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Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation3.12 Word Problems - Age - 4

 Example: Mike is 7 years younger than Jill. Together, there ages add up to 19. Find both their ages. Solution: Let x = Jill's age Then x - 7 = Mike's age Their ages add up to 19 x + (x-7) = 19 x + x - 7 = 19 2x - 7 = 19 2x - 7 + 7 = 19 + 7 2x = 26 x = 13 x - 7 = 13 - 7 = 6 Verification: x + (x-7) = 13 + (13-7) = 13 + 6 = 19 Jill is 13 and Mike is 6 year old. Example: Harold's age is four times the sum of the ages of his two daughters. Six years hence, his age will be double the sum of their ages. Find Harold's age. Solution: Let the sum of the ages of his two daughter be 'x' years. Harold's age = 4x Six years hence, the sum of the ages of the two daughters = x + 6 + 6 = x + 12 Harold's age = 4x + 6 4x + 6 = 2(x + 12) 4x + 6 = 2x + 24 2x = 24 - 6 = 18 x = 9. Harold's age = 4x = 36 years Directions: Solve the following word problems.
 Q 1: Ben's age is three times Molly's. If 20 is added to Molly's age and 20 is subtracted from Ben's, their ages will be equal. How old is each now?Molly's age is 30 and Ben's age is 60Molly's age is 20 and Ben's age is 60Molly's age is 20 and Ben's age is 80 Q 2: One painting is three times as old as the second painting. One hundred years from now the first painting will be twice as old. How old is the first painting? First painting is 300 years and second painting is 100 yearsFirst painting is 100 years and second painting is 200 yearsFirst painting is 400 years and second painting is 200 years Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only!

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