If # = +, @ = ×, % = −, then (5 # 2) @ 3 % 4 = ?

Correct — B.
Rule: swap # → +, @ → ×, % → −, and keep the brackets.

(5 # 2) @ 3 % 4 becomes (5 + 2) × 3 − 4

Bracket: 5 + 2 = 7
Multiply: 7 × 3 = 21
Subtract: 21 − 4 = 17 → option (b).

Why the others are wrong

• (a)
  20 — 20 comes out of neither reading. Brackets first gives 21 − 4 = 17, and dropping the bracket gives 5 + 2 × 3 − 4 = 7.
• (c)
  18 — 18 would need the last step to subtract 3. The % stands before 4, so the step is 21 − 4 = 17.
• (d)
  21 — 21 stops at (5 + 2) × 3 and drops the last step. The % 4 still applies as minus 4, giving 17.

Here each symbol stands for one ordinary sign, so the item tests order of operations more than decoding.

The bracket around 5 # 2 forces the addition first. Without it, multiplication would come first: 5 + 2 × 3 − 4 = 7.

So swap the symbols, keep the brackets exactly where they are, and work outwards: bracket, then ×, then −.

Key facts

• In this item # is +, @ is × and % is −.
• Brackets are worked before × and ÷.
• (5 + 2) × 3 − 4 = 17, while 5 + 2 × 3 − 4 = 7.

Common traps

• Dropping the brackets after swapping symbols, which gives 5 + 2 × 3 − 4 = 7.
• Stopping at 7 × 3 = 21 and missing the % 4.

No directly related past PYQ was found.