Correct Answer: 6 and 4

Solution : Given:
18 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112

Interchange the numbers given in the options one by one with the original numbers in the given equation.

First option: 18 and 11
18 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112
⇒ 11 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 18 = 112
= 11 + 72 + 9 x 4 + 3 – 18 = 112
= 11 + 72 + 36 + 3 – 18 = 112
= 122 – 18 = 104 ≠ 112
Second option: 12 and 4
18 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112
⇒ 18 + (4 x 6) + (63 ÷ 7) x 12 + (27) ^⅓ – 11 = 112
=18 + 24 + 9 x 12 + 3 – 11 = 112
=18 + 24 + 108 + 3 – 11 = 112
=153 – 11 = 142 ≠ 112
Third option: 6 and 4
18 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112
⇒ 18 + (12 x 4) + (63 ÷ 7) x 6 + (27) ^⅓ – 11 = 112
=18 + 48 + 9 x 6 + 3 – 11 = 112
=18 + 48 + 54 + 3 – 11 = 112
=123 – 11 = 112 = R.H.S
Fourth option: 18 and 12
18 + (12 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112
⇒ 12 + (18 x 6) + (63 ÷ 7) x 4 + (27) ^⅓ – 11 = 112
= 12 + 108 + 9 x 4 + 3 – 11 = 112
= 12 + 108 + 36 + 3 – 11 = 112
= 159 – 11 = 148 ≠ 112

So, only the third option satisfies the R.H.S. of the given equation. Hence, the third option is correct.