Age problems involve translating relationships about people's ages — either in the present, past, or future — into algebraic expressions or equations.
Common Phrases in Age Problems:
"x years ago" → Subtract: $x - \text{years}$
"in x years" → Add: $x + \text{years}$
"Twice as old" → Multiply: $2x$
"Half as old" → Divide: $\frac{x}{2}$
"The difference in age" → Subtract one from the other
Typical Structure:
Let one person's age be a variable (e.g., $x$), and express the others in terms of $x$.
$$
\text{Let John's age be } x.
\text{Then, his sister's age (3 years younger) is } x - 3
$$
Problem:
Itadori is 4 years younger than Megumi. Twenty years ago, Megumi's age was 13 years more than half the age of Itadori. How old are they now?
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Problem:
When I am as old as my father is now, I shall be five times as old as my son is now. By then, my son will be eight years older than I am now. The combined ages of my father and me is 100 years. How old is my son?
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Problem:
Six years ago, Tom was five times as old as Ana. In five years, Tom will be three times as old as Ana. What is the present age of Ana?
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Problem:
If Luna was four times as old as Lina eight years ago, and if Luna will be twice as old as Lina eight years hence, how old is Lina now?
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Problem:
Rumi is three times as old as Zoey. Three years ago, she is four times as old as Zoey. Find the sum of their ages. Ans. 36
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Exam Generator Problems
Additional board-style practice items for this topic.
Six years ago, Kazuha was five times as old as Xiao. In five years, Kazuha will be three times as old as Xiao. What is the present age of Xiao?
Answer:
17
15
16
14
Let Xiao's present age be $x$ and Kazuha's present age be $k$. Six years ago: $k-6=5(x-6)$ In five years: $k+5=3(x+5)$ Solving gives $x=17$. $\boxed{17}$
When I am as old as my father is now, I shall be five times as old as my son is now. By then, my son will be eight years older than I am now. The combined age of my father, son, and me is 100 years.
Itadori is 4 years younger than Megumi. Twenty years ago, Megumi's age was 13 years more than half the age of Itadori. How old are they now?
Itadori's age:
38
42
40
46
Megumi's age:
42
38
46
40
Part 1.
Let Itadori's age be $i$ and Megumi's age be $m$. Since Itadori is 4 years younger: $m=i+4$ Twenty years ago: $m-20=\frac{1}{2}(i-20)+13$ Substitute $m=i+4$: $i-16=\frac{i}{2}+3$ $i=38$ $\boxed{38}$
Part 2.
Megumi is 4 years older than Itadori: $m=38+4$ $\boxed{42}$
What time between 2 and 3 o'clock will the angle between the hands of the clock be bisected by the line connecting the center of the clock and the 3 o'clock mark?
Peter's age 13 years ago was 1/3 of his age 7 years hence. How old is Peter?
Answer:
23
21
15
27
Let Peter's present age be $x$. Thirteen years ago his age was one-third of his age seven years hence: $x-13=\frac{1}{3}(x+7)$ $3x-39=x+7$ $2x=46$ $\boxed{x=23}$
The sum of the parents' ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages if four times the sum of their children's ages. In fifteen years, the sum of the parents' ages will be equal to the sum of their children's ages. How many children were in the family?
Answer:
5
2
3
4
Let $P$ = sum of parents' ages (2 parents), $C$ = sum of children's ages, $n$ = number of children. Condition 1 (now): $P = 2C$ — (1) Condition 2 (5 years ago): $(P-10) = 4(C-5n)$ — (2) Condition 3 (15 years from now): $(P+30) = (C+15n)$ — (3) From (1) into (2): $2C-10=4C-20n \Rightarrow C=10n-5$ From (1) into (3): $2C+30=C+15n \Rightarrow C=15n-30$ Setting equal: $10n-5=15n-30 \Rightarrow 5n=25$ $\boxed{n = 5 \text{ children}}$
The price of houses increases at 10% per year. A house is currently priced at P8,000,000. After how many years will its price be double what it is now?
7.93
6.87
7.27
8.32
Use compound growth: $P=P_0(1.10)^t$. For the price to double: $2P_0=P_0(1.10)^t$, so $2=(1.10)^t$. Taking logarithms: $t=\frac{\log 2}{\log 1.10}=7.27$ years. $\boxed{7.27}$
