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### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation2.8 Word Problems - Age

 Steps: 1. Read the problem carefully and note down what is given and what is required. 2. Select a letter say x or y or z to represent the unknown quantity asked for. 3. Represent the word statements of the problems in the symbolic language step by step. 4. Look for quantities which are equal as per conditions given and form an equation. 5. Solve the equation. 6. Verify the result for making sure that your answer satisfies the requirements of the problems. Example: Five years ago the age of a father was twice the age of his son. The sum of their present ages is 55 years. Find their present ages. Solution: The age of the father, 5 years ago = x years Age of the father 5 years ago = 2x years Present ages of son and father respectively are (x+5) years and (2x+5)years. Given that, sum of their present ages = 55 years Therefore (x+5)+(2x+5) = 55 3x + 10 = 55 3x = 55 - 45 3x = 45 x = 45/3 x = 15 Verification: Present age of son = x + 5 = 15 + 5 = 20 years Present age of father = 2x + 5 = 2*15 + 5 = 30 + 5 = 35 years The sum of their present ages = 20 + 35 = 55 years. Directions: Solve the following word problems. Also write at least ten word problem examples of your own.
 Q 1: The present age of Adam is three times that of his son. Six years ago, the age of Adam was four times that of his son. Find the ratio of their ages 6 years later.5:26:36:45:6 Q 2: Kate's age 13 years ago is half her age 13 years later. Find her present age.39491629 Q 3: The present age of Nancy is one fourth that of her mother Christina. 20 years later the age of Nancy will be half the age of her mother. Find their present age.Nancy 10 years Christina 40 yearsNancy 15 years Christina 35 yearsNancy 10 years Christina 20 yearsNancy 15 years Christina 30 years Q 4: George aged 50 years, has a daughter Jessica of age 20 years. How many years ago was the age of George three times that of his daughter.2616545 Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!