If 2 times the mother's age is 26 years more than 4 times her daughter's age, and 3 times the daughter's age is 7 years less than the mother's age, then what is the difference (in years) between the ages of the mother and the daughter?
This problem requires us to find the age difference between a mother and her daughter based on two given conditions relating their ages. We can solve this using algebra by setting up and solving a system of equations.
Let's represent the mother's current age as '$M$' years and the daughter's current age as '$D$' years.
We are given two conditions. Let's translate them into mathematical equations:
$2M = 4D + 26$
$3D = M - 7$
We now have a system of two linear equations with two variables:
$2M = 4D + 26$
$3D = M - 7$
Let's simplify the first equation by dividing both sides by 2:
$M = 2D + 13 \quad (\text{Equation 1 simplified})$
Now, we can use substitution. Substitute the expression for '$M$' from the simplified Equation 1 into Equation 2:
$3D = (2D + 13) - 7$
Simplify the equation:
$3D = 2D + 6$
Subtract '$2D$' from both sides to solve for '$D$':
$3D - 2D = 6$
$D = 6$
So, the daughter's age is 6 years.
Now, substitute the value of '$D$' back into the simplified Equation 1 to find the mother's age:
$M = 2(6) + 13$
$M = 12 + 13$
$M = 25$
So, the mother's age is 25 years.
The question asks for the difference between the mother's age and the daughter's age.
Age Difference = Mother's Age - Daughter's Age
Age Difference = $M - D$
Age Difference = $25 - 6$
Age Difference = $19$
The difference in age between the mother and the daughter is 19 years.
If 2 times a son's age is added to his father's age, the sum is 36 years. If 2.5 times the father's age is added to the son's age, the sum is 72 years. What is the father's age (in years)?
Two times the present age of P exceeds the present age of Q by 41 years. After 7 years, nine times Q's age will be 24 years less than six times P's age at that time. What is the present age (in years) of Q?