Age Word Problems

 

1. Sid Upp’s father is four times as old as Tip. Five years ago he was seven times as old. Find the absolute value of the difference of the ages now.

Solution

Let x = Sid’s age now   (smaller number)

4x = father’s age now

The problem tells you about their ages at another time. Five years ago your age would be 5 less than your age now.

So

x – 5 = Sid’s age 5 years ago

4x – 5 = father’s age 5 years ago

Five years ago he was seven times as old (as he was)

4x – 5 = 7(x – 5)

4x – 5 = 7x – 35

4x = 7x – 30

-3x = -30

Divide both sides of the equation by 3

x = 10

So,

Sid is 10 and the father is 40.

 

2. Abbott is 6 years older than Costello. Six years ago Abbott was twice as old as Costello. How old is each now?

Solution

Let x = Costello’s age now   (smaller number)

x + 6 = Abbott’s age now

Then

x – 6 = Costello’s age 6 years ago

(x + 6) – 6 = Abbott’s age 6 years ago

Six years ago she was twice as old as him.

Equation

(x + 6) – 6 = 2(x – 6)

x = 2x – 12

-x = -12

x = 12 and x + 6 = 18

So,

Costello is 12 and Abbott is 18.

 

3. Willie’s Father is 26 years older than Willie. In 10 years, the sum of their ages will be 80. What are their present ages?

Solution

Now

Let x = Willie’s age now.

Let x + 26 = Father’s age now.

In 10 Years

x + 10 = Willie’s age in 10 years

(x + 26) + 10 = Father’s age in 10 years

Equation

x + 10 + (x + 26) + 10 = 80

2x + 46 = 80

x = 17 and x + 26 = 43

So,

Willie is 17 now.

Willie’s father is 43 now.

 

4. Tess T. Fye is twice as old as Inda Cates. If 8 is subtracted from Inda Cate’s age and 4 is added to Tess T. Fye’s age, Tess T. Fye will then be four times as old as Inda Cates. Find their ages.

Solution

Let x = Inda Cates age now.

Let 2x = Tess T. Fye’s age now.

Then

x – 8 = 8 subtracted from Inda Cate’s age.

2x + 4 = 4 added to Tess T. Fye’s age.

Equation:

2x +  4 = 4(x – 8)

2x + 4 = 4x – 32

-2x = -36

x = 18

2x = 36

So,

Tess T. Fye’s age is 36 years.

Inda Cate’s age is 18 years.

5. A man is four times as old as his son. In 3 years, the father will be three times as old as his son. How old is each now?

Solution

Let x = son’s age now.

Let 4x = Father’s age now.

In 3 Years:

x + 3 = son’s age in 3 years

4x + 3 = father’s age in 3 years

In 3 years, the father will be three times as old as his son.

Equation:

4x + 3 = 3(x + 3)

4x + 3 = 3x + 9

x = 6 and 4x = 24

So,

The father’s age now is 24 years.

The son’s age now is 6 years.

6. Robin Banks is 8 years older than Willie Catchup. Twenty years ago Robin was three times as old as Willie. How old is each now?

Solution
Let w = Willie’s age now.

Let w + 8 Robin’s age now.

Then

w – 20 = Willie’s age 20 years ago.

(w + 8) – 20 = Robin’s age 20 years ago.

Equation

(w + 8) – 20 = 3(w – 20)

w -12 = 3w – 60

-2w = -48

w = 24 and w + 8 = 32

So,

Willy is 24 and Robin is 32.

7. Bill Dupp’s is twice as old as Doug Upp. If 16 is added to Doug’s age and 16 is subtracted from Bill’s age, their ages will then be equal. What are their present ages?

Solution

Doug’s age = x

Bill’s age = 2x

Bill Dupp is twice as old as Doug Upp.

Then

2x – 16 = Bill’s current age

x + 16 = Doug’s current age

If 16 is added to Doug’s age and 16 is subtracted from Bill’s age they will be equal.

Equation

2x – 16 = x + 16

2x = x + 32

x = 32

So,

Doug Dupps is 32 and Bill Dupp is 64.

 

8. In 4 years Calvin’s age will be the same as Hobbes’s age now. In 2 years, Hobbes will be twice as old as Calvin. Find their ages now.

Solution

Let x = Hobbes’s age now.

x – 4 = Calvin’s age now.

In 2 years time, Hobbes’s age is x + 2

In 2 years time, Calvin’s age is (x – 4) + 2

The problem tells you about their ages at another time. Two years later your age would be 2 more than your age now.

Equation

x + 2 = 2[(x – 4) + 2]

x + 2 = 2x – 8 + 4

x + 2 = 2x – 4

x = 6 and x – 4 = 2

So,

So Calvin’s age is 2 and Hobbes’s age is 6.

 

9. Justin Case is twice as old as her daughter Jewell. Ten years ago the sum of their ages was 46 years. How old is Justin?

