The mother's age is 4 times her daughter's age. Four years hence, the age of the mother will be one year more than three times her daughter's age. The present ages in years of daughter and mother are, respectively?
Answer (2)
To solve the problem about the ages of the mother and daughter, let's use algebra.
Define Variables:
Let the daughter's present age be x years.
Therefore, the mother's present age is 4 x years because she is four times older than her daughter.
Future Ages in 4 Years:
In four years, the daughter's age will be x + 4 .
In four years, the mother's age will be 4 x + 4 .
Formulate an Equation:
According to the problem, in four years, the mother's age will be one year more than three times the daughter's age.
This can be written as: 4 x + 4 = 3 ( x + 4 ) + 1
Simplify and Solve the Equation:
Start by expanding the right side: 3 ( x + 4 ) = 3 x + 12
Add 1 to it: 3 x + 12 + 1 = 3 x + 13
Substitute back into the equation: 4 x + 4 = 3 x + 13
Rearrange the terms to isolate x : 4 x − 3 x = 13 − 4 x = 9
Determine the Ages:
The daughter's present age is x = 9 years.
The mother's present age is 4 x = 4 × 9 = 36 years.
Therefore, the present ages of the daughter and the mother are 9 years and 36 years, respectively.
The daughter's current age is 9 years, while the mother's current age is 36 years. This is determined through algebraic equations based on their age relationship and future age conditions. Thus, their ages reflect the given problem statements accurately.
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