Which two numbers should be interchanged to make the given equation correct?
72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54
(
Note:
Interchange should be done for an entire number and not for individual digits of the number.)
- AOption (a)
6 and 3
- BOption (b)
2 and 3
- COption (c)
18 and 24
- DOption (d)
6 and 2
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Correct answer
2 and 3
Why this answer is correct
Given: 72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54
According to the question, after interchanging the numbers:
Option 1) 6 and 3
72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54 ⇒ 72 + (18 ÷ 3) + (9 × 2) – (24 × 6) = 54
⇒ 72 + (18 ÷ 3) + (9 × 2) – (24 × 6) = 54
⇒ 72 + 6 + (9 × 2) – (24 × 6) = 54
⇒ 72 + 6 + 18 – 144 = 54
⇒ 90 – 144 = 54
⇒ -54 ≠ 54 (LHS ≠ RHS)
Option 2) 2 and 3
72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54 ⇒ 72 + (18 ÷ 6) + (9 × 3) – (24 × 2) = 54
⇒ 72 + (18 ÷ 6) + (9 × 3) – (24 × 2) = 54
⇒ 72 + 3 + (9 × 3) – (24 × 2) = 54
⇒ 72 + 3 + 27 – 48 = 54
⇒ 102 – 48 = 54
⇒ 54 = 54 (LHS = RHS)
Option 3) 18 and 24
72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54 ⇒ 72 + (24 ÷ 6) + (9 × 2) – (18 × 3) = 54
⇒ 72 + (24 ÷ 6) + (9 × 2) – (18 × 3) = 54
⇒ 72 + 4 + (9 × 2) – (18 × 3) = 54
⇒ 72 + 4 + 18 – 54 = 54
⇒ 94 – 54 = 54
⇒ 40 ≠ 54 (LHS ≠ RHS)
Option 4) 6 and 2
72 + (18 ÷ 6) + (9 × 2) – (24 × 3) = 54 ⇒ 72 + (18 ÷ 2) + (9 × 6) – (24 × 3) = 54
⇒ 72 + (18 ÷ 2) + (9 × 6) – (24 × 3) = 54
⇒ 72 + 9 + (9 × 6) – (24 × 3) = 54
⇒ 72 + 9 + 54 – 72 = 54
⇒ 135 – 72 = 54
⇒ 63 ≠ 54 (LHS ≠ RHS)
Hence, "Option 2" is the correct answer.
