If 6 $ 2 = 20 and 4 $ 3 = 19, then 5 $ 2 = ?

Correct — B. Rule: a $ b = a × b + a + b (the product, plus both numbers).

6 $ 2 = 12 + 6 + 2 = 20 ✓
4 $ 3 = 12 + 4 + 3 = 19 ✓

5 $ 2 = 5 × 2 + 5 + 2
= 10 + 7 = 17 → option (b)

Why the others are wrong

• (a)
  18 — 18 is (5 + 1) × (2 + 1), the same rule without its final − 1. That shortcut gives 7 × 3 = 21 for the first example, not the printed 20.
• (c)
  16 — 16 would need 6 added to the product 10, but the added part is a + b = 5 + 2 = 7. The examples fix it: 6 + 2 = 8 lifts 12 to 20.
• (d)
  15 — 15 is 5 × 2 + 5, adding only the first number. That version gives 6 × 2 + 6 = 18 for the first example, not the printed 20.

A made-up operator hides a rule built from the two numbers. Test the simplest combinations first: sum, product, squares.

The two examples here share a product, 6 × 2 = 12 = 4 × 3. So the gap between 20 and 19 must come from something else, and 6 + 2 = 8 against 4 + 3 = 7 accounts for it exactly.

That gives product plus sum. The same rule can be written (a + 1)(b + 1) − 1.

Key facts

• 6 × 2 + 6 + 2 = 20 and 4 × 3 + 4 + 3 = 19.
• 5 × 2 + 5 + 2 = 17.
• a × b + a + b equals (a + 1)(b + 1) − 1.

Common traps

• Stopping at a rule that fits one example: 3a + b gives 20 for 6 $ 2 but 15 for 4 $ 3.
• Writing (a + 1)(b + 1) and forgetting the − 1, which gives 18.

No directly related past PYQ was found.