Two natural numbers differ by 2 and the sum of their squares is 202. The sum of the numbers is :

1. 14

2. 16

3. 18

4. 20

Let the two natural numbers be x and x + 2.

Given,

Sum of the squares of numbers is 202.

⇒ x² + (x + 2)² = 202

⇒ x² + x² + 4 + 4x = 202

⇒ 2x² + 4x + 4 - 202 = 0

⇒ 2x² + 4x - 198 = 0

⇒ 2x² + 22x - 18x - 198 = 0

⇒ 2x(x + 11) - 18(x + 11) = 0

⇒ (2x - 18)(x + 11) = 0

⇒ (2x - 18) = 0 or (x + 11) = 0     [Using zero-product rule]

⇒ 2x = 18 or x = -11

⇒ x = $\dfrac{18}{2}$ or x = -11

⇒ x = 9 or x = -11.

Since, the number required is natural number, thus x ≠ -11,

⇒ x + 2 = 9 + 2 = 11

Sum of the two numbers = 9 + 11 = 20.

Hence, option 4 is the correct option.