Correct option is B
Logic:
Symbol | × | # | @ | $ | & |
Means | - | + | = | ÷ | × |
Option a) : (6#2)$2@(2&8)X1
After changing the symbols from given information the equation is as follows:
=> (6 + 2) ÷ 2 =(2 × 8)- 1
=> 8 ÷ 2= 16 - 1
=> 16 =16 - 1
=> 16 = 15 → False
LHS ≠ RHS
Option b) : (15X4)&3@(60$2)#3
After changing the symbols from given information the equation is as follows:
=> (15 - 4) × 3 = (60 ÷ 2)+ 3
=>11 × 3=30 + 3
=> 33 = 33 → True
LHS = RHS
Option c) : (81X4)&3@4#16X3
After changing the symbols from given information the equation is as follows:
=> (81 - 4) × 3 = 4 + 16 - 3
=> 77 × 3= 4 + 16 - 3
=> 231 =4 + 16- 3
=> 231 =20 - 3
=> 231 = 17 → False
LHS ≠ RHS
Option c) : (81X4)&3@8$15&7
After changing the symbols from given information the equation is as follows:
=> (81 - 4) × 3 = 8 ÷ 15 × 7
=> 77 × 3 =8 ÷ 15 × 7
=> 77 × 3=0.533 × 7
=> 231= 3.733 → False
LHS ≠ RHS
∴ Here '(15X4)&3@(60$2)#3' is true after changing the symbols.
Thus, the correct answer is (b) "(15X4)&3@(60$2)#3".
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