Saturday, March 15, 2008

Age Problem 1 Solution

For a non-algebraic approach:

If Steves current age was 20 then 20 years ago he would be zero, so he has to be at least 21 years old. This would make Bob 42 and Steve 21. 20 years ago they would be 22 and 1, which is 22 times as old and not 6 times. Were not that far off though. So go up by increments. Try Steve at 22 and Bob 44-- 20 years ago would be 2 and 24 which is 12 times-- at 23 and 46 we can see than 20 years ago they would be 3 and 26 which is over 8 times were close-- and at 24 and 48 20 years ago would make 4 and 28 which is 7 times and really close.. at 25 and 50 20 years ago we would have 5 and 30 which IS 6 times older and what were looking for. So, Steve is 25 and Bob is 50...


For the algebraic approach we know that if something is twice as old, big, fast, etc as something else it is 2x more than x. So in this problem Bob is 2x as old as x Steve is.

Twenty years ago would be :

2x - 20 for Bob

and

x - 20 for Steve

but, 20 years ago Bob was 6 times older so

2x - 20 = 6(x - 20)

2x - 20 = 6x - 120

4x = 100

x = 100/4

x = 25

So x or Steve is 25 and Bob is twice or 2(25): 50 years old.

6 comments:

  1. Woow! thanks for the algebra problem now i can answer my assignment given by my math teacher.

    ReplyDelete
  2. Bob is 50 yrs. old and Steve is 25 yrs. old.

    ReplyDelete
  3. their ages are interchanged. steve's age is x. since x is 25, so steve is 25. and bob's is 2x, so 2(25)=50

    ReplyDelete
  4. i'm still confused and did not get it. :(

    ReplyDelete
  5. past present
    Bob 2x-20 2x
    steve 6(x-20) x

    2x-20 = 6(X-20)
    2x -20 = 6x - 120
    120 - 20 = 6x - 2x
    x = 25

    ReplyDelete
  6. Wow! thanks for the algebra problem now i can answer my assignment given by my math teacher.

    ReplyDelete


Interested in improving you brain the fun way? Check out BrainGamesFun.com