Question 464882: The sum of the ages of a married couple and their four children is 40 years over the century mark. Ten years ago, the sum f the ages of the couple, the eldest child and the second cild is 5 decades and 5 years less than the sum of all (the six of them) their ages now. The fatherīs age is 10 times the age of the youngest, while the motherīs age is 8 times that of the youngest. The difference between the age of the second cild and the sum of the ages of the third and the last child is zero, while the difference between the ages of the first child and the sum of the ages of the second and last child is also zero. How old is the father now?
Answer by solver91311(24713)   (Show Source):
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Let  represent the father's age. Let  represent the mothers age, and let    and  represent the ages of child 1, 2, 3, and 4.
The first thing we know is that:
 (Eq 1)
The next thing we are told is something about the situation 10 years ago:
Which can be simplified to:
 (Eq 2)
Then we are told that
 (Eq 3)
and
 (Eq 4)
and
\ =\ 0) (Eq 5)
and finally:
\ =\ 0) (Eq 6)
Notice the handy (and very necessary) fact that we have 6 equations to match our 6 variables.
Start by combining Equations 1 and 2. Multiply (2) by -1 and add the result to (1) with the following result:
 (Eq 7)
Which can be written as:
 (Eq 8)
Substituting the value of  from Equation 7 into Equation 5, we can deduce that:
 (Eq 9)
Using the value of  from Equation 9 in Equation 6, we can deduce that
 (Eq 10)
Notice that we now have an expression for each of the variables   and  in terms of , (Equations 3, 4, 10, 9, and 8) so make all of the appropriate substitutions:
The rest is simply solving the single variable equation for the value of to get the youngest child's age and then multiplying by 10 to get the father's age.
John
My calculator said it, I believe it, that settles it
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