Question

Ten years ago, A was four times as old as B and after 10 years A will be twice as old as B find their present ages​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let the present age of A be x

And present age of B be y

Ten years ago:-

Age of A = x - 10

Age of B = y - 10

Given, A was four times as old as B

$\sf \longrightarrow x - 10 = 4(y - 10)$

$\sf \longrightarrow x - 10 = 4y - 40$

$\sf \longrightarrow x - 4y = - 40 + 10$

$\sf \longrightarrow x - 4y = - 30......(i)$

Ten years later:-

Age of A = x + 10

Age of B = y + 10

Given, A will be twice as old as B

$\sf \longrightarrow x + 10 = 2(y + 10)$

$\sf \longrightarrow x + 10 = 2y + 20$

$\sf \longrightarrow x - 2y= 20 - 10$

$\sf \longrightarrow x - 2y = 10......(ii)$

Equation (ii) - (i)

$\sf \longrightarrow x - 2y - (x - 4y) = 10 - (-30)$

$\sf \longrightarrow x - 2y - x + 4y= 10 + 30$

$\sf \longrightarrow 2y= 40$

$\sf \longrightarrow y = 20$

Substituting y = 20 in equation (ii)

$\sf \longrightarrow x - 2y = 10$

$\sf \longrightarrow x - 2(20) = 10$

$\sf \longrightarrow x - 40 = 10$

$\sf \longrightarrow x = 10 + 40$

$\sf \longrightarrow x = 50$

Therefore:-

Present age of A = 50 years

Present age of B = 20 years