Question
Download Solution PDFWhich two numbers, from amongst the given options, should be interchanged to make the given equation correct?
\((6)^3 \div12+ [(\sqrt{81}) \times 4] - (28 \div 2) + 24 = 43\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven : \((6)^3 \div 12+ [(\sqrt{81}) × 4] - (28 \div 2) + 24 = 43\)
According to the question, after interchanging the numbers:
So,
- Option - (1) : 12 and 24
\((6)^3 \div 24 + [(\sqrt{81}) × 4] - (28 \div 2) + 12 = 43\)
216 ÷ 24 + (9 × 4) - (28 ÷ 2) + 12 = 43
216 ÷ 24 + 36 - 14 + 12 = 43
9 + 36 + 12 - 14 = 43
57 - 14 = 43
43 = 43 (LHS = RHS)
- Option - (2) : 6 and 24
\((24)^3 \div 12 + [(\sqrt{81}) × 4] - (28 \div 2) + 6 = 43\)
13824 ÷ 12 + (9 × 4) - (28 ÷ 2) + 6 = 43
13824 ÷ 12 + 36 - 14 + 6 = 43
1152 + 36 + 6 - 14 = 43
1194 - 14 = 43
1180 ≠ 43 (LHS ≠ RHS)
- Option - (3) : 81 and 4
\((6)^3 \div 12 + [(\sqrt{4}) × 81] - (28 \div 2) + 24 = 43\)
216 ÷ 12 + (2 × 81) - (28 ÷ 2) + 24 = 43
216 ÷ 12 + 162 - 14 + 24 = 43
18 + 162 + 24 - 14 = 43
204 - 14 = 43
190 ≠ 43 (LHS ≠ RHS)
- Option - (4) : 28 and 24
\((6)^3 \div 12 + [(\sqrt{81}) × 4] - (24 \div 2) + 28 = 43\)
216 ÷ 12 + (9 × 4) - (24 ÷ 2) + 28 = 43
216 ÷ 12 + 36 - 12 + 28 = 43
18 + 36 + 28 - 12 = 43
82 - 14 = 43
68 ≠ 43 (LHS ≠ RHS)
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