Question

Sumita is twice as old as Ashima. If 6 years is subtracted from Ashima's age and 4 years added to Sumita's age than Sumita will be four times that of Ashima's age. Then their ages are

• A. 14
• B. 28
• C. 12
• D. 26

Answer 1

Answer: (C), (D)

Let Ashima's present age be $x$ years and Sumita's present age be y years From question
$y=2 x$....(i)
Also,
$ (y+4)=4(x-6) $
$ \therefore 2 x+4=4 x-24 $
$\text { [From (i) } y=2 x \text { ] } $
$ \therefore 2 x=28 $
$ \therefore x=14$
$ \text { And } \mathrm{y}=28 $
$\therefore 2 \text { years ago, Ashima's age }=x-2= 12 \text { years }$
$ 2 \text { years ago Sunita's age }=y-2 $
$=28-2$
$ =26 \text { years } $