# PROBLEMS ON AGES

Problem 1 :

Four times of Liam's age 3 years ago is equal to twice his age 6 years hence. Find the present age of Liam.

Solution :

Let x be the present age of Liam.

Then,

Liam's age 3 years ago = x - 3

Liam's age 6 years hence= x + 6

Given : Four times of Liam's age 3 years ago is equal to twice his age 6 years hence.

4(x - 3) = 2(x + 6)

Using Distributive Property,

4x - 12 = 2x + 12

Subtract 2x from both sides.

2x - 12 = 12

2x = 24

Divide both sides by 2.

x = 12

The present age of Liam is 12 years.

Problem 2 :

The age of a man is three times  the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man.

Solution :

Let 'x' be the present age of the man and 'y' be the sum of the present ages of two sons.

Given : Present age of the man is 3 times the sum of the ages of 2 sons.

x = 3y ----(1)

Given : 5 years hence, age of the man will be double the sum of the ages of his two sons.

x + 5 = 2(y + 5 + 5)

(Here, 5 is added twice to y. Because, there are two sons in y)

x + 5 = 2(y + 10)

x + 5 = 2y + 20

From (1), we can plug x  =  3y.

3y + 5 = 2y + 20

y = 15

Substitute y = 15 in (1).

x = 3(15)

x = 45

So, the present age of the man is 45 years.

Problem 3 :

Present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.

Solution :

Let 'x' be the present age of the son and 'y' be the present age of the father.

Given : Present age of a father is 3 years more than three times the age of his son.

y = 3x + 3 ----(1)

Given : Three years hence, father's age will be 10 years more than twice the age of the son.

y + 3 = 2(x + 3) + 10

y + 3 = 2x + 6 + 10

y + 3 = 2x + 16

From (1), substitute y = 3x + 3.

3x + 3 + 3 = 2x + 16

3x + 6 = 2x + 16

x = 10

Substitute x = 10 in (1).

y = 3 ⋅ 10 + 3

y = 33

So, the present age of the man is 33 years.

Problem 4 :

The ratio of the age of a man and his wife is 4 : 3. After 4 years this ratio will become 9 : 7. At the time of marriage, if the ratio was 5 : 3, how many years ago were they married ?

Solution :

Given : The ratio of the present ages of a man and his wife is 4 : 3.

Then, the present age of the man is 4x and his wife is 3x.

Given : After 4 years this ratio will become 9 : 7.

(4x + 4) : (3x + 4) = 9 : 7

(4x + 4)/(3x + 4) = 9/7

7(4x + 4) = 9(3x + 4)

28x + 28 = 27x + 36

x = 8

So, the present age of the man is

= 4x

= 4 ⋅ 8

= 32 years

The present age of his wife is

= 3x

= 3 ⋅ 8

= 24 years

Let us assume that they got married before 't' years from now.

Given : At the time of marriage, if the ratio was 5 : 3.

(32 - t) : (24 - t) = 5 : 3

(32 - t)/(24 - t) = 5/3

3(32 - t) = 5(24 - t)

96 - 3t = 120 - 5t

2t = 24

t = 12

So, they got married 12 years before.

Problem 5 :

John's age after six years will be three seventh of his father's age. Ten years ago the ratio of their ages was 1 : 5 . What is John's father's age at present ?

Solution :

Given : Ten years ago the ratio of their ages was 1 : 5.

Then, ten years ago,

age of John = x

age of his father = 5x

At present,

age of John = x + 10

age of his father = 5x + 10

After six years,

age of John = x + 10 + 6 = x + 16

age of his father = 5x + 10 + 6 = 5x + 16

Given : John's age after six years will be three seventh of his father's age.

(x + 16) = (3 / 7)(5x + 16)

7(x + 16) = 3(5x + 16)

7x + 112 = 15x + 48

64 = 8x

8 = x

At present, John's father's age is

5x + 10

Substitute x = 8.

= 5 ⋅ 8 + 10

= 40 + 10

= 50

So, the present age of John's father is 50 years.

Problem 6 :

The total age of P and Q is 12 years more than the total age of Q and R. R is how many year younger to P ?

Solution :

From the given information, we have

P + Q = 12 + Q + R

When we rearrange the above equation, we get

P - R = 12 + Q - Q

P - R = 12

From the above equation, it is clear that R is 12 years younger to P.

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