EIqB1 MILDRED"'OHNSON nM JOHNSON == How to Solve Word Problems in Algebra A Solved Problem Approach Second Edition Mildred johnson Late. We present an approach for automatically learning to solve algebra word problems. Our algorithm reasons across sentence boundaries to construct and solve a. Word Problem Examples: 1. Age: Abigail is 6 years older than Jonathan. Six years ago she was twice as old as he. How old is each now? Solution.
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Hcference Book: HOW TO SOLVE WORD PROBLEMS IN ALGEBRA: A solved problem approach by Mildred Johnson. How to start to work a word problem: 1. Check: Return to the original word problem and see whether these amounts satisfy the conditions . Seeing a pattern can help you solve algebra problems. (Mathematics) I. Title: Five hundred one math word problems. II. Title: Five Algebra Success in 20 Minutes a Day, Algebra Questions, Geometry Success in 20 pose of drill and skill practice is to make you proficient at solving problems.
Ken Arellano. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of , no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. Walker, and the production supervisor was Elizabeth ,. Printed and bound by R. McGraw-Hill books are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training sessions.
Or contact your local bookstore. Most textbooks in algebra do not have adequate explanations and examples for the stu- dent who is having trouble with them. This book's purpose is to give the student detailed instructions in procedures and many completely worked examples to follow. All major types of word problems usually found in algebra texts are here. Emphasis is on the mechanics of word-problem solving because it has been my experience that students having dif- ficulty can learn basic procedures even if they are unable to reason out a problem.
This book may be used independently or in conjunction with a text to improve skills in solving word problems. The problems are suitable for either elementary- or intermediate- level algebra students.
A supplementary miscellaneous prob- lem set with answers only is at the end of the book, for drill or testing purposes. In it you will find many examples of the basic types of word problems completely worked out for you.
Learning to work problems is like learning to play the piano. First you are shown how. Then you must practice and practice and practice. Just reading this book will not help unless you work the problems. The more you work, the more confident you will become. After you have worked many problems with the solutions there to gUide you, you will find miscellaneous problems at the end of the book for extra practice.
MATH 101 - Week 11 Section 1.pdf - College Algebra Solving...
You will find certain basic types of word problems in almost every algebra book. You can't go out and use them in daily life, or in electronics, or in nursing. But they teach you basic procedures which you will be able to use elsewhere.
This book will show you step by step what to do in each type of problem. Let's learn how it's done! How do you start to work a word problem? Read the problem all the way through quickly to see what kind of word problem it is and what it is about. Look for a question at the end of the problem.
This is often a good way to find what you are solving for. Sometimes two or three things need to be found. You let x equal what you are trying to find. What you are trying to find is usually stated in the question at the end of the problem.
You must show and label what x stands for in your problem, or your equation has no meaning. You'll note in each solved problem in this book that x is always labeled with the unit of measure called for in the problem inches, miles per hour, pounds, etc.
That's why we don't bother repeating the units label for the answer line. If you have to find more than one quantity or unknown, try to determine the smallest unknown. This unknown is often the one to let x equal. Go back and read the problem over again. This time read it one piece at a time. Simple problems generally have two statements. One statement helps you set up the unknowns, and the other gives you equation informa- tion.
Translate the problem from words to symbols one piece at a time. Here are some examples of statements translated into algebraic language, using x as the unknown. From time to time refer to these examples to refresh your memory as you work the problems in this book. Statement Algebra 1. Twice as much as the unknown 2x 2. Two less than the unknown x-2 3. A number decreased by 7 x-7 6. Ten decreased by the unknown 10 - x 7. Dan's age x 10 years ago x - 10 9.
Number of cents in x quarters 2Sx Number of cents in 2x dimes 10 2x Separate 17 into two parts x and 17 - x Distance traveled in x hours at SOx SO mph Interest on x dollars for 1 year O. OSx at 5 percent Distance traveled in 3 hours atx mph Product of a number and 3 3x Four times as much 4x No unit labels such as feet, degrees, and dollars are used in equations. In this book we have left these labels off the answers as well.