Solution

Let x = Jewell’s age
2x = Justin’s age

The problem talked about their age at another time which was ten years ago. So, that would be less than 10 because it was ten years ago.
x – 10 = Jewell’s age ten years ago
2x – 10 = Justin’s age ten years ago.

Ten years ago the sum of their ages was 46.

Equation

x – 10 + 2x – 10 = 46

3x – 20 = 46

3x = 66

Answer

x = 22

So, 

Jewell’s age is 22 and Justin’s is 44.

10. A Roman statue is three times as old as a Florentine statue. One hundred years from now the Roman statue will be twice as old. How old is the Roman statue?

Solution

Now

Let the Florentine statue = x

Let the Roman statue = 3x

In 100 years

The age of the Florentine statue = x + 100

The age of the roman statute =  3x + 100

Equation

3x + 100 = 2(x + 100)

3x + 100 = 2x + 200

x = 100

3x = 300

So

The age of the Florentine statue is 100 now.

The age of the roman statute is 300 now.

11. Butch Err’s age in 20 years will be the same as Janet’s age is now. Ten years from now, Janet’s age will be twice Butch’s. What are their present ages?

Solution

Now

Let Butch’s age= X+20

Let Janet’s age= X

10 Years From Now

Let Butch’s age = X + 30

Let Janet’s age = X + 10

Equation

x + 30 = 2(x + 10)

x + 30 = 2x + 20

10 = x

So,

Butch Err’s age now is 10 years.

Janet’s age now is 30 years.

 

12. Moe Tell’s age is three times Fran Tick’s. If 20 is added to Fran’s age and 20 is subtracted from Moe’s, their ages will be equal. How old is each now?

Solution

Let x = Fran Tick’s age now.

Let 3x = Moe Tell’s age now.

Then

Moe’s age = 3x – 20

Fran’s age = x + 20

If 20 is added to Fran’s age and 20 is subtracted from Moe’s then their ages will be equal.

Equation

3x – 20 = x + 20

3x= x + 40

2x = 40

x = 20 and 3x = 60

So,

Moe Tell’s age is 60 and Fran Tick’s age is 20 years old.

 

 

13. Andy is twice as old as Kate, In 6 years, their ages will total 60. How old is each now?

Solution

Let a = Andy’s age now.

Let k = Kate’s age now.

Then

a = 2k

 

(a+6) + (k+6) = 60

 

Substitute a for 2k

 

Equation

 

(2k + 6) + (k+6) = 60

 

Combine Like Terms

 

3k + 12 = 60

 

3k = 48

 

k = 16

 

a = 32

 

So,

 

Andy is 32 years old and Kate is 16 years old.

 

14. Mrs. Wang is 23 years older than her daughter. In 5 years, their ages will total 63. How old are they now?

Solution

Now

Let w = Mrs. Wang’s age now.

Let d = daughter’s age now.

w = d + 23

d + 5 + w + 5 = 63 or d + w + 10 = 63 or d + w = 53

Solve
d + w + = 53

w = d + 23

d + (d + 23) = 53

2d + 23 = 53

2d = 30

d = 15 and w = 38

So

The daughter is 15 and Mrs. Wang is 38.

15. Matthew is 3 times as old as Jenny. In 7 years, he will be twice as old as she will be then. How old is each now?

 

Solution

Now

Let Matthew’s age = 3j

Let Jenny’s age = j

7 Years From Now

3j + 7 = 2(j + 7)

Equation

3j + 7 = 2(j + 7)

3j + 7 = 2j + 14

j + 7 = 14

j = 7

So,

Matthew’s age is 21 and Jenny’s age= 7

16. Juan is 8 years older than his sister. In 3 years, he will be twice as old as she will be then. How old are they now?

Solution

Let j = Juan’s age now.

Let s = sister’s age now.

Then

j  = s + 8

 

j + 3 = 2(s + 3)

 

Substitute s + 8 for j

 

Equation

 

(s + 8) + 3 = 2(s + 3)

 

s + 11 = 2s + 6

 

-s + 11 = 6

 

-s = -5

 

s = 5

 

So,

 

Juan is 13

 

His sister is 5

 

17. Melissa is 24 years Younger than Joyce. In 2 years, Joyce will be 3 times as old as Melissa will be then. How old are they now?

Solution

Now

Let m = Melissa’s age now

Let j = Joyce’s age now

Then

j = m + 24

j + 2 = 3 (m + 2)

Substitute  m + 24 for j

Equation

(m + 24) + 2 = 3 (m + 2)

m + 22 = 3m + 2

-2m = -20

m = 10

So,

Joyce is 34 years old

Melissa is 10 years old

 

18. Tom is 4 years Older than Jerry. Nine years ago Tom was 5 times as old as Jerry was then. How old is each now?

 

Solution

Let t = Tom’s age now.

