SOLUTION: Jasmine is nw half as old as her father.12 years ao, the father was 3 times as old as jasmine.find their ages.

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Question 1062471: Jasmine is nw half as old as her father.12 years ao, the father was 3 times as old as jasmine.find their ages.

Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!
"nw", what?

j Jasmine
f Father






-







Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Jasmine is nw half as old as her father.12 years ao, the father was 3 times as old as jasmine.find their ages.
Let Jasmine's age be J
Then father's age is: 2J
We then get: 2J - 12 = 3(J - 12)
2J - 12 = 3J - 36
2J - 3J = - 36 + 12
- J = - 24
J, or Jasmine is:
Father's age:
It's as simple as that.....nothing COMPLEX and/or CONFUSING!
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
Jasmine is now half as old as her father. 12 years ago, the father was 3 times as old as jasmine.find their ages.
~~~~~~~~~~~~~~~~~~~~~

F = 2J,              (1)   ("Jasmine is now half as old as her father")
F - 12 = 3(J-12).    (2)   ("12 years ago . . .")

From (1), substitute F = 2J into (2) replacing F. You will get the single equation for J:

2J - 12 = 3(J-12)  --->  2J - 12 = 3J - 36  --->  J = 36 - 12 = 24.

Answer.  Jasmin is 24 years old now. The father is 2*J = 2*24 = 48 years old now.

Solved.


There is a bunch of lessons on age word problems
    - Age problems and their solutions
    - A fresh formulation of a traditional age problem
    - Really intricate age word problem
    - Selected age word problems from the archive
    - Age problem for the day of April, 1
    - OVERVIEW of lessons on age problems
in this site.

Read them and become an expert in solving age problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Age word problems".



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