A boy is 10 years older than his brother. In 4 years he will be twice as old an his brother. Find the present age of each. 2. A father is 4 times as old as his son. In 20 years the father will be twice as as his son. Find the present age of each.
The Correct Answer and Explanation is :
Let’s solve each of these problems step-by-step.
Problem 1: The Boy and His Brother
- Define Variables:
- Let the present age of the brother be ( x ).
- Then, the age of the boy is ( x + 10 ) (since he is 10 years older).
- Set Up the Equation:
- In 4 years, the brother’s age will be ( x + 4 ).
- In 4 years, the boy’s age will be ( (x + 10) + 4 = x + 14 ).
- According to the problem, in 4 years, the boy will be twice as old as his brother:
[
x + 14 = 2(x + 4)
]
- Solve the Equation:
- Expand the equation:
[
x + 14 = 2x + 8
] - Rearranging gives:
[
14 – 8 = 2x – x \quad \Rightarrow \quad 6 = x
]
- Find Present Ages:
- Brother’s age: ( x = 6 ) years
- Boy’s age: ( x + 10 = 16 ) years
Problem 2: The Father and His Son
- Define Variables:
- Let the present age of the son be ( y ).
- Then, the father’s age is ( 4y ) (since he is 4 times as old as his son).
- Set Up the Equation:
- In 20 years, the son’s age will be ( y + 20 ).
- In 20 years, the father’s age will be ( 4y + 20 ).
- According to the problem, in 20 years, the father will be twice as old as his son:
[
4y + 20 = 2(y + 20)
]
- Solve the Equation:
- Expand the equation:
[
4y + 20 = 2y + 40
] - Rearranging gives:
[
4y – 2y = 40 – 20 \quad \Rightarrow \quad 2y = 20 \quad \Rightarrow \quad y = 10
]
- Find Present Ages:
- Son’s age: ( y = 10 ) years
- Father’s age: ( 4y = 40 ) years
Summary of Ages:
- The boy is 16 years old, and his brother is 6 years old.
- The father is 40 years old, and his son is 10 years old.
This methodical approach of defining variables, setting up equations based on the relationships described in the problem, and systematically solving for the variables ensures clarity and accuracy in finding the present ages.