10 years ago, a man’s age was 5 times of his son’s age. 2 years hence, twice his age will be equal to 4 times the age of his son. What is the present age (in years) of the son?

The correct option is 14.

Let the present age of the son be s years and the present age of the father (man) be f years.

Step 1: Formulate the equation for 10 years ago.
10 years ago:
Son's age = s - 10
Father's age = f - 10

According to the given condition, 10 years ago, the man's age was 5 times his son's age:
$f-10=5\×(s-10)$
$f-10=5s-50$
$f=5s-40$    --- (Equation 1)

Step 2: Formulate the equation for 2 years hence (in the future).
2 years from now:
Son's age = s + 2
Father's age = f + 2

According to the given condition, twice the man's age will be equal to 4 times the age of his son:
$2\×(f+2)=4\×(s+2)$
Divide both sides by 2:
$f+2=2\×(s+2)$
$f+2=2s+4$
$f=2s+2$    --- (Equation 2)

Step 3: Solve for the present age of the son (s).
Equating Equation 1 and Equation 2:
$5s-40=2s+2$
$5s-2s=2+40$
$3s=42$
$s=\frac{42}{3}=14$

Thus, the present age of the son is 14 years.