SOLUTION: John is 43 years old and his daughter Susan is 7. In n-years, John will be four times as old as Susan is at that point. Find n and the ages of John and Susan after n-years.
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-> SOLUTION: John is 43 years old and his daughter Susan is 7. In n-years, John will be four times as old as Susan is at that point. Find n and the ages of John and Susan after n-years.
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Question 1174209: John is 43 years old and his daughter Susan is 7. In n-years, John will be four times as old as Susan is at that point. Find n and the ages of John and Susan after n-years.
Write the equation for their ages in n years
43 + n = 4*(7 + n).
Now solve it
43 + n = 28 + 4n
43 - 28 = 4n - n
15 = 3n
n = 15/3 = 5.
Answer. a) n= 5: in 5 years.
b) At that time, they will be 48 (the father) and 12 (Susan) years old.
The other tutor showed a typical formal algebraic solution; you should certainly understand that method and know how to use it.
Here is a less formal path to the solution using logical reasoning. Solving a problem like this can give you very beneficial mental exercise.
(1) The difference in their ages is (and always will be) 43-7 = 36 years.
(2) When John is 4 times as old as his daughter, the difference in their ages will be 3 times her age. (If we want to use a bit of very basic algebra here, we are saying 4x-x = 3x.)
(3) So when John is 4 times as old as his daughter, 3x=36 so x=12.
That means he will be 4 times as old as her when she is 12 and he is 4*12 = 48.
Since their ages are now 43 and 7, that will be 5 years from now.
ANSWERS: 5 years from now their ages will be 12 and 48.
