Question

A is twice as old as B. Ten years ago he was four times as B. what are their present ages​

Answer 1

Solution :

Let the age of B be x yrs.

Then, the age of A = 2x yrs

Ten years ago :

The age of B = (x-10) yrs

The age of A = (2x-10) yrs

Atq,

$\: \: \: \: \: \: \: 4(x - 10) = 2x - 10\\= > 4x - 40 = 2x - 10 \\ = > 4x - 2x = - 10 + 40 \\ = > 2x = 30\\ = > x = \frac{30}{2} \\ = > x = 15$

Present age of B = 15 yrs

Present age of A = (2×15) = 30 yrs.

Answer 2

Step-by-step explanation:

let the age of A be x and B be y

acc. to question

x=2y

x-2y=0.....eq.1

ten years ago. x-10=4(4-10)

x-10=4y-40

x-4y=-40+10

x-4y=-30.......eq.2

x=2y......eq3 substitute in eq.2

2y-4y=-30

-2y=-30

2y=30

y=15

fro eq.3

x=2y

x=2(15)

x=30

so age of A is 30

and age of B is 15