Using a problem solving technique on a word problem

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cbarker12

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Dear Everyone,

Disclaimer: This question is not a homework question. I am using a outdate book (meaning that the book, College Algebra by Richard Heineman, is not taught out of) to use this technique (as you will read below).

I was given a document about problem solving technique by my previous tutoring supervisor and Intermediate Algebra instructor at my undergraduate university. The technique by using some transitional words (let, then, but, so, therefore) in the solution in order for the students not to be overwhelm and confused on the story problems. For instead,
"Suppose in designing a house, the living room is to be thrice as long as it is wide. The total area of the room is 507 square feet. What should be the length of the room?"

Solution to this example:
Let [MATH]w[/MATH] be the width. Then the length is [MATH] 3w[/MATH], and the area will be [MATH] (3w)(w)=3w^2[/MATH]. But we know the total area is 507. So [MATH]3w^2=507[/MATH]. After we divide 3 and take the positive root on both side of the equation, we discovered [MATH]w=13[/MATH]. Therefore (or any synonym), the length of the room is 39 feet because [MATH] 3 \cdot 13=39[/MATH].

Here is my problem, Exercise 15 problem 5: A vending machine contains $4.30 consisting of nickels and dimes. How many of each if the total number of coins is 60?
I have notes some facts in the problem: There is 4.30 dollars. We know that 2 nickels is equivalent to one dime. We have something with 60 coins.
But I am struck on how to set it up.

Here is what I think the solution should look like:

Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then the machine have a total of $4.30. But the machine has 60 coins all together. So....
 
Last edited:
cbarker12 said:
Dear Everyone,

Disclaimer: This question is not a homework question. I am using a outdate book (meaning that the book, College Algebra by Richard Heineman, is not taught out of) to use this technique (as you will read below).

I was given a document about problem solving technique by my previous tutoring supervisor and Intermediate Algebra instructor at my undergraduate university. The technique by using some transitional words (let, then, but, so, therefore) in the solution in order for the students not to be overwhelm and confused on the story problems. For instead,
"Suppose in designing a house, the living room is to be thrice as long as it is wide. The total area of the room is 507 square feet. What should be the length of the room?"

Solution to this example:
Let [MATH]w[/MATH] be the width. Then the length is [MATH] 3w[/MATH], and the area will be [MATH] (3w)(w)=3w^2[/MATH]. But we know the total area is 507. So [MATH]3w^2=507[/MATH]. After we divide 3 and take the positive root on both side of the equation, we discovered [MATH]w=13[/MATH]. Therefore (or any synonym), the length of the room is 39 feet because [MATH] 3 \cdot 13=39[/MATH].

Here is my problem, Exercise 15 problem 5: A vending machine contains $4.30 consisting of nickels and dimes. How many of each if the total number of coins is 60?
I have notes some facts in the problem: There is 4.30 dollars. We know that 2 nickels is equivalent to one dime. We have something with 60 coins.
But I am struck on how to set it up.

Here is what I think the solution should look like:

Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then the machine have a total of $4.30. But the machine has 60 coins all together. So....
Click to expand...
I am not a big fan of this method. It makes total sense after someone else solves the problem. But is it obvious to you what comes after