Question

The difference between the ages of two cousins is 10 years. 15 years ago , if elder one was twice as old as younger one , find their present ages​

Answer 1

Let the age of one cousin be x

Age of another cousin=x+10

15 years ago,

younger one's age=x-15

elder one's age=2(x-15)

ATQ,

2(x-15) = x+10-15

2x-30= x-5

2x-x=30-5

x=25

elder one's age=25+10 =35 years

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Answer 2

Answer:

The Elder sibling is 35 years old and younger one is 25 years old.

Step-by-step explanation:

Given :

Difference between their ages = 10 years

15 years ago = elder one was twice old as younger one

To find :

Their present ages

Solution :

Let the present ages be -

• Younger one as y
• Elder one as (y + 10)

Ages 15 years ago -

• Younger as (y – 15)
• Elder as (y + 10) – 15

________________________________

According to the question, 15 years ago, elder one was twice the age of the younger.

$\longrightarrow$ [(y + 10) – 15] = 2(y – 15)

$\longrightarrow$ y – 5 = 2y – 30

$\longrightarrow$ y – 2y = –30 + 5

$\longrightarrow$ –y = –25

$\longrightarrow$ y = 25

Younger one = 25 years old.

________________________________

★ Age of the elder one -

$\longrightarrow$ y + 10

$\longrightarrow$ 25 + 10

$\longrightarrow$ 35 years

Elder one = 35 years old.

$\therefore$ The Elder sibling is 35 years old and younger one is 25 years old.