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Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation
3.12 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: 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 100 years and second painting is 200 years
First painting is 300 years and second painting is 100 years
First painting is 400 years and second painting is 200 years

Q 2: Sam's age in 20 years will be the same as Tim's age is now. Ten years from now, Tim's age will be twice Sam's. What are their present ages?
Tim's age is 40 and Sam's age is 20
Tim's age is 20 and Sam's age is 10
Tim's age is 30 and Sam's age is 10

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Question 4: This question is available to subscribers only!


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