Algebra Problem Solving Strategies

There are ways to solve word problems and the following is two strategies for solving an algebra problem.

Problem

10 years ago Bob was three times as old as Sallie was, 10 years from now Bob will be twice as old.

What is Bob’s present age?

a. 30 years

b. 50 years

c. 60 years

d. 70 years

e. 80 years

Solving Problem Technique #1

We know that Bob’s age 10 years ago is three times Sallie’s age.

i.e. (b − 10) = 3(s − 10) where ‘b’ and ‘s’ are present ages of Bob and Sallie respectively.

Therefore, (b – 10) = multiple of 3.

From the options, subtract 10 to get the age of Bob 10 years ago and see which

of these is a multiple of 3.

a. 30 => 30 – 10 = 20 (Not a multiple of 3)

b. 50 => 50 – 10 = 40 (Not a multiple of 3)

c. 60 => 60 – 10 = 50 (Not a multiple of 3)

d. 70 => 70 – 10 = 60 (is a multiple of 3)

e. 80 => 80 – 10 = 70 (Not a multiple of 3)

Since, only option ‘d’ satisfies the conditions, 70 years is the present age of Bob.

(Ans: d)

Solving Problem Technique #2

Let the present age of Bob be ‘b’ years.

Therefore, Bob’s age 10 years ago = (b – 10) years

Let the present age of Sallie be ‘s’ years.

Therefore, Sallie’s age 10 years ago = (s – 10) years

Equation 1:   (b − 10) = 3(s − 10)

Similarly Bob’s age 10 years from now = (b + 10)

And Sallie’s age 10 years from now = (s + 10)

Equation 2:   (b + 10) = 2(s + 10)

From equations 1 and 2 we get:

b − 10 = 3s − 30 and b + 10 = 2s + 20

b − 3s = −20 and b − 2s = 10

Solving simultaneously, we get:

s = 30 and b = 70

Therefore, Bob’s present age = 70 years.

Resources

12 Tips for Solving Word Problems

Math Problem Solving Strategies

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