Let the number of 100 cheques, 250 cheques and 500 cheques be x,y and z respectively. We need to find the maximum value of z. x+y+z=100...(1) 100x+250y+500z=15250 2x+5y+10z=305...(2) 2x+2y+2z=200...(1) Eqs(2)−(1), we get 3y+8z=105 At z=12,x=3 Therefore, maximum value z can take is 12 .