SOLUTION: Mrs. Andrew’s is two years less than four times as old as her daughter Janelle. In six years Janelle will be three-eighths as old as her mother. How old is each one now?
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Question 1175640: Mrs. Andrew’s is two years less than four times as old as her daughter Janelle. In six years Janelle will be three-eighths as old as her mother. How old is each one now? Found 2 solutions by greenestamps, Boreal:Answer by greenestamps(13200) (Show Source):
There are dozens of ways to set this problem up for solving.
I will show you a path to the solution that is a bit unusual but that "works" well for me.
You will probably get responses from other tutors that show more traditional algebraic solution methods.
"In six years Janelle will be three-eighths as old as her mother."
With that statement, most ways of setting up the problem would involve working with fractions. I find it easier to avoid fractions, as the computations are a bit easier and faster without them.
So my start on the problem is this:
Let the ages of Mrs. Andrew and her daughter 6 years from now be 8x and 3x.
Then their current ages are 8x-6 and 3x-6.
Mrs. Andrew's age is 2 less than 4 times Janelle's age:
ANSWERS:
Mrs. Andrew's age is 8x-6 = 40-6 = 34
Janelle's age is 3x-6 = 9
You can put this solution on YOUR website! daughter is x
Mrs. Andrew is 4x-2
in 6 years, Janelle is x+6
x+6=(3/8)(4x+4)
That is the set up.
x+6=1.5x+1.5
-.5x=-4.5
x=9
Janelle is 9 and her mother is 34, which is the answer
In 6 years, she will be 15 and her mother 40--15 is 3/8 of 40.
