Sep 21, 2009

Age Problems - Equations in single variable - Algebra

This worksheet is based on age word problems which come under equations in one variable. Intermediate Algebra, Algebra – I and Algebra – II students can practice these questions.
1. Miguel is 46 years old. He is 4 years older than thrice his son’s age. Find the age of his son.
Solution:
Age of Miguel = 46 years
Age of Son       = x years
So, 46               = 4 +3x
Subtracting 4 on both sides,
                  42   =3x
Dividing by 3 on both sides,
                     X =14
Therefore, the Son is 14 years old.
2. Pete, Bryan and Philip are cousins. Pete’s age is one-third of Bryan and Philip is five years elder than Bryan. If the sum of the age of the cousins is 40, find the ages of each.
Solution:
From the statement,
Age of Bryan= x
Age of Pete   =x/3
Age of Philip = x+5
Sum of the ages=40.
X/3 + x+x+5= 40
(7x+15)/3 =40
Multiplying by 3 on both sides,
7x +15=120
Subtracting 15 on both sides,
7x=105
Dividing by 7 on both sides,
X=15
Therefore, age of Pete=15/3=5
Age of Bryan                 =15
Age of Philip                  =15+5=20
3. Ana is 5 years more than Jack. The sum of their ages is 29. Find the ages of Ana and Jack.
4. Tina is 3 years younger than Tom. The ages are in the ratio 2:3. Find their ages.
5. George is 8 years more than Christopher and Ford is 2 years younger than Christopher. The sum of their ages is 60. Find the ages of George, Christopher and Ford.
6. Devon is 12 years old. His age is 2 more than half the age of Steven. Find the age of Steven.
7. Mrs. Smith is 8 years more than twice the age of his son. The ages of son is 12.Find the age of mother and find the difference between their ages.
8. Kate is 12 years old. His age is 4 times the age of Robbie. Find the age of Robbie.
9. The sum of present age of Abe and the age before 7 years is 31. Find the present age of Abe. What will be his age after 7 years?
10. George is 8 years more than Christopher and Ford is 2 years younger than Christopher. The sum of their ages is 60. Find the ages of George, Christopher and Ford.
11. Margaret is 8 years more than twice the age of his son. The age of son is 12. Find the age of mother and find the difference between their ages.
12. Miguel is 46 years old. He is 6 years older than thrice his son’s age. Find the age of his son.
13. Nick and Kane were born on consecutive years and on same date. Nick is younger.  The sum of their ages is 11. Find the age of the brothers.
14. The ages of Ashley and Mary are in the ratio 4:7. The sum of their ages is 22. Find the ages of Ashley and Mary.
15. The ages of Peter, Margaret and Jack are consecutive odd numbers. The sum of the ages of Peter and Margaret equals Jack’s age before 5 years. Find the ages of Peter, Margaret and Jack.
16. The sum of the ages of Tony and Teddy equals thrice their mother’s age. Tony is two years younger than Teddy. If the age of the mother is 42, find the ages of Tony and Teddy?
17. At present Allen is 10 years younger than Scott. Five years from now, Scott’s age will be 2 times the present age of Allen. Find the present age of Allen and Scott.
 18. The present ages of Lewis and Brown are in the ratio 1:2. Three years from now, the ages will be in the ratio 3:5. Find the present ages of Lewis and Brown.

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Posted by Mr. Math at 4:37 PM
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15 comments:

  1. Great Website..Thank You for the help provided !
    :)Regards,
    Well-wisher
    ReplyDelete
  2. thank you so much..it helps me a lot..
    can i ask a favor?
    can you show me how to solve this..it's confusing..and also show me an example in ratio form..

    Two times the father's age is 8 more than six times his son's age. Ten years ago, the sum of their ages was 44. The age of the son is:

    thank you so much..!:)
    ReplyDelete
  3. Son's age = x

    2 * Father's age = 8 + 6x

    Divide by 2 on both sides

    Father's age = 4 + 3x

    Son's age (10 years ago) = x - 10 ....(1)
    Father's age (10 years ago) = (4+3x)-10
    = -6+3x .... (2)
    Add equation (1) and (2, set it equals to 44 and solve for x.
    ReplyDelete
  4. In four years, Donna's grandmother will be eight times as old as Donna last year. When Donna's present age is added to his grandmother's present sge, the total is 72. How old is each one now?
    ReplyDelete
  5. this helped me a lot!

    thanks!

    ..........♥..............
    ReplyDelete
  6. will you please help me solve the problem number 4..thank you
    ReplyDelete
  7. ..it makes our assignments clearer!!
    ReplyDelete
  8. thank you i used it in my career examination
    ReplyDelete
  9. i didnt get why the 2nd one becomes (7x+15)/3?????
    ReplyDelete
  10. x/3 + x + (x+5) = 40

    Multiplying every term by 3, we have

    (x/3)*3 + x * 3 + (x+5) * 3 = 40 * 3

    x + 3x + 3x + 15 = 120

    7x + 15 = 120

    or

    x/3 + x + (x+5)

    This can be written as

    x/3 + x/1 + (x+5)/1

    Denominators are 3,1,1

    So, LCM = 3

    (x + x*3 + (x+5)*3)/3

    (x+3x+3x+15)/3

    (7x+15)/3

    Hope you understand this.
    ReplyDelete
  11. Thank You help a lot for my exam =)
    ReplyDelete
  12. ,,great posts.. big help to math teachers.. tnx lots
    ReplyDelete
  13. Can you please explain to me question 17 on the page on algebraic age problems? Thanks a lot!
    ReplyDelete
  14. There was an error with question 17 and is corrected now. Please let me know in case of any doubt.
    ReplyDelete
  15. Pls. answer this.

    Ana is three times as old as Rain. Four years from now she will be twice as old as rain. How old are they now? I need the Representation and the Solution. Thanks
    ReplyDelete

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