Question

Rakesh is half his father`s age now. Ten years ago , he was 20 years younger to his father.

What are their present ages ? Pls explain in detail

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Answer 1

Answer:

The present age of the son is 20 years and present age of father is 40 years.

Step-by-step explanation:

Let us assume, the present age of son and father be x years and y years respectively.

Given:

x = 1/2y -----------1

Also given:

x  - 10 = y – 10 – 20

y – x = 20 -----------2

Substitute the value of x from equation 1 in equation 2

y – 1/2y = 20

y = 40 years

Therefore, x = 1/2 * 40 = 20 years

The present age of the son is 20 years and present age of father is 40 years.

Pls mark brainliest

Answer 2

Answer:-

$\red{\bigstar}$ Rakesh's Present age

$\large\leadsto\boxed{\sf{20 \: years}}$

$\red{\bigstar}$ Rakesh's Father's Present age

$\large\leadsto\boxed{\sf{40 \: years}}$

• Given:-

Rakesh is half his father's age

10 years ago, he was 20 years younger to his father.

• To Find:-

Present ages of Rakesh and his father = ?

• Solution:-

Let the age of Rakesh be = x

Let the age of Rakesh's father be = y

• ATQ:-

→ $\sf{x = \dfrac{1}{2}y}\: \: \: \: \: \: \longrightarrow[eqn.1]$

Also,

→ $\sf{x - 10 = y - 10 - 20}$

→ $\sf{x - 10= y -30}$

→ $\sf{x - y = -30 + 10}$

→ $\sf{x - y = -20}\: \: \: \: \: \: \longrightarrow[eqn.2]$

• Substituting the value of x in [eqn.2]:-

→ $\sf{\dfrac{1}{2}y - y = -20}$

→ $\sf{\dfrac{y-2y}{2} = - 20}$

→ $\sf{\dfrac{-y}{2} = - 20}$

→ $\sf{-y = -20 \times 2}$

→ $\sf{-y = -40}$

→ $\large\boxed{\sf{y= 40}}$

• Substituting the value of y in eqn.1:-

→ $\sf{x = \dfrac{1}{2} \times 40}$

→ $\large\boxed{\sf{x = 20}}$

Therefore,

• Rakesh's Present age =

$\large\boxed{\sf{20 \: years}}$

• Rakesh's Father's Present age =

$\large\boxed{\sf{40 \: years}}$