SOLVING WORD PORBLEMS INVOLVING AGES

Problem 1 :

In 8 years Jamie will be three times as old as she is now. How old is she now?

Solution:

Let x be the present age.

Age after 8 years = 3 × present age

x + 8 = 3x

2x = 8

x = 8/2

x = 4 years

Problem 2 :

Mattie is twice as old as Pat. Jon is twice as old as Mattie. In 4 years the sum of their ages will be 61. How old is each person?

Solution:

Let pat's age be x years.

Mattie's age = 2x

Jon's age = 2(2x) = 4x

Sum of their ages after 4 years

(x + 4) + (2x + 4) + (4x + 4) = 61

7x + 12 = 61

7x = 61 - 12

7x = 49

x = 7

So, 2x = 14 and 4x = 28

Pat's age = 7 years

Mattie's age = 14 years

Jon's age = 28 years

Problem 3 :

The difference between Jon’s age and Sue’s age is 18 years. If Jon is six years more than five times Sue’s age, how old is each person?

Solution:

Let x be the Sue's age.

Jon's age = 5x + 6

5x + 6 - x = 18

4x + 6 = 18

4x = 12

x = 3 years

Jon's age = 5(3) + 6

= 15 + 6

= 21 years

So, Sue's age = 3 years

Jon's age = 21 years

Problem 4 :

The Smith family has a set of triplets and a set of twins. If the triplets are four years younger than the twins, and the sum of their ages is 68, how old are the triplets?

Solution:

Let's the age of the twins as 't'

and the age of the triplets as 't - 4'.

2t + 3(t - 4) = 68

2t + 3t - 12 = 68

5t - 12 = 68

5t = 80

t = 80/5 

t = 16 years

Age of the triplets = t - 4

= 16 - 4

= 12 years

Problem 5 :

Jamie is 3 years less than three times Jessie’s age. In five years the sum of their ages will be 59. How old are they now?

Solution:

Let represent Jessie's age x.

Jamie's age = 3x - 3

In 5 years, Jessie's age will be x + 5

and Jamie's age will be 3x - 3 + 5.

x + 5 + 3x + 2 = 59

4x + 7 = 59

4x = 52

x = 52/4

x = 13

Jessie's age = 13 years

Jamie's age = 3x - 3

= 3(13) - 3

= 39 - 3

= 36 years

Problem 6 :

In five years the sum of Don’s and Dan’s ages will be 34. Don’s age now is 8 years less than three times Dan’s age. How old is Don?

Solution:

Let 'Do' be the Don's and 'D' be the Dan's.

In five years the sum of Don’s and Dan’s ages will be 34.

D + 5 + Do + 5 = 34

D + Do = 24

Don's age is 8 years less than three times Dan's age.

Do = 3D - 8

D + 3D - 8 = 24

4D - 8 = 24

4D = 32

D = 8 years

Substitute D = 8 into Do = 3D - 8

Do = 3(8) - 8

Do = 24 - 8

Do = 16 years

So, Dan's age = 8 years

Don's age = 16 years

Problem 7 :

The total of the ages of three students is 38. If the youngest student is 7 years younger than the oldest student, and the middle student is 4 years older than the youngest student, how old is each student?

Solution:

Let's the youngest student is x years old

the oldest student is (x + 7) years old

the middle student is (x + 4) years old

x + x + 7 + x + 4 = 38

3x + 11 = 38

3x = 27

x = 9

So, the youngest student is 9 years old

the oldest student is (x + 7) = (9 + 7) = 16 years old

the middle student is (x + 4)  = (9 + 4) = 13 years old.

Problem 8 :

A man is 35 years old, and his son is 7 years old. In how many years will the father be 3 times as old as his son will be then?

Solution:

Given, father age = 35 years

son age = 7 years

35 + x = 3(7 + x)

35 + x = 21 + 3x

35 - 21 = 3x - x

14 = 2x

x = 7 years

Problem 9 :

Jan has 6 more dimes than quarters. If the value of the dimes is the same as the value of the quarters, how many dimes and quarters does Jan have?

Solution:

Let the number of quarters be q.

The number of dimes is q + 6.

The value of 1 quarter is 0.25q.

The value of 1 dime = 0.10(q + 6)

0.25q = 0.10(q + 6)

0.25q = 0.10q + 0.60

0.15q = 0.60

q = 4

The number of dimes = 4 + 6 = 10

So, quarter = 4 

Dimes = 10

Problem 10 :

I have 7 more nickels than dimes. If the value of the nickels and dimes is the same, how many of each coin do I have?

Solution:

Let the number of dimes be x.

Number of nickels = x + 7

The value of dimes = 0.10x

The value of nickels = 0.05(x + 7)

0.10x = 0.05(x + 7)

0.10x = 0.05x + 0.35

0.05x = 0.35

x = 7

So, Number of dimes = 7

Number of nickels = 14

Problem 11 :

Alex broke open his piggy bank. He had 30 more dimes than nickels and 8 more quarters than nickels. The value of the quarters is exactly the same as the value of the sum of the dimes and nickels. How many of each coin does Alex have?

Solution:

Alex had 30 more dimes then nickels.

D = N + 30

Alex had 8 more quarters than nickels.

Q = N + 8

The value of the quarters is exactly the same as the value of the sum of the dimes and nickels.

0.25Q = 0.05N + 0.10D

0.25(N + 8) = 0.05N + 0.10(N + 30)

0.25N + 2 = 0.05N + 0.10N + 3

0.25N - 0.15N = 3 - 2

0.1N = 1

N = 1/0.1

N = 10

D = N + 30

D = 10 + 30

D = 40

Q = N + 8

Q = 10 + 8

Q = 18

So, 10 nickels, 40 dimes and 18 quarters.