, y and z are all non negative integers from 0 to 100. If we see equation

3y+8z = 105, we will notice that for max value of z, the value of y should be minimum. So, we will hit and try the possible value of y starting from 0, for max value of z.

z = (105 – 3y) / 8

Here y = 0, 1, 2 will not result an integer value of z, but y = 3 will. So, when y = 3, then z = 12.

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