Some Algebra Word Problems – Part II

(edited 5/31/2017)

Some Algebra Word Problems – Part II

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.

Here are five more problems.  For the previous installment, click this link:  http://edspi31415.blogspot.com/2017/05/some-algebra-word-problems.html

What size is the Garden?

Problem:  

The length of a rectangular garden is 5 feet greater than its width.  The garden’s area is 300 square feet.  Find the dimensions of the garden.  [pg. 452, number 82 – Blitzer]

Discussion:

The area of a rectangle is:

Area = Length * Width

In this case, Width = x, Length = x + 5, and Area = 300, with width and length in feet, area in square feet.

Then:

(x + 5) * x = 300

x^2 + 5 * x = 300

What we have is a quadratic equation.  Subtract 300 from both sides yields

x^2 + 5*x – 300 = 0

We can use the quadratic equation or try to factor.  I’m going to factor:

(x – 15) * (x + 20) = 0

Which leads to x = 15 and x = -20.   In this case, negative lengths do not make sense, hence the solution we are looking for x = 15.

To conclude, the Width is 15 feet, while the Length is 20 feet (15 + 5). 

Trees and Triangles

Problem: 

A person who is 5 feet tall is standing 80 feet from a tree.  At the time, the tree casts an 86-foot shadow, while the person casts a 6-foot shadow.  What is the tree’s height?  [pg. 527, number 32 – Blitzer]

Discussion:

We are working with similar triangles.  Please see the diagram.  The ratio of height/length is:

x/86 = 5/6

x = 86 * 5/6

x = 71.666666

The height of the tree is 71.666666 feet. 

* I previously had the incorrect answer of 70.83333.  I thank Gianfranco Cazzaro for alerting me on my mistake.  The answer is now correct.  - EWS 5/31/2017

From Bangkok to Phnom Penh

Problem:

On a map, Bangkok are given coordinates (-115, 170) and Phnom Penh has the coordinates (65, 70).  Units are in miles.  How long will it take a plane flying 400 miles an hour (mph) on a direct flight between Bangkok and Phnom Penh? Round off the answer to the nearest tenth of an hour.  Assume the flight takes place in a straight line.  [pg. 750, number 121 – Blitzer]

Discussion:

We can use the following two formulas to help us solve the problem:

Distance between points (x1, y1) and (x2, y2):

Distance = √( (x1 – x2)^2 + (y1 – y2)^2 )

Relationship between Distance, Speed, and Time:

Time = Distance / Speed

Let Point 1 (x1, y1) represent Bangkok (-115, 170) and Point 2 represents Phnom Penh (65, 70).  Hence, Distance =√( (-115 – 65)^2 + (170 – 70)^2 ) , and Speed = 400. 

Then:

Time = √( (-115 – 65)^2 + (170 – 70)^2 ) / 400

Time = √( 180^2 + 100^2 ) / 400

Time ≈ 0.5

The plane will take about a half an hour to travel from Bangkok to Phnom Penh.

Making It Rain

Problem:  

A rain gutter (as shown above) is made from aluminum sheet 20 inches wide.  Edges that are x inches wide are turned up on each side to form the gutter.  Find x to allow a cross sectional area of 13 square inches.  Round off the answer to the nearest tenth of an inch.  [pg. 764, number 81 – Blitzer]

Discussion:

If you fold the edges up, you get a rectangle with sides x in. and 20 – 2*x in.  We want to find x such that the area is 13 in^2.  Then:

x * (20 – 2 * x) = 13

20 * x – 2 *x^2 = 13

Bringing all terms to one side:

0 = 2*x^2 – 20 *x + 13

We have a quadratic equation.  This time, we’ll use the quadratic formula:

x = ( 20 ± √(20^2 – 4 * 2 * 13) ) / (2 * 2)

With a calculator the results are (results rounded to tenths):

x ≈ 9.3 in and x ≈ 0.7 in.

There is no specification which measure is preferable, so both answers are acceptable. 

Plot Enclosure

Problem:  

You have 600 feet of fencing to enclose a rectangular plot (see figure).  The side facing the river will not be enclosed.  What is the largest area that can be enclosed?  [pg. 782, number 65 – Blitzer]

The area of the enclosure is:

A = x * (600 – 2*x)

A = 600 * x – 2 * x^2  (I)

The equation is in the form of a quadratic equation y = a*x^2 + b*x + c.  In general, if a < 0, the quadratic equation will have a maximum at x = -a / (2*b).

From (I), a = -2, b = 600, and c = 0.  Since a < 0, we will have a maximum at x = -600 / (2 * -2) = 150.

The sides will have 150 feet by 300 feet.   The maximum area is 150 ft * 300 ft = 45,000 ft^2.

Eddie

This blog is property of Edward Shore, 2017.

Some Algebra Word Problems

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.