Lesson 120: Age Word Problems

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Presentation transcript:

Lesson 120: Age Word Problems

Most age word problems discuss the present ages of two or more people and their ages at some given time in the past and/or in the future. The key to these problems is the proper choice of unknowns. Subscripted variables are very helpful.

Example: A man is 13 times as old as his son Example: A man is 13 times as old as his son. In 10 years he will be 3 times as old as his son will be. How old are they now?

Answer: M = 13S M + 10 = 3(S + 10) S = 2 M = 26 N N N N N

Example: Five years ago Brenda was 4/5 as old as Layton Example: Five years ago Brenda was 4/5 as old as Layton. Ten years from now she will be 7/8 as old as Layton. How old is each now?

Answer: B – 5 = 4/5(L – 5) B + 10 = 7/8(L + 10) B = 25 L = 30 N N N N N

Example: Thirty years ago Barbie was 1 year older than twice Mary’s age then. Twenty years ago Mary was 4/5 as old as Barbie was then. How old is each woman now?

Answer: B – 30 – 1 = 2(M – 30) M – 20 = 4/5(B – 20) M = 32 B = 35 N N N N N

HW: Lesson 120 #1-30