Question 1194746: The average age of a family of eight is 30 years. The average age of the six children is 19 years. If the mother is four years younger than the father. Calculate the age of the father.
Found 3 solutions by josgarithmetic, MathTherapy, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
The average age of a family of eight is 30 years. The average age of the six children is 19 years. If the mother is four years younger than the father. Calculate the age of the father.
With average age of the 8 members being 30, ages of all 8 members TOTAL 8(30) = 240 years
With average age of the 6 children being 19, the ages of ALL 6 children TOTAL 6(19) = 114 years
Therefore, mother's and father's ages TOTAL 240 - 114 = 126 years
Let father's age be F
Then mother's is, F - 4
Since mother's and father's ages sum to 126 years, we get: F + F - 4 = 126
2F = 130
Father, or 
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
You have received two responses to your post, one showing setting up the problem in a very formal algebraic way, the other using an easier and less formal solution.
Here is another way you can solve problems like this that involve averages of numbers that are relatively close together.
The average age of all eight family members is 30; the average age of the 6 children is 19. That means the sum of the ages of all 6 children is, all together, 6(30-19) = 6(11) = 66 years below average.
To balance that, the sum of the ages of the mother and father must be 66 years above the average.
Since the mother is 4 years younger than the father, a bit of mental arithmetic or logical reasoning, or perhaps some basic algebra, determines that the father is 35 years above the average and the mother is 31 years above the average.
So the father is 30+35 = 65 years old, and the mother is 30+31 = 61 years old.
ANSWER: The father's age is 65 years.
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