Let j = Jerry’s age now.

Then

t = j + 4

t + 9 = 5(j + 9)

Substitute (j + 4) for t.

Equation

(j + 4) – 9 = 5 (j – 9)

j – 5 = 5j – 45

j = 5j – 40

– 4j = – 40

– j = – 10

j = 10

So,

t = 14

j = 10

Tome is 14 and Jerry is 10 now.

 

19. Kathy is 6 years younger than Bill. Twelve years ago, Bill was twice as old as Kathy Was then. How old are they now?

Solution

Let k = Kathy’s age now.

Let b = Bill’s age now.

Then

k + 6 = b

2(k – 12) = b – 12

Substitute (k + 6) for b.

Equation

 

2(k – 12) = (k + 6) – 12

2k – 24 = k – 6

k – 24 = -6

k = 18

So,

k = 18

b = 24

Kathy is 18 and Bill is 24 now.

12 years ago Kathy was 6 and Bill was 12

 

20. Dr. Garcia is twice as old as his son. Twenty years ago, he was 4 times as old as his Son was then How old are they now?

Solution

Let g = Dr. Garcia’s age now.

Let s = son’s age now.

Then

2s = g

g – 20 = 4(s-20)

Substitute 2s for g.

Equation

2s – 20 = 4(s-20)

2s – 20 = 4s-80

-20 = 2s – 80

60 = 2s

s = 30

So,

s = 30

g = 60

The son is 30 now and Dr. Garcia is 60 now.

20 years ago the son was 10 and Dr. Garcia was 40

21. Mr. Klinker is 35 and his daughter is 10. In how many years will Mr. Klinker be twice as old as his daughter?

Solution

Let k = Mr. Klinker’s age now.

Let d = Daughter’s age now.

Then

k + x = 2( d + x)

Replace k with 35 and d with 10

Equation

35 + x = 20 + 2x

35 + x = 20 + 2x

35 = 20 + x

15 = x

x = 15

So,

In 15 years, Mr. Klinker will be 50, and his daughter will be 25, thus making Mr. Klinker twice his daughter’s age.

 

22. George is 7 and his mother is 37. In how many years will his mother be 3 times as old as he is?

Solution

Let g = George’s age now.

Let m = mother’s age now.

Then

g + x = 3(m + x)

Replace g with 7, and m with 37.

Equation

37 + x = 3(7+x)

37 + x = 21 + 3x

37 = 21 + 2x

16 = 2x

x = 8

So,

In 8 years, George’s mother will be 3 times older than him.

 

23. Pete is 14 and his grandfather is 54. How many years ago was his grandfather 6 times as old as Pete?

Solution

Let p = Pete’s age now.

Let g = grandfather’s age now.

Then

6(p – x) = g – x

Replace p with 14, and g with 54.

Equation
6(14 – x) = 54 – x

84 – 6x = 54 – x

84 = 54 + 5x

30 = 5x

x = 6

So,

6 years ago Pete was 8 and the grandfather was 48.

 

24. Dorothy is 14 years younger than Rita. Ten years ago, Rita was 3 times as old as Dorothy was then. How old is each now?

Solution

Let d = Dorothy’s age now

Let r = Rita’s age now

Then

d + 14 = r

r – 10 = 3(d – 10)

Replace r with d + 14

Equation

r – 10 = 3(d – 10)

d + 14 – 10 = 3(d – 10)

d + 4 = 3d – 30

4 = 2d – 30

2d = 34

d = 17

So,

Dorothy is 17 and Rita is 31.

 

10 years ago Dorothy was 7 and Rita was 21.

 

25. Ms. Ford is 48 and Ms. Lincoln is 35. How many years ago was Ms. Ford exactly twice as old as Ms. Lincoln?

Solution

Let f = Ms. Ford’s age now

Let l = Ms. Lincoln’s age now

Then

2(f –  x) = l – x

Replace f with 48, and l with 35.

Equation

f –  x = 2(l – x)

48 –  x = 2(35 – x)

48 – x = 70 – 2x

48 + x = 70

 

x = 22

 

So,

In 22 years ago , Ms. Ford’s was 26 and Ms. Lincoln’s was 13.

 

 

26. Steve is 5 times as old as Janis. In 12 years, he will be twice as old as she will be then. How old are they now?

Solution

Now

Let s = Steve’s age now

Let j = Janis’s age now

s = 5j

Then (In 12 Years)

(s + 12) = 2(j + 12)

Solve

s = 5j

(s + 12) = 2(j + 12)

(5j + 12) = 2(j + 12)

5j + 12 = 2j + 24

3j = 12

j = 4 so s = 32

So,

Janis’s age is 4 and Steve’s age is 32.

 

27. Mary is 4 years older than Toni. Sam is twice as old as Mary. The sum of their three ages is 8 times Toni’s age. How old are they?.

