Search 72,588 tutors
FIND TUTORS

Algebra Word Problems: Relative Comparisons

Solving Word Problems with Proportions and Relative Comparisons
These word problems are set-up where the dependent variable is not provided as is, but rather as a part of an operation. You will have to set-up each side of the equality with its own operations.
Example 1:
“Shelley finished x number of her math homework problems before dinner. Had she finished 3 more, she would have finished half her math homework. Write an equation which represents the relationship between y, total problems and x, number of problems Shelley completed.”
This isn’t set-up in the same way as problems presented in previous entries because there isn’t a defined rate of change right away. So, it will be set-up this way with one variable on each side of the equality. You're already given the variables to use in the problem.
Proportion of completed problems = proportion of total problems.
“3 more than completed problems” = “half her math homework” (half total problems)
(x + 3) = (1/2)*(y)
Or
x + 3 = y/2
If you’re asked in the problem to isolate – usually if the problem wants you to graph or establish the rate of change - you will isolate y by multiplying each side by 2. You get:
2x + 6 = y.

Example 2:
“On Thursday Brenda’s schedule allows for 60 more minutes of training than any other day of the week. One round of strength routine takes 8 minutes. One round of endurance training takes 12 minutes. Write an equation which represents the time available to Brenda to spend on each type of routine on Thursday compared to any given day. Assume all other days she can train the same amount of time.”
So, let’s start by doing a comparison of how much time you can spend on any day to Thursday. Thursday is “60 more than any other day” or 60 + d. d represents the amount of time she can train any other day in minutes.
So, you get
“Amount of Training on Thursday” = Time spent on Strength + Time spent on Endurance.
60 + d = 8 minutes per session + 12 minutes per session.
60 + d = 8s + 12e

Example 3:
“Jennifer and Alex work at a call center for a major company, but in different departments. Monday was a particularly busy day for Jennifer’s department. If Jennifer had received 10 fewer calls, she would have answered 40% more calls than Alex.”
Set-up each side as:
“10 fewer calls” than actual number for Jennifer = “40% more” than actual calls for Alex.
Translate each side into symbols:
j – 10 = a + 0.40a
Note: The problem indicates 40% more calls, not 40 more calls. That's why we use 0.40a and not simply add 40 to the right side.

Seattle tutors