the difference of the present age of two sisters is 5 year. after 4 year ago if the product of their age become 104 find their present age.
Answers
Given data : The difference of the present age of two sisters is 5 year. After 4 years if the product of their ages become 104.
Solution :
A/C to given data; Let, age of the first sister be x years and age of second sister be y years.
Hence,
➜ x - y = 5 years -----{1}
➜ x = (5 + y) years ----{2}
Now, after four years a/c to given data;
➜ x * y = 104 ----{3}
from eq. {1}
➜ (5 + y) * y = 104
➜ 5y + y² = 104
➜ y² + 5y - 104 = 0
➜ y² + 13y - 8y - 104 = 0
➜ y(y + 13) - 8(y + 13) = 0
➜ (y + 13) (y - 8) = 0
Now,
➜ y + 13 = 0 or y - 8 = 0
➜ y = - 13 or y = 8
Here, we know that age of person is never negative,
∴ y ≠ - 13 and y = 8 years
Now put value of y in eq. {1}
➜ x = (5 + y) years
➜ x = (5 + 8) years
➜ x = 13 years
Answer : Hence, the present age of first sister is 13 years and second sister is 8 years.
{Verification: Put value of x and y in eq. {3}
➜ x * y = 104
➜ 8 * 13 = 104
➜ 104 = 104
Similarly put value of x and y in eq. {1}
➜ x - y = 5
➜ 13 - 8 = 5
➜ 5 = 5
Hence it verified.}
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