# Age Word Problems – Solving Age Related Math Problems

Age word problems come in a variety of forms. Learn how to solve age word problems with 2 variables, 3 variables as well as age related problems involving past, present and future ages.

###### Age Word Problems (2 variable)

Bruce is 4 times as old as Benjamin and is also 6 years older than Benjamin. How old is Bruce? How old is Benjamin?

Let A = Bruce and B = Benjamin

$A = 4B$ A is four times older than B

$A = B + 6$ A is 6 years older than B or we can rewrite this as$B = A - 6$

Now, we use substitution to solve for a single variable. Let's solve for Bruce's age first (A).

$A = 4(A-6)$

$A = 4A-24$

$24 = 3A$

$A = 8$

so Bruce = 8 years old and Benjamin = 2 years old Age Word Problems (3 variable)

###### Age Word Problems (3 variable)

John is 3 years older than Jim. Jim is 4 years less than twice David’s age. How old are
the three boys if their ages add up to 35?

Let N = John and let M = Jim and let D = David

$N + M + D = 35$

$N = M + 3$

$M = 2D - 4$ which can be written as $D = \frac{M + 4}{2}$

Now, we use substitution to solve for a single variable. Let's solve for Jim's age first (M).

$(M + 3) + M + (\frac{M + 4}{2}) = 35$

simplifying...

$2M+ (\frac{M + 4}{2}) = 32$

$2M+ \frac{M}{2} + 2 = 32$

$2M+ \frac{M}{2} = 30$

$\frac{4M + 2M}{2} = 30$

$\frac{5M}{2} = 30$

$5M = 60$

$M = 12$ so Jim = 12 years old and John = 15 Years old and David = 8 years old
###### Age Word Problems (past, present and future)

In January of the year 2000, JOHN was one more than eleven times as old as his son WILLIAM. In January of 2009, JOHN was seven more than three times as old as WILLIAM. How old was WILLIAM in January of 2000?

Let J=John and let W=William

In 2000: $J = 11W + 1$

In 2009: $J+9 = 3(W + 9) + 7$ (we add 9 years to both John and William)

Simplifying...

$J = 3W + 27 + 7 - 9$

$J = 3W + 25$

Substituting...

$3W+25=11W+1$

Simplifying...

$8W = 24$

$W = 3$ William = 3 years old in 2000

In three more years, Miguel’s grandfather will be six times as old as Miguel was last year. When Miguel’s present age is added to his grandfather’s present age, the total is 68. How old is each one now?

Let M = Miguel and let G = Grandfather

$M + G = 68$

$G+3=6(M-1)$ add 3 for G and subtract 1 from M
Simplifying...

$G = 6M-6-3$

$G = 6M-9$

Substituting...

$M + 6M-9 =68$

$7M=77$

$M=11$ Miguel = 11 years old

$11+G=68$

$G=57$ Grandfather = 57 years old

Fran is now twice as old as Pauline. Five years ago Fran was three times as old as Pauline. Find their present ages.

Let F = Fran and let P = Pauline

$F = 2P$

$F-5 = 3(P-5)$ rewrite as $F= 3P-10$

Substituting...

$3P-10=2P$

$P=10$ Pauline = 10 years old

$F=2(10)$ so Fran = 20 years old



John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now?

Let F = John's Father, Let J = John, Let A = Alice

$(F+2) + (J+2) + (A+2)=58$ or $F+J+A=52$

$F=5J$

$J=2A$ or $A=\frac{J}{2}$

We have 3 variables and 3 equations... Now, we can substitute and simplify...

Let's isolate and solve for J...

$5J+J+\frac{J}{2}=52$

$6J+\frac{J}{2}=52$

$\frac{13J}{2}=52$

$13J=104$

$J=8$ John = 8 years old