a family has 3 children the eldest is 8 years older than the middle child and the youngest is 12 years younger than the eldest child of the total of their age is 88 how old is each child?
If their ages are a < b < c then
c = b+8
a = c-12
a+b+c = 88
...
E-8=M
Y+12=E
E+Y+M=88
putting it in standard form..
E-M+0y=8
E-oM-y=12
E+M+Y=88
add the last two
2E+M=100 add to the first equation
3E=108
E=36
M=28
Y=24
To solve this problem, let's use algebra. Let's assume the middle child's age is "x" years.
According to the given information:
- The eldest child is 8 years older than the middle child. So, the eldest child's age is (x + 8) years.
- The youngest child is 12 years younger than the eldest child. So, the youngest child's age is (x + 8 - 12) years, which simplifies to (x - 4) years.
The sum of their ages is 88. Therefore, we can form an equation:
x + (x + 8) + (x - 4) = 88
Now, let's solve this equation step by step:
Combining like terms:
3x + 4 = 88
Subtracting 4 from both sides:
3x = 88 - 4
3x = 84
Dividing both sides by 3:
x = 84 / 3
x = 28
So, the middle child is 28 years old.
Substituting this value back into our expressions for the eldest and youngest children's ages:
- The eldest child's age is (28 + 8) = 36 years.
- The youngest child's age is (28 - 4) = 24 years.
Therefore, the ages of the children are:
Middle child: 28 years
Eldest child: 36 years
Youngest child: 24 years.