Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the present age be x then 2nd's age= x+10

15 years ago'

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

2x-30=x-5

2x-x=30-5

x=25 years

x+10=35 years

Answer 2

Answer:

• Present age of elder one = 35 years

• Present age of younger one = 25 years

Step-by-step explanation:

Given:

• The difference between the ages of two cousins is 10 years.
• 15 years ago, the elder one was twice as old as the younger one.

To find:

• Their present ages.

Solution:

Let the present age of elder one be x years and the present age of younger one be y years.

★ According to the 1st condition,

• The difference between the ages of two cousins is 10 years.

$\implies$ x-y=10

$\implies$ x = 10+y..........(i)

★ According to the 2nd condition,

• 15 years ago, the elder one was twice as old as the younger one.

15 years ago,

• Age of elder one = (x-15) years
• Age of younger one = (y-15) years

$\implies$ x-15=2(y-15)

$\implies$ 10+y-15=2y-30

$\implies$ y-5 = 2y -30

$\implies$ y - 2y = -30 + 5

$\implies$ -y = -25

$\implies$ y = 25

• Present age of younger one = 25 years.
• Present age of elder one = 10+25 = 35 years

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