The Smith family has 13 children, all born exactly 2 years apart. How old will the second-youngest child be when the oldest child is five times the age of the youngest child?
ages then:
n, n-2, n-4, n-6, n-2(13-1)
so last kid is n-24
5*(n-24)=n
n=5*6=30
youngestg will be 30-2(12)=6
To find the age of the second-youngest child when the oldest child is five times the age of the youngest child, we can approach this problem step by step.
Let's assume the age of the youngest child is "x" years. Since the Smith family has 13 children, the age of the oldest child would be x + 24 (2 years apart for each child, so 12 times 2 = 24).
According to the problem, the oldest child will be five times the age of the youngest child when the youngest child is x years old. Mathematically, this can be written as:
x + 24 = 5x
To solve this equation for x, we need to isolate x on one side of the equation. Let's begin the process:
Subtract x from both sides:
24 = 4x
Divide both sides by 4:
6 = x
So, the youngest child is 6 years old.
Now, to find the age of the second-youngest child when the oldest child is five times the youngest child's age, we subtract 2 (since the children are born exactly 2 years apart) from the age of the oldest child.
The age of the oldest child = x + 24 = 6 + 24 = 30 years
The age of the second-youngest child = 30 - 2 = 28 years.
Therefore, the second-youngest child will be 28 years old when the oldest child is five times the age of the youngest child.