One number is two times another. A man is 3 years older than twice his son's age. Represent two numbers whose sum is Represent interest income. Represent the amount of acid in quarts. A woman drove for 5 hours at a uniform rate per hour.
Represent the distance traveled. A girl had two more dimes than nickels. Represent how much money she had in cents. Remove parentheses first. Remove fractions by multiplying by the lowest common denominator LCD. Decimals should be removed from an equation before solving.
Multiply by a power of 10 large enough to make all decimal numbers whole numbers. If you multiply by 10, you move the decimal point in all terms one place to the right. If you multiply by , you move the decimal point in all terms two places to the right. In this book they are grouped into general types. That way you can look up each kind of problem for basic steps if you are using an algebra text.
The unknown is a whole number, not a frac- tion or a mixed number. It is almost always a positive number. In some problems the numbers are referred to as integers, which you may remember are positive numbers, negative numbers, or zero. One number is twice the other. What are the numbers? Steps 1. Read the problem. It is about numbers. The question at the end asks, "What are the numbers? Always label x as care- fully as you can.
In this problem, that means just the label "smaller number," since no units of measure are used in this problem. Read the problem again, one piece at a time. The first line says you have two numbers. So far, you have one number represented by x. You know you have to represent two unknowns because the problem asks you to find two num- bers.
So read on. Next the problem states that the sum of the two numbers equals Most of the time it is good to save a sum for the equation statement if you can. The next statement says "one number is twice the other.
II Here is a fact for the second unknown! Now that both unknowns are represented, we can set up the equation with the fact which has not been used.
The sum of the numbers is Be sure you have answered the question completely; that is, be sure you have solved for the unknown or unknowns asked for in the problem. Three times the first is 5 more than twice the second.
First carefully read the problem. Reread the question at the end of the problem: Note the individual facts about the numbers: This problem is different, you say. It sure is! Since state- ment c tells you to do something with the numbers, you have to represent them first. So, let's see another way to use a total or sum. It is like saying, liThe sum of two numbers is One number is 6. What is the other number? That is why if the sum is SO and one number is x, we subtract x from SO and let SO - x represent the second number or unknown.
If you know the sum total , you can subtract one part x to get the other part. Let's get back to the problem. Now put it together: For example, "5 less than x" translates as x - 5. It's good to be consistent and put the " more than" and "less than" quantities on the right: The sum of the numbers is SO.
What is the question what are you looking for? The prob- lem really tells you to find two numbers whose sum is Separate 71 into two parts. One part exceeds the other by 7. The larger minus the smaller equals the difference, and the larger exceeds the smaller by 7. If we showed you all of them, you would be utterly confused, so if your method always works, you are probably correct. It is like saying: The words translate into symbols in the same order as you read the statement. These are consecutive integers.
They are gener- ally qualified as positive. They may be 1. Consecutive integers. Consecutive integers are, for example, 21, 22, The difference between consecutive integers is 1. Represent three consecutive integers. Consecutive even integers. Consecutive even integers might be 2, 4, 6. The difference between consecutive even integers is 2. The first integer in the sequence has to be even, of course. Represent three consecutive even integers. Consecutive odd integers. An example of consecutive odd integers is 5, 7, 9.
Here we find the difference is also 2. But the first integer is odd. The difference is that x represents an even integer in one and an odd integer in the other. The numbers 28, 29, and 30 are consecutive integers.
The sum of 28, 29, and 30 is The largest is three times the smallest. The numbers 2, 4, 6 are even, consecutive inte- gers. Always be sure that all the unknowns at the beginning of the problem are solved for at the end.
Find the integers. The sum of 13, 15, 17, and 19 is Now it is time to try some problems by yourself.
Read the problem all the way through. Find out what you are trying to solve for. If possible, let x represent the smallest unknown. Read the problem again, a small step at a time, translating words into algebraic statements to set up the unknowns and to state the equation.
Supplementary Number Problems I. There is a number such that three times the number minus 6 is equal to Find the number. The sum of two numbers is The larger number is I less than twice the smaller number. Find the numbers. Separate 90 into two parts so that one part is four times the other part. The sum of three consecutive integers is There are two numbers whose sum is Three times the smaller number is equal to 19 more than the larger number.
