In a quantitative aptitude problem on ages, the present age of Dr Pandey is four times the present age of his son. After 10 years, the age of Dr Pandey will be exactly twice the age of his son at that time. What is the present age, in years, of the son of Dr Pandey?

Introduction / Context:
This is a standard aptitude question on ages where present ages are connected to future ages using simple linear relationships. Such problems test whether you can convert English statements like “four times” and “twice the age after 10 years” into algebraic equations, and then solve them correctly. Mastering this pattern is very useful for bank exams, SSC exams and other quantitative reasoning tests.

Given Data / Assumptions:

• The present age of Dr Pandey is four times the present age of his son.
• After 10 years, the age of Dr Pandey will be exactly twice the age of his son at that time.
• We are asked to find the present age of the son in years.
• All ages are measured in whole years and are positive.

Concept / Approach:
The idea is to assume a variable for the unknown present age of the son and then write expressions for present and future ages. Phrases like “four times” translate to multiplication by 4, and “after 10 years” means we add 10 to the present age. The statement that one age becomes “twice” another age after a fixed time becomes an equation which we can solve using basic algebra. Once the value of the variable is obtained, we interpret it as the required present age of the son.

Step-by-Step Solution:
Step 1: Let the present age of the son be x years. Step 2: Then the present age of Dr Pandey is 4x years, because it is four times the age of his son. Step 3: After 10 years, the son will be x + 10 years old. Step 4: After 10 years, Dr Pandey will be 4x + 10 years old. Step 5: According to the question, after 10 years the age of Dr Pandey will be twice the age of his son, so we form the equation 4x + 10 = 2 * (x + 10). Step 6: Expand the right side: 4x + 10 = 2x + 20. Step 7: Rearrange to bring variables to one side: 4x - 2x = 20 - 10, so 2x = 10. Step 8: Divide by 2 to get x = 5. Step 9: Therefore the present age of the son is 5 years.

Verification / Alternative check:
We can quickly verify the result by substituting back into the story. If the son is 5 years old now, Dr Pandey is 4 * 5 = 20 years old. After 10 years, the son will be 5 + 10 = 15 years old and Dr Pandey will be 20 + 10 = 30 years old. At that time, 30 is exactly twice 15, which perfectly matches the condition in the question, so the solution is consistent.

Why Other Options Are Wrong:
Option A (4 years) would make the father 16 years old now and 26 after 10 years, while the son would be 14 after 10 years. Then 26 is not twice 14, so this is incorrect.
Option C (6 years) would give present ages 6 and 24. After 10 years, they become 16 and 34, and 34 is not twice 16, so this option fails.
Option D (8 years) would give present ages 8 and 32. After 10 years, the ages would be 18 and 42, and 42 is not twice 18, so this is also wrong.

Common Pitfalls:
A common mistake is to confuse present ages with future ages and directly apply the “twice” relation on present ages. Another frequent error is to forget that both father and son age by the same number of years, in this case 10 years. Carefully translating each line of the statement into algebra avoids these mistakes.

Final Answer:
The correct present age of the son of Dr Pandey is 5 years.