A father who is 42 years old has a son and a daughter. The daughter is three times as old as the son. In 10 years, the sum of all their ages will be 100 years. How old are the two siblings at present?
D = 3S
(D+10)+(S+10) + (42+10) = 100
Substitute 3S for D in the second equation and solve for S. Insert that value into the first equation to solve for D. Check by putting both values into the second equation.
To solve this problem, let's work through it step by step:
Step 1: Define the variables:
Let's denote the son's age as S.
The daughter's age would then be 3S (since the daughter is three times the age of the son).
And the father's age is given as 42.
Step 2: Determine the future sum of their ages:
In 10 years, the son's age will be S + 10.
The daughter's age will be 3S + 10.
And the father's age will be 42 + 10.
Step 3: Write an equation based on the given information:
The sum of their ages in 10 years will be 100, so we can write the equation as:
(S + 10) + (3S + 10) + (42 + 10) = 100.
Step 4: Simplify and solve the equation:
Combining like terms, we get:
S + 10 + 3S + 10 + 42 + 10 = 100.
4S + 72 = 100.
Subtracting 72 from both sides, we have:
4S = 28.
Dividing both sides by 4, we get:
S = 7.
Step 5: Find the ages of the daughter and siblings:
Now that we know the son's age is 7, we can find the daughter's age:
Daughter's age (3S) = 3 * 7 = 21.
So, the son is 7 years old, and the daughter is 21 years old.