Question

Ten years later, A will be twice as old as B and 3 years ago, A was 3 times as old as B. What are their present ages

Answer 1

Answer:

Step-by-step explanation:

Let the present age of A be x and B be y.

After ten years, age of A will be (x+10) and B will be (y+10)

ATQ, age of A = twice age of B

                x+10 = 2(y+10)

            => x-2y = 10       ................(1)

Also 3 years ago, age of A be (x-3) and age of B be (y-3)

ATQ, age of A = thrice the age of B

                  x-3 = 3(y-3)  

              => x-3y = -6     .................(2)

Equating equation (1) and (2)

(x-2y)-(x-3y) = 10-(-6)

=> x-2y-x+3y = 10+6

=> y = 16

Now, x-2y = 10

=> x = 10+2(16)

=> x = 42

Hence, present age of A is 42 years and B is 16 years.

Answer 2

Answer:

Step-by-step explanation:

Let the present age of A be x and B be y.

After ten years, age of A will be (x+10) and B will be (y+10)

ATQ, age of A = twice age of B

                x+10 = 2(y+10)

            => x-2y = 10       ................(1)

Also 3 years ago, age of A be (x-3) and age of B be (y-3)

ATQ, age of A = thrice the age of B

                  x-3 = 3(y-3)  

              => x-3y = -6     .................(2)

Equating equation (1) and (2)

(x-2y)-(x-3y) = 10-(-6)

=> x-2y-x+3y = 10+6

=> y = 16

Now, x-2y = 10

=> x = 10+2(16)

=> x = 42

Hence, present age of A is 42 years and B is 16 years

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