Saturday, May 20, 2017

Some Algebra Word Problems

Some Algebra Word Problems

Source:  Blitzer, Robert  Introductory & Intermediate Algebra for College Students 3rd Edition Pearson, Prentice Hall:  Upper Saddle River, New Jersey 2009  ISBN-13: 978-0-13-602895-6

Problems and diagrams are rewritten.  No claims of profit are made.

A Simple Equation for Heart Beat

Problem:
Write a formula that models that the heart rate, H, in beats per minute, for a person who is a years old and likes to achieve 9/10 of maximum (90% of maximum) heart rate during exercise.  Find the heart during exercise for a 40 year old. 

Information:  The book gives the maximum heart rate in beats per minute as 220 – a.  [pg. 29, number 127 - Blitzer]

Discussion:
The maximum heart rate is given by 220 – a.  Since we want to achieve 9/10 of the maximum, multiply the percentage by the heart rate.  An appropriate function is:

H = 9/10 * (220 – a)

At 40 years old, let a = 40 and H = 9/10 * (220 – 40) = 9/10 * 180 = 162

The maximum rate would be 162 beats per minute.

Doubling the Sphere

Problem: 
What happens to the volume of a sphere if its radius is doubled? [pg. 171 – number 89 – Blitzer]

Discussion:
The volume of a sphere is V = 4/3 * π * r^3, where r is the radius.

If we double the radius (sub 2*r for r), then the volume would be:
V = 4/3 * π * (2*r)^3
V = 4/3 * π * 8 * r^3   (I)
Simplifying:
V = 32/3 * π * r^3

Effectively, doubling the radius increases the sphere’s volume eight times. (see (I))

What Angle Are Talking About?

Problem:  (diagram page 171 number 91)
Find the measure of angle x as indicated by the figure below.  [pg. 171, number 91 - Blitzer]



Discussion:
Assume all angles are measured in degrees.  From the diagram above, label angle A = 2 * x and angle B = 2 * x + 40. 
Observe that A + B = 180°.  We will use this to solve for x.

A + B = 180
(2*x) + (2*x + 40) = 180
4*x + 40 = 180
4*x + 40 – 40 = 180 – 40
4*x = 140
4*x/4 = 140/4
x = 35

The measure of x is 35°.

The Sum and Difference of Two Numbers

Problem:
The sum of two numbers, x and y, is 28.  The difference between the numbers is 6.  What are x and y?  [pg. 287, number 85 – Blitzer]

Discussion:
The goal is to find the values of the two unknowns, x and y. 

The sum of x and y is 28.  Sum means addition.  Hence, x + y = 28.

The difference is 6, and difference means subtraction.  Since no variable is specified, let’s assume x is greater than y.   The equation would be x – y = 6.

Given both statements are true, we have a pair of simultaneous equations:

(I)   x + y = 28
(II) x – y = 6

Now it is a matter of solving for both x and y.   Note how the equations are marked above as (I) and (II).

First add equations (I) and (II) to get:

2 * x = 34
2 * x / 2 = 34 / 2
x = 17

I am going to substitute x = 17 in equation (I) and solve for y (doing this in (II) would work too).

17 + y = 28
17 + y – 17 = 28 – 17
y = 11

Check:  17 + 11 = 28 and 17 – 11 = 6

Yes, it checks out.  The solution is x = 17 and y = 11.

The Cost of Competing Telephone Plans

Problem: 
You are choosing between two-long distance telephone plans.  Plan A has a monthly fee of $15 with a charge of $0.08 per minute for all-long distance calls.  Plan B has a monthly fee of $5 with a charge of $0.10 per minute for long-distance calls.  Determine the number of minutes when the cost of the two plans will be the same.  What will be the cost?  [pg. 301, number 17a – Blitzer]

Discussion:
Let x be the number of minutes.  We can describe the two plans as equations:

Plan A:  C = 15 + 0.08*x  
Plan B:  C = 5 + 0.10*x

Note that I assigned C as the cost.  Also, both equations take the form of C = fixed cost + variable cost * x.

The problem asks us how many minutes will make both plans have the same cost.  For this, equate the costs of both plans A and B.

15 + 0.08 * x = 5 + 0.10 * x
15 + 0.08 * x – 5 = 5 + 0.10 * x – 5
10 + 0.08 * x = 0.10 * x
10 + 0.08 * x – 0.08 * x = 0.10 * x – 0.08 * x
10 = 0.02 * x    (note that 0.02 = 2/100)
10 * 100/2 = 2/100 * x * 100/2
500 = x

As a check:  15 + 0.08 * 500 = 55 and 5 + 0.10 *500 = 55

The amount of minutes of long distance calls that will make both plans cost the same is 500 minutes, and the cost will be $55.00.


Please let me know if you find this helpful, especially students who are in algebra classes.  This is something different I want to do with the blog in addition to all the calculator programming. 

Eddie



This blog is property of Edward Shore, 2017.

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