There are three consecutive odd integers. Three times the largest is seven times the smallest. The sum of four consecutive even integers is There are three consecutive integers. The sum of the first two is 35 more than the third. A foot-long board is to be cut into two parts.
The longer part is I foot more than twice the shorter part. Mahoney went shopping for some canned goods which were on sale. She bought three times as many cans of tomatoes as cans of peaches and two times as many cans of tuna as cans of peaches. If Mrs.
The first side of a triangle is 2 inches shorter than the second side. The third side is 5 inches longer than the second. In the afternoon, Kerrie and Shelly rode their bicycles 4 miles more than three times the distance in miles they rode in the morning on a trip to the lake.
Patton and their daughter Carolyn own three cars. Carolyn drives I0 miles per week farther with her car than her father does with his. Patton drives twice as many miles per week as Mrs. There were , people who attended a rock festival. In a 3-digit number, the hundreds digit is 4 more than the units digit and the tens digit is twice the hundreds digit. If the sum of the digits is 12, find the 3 digits.
Write the number. Solutions to Supplementary Number Problems I. The sum is The sum of the first two is the third plus Because most of you understand the time, rate, and distance relationship, let's learn a method for setting up time, rate, and distance problems.
First, a little review. If you travel for 2 hours at SO mph to reach a destination, you know I hope that you would have traveled miles. It is convenient to have a dia- gram in time, rate, and distance problems. There are usually two moving objects. Sometimes there is one moving object traveling at two different speeds at different times.
Show in a small sketch the direction and distance of each movement. Then put the information in a simple diagram. For example, one train leaves Chicago for Boston and at the same time another train leaves Boston for Chicago on the same track but traveling at a different speed.
The trains travel until they meet. Your sketch would look like this: First train Second train Chicago ' Or suppose two trains leave the same station at different times traveling in the same direction. One overtakes the other. The distances are equal at the point where one over- takes the other. Nemuth leaves New York for a drive to Long Island. Later he returns to New York. New York Long Island Distances are equal.. The diagram for your time, rate, and distance informa- tion might look like this: You should fill in the appropriate information from each problem.
Rate is in miles per hour mph and time is in hours. Distance will therefore be in miles. The basic rela- tionship is time multiplied by rate equals distance.
Two hours later a passenger train leaves the same station for Chicago traveling at 60 mph. How long will it be before the passenger train overtakes the freight train?
Read the problem through, carefully. The question at the end of the problem asks "how long? This question is your unknown. Second, make a diagram to put in information: Now, read the problem again from the beginning for those individual steps.
It says the freight train traveled at 40 mph, so fill in the box for the rate of the freight train: Continue reading: You will always know either both rates or both times in the beginning problems. The question asked is, "How long before the passenger train overtakes the freight train? But when you use the time, rate, and distance table, it helps to put down known facts first because the problem tells you either both the times or both the rates.
Now put in x for the time for the passenger train. It's the unknown to be solved for. By the way, did you notice by your sketch that the distances are equal when one train over- takes the other? This fact is very important to remember. Fill in both rates or both times. The other two will be filled in as unknowns. You will multiply time by rate to get distance and put this product in the dis- tance box to represent distance.
Later you may have prob- lems where each distance is given, but this type of problem results in fractions. For now, the distances will not be given. Back to the problem. You have filled in both rates and one time. You must represent the time for the freight train in terms of x because you know both rates, and therefore both times are unknown. The problem stated that the passenger train started 2 hours after the freight train, so the freight train took 2 hours longer.
This one had the distances equal. That is, the trains traveled the same distance because they started at the same place and traveled until one caught up with the other. This fact was not stated. You have to watch for the relationship. We have two distances in the table above. Set these two distances equal for your equation: There was only one unknown asked for, but you had to use two to work the problem. Also, you could have used x for the freight train's time and x - 2 for the passenger train's time.
Always be very careful that you have answered the question. We have seen one object overtake another, starting from the same place at a different time. Here is another type. At the same time, another car leaves Los Angeles for San Francisco traveling 60 mph. If it is miles between San Francisco and Los Angeles, how long before the two cars meet, assuming that each maintains its average speed?
