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Method we are given that, 2A8 + 3B5 +5C8 = 1111 Let us solve by following method : (200 + A0 + 8) + (300 + B0 + 5) + (500 + C0 + 8)=1111 After adding all 100 and one''s unit we get, (1000+21) + (A+B+C)10=1111 Taking the total 1021 on another side, (A+B+C)10 = 1111-1021 ,we will get 90 then dividing 90 / 10 = 9 Thus max value i.e 9 will be given to B, whereas A =0, C=0 208 +395+508 =1111 Nandini 06-16-2019 11:09 AM

Discuss I want solution Roshan sharma 10-8-2018 12:52 PM

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