A boy is 42 years younger than his father .In 8 years' times , he will be 1/4 times as old as his father .Find their present ages.
(Answer is boy= 6, father= 48)
correct
I just did this in the previous post
To solve this problem, we can make use of algebraic equations.
Let's assume the boy's present age is represented by 'b' and the father's present age is represented by 'f'.
We are given two pieces of information. Firstly, the boy is 42 years younger than his father, so we can write:
b = f - 42 -- Equation 1
Secondly, in 8 years' time, the boy will be 1/4 times as old as his father. So, in 8 years, the boy's age will be (b + 8) and the father's age will be (f + 8). This can be expressed as:
(b + 8) = (1/4) * (f + 8) -- Equation 2
Now, we have two equations with two variables (b and f). We can solve this system of equations to find their present ages.
First, let's substitute the value of 'b' from Equation 1 into Equation 2:
(f - 42 + 8) = (1/4) * (f + 8)
Simplifying this equation:
f - 34 = (1/4) * f + 2
Multiplying both sides by 4 to eliminate the fraction:
4f - 136 = f + 8
Bringing all terms with 'f' to one side:
4f - f = 8 + 136
Simplifying:
3f = 144
Dividing both sides by 3:
f = 48
Now that we have the value of 'f', we can substitute it back into Equation 1 to find 'b':
b = 48 - 42
b = 6
Therefore, the boy's present age is 6 and the father's present age is 48.