SOLUTION: the sum of the ages of the parents and the three children is decades over a century.
The father is twice as old as the eldest child.
When the eldest child was born, the sum of
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Question 758092: the sum of the ages of the parents and the three children is decades over a century.
The father is twice as old as the eldest child.
When the eldest child was born, the sum of the parents age is 54.
When the youngest child was born, the sum of the parents age was 70.
In 38 years, the sum of the parents ages will be equal to the sum of the childrens ages.
How old is the second child?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
the sum of the ages of the parents and the three children is decades
over a century.
:
f, m, a, b, c; (a is the oldest child)
The father is twice as old as the eldest child.
f = 2a
:
When the eldest child was born, the sum of the parents age is 54.
(f-a) + (m-a) = 54,
f + m - 2a = 54
we know that f = 2a
2a + m - 2a = 54
m = 54 is mom's age
:
When the youngest child was born, the sum of the parents age was 70.
(f-c) + (m-c) = 70,
f + m - 2c = 70
:
Use elimination with these two equations
f + m - 2c = 70
f + m - 2a = 54
---------------subtraction eliminates f and m
+2a - 2c = 16
simplify
a - c = 8
or
c = a - 8
:
In 38 years, the sum of the parents ages will be equal to the sum
of the children's ages.
(f+38) + (m+38) = (a+38) + (b+38) + (c+38)
combine the values
f + m + 76 = a + b + c + 114
subtract 76 from both sides
f + m = a + b + c + 38
Replace f with 2a
2a + m = a + b + c + 38
subtract a from both sides and we have
a + m = b + c + 38
replace c with (a-8)
a + m = b + a - 8 + 38
subtract a from both sides
m = b + 30
We know that mom is 54
54 = b + 30
b = 54 - 30
b = 24 is the 2nd child's age
:
:
:
Check this mess out by assuming the total of their present ages is
an even 10 (a decade) over 100 and a=30
f = 60 (2a)
m = 54
a = 30
b = 24 (m-30)
c = 22 (a-8)
-------
tot: 190
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