a woman is four times older than her daughter five years ago the product of their age was 175 find their present age
To solve this problem, let's break it down step by step.
Let's assign variables to the present ages of the woman and her daughter. Let's say the woman's present age is W, and her daughter's present age is D.
According to the problem, "a woman is four times older than her daughter." In other words, the woman's age is four times the age of her daughter. Mathematically, this can be expressed as W = 4D.
Now, let's consider the information "five years ago." This means we need to subtract 5 from the present ages of both the woman and her daughter. Therefore, the woman's age five years ago would be W - 5, and the daughter's age five years ago would be D - 5.
According to the problem, "the product of their age was 175." This means that if you multiply the age of the woman five years ago with the age of her daughter five years ago, the result would be 175. Mathematically, this can be expressed as (W - 5) * (D - 5) = 175.
Now that we have two equations, we can solve them simultaneously. Substituting the value of W from the first equation into the second equation, we get (4D - 5) * (D - 5) = 175.
Expanding the equation, we have 4D^2 - 20D - 5D + 25 = 175.
Simplifying further, we get 4D^2 - 25D + 25 - 175 = 0.
Combining like terms, we have 4D^2 - 25D - 150 = 0.
Now we can solve this quadratic equation for the value of D (daughter's present age) using factoring, completing the square, or using the quadratic formula.
If we factor the equation, we have (D - 10)(4D + 15) = 0.
Setting each factor equal to zero, we get D - 10 = 0 or 4D + 15 = 0.
Solving each equation, we find D = 10 or D = -15/4.
Since the daughter's age cannot be negative, we can conclude that the daughter's present age (D) is 10.
Now, substituting this value into the first equation W = 4D, we have W = 4 * 10, which means the woman's present age (W) is 40.
Therefore, the woman is 40 years old, and her daughter is 10 years old.