CE 111-
BASIC CE MATH
Lesson 6
Algebra
Age-Related Problems
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Mathematical logic lectures and sample problems, algebra, and calculus., Lecture notes of Mathematics
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CE 111-
BASIC CE MATH
Lesson 6 Algebra Age-Related Problems
LECTURE
OBJECTIVES
The primary objective of this lecture is to learn about verbal problems in algebra.
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ALGEBRA
If = present age of a person
− 3 = age of the person 3 years ago
+ 5 = age of the person 5 years from now or 5 years hence
ALGEBRA
Note:
The difference of the ages of two persons is constant at any time.
If A = present age of Albert and B = present age of Bryan
Sum of their ages 4 years ago = (A - 4) + (B - 4) Sum of their ages 2 years hence = (A + 2) + (B + 2) Difference of their ages = A - B
ALGEBRA
ALGEBRA
Example:
Alfred is four times as old as his nephew Franco. 5 years ago, the sum of their ages is equal to the present age of Alfred. How old is each?
Solution:
ALGEBRA
Example:
The sum of the parents' ages is twice the sum of their children’s ages. Five years ago, the sum of the parents' ages was four times the sum of their children’s ages. In fifteen years, the sum of the parents' ages will be equal to the sum of their children’s ages. How many children were in the family?
Solution:
ALGEBRA
SEATWORK # 7
- Mary is 24 years old. Mary is twice as old as Anna was when Mary was as old as Anna is now. How old is Anna now?
- Albert is as old as Bryan will be when Albert is twice as old as Bryan was when Albert's age was half the sum of their present ages. Bryan is as old as Albert was when Bryan was half the age, he will be ten years from now. How old are Albert and Bryan?
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