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\(\textbf{1)}\) In 10 years, Steve will be twice as old as he is today. How old is Steve today? Steve’s age is 10 years old
\(\,\,\,\,\,\)Let \(x\) = Steve’s age today
\(\,\,\,\,\,\)In 10 years: \(x + 10\)
\(\,\,\,\,\,\)Set up equation: \(x + 10 = 2x\)
\(\,\,\,\,\,\)Solve: \(x + 10 = 2x\)
\(\,\,\,\,\,\)\(10 = x\)
\(\,\,\,\,\,\)So, Steve is \(10\) years old
\(\,\,\,\,\,\)Let \(x\) = Steve’s age today
\(\,\,\,\,\,\)In 10 years: \(x + 10\)
\(\,\,\,\,\,\)Set up equation: \(x + 10 = 2x\)
\(\,\,\,\,\,\)Solve: \(x + 10 = 2x\)
\(\,\,\,\,\,\)\(10 = x\)
\(\,\,\,\,\,\)So, Steve is \(10\) years old
\(\textbf{2)}\) 10 years ago, Jessica’s age was 1/3 the age she will be in 6 years. How old is Jessica now? Jessica’s age is 18 years old
\(\,\,\,\,\,\)Let \(x\) = Jessica’s age today
\(\,\,\,\,\,\)10 years ago: \(x – 10\)
\(\,\,\,\,\,\)In 6 years: \(x + 6\)
\(\,\,\,\,\,\)Set up equation: \(x – 10 = \frac{1}{3}(x + 6)\)
\(\,\,\,\,\,\)Multiply both sides by 3: \(3(x – 10) = x + 6\)
\(\,\,\,\,\,\)\(3x – 30 = x + 6\)
\(\,\,\,\,\,\)\(2x = 36\)
\(\,\,\,\,\,\)\(x = 18\)
\(\,\,\,\,\,\)So, Jessica is \(18\) years old
\(\,\,\,\,\,\)Let \(x\) = Jessica’s age today
\(\,\,\,\,\,\)10 years ago: \(x – 10\)
\(\,\,\,\,\,\)In 6 years: \(x + 6\)
\(\,\,\,\,\,\)Set up equation: \(x – 10 = \frac{1}{3}(x + 6)\)
\(\,\,\,\,\,\)Multiply both sides by 3: \(3(x – 10) = x + 6\)
\(\,\,\,\,\,\)\(3x – 30 = x + 6\)
\(\,\,\,\,\,\)\(2x = 36\)
\(\,\,\,\,\,\)\(x = 18\)
\(\,\,\,\,\,\)So, Jessica is \(18\) years old
\(\textbf{3)}\) Michael is twice as old as his brother. Five years ago he was 3 times as old as his brother. How old is Michael now? Michael’s age is 20 years old
\(\,\,\,\,\,\)Let \(x\) = Michael’s age
\(\,\,\,\,\,\)Let \(y\) = his brother’s age
\(\,\,\,\,\,\)Given: \(x = 2y\)
\(\,\,\,\,\,\)Five years ago: \(x – 5 = 3(y – 5)\)
\(\,\,\,\,\,\)Substitute \(x = 2y\) into the second equation: \(2y – 5 = 3(y – 5)\)
\(\,\,\,\,\,\)\(2y – 5 = 3y – 15\)
\(\,\,\,\,\,\)\(10 = y\)
\(\,\,\,\,\,\)\(x = 2(10) = 20\)
\(\,\,\,\,\,\)So, Michael is \(20\) years old
\(\,\,\,\,\,\)Let \(x\) = Michael’s age
\(\,\,\,\,\,\)Let \(y\) = his brother’s age
\(\,\,\,\,\,\)Given: \(x = 2y\)
\(\,\,\,\,\,\)Five years ago: \(x – 5 = 3(y – 5)\)
\(\,\,\,\,\,\)Substitute \(x = 2y\) into the second equation: \(2y – 5 = 3(y – 5)\)
\(\,\,\,\,\,\)\(2y – 5 = 3y – 15\)
\(\,\,\,\,\,\)\(10 = y\)
\(\,\,\,\,\,\)\(x = 2(10) = 20\)
\(\,\,\,\,\,\)So, Michael is \(20\) years old
\(\textbf{4)}\) In 12 years, Rachel will be three times as old as she is today. How old is Rachel today? Rachel’s age is 6 years old
\(\,\,\,\,\,\)Let \(x\) = Rachel’s age today
\(\,\,\,\,\,\)In 12 years: \(x + 12\)
\(\,\,\,\,\,\)Set up equation: \(x + 12 = 3x\)
\(\,\,\,\,\,\)\(12 = 2x\) \(x = 6\)
\(\,\,\,\,\,\)So, Rachel is \(6\) years old
\(\,\,\,\,\,\)Let \(x\) = Rachel’s age today
\(\,\,\,\,\,\)In 12 years: \(x + 12\)
\(\,\,\,\,\,\)Set up equation: \(x + 12 = 3x\)
\(\,\,\,\,\,\)\(12 = 2x\) \(x = 6\)
\(\,\,\,\,\,\)So, Rachel is \(6\) years old
\(\textbf{5)}\) 8 years ago, Sam’s age was 1/4 the age he will be in 10 years from now. How old is Sam now? Sam’s age is 14 years old
\(\,\,\,\,\,\)Let \(x\) = Sam’s age today
\(\,\,\,\,\,\)8 years ago: \(x – 8\)
\(\,\,\,\,\,\)In 10 years: \(x + 10\)
