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    Which two number should be interchanged to make the given equation correct?​21×4+(84÷7)×2−15×5+42=10521 \times 4 + (84 \div 7) \times 2 - 15 \times 5
    Question

    Which two number should be interchanged to make the given equation correct?

    21×4+(84÷7)×215×5+42=10521 \times 4 + (84 \div 7) \times 2 - 15 \times 5 + 42 = 105

    (Note: Interchange should be done of entire number and not individual digits of a given number.)

    A.

    4 and 2

    B.

    84 and 42

    C.

    4 and 5

    D.

    7 and 21

    Correct option is B

    Given: ​21×4+(84÷7)×215×5+42=10521 \times 4 + (84 \div 7) \times 2 - 15 \times 5 + 42 = 105 
    21×2+(84÷7)×415×5+42=105 42+12×475+42=105 42+4875+42=105 5710521 \times 2+ (84 \div 7) \times 4 - 15 \times 5 + 42 = 105\\ \ \\42+12\times4-75+42=105\\ \ \\42 +48 -75+42= 105\\ \ \\57\not=105
    By option (a): Inter  change 4 and 2By option (b) Inter  change 84 and 42
    21×4+(42÷7)×215×5+84=105 84+6×275+84=105 84+1275+84=105 18075=105 105=10521 \times 4 + (42 \div 7) \times 2 - 15 \times 5 + 84= 105\\ \ \\84+6\times2- 75 +84= 105\\ \ \\84 +12-75 +84= 105\\ \ \\ 180-75= 105 \\ \ \\\bf105 = 105​​

    Thus, correct option is (b).

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