sunita is twice as old as ashima.if six years is subtracted from ashima age and four year added to sunita age,then sunita will be four times ashima's age.how old were they two years ago?
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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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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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