\(\,\,\,\,\,\)Set up equation: \(x – 8 = \frac{1}{4}(x + 10)\)
\(\,\,\,\,\,\)Multiply both sides by 4: \(4(x – 8) = x + 10\)
\(\,\,\,\,\,\)\(4x – 32 = x + 10\) \(3x = 42\)
\(\,\,\,\,\,\)\(x = 14\)
\(\,\,\,\,\,\)So, Sam is \(14\) years old
\(\,\,\,\,\,\)Let \(x\) = Sam’s age today
\(\,\,\,\,\,\)8 years ago: \(x – 8\)
\(\,\,\,\,\,\)In 10 years: \(x + 10\)
\(\,\,\,\,\,\)Set up equation: \(x – 8 = \frac{1}{4}(x + 10)\)
\(\,\,\,\,\,\)Multiply both sides by 4: \(4(x – 8) = x + 10\)
\(\,\,\,\,\,\)\(4x – 32 = x + 10\) \(3x = 42\)
\(\,\,\,\,\,\)\(x = 14\)
\(\,\,\,\,\,\)So, Sam is \(14\) years old
\(\textbf{6)}\) Emma is two times as old as her sister. Ten years ago, she was 4 times as old as her sister. How old is Emma now? Emma’s age is 30 years old
\(\textbf{7)}\) In 18 years, Nick will be 3 times as old as he is today. How old is Nick today? Nick’s age is 9 years old
\(\textbf{8)}\) 14 years ago, Jake’s age was 1/5 the age he will be in 10 years. How old is Jake now? Jake’s age is 20 years old
\(\textbf{9)}\) Max is 3 times as old as his sister. In ten years, he will be 5/3 times as old as his sister. How old is Max now? Max’s age is 15 years old
\(\textbf{10)}\) John is 10 years younger than Morgan. Four years ago John was 3/8 as old as Morgan. How old are John and Morgan now? John’s age is 10 years old Morgan’s age is 20 years old
\(\textbf{11)}\) John is 15 years older than Karen. Two years ago, John was 2 times as old as Karen. How old is John? John’s age is 32 years old
\(\textbf{12)}\) When I was 4, my brother was half my age. Now I am 20, and my sister is half my age. When was my sister half my brother’s age? 2 years ago
See Related Pages\(\)
\(\bullet\text{ Word Problems- Linear Equations}\)
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\(\bullet\text{ Word Problems- Averages}\)
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\(\bullet\text{ Word Problems- Consecutive Integers}\)
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\(\bullet\text{ Word Problems- Distance, Rate and Time}\)
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\(\bullet\text{ Word Problems- Break Even}\)
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\(\bullet\text{ Word Problems- Ratios}\)
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\(\bullet\text{ Word Problems- Age}\)
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\(\bullet\text{ Word Problems- Mixtures and Concentration}\)
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In Summary
Word problems involving ages are a common type of math problem that students encounter in school. These problems ask students to solve for an unknown age using information about the ages of different people. These problems often include information about the present age and the age at a specific point in the past or future. These are an important part of math education because they help students develop problem-solving skills and apply their knowledge of math concepts to real-world situations.
Word problems are typically covered in math classes for elementary and middle school students. These problems may be introduced in earlier grades as part of a lesson on basic addition and subtraction, and then become more complex as students progress.