Hence the use of clue words, although designed to be helpful to the students and to serve as a crutch, is actually a disservice to the student since it does not convey to the student that there are no short-cut to doing verbal problems, rather the student must think!
There are methods, however, that the student can use to help break down the problem and represent the various elements. The following example shows how this can be done with one such problem using Hands-On Equations. This problem is taken from the Hands-On Equations Verbal Problems Book. Using Hands-On Equations, this problem is accessible to students as early as the 4th grade.
Theresa could purchase four small gifts and a $3 doll for the same price as three of the same small gifts and one $5 doll. What was the price of each of the small gifts?
Solution:
We let the blue pawn represent the price of each of the small gifts. The price of four of the small gifts would therefore be represented by 4 blue pawns. The $3 doll would be represented by a red 3 cube. And likewise for the other side.
The setup for the problem therefore looks as follows:
From here, we can use legal moves (remove three blue pawns from each side) to simplify the setup.
From this simplified setup we can see that the blue pawn is worth 2. Hence, The cost of each small gift is $2.
Check: $11=$11
ATTENTION TEACHER OR HOME SCHOOLER:
Additionally, if you are a teacher in grades 3 to 8 you may wish to attend a Making Algebra Child's Play workshop this season, In this workshop, you will learn how to use the Hands-On Equations program to solve equations, and also how to apply the concepts to verbal problems.
If you have already attended a Making Algebra Child's Play workshop, or are already using Hands-On Equations in your classroom or in your math program, and you are teaching in grades 6 and up, we encourage you to consider attending the Day2 Hands-On Equations Verbal Problems Workshop. In this workshop, you will review the key ideas of Hands-On Equations and you will also see how to apply these ideas to solve a wide variety of consecutive integer, age, coin and distance problems, including rowing up and downstream! This workshop will also be of interested to teachers of the gifted grades 2 and up, and teachers of low-achieving high school students.
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Kira Brennan, age 8 presented the solution below to the above problem:
"When I saw the picture, I saw that each present could be a blue pawn, and the doll could be a block (cube). So I put four blue pawns and a red 3 cube on the left hand side, and three pawns and the 5 cube on the right side. I guessed then that each present costs $2, but I took three pawns off each side anyway, and I could see you have to add $2 to 3 to equal $5 on the other side. Also, I counted 2-4-6-8-11 on the left, and 2-4-6-11 on the right, 11 equals 11, so each present must cost $2. It's harder if you just set up the equation, I think, but it was easy after I drew the picture."
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This problem is appropriate for students in grades 5 and up who have had Level I of Hands-On Equations
Pedro’s dad is three times Pedro’s age. In 10 years, Pedro’s dad will be twice as old as Pedro will be then. How old is each now?
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