Solution

Now

Let m = Mary’s age now.

Let t = Toni’s age now.

Let s = Sam’s age now.

t + 4 = m

s = 2m

m + t + s = 8t

Replace m with (t + 4)

Replace s with 2m and then replace the m in 2m with (t + 4)

Equation

(t + 4) + t + 2m = 8t

(t + 4) + t + 2(t+ 4) = 8t

t + 4 + t + 2t + 8 = 8t

4t + 12 = 8t

4t = 12

t = 3

So,

Toni is 3

Mary is 7

Sam is 14

3 + 7 + 14 = 8(3)

24 = 24

 

 

28. Larry is 8 years older than his sister. In 3 years, he will be twice as old as she is now. How old are they now?

Solution

Now

Let l = Larry’s age now

Let s = sister’s age now

s + 8 = l

2(s + 3) = l + 3

Replace l with s + 8

2(s + 3) = l + 3

2(s + 3) = (s + 8) + 3

2s + 6 = s + 11

s + 6 = 11

s = 5

So,

Larry is 13 and his sister is 5

In 3 years Larry will be 16 and his sister will be 8.

 

 

29. Barry is 8 years older than his sister. In 3 years, he will be twice as old as she will be then. How old is each now?

 

Solution

Now

Let b = Barry’s age now.

Let s = sister’s age now

b = 8 + s

Then (In 3 Years)

(b + 3) = 2(s + 3)

Solve

b = 8 + s

(b + 3) = 2(s + 3)

s + 11 = 2s + 6

11 = s + 6

s = 5

So,

Barry is 13 and his sister is 5.

 

30. Jennifer is 6 years older than Sue. In 4 years, she will be twice as old as Sue was 5 years ago. Find their ages now.

 

Solution

Now

Let j = Jennifer’s age now.

Let s = Sue’s age now

j = s + 6

Then (In 4 Years)

j + 4 = 2(s – 5)

Solve

(s + 6) + 4 = 2(s – 5)

s + 10 = 2s – 10

10 = s – 10

20 = s

s = 20

So,

Sue is 20 and Jennifer is 26.

 

31. Adam is 5 years younger than Eve. In 1 year, Eve will be three times as old as Adam was 4 years ago. Find their ages now.

Solution

Now

Let a = Adam’s age now.

Let e = Eve’s age now

a + 5 = e

Then (In 1 Years)

3(a – 4) = e + 1

Solve

3(a – 4) = e + 1

Replace e with (a + 5)

3(a – 4) = a + 5 + 1

3a – 12 = a + 6

2a – 12 = 6

2a = 18

a = 9

So,

Adam is 9 and Eve is 14

In one year Eve will be 15 and 4 years ago Adam was 5.

 

 

32. Jack is twice as old as Jill. In 2 years, Jack will be 4 times as old as Jill was 9 years ago. How old are they now?

Solution

Now

Let j = Jack’s age now.

Let x = Jill’s age now

j = 2x

Then (In 2 Years)

j + 2 = 4(x – 9)

Solve

j + 2 = 4(x – 9)

(2x) + 2 = 4(x – 9)

2x + 2 = 4x – 36

2 = 2x – 36

38 = 2x

x = 19

So,

Jill is 19 and Jack is 38.

 

33. Four years ago, Katie was twice as old as Anne was then. In 6 years, Anne will be the same age that Katie is now. How old is each now?

Solution

Now

Let k = Katie’s age now.

Let a = Anne’s age now

k – 4 = 2(a – 4)

k = a + 6

Solve

k – 4 = 2(a – 4)

Replace k with (a + 6)

(a + 6) – 4 = 2(a – 4)

a + 2 = 2a – 8

2 = a – 8

a = 10

So,

Anne is 10 and Katie is 16

In 6 years Anne will be 16 and Katie is 16 now.

4 years ago Anne was 6 and Katie was 12

 

34. Five years ago. Tom was one third as old as his father was then. In 5 years Tom will be half as old as his father will be then. Find their ages now.

Solution

Now

Let t = Tom’s age now.

Let f = Father’s age now

3(t – 5) = f – 5

2(t + 5) = f + 5

Solve

2 equations

3(t – 5) = f – 5

2(t + 5) = f + 5

First equation

3(t – 5) = f – 5

3t – 15 = f – 5

Second equation

2(t + 5) = f + 5

2t + 10  = f + 5

Solve second equation for f

2t + 10  = f + 5

2t + 5 = f

Substatute f form second equation for f in first equation

3t – 15 = (2t + 5) – 5

Solve

3t – 15 = 2t

t = 15

So,

Tom is 15 and his Father is 35.

Five years ago Tom was 10 and his Father was 30. Tom will be 1/3 his father’s age.

In 5 years Tom will be 20 and his Father will be 40. Tom will be half his father’s a