Question

The difference between the ages of two sisters is 8 years. 12 years ago, if the older one was twice as old as the younger one, find their present ages.​

Answer 1

Answer:

• Their present age are 20 & 28 years respectively .

Step-by-step explanation:

According to the Question

It is given that,

• Difference between the ages of two sisters is 8 years. 12 years ago

• The older one was twice as old as the younger one.

Let the age of younger one be x years

Then,

older one be (x + 8 ) years

According to the given statement

12 years ago older one was twice as old as the younger's one .

→ ( x + 8) - 12 = 2 × ( x - 12)

→ (x +8) -12 = 2x - 24

→ x + 8 -12 = 2x -24

→ x - 4 = 2x - 24

→ x - 2x = -24 + 4

→ - x = -20

→ x = 20 years

Now,

Older one be ( x + 8) = 20 + 8 = 28 years .

• Hence, their present ages are 20 years & 28 years respectively .

Answer 2

Answer :

• Present age are 20 years and 28 years

Given :

• The difference between the ages of two sisters is 8 years
• 12 years ago, If the older one was twice as old as the younger one

To find :

• Present ages

Solution :

• Let the elder sister age be x

Given, the difference between the ages of two sisters is 8 years so,

• Let the younger sister age be x + 8

Given,

12 years ago, if the older one was twice as old as the younger one so,

• Elder age be x - 12

Substituting the value:

⇢ (x + 8) - 12 = 2 × (x - 12)

⇢ (x + 8) - 12 = 2x - 24

⇢ x + 8 - 12 = 2x - 24

⇢ x - 4 = 2x - 24

⇢ x - 2x = -24 + 4

⇢ x = 24 - 4

⇢ x = 20

Finding the present age :

⇢ Elder sister = x

⇢ Elder sister age = 20 Years

⇢ Younger age = x + 8

⇢ Younger age = 20 + 8

⇢ Younger sister age = 28 years.

Hence , Present age are 20 years and 28 years