Find out if you know both speeds or both times. In this problem we find that both speeds are given. What is the unknown? The question is, "How long before the two cars meet? But which shall we let x equal? They leave at the same time and meet at the same time, so the traveling times must be equal if nei- ther one stops.
Watch for the times being equal when two objects start at the same time and meet at the same time. What fact is known about distance? The miles is the total distance, so here is equation information. Remember, don't put distance information in the distance box unless you have to for example, when the exact distance each object travels is given. Solution First, draw a sketch of movement: Los Angeles And remember, time multiplied by rate equals distance in the table.
You know that the total distance is miles, so add the two distances in the diagram and set them equal to miles for the equation. So it isn't a jet! At what time will they be miles apart? How far has each traveled? You know both speeds. The times must be unknowns and they are equal or the same because the planes leave at the same time and travel until a certain given time when they are miles apart.
The total distance is miles. It stands for traveling time in hours!
Algebra Word Problems
The rate is in miles per hour, so time must be in hours in all time, rate, and dis- tance problems. New York.. Always check the ques- tion, especially on time, rate, and distance problems.
The problem asked what time they would be miles apart. Since the problem says they left at 10 A. So the answer is Derbyshire makes a business trip from his house to Loganville in 2 hours. One hour later, he returns home in traffic at a rate 20 mph less than his rate going. If Mr. Derbyshire is gone a total of 6 hours, how fast did he travel on each leg of the trip? This problem does not give us the rates, but it does give us the time both ways.
It gives the time to Loganville as 2 hours. The total time is 6 hours, but he did not travel dur- ing one of those hours. Deducting this hour leaves only 5 hours of total traveling time. That means that if it took 2 hours to travel to Loganville, it must have taken 3 hours to travel back.
The rates are both unknown. We can let x equal either one. Just be careful that the faster rate goes with the shorter time. If x equals the rate going, then x - 20 equals the rate returning. Either way is correct. Solution Sketch: To Loganville..
The distance to Loganville equals the dis- tance back home.
One day Jerry left camp on his motorcycle to go to the village. Ten minutes later Jake decided to go too. If Jerry was traveling 30 mph and Jake traveled 3S mph, how long was it before Jake caught up with Jerry?
We know both speeds. We know that Jake traveled for 10 minutes less than Jerry. But minutes cannot be used in time, rate, and distance problems because rate is in miles per hour. They travel the same distance so the distances are equal. Also, be sure to answer the question and solve for x - l, Jake's time. Can you guess why this problem is here?
You must be very careful to have time always in hours or fractions of an hour because rate is in miles per hour. Basically, our time, rate, and distance problems look like the following: Distances equal Distances equal Times equal, total distance given miles.. One is SO miles ahead of the other on the same road. The one in front is traveling 60 mph while the second car is traveling 70 mph. How long will it be before the second car overtakes the first car?
The speeds are given. The times are the unknowns and are equal. The distances are not equal; this inequality is the tricky variation we mentioned above. But if we add SO miles to the distance of one, it will equal the distance of the other. So each takes x hours. The difficulty is that the two cars do not start from the same pOint, so their distances are not equal. The sketch shows that you have to add SO miles to the distance the first car travels to make it equal to the dis- tance the second car travels.
Of course, the equation could be: If you can see the relationship better one way than another, always do the problem that way. The only problems of this type which we can solve are those where the objects move directly with or against the wind or water. The plane must have a direct headwind or tailwind, and the boat must be going upstream or downstream. In this type of problem the plane's speed in still air would be increased by a tailwind or decreased by a headwind to determine how fast it actu- ally covers the ground.
For example, a plane flies mph in still air. This is called airspeed. If there is a mph headwind blowing, it would decrease the speed over the ground by 20 mph, so the ground speed of the plane would be - 20 or mph.
The ground speed is the rate in time, rate, and dis- tance problems. A headwind reduces the speed of the plane by the velocity of the wind.
A tailwind increases the speed of the plane over the ground by the velocity of the wind. A plane with an airspeed speed in still air of mph with a mph tailwind actually travels over the ground ground speed at mph. A current affects a boat in the same way.
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This PDF book incorporate commoncoresheet perimeter conduct. The teaching of heuristics  and other problem-solving strategies to solve non-routine problems. Methods[ edit ] The method or methods used in any particular context are largely determined by the objectives that the relevant educational system is trying to achieve. Methods of teaching mathematics include the following: Classical education: the teaching of mathematics within the quadrivium , part of the classical education curriculum of the Middle Ages , which was typically based on Euclid's Elements taught as a paradigm of deductive reasoning.
In "Number Bingo," players roll 3 dice, then perform basic mathematical operations on those numbers to get a new number, which they cover on the board trying to cover 4 squares in a row. Computer-based math an approach based around use of mathematical software as the primary tool of computation. Computer-based mathematics education involving the use of computers to teach mathematics. Mobile applications have also been developed to help students learn mathematics.
Starts with arithmetic and is followed by Euclidean geometry and elementary algebra taught concurrently. Requires the instructor to be well informed about elementary mathematics , since didactic and curriculum decisions are often dictated by the logic of the subject rather than pedagogical considerations. Other methods emerge by emphasizing some aspects of this approach. Exercises : the reinforcement of mathematical skills by completing large numbers of exercises of a similar type, such as adding vulgar fractions or solving quadratic equations.
Historical method: teaching the development of mathematics within an historical, social and cultural context. Provides more human interest than the conventional approach. Adopted in the US as a response to the challenge of early Soviet technical superiority in space, it began to be challenged in the late s.
Solving Word Questions
The New Math method was the topic of one of Tom Lehrer 's most popular parody songs, with his introductory remarks to the song: " The problems can range from simple word problems to problems from international mathematics competitions such as the International Mathematical Olympiad. Problem solving is used as a means to build new mathematical knowledge, typically by building on students' prior understandings.
Recreational mathematics : Mathematical problems that are fun can motivate students to learn mathematics and can increase enjoyment of mathematics. Relational approach: Uses class topics to solve everyday problems and relates the topic to current events.
Rote learning : the teaching of mathematical results, definitions and concepts by repetition and memorisation typically without meaning or supported by mathematical reasoning. A derisory term is drill and kill. In traditional education , rote learning is used to teach multiplication tables , definitions, formulas, and other aspects of mathematics.
Content and age levels[ edit ] Different levels of mathematics are taught at different ages and in somewhat different sequences in different countries. Sometimes a class may be taught at an earlier age than typical as a special or honors class.
Elementary mathematics in most countries is taught in a similar fashion, though there are differences. Most countries tend to cover fewer topics in greater depth than in the United States. Mathematics in most other countries and in a few U.
Students in science-oriented curricula typically study differential calculus and trigonometry at age 16—17 and integral calculus , complex numbers , analytic geometry , exponential and logarithmic functions , and infinite series in their final year of secondary school. Probability and statistics may be taught in secondary education classes. Science and engineering students in colleges and universities may be required to take multivariable calculus , differential equations , and linear algebra.
Applied mathematics is also used in specific majors; for example, civil engineers may be required to study fluid mechanics ,  while "math for computer science" might include graph theory , permutation , probability, and proofs.
Standards[ edit ] Throughout most of history, standards for mathematics education were set locally, by individual schools or teachers, depending on the levels of achievement that were relevant to, realistic for, and considered socially appropriate for their pupils. In modern times, there has been a move towards regional or national standards, usually under the umbrella of a wider standard school curriculum.
In England , for example, standards for mathematics education are set as part of the National Curriculum for England,  while Scotland maintains its own educational system. Ma summarised the research of others who found, based on nationwide data, that students with higher scores on standardised mathematics tests had taken more mathematics courses in high school.In order to understand them, you must under- stand what each digit in a number means.
Assume simple interest compounded yearly. How long did it take Cheryl to finish the first car alone? Quadratic Formula. Jennifer went to the drugstore. Simplifying Eq. The distances are equal at the point where one over- takes the other.
Later he returns to New York. The unknown is a whole number, not a frac- tion or a mixed number. The perimeter of a right triangle is 60 cm.