Question

A and B are friends and their ages differ by 2 yr. A's father, D is twice as old as A and B is twice as old as his sister C. The ages of D differ by 40 years. Find the age of A and B.​

Answer 1

Answer:

Let the Age of A = x

Age of B = x+2

since, A's father D is twice as old as A

So, Age of D = 2×x = 2x

and since B is twice as old as his sister

So, Age of C = age of B / 2 = (x+2)/2

According to the question,

The Ages of D differ by 40yrs.

it means,

Age of D - age of C = 40

=> 2x - [(x+2)/2] = 40

=> 2x - (x+2)/2 = 40

=> 4x/2 - (x+2)/2 = 40

=> (4x - x + 2)/2 = 40

=> 3x+2/2 = 40

=> 3x + 2 = 40 × 2

=> 3x = 80 - 2

=> x = 78/3

= 26

So, Age of A = x = 26yr.s

Age of B = x + 2 = 26 + 2 = 28yr.s

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Answer 2

A's age =26 years and B's age =24 years

Step-by-step explanation:

Let the age of A = x years

and the age of B = y years respectively.

From the above question, it is given that ,

x-y = 2 years..........(1)

D is twice as A, D= 2x and 2C = B, C= B/2=y/2

it is clear that, D is older than C

So, 2x-y/2= 40( given)

=>4x-y = 80.............(2)

Now subtracting eq(1) from eq(2)..

we get,

=> 4x-y-x+y = 80-2

=> 3x= 78

=>x = 78/3= 26

Now putting the value of x in eq(1)

26-y = 2

=>y = 26-2 = 24

Hence, A's age =26 years and B's age =24 years

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