This is the third part of the Solving Age Problems Series. In this part, we will solve age problems with a variety of formats and difficulty that are not discussed in the first two parts. We have already solved six problems in the first and second part, so we start with the seventh problem.
Bill is four times as old as Carol. One fifth of Bill’s age added to one half Carol’s age is equal to years. How old are both of them?
Bill is older than Carol and he is four times older. This means that if Carol is years old, then Bill is years old. Now, one fifth of Bill’s age is and one half of Carol’s age is . Add these together and you get . Now, we have an equation.
Let be Carol’s age and be Bill’s age.
Simplifying, we have
Since we have a fraction, we can eliminate the denominator by multiplying everything with the least common multiple of and which is . Multiplying both sides of the equation by , we have
This means that Carol is and Bill is .
Bill is and Carol is . Yes, Bill is four times as old as Carol. One fifth of is . One half of is and . So, we are correct.
When a really smart math kid was asked about his age, he said:
“I am one fifth as old as my mother. In six years, I will be one-third as old.”
How old is the kid and his mother?
The kids is one fifth as old as his mother. So, if the mother is years old, then the kid is Six years from now, the ages of the mother and the kid respectively are and as shown in the table below.
As the kid said, in years, his age will be a third of his mother. This means that if we multiply his age by $latex3$, then it will equal the age of his mother. In equation form, we have
Now, we write the solution.
Let be the mother’s age and be the kid’s age.
We simplify the right hand side by Distributive Property. This gives us
Now, to eliminate the fraction, we multiply both sides of the equation by .
Again, by distributive property, we have
Putting all the x’s on the left hand side and all the numbers on the right hand side, we have
So, the mother and and the kid is . A smart kid indeed, giving problems such as this at age 6.
Left as an exercise.
Donna is years older than Demi. One fifth of Donna’s age a year ago added to three fourth of Demi’s age is equal to Demi’s age. How old is Donna?
Demi is years old and Donna is . Now, Donna’s age a year a go is which is equal to . How, one fifth of Donna’s age a year ago is and one fourth of Demi’s age is .
Now, these ages if added equal’s Donna’s age which is . Therefore, the equation is
Let be Demi’s age and be Donna’s age
Simplifying, the left hand side by distributive property, we have
Now, to eliminate the fractions, we multiply both sides of the equation by the least common multiple of and which is . This will result to
Therefore, Demi is and Donna is .
Check: Left as an exercise.