Question

sunita is as twice as old as ashima..if six years is subtracted from ashima's age and 4 years added to sunitas age,then sunita will be four times ashimas age..how old were they two years ago ( ans - sunita 28 , ashima 14)

i an currentl 5 times as old as my son ...fin 6 years time i will be three times as old as he will be then...what are our ages now( ans 30years,6years)


plzz solve these for me and get thumps up..!!

Answer 1

Let the age of Sunita be x and the age of Ashima be y.

Sunita is twice as old as ashima

⇒ x = 2y

⇒ x - 2y = 0 ------------------ (1)

Six years subtracted from Ashima's age = (y - 6)

Four years added to Sunita's age = (x + 4)

⇒ (x + 4) = 4(y - 6)

⇒ (x + 4) = 4(y - 6)

⇒ x + 4 = 4y - 24

⇒ x - 4y = - 28 ------------------ (2)

Solving equations (1) and (2), we get x = 28 and y = 14.

Therefore, present age of Sunita is 28 years and the present age of Ashima is 14 years.

Age of Sunita and Ashima two years ago was 26 years and 12 years respectively.

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