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.
Answer is 12.