Correct option is A
Given:
25÷10−{47×31}×56+314×109−{51÷251}
BODMAS Rule:
Operation preference wiseBracketsOrders,ofDivisionMultiplicationAdditionSubtractionSymbol[],,()²(power),√(root),of÷×+−
Solution:
25÷10−{47×31×56+314×109−(51÷251)}⟹25÷10−(127×56+314×109−5)⟹1025−(25−107+521−5)⟹1067−107−5⟹1067−1057⟹1010⟹1
Hence, option (a) is the correct answer.