**Some Algebra Word Problems**

Source: Blitzer, Robert

*Introductory & Intermediate Algebra for College Students*3^{rd}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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