👴👦 Let’s solve the problem! 🤔

Let the present age of the father be F years. 📆

Let the present age of the son be S years. 📆

10 years ago, the father’s age was F – 10 years. 🕰️

10 years ago, the son’s age was S – 10 years. 🕰️

According to the problem, 10 years ago, the father’s age was thrice the son’s age. 📝

So, F – 10 = 3(S – 10)

F – 10 = 3S – 30

F = 3S – 20

Now, let’s consider the situation 10 years hence. 🔮

10 years hence, the father’s age will be F + 10 years. 🕰️

10 years hence, the son’s age will be S + 10 years. 🕰️

According to the problem, 10 years hence, the father’s age will be twice the son’s age. 📝

So, F + 10 = 2(S + 10)

F + 10 = 2S + 20

F = 2S + 10

Now we have two equations: 📝

F = 3S – 20 … (1)

F = 2S + 10 … (2)

Equating (1) and (2), we get: 🤔

3S – 20 = 2S + 10

3S – 2S = 10 + 20

S = 30

Now, substitute S = 30 in equation (1) or (2) to find F. 📝

F = 3S – 20

F = 3(30) – 20

F = 90 – 20

F = 70

So, the present age of the father is 70 years. 📆

And, the present age of the son is 30 years. 📆

The ratio of their present ages is: 📊

F : S = 70 : 30

= 7 : 3

The final answer is: 🎉 The ratio of their present ages is 7 : 3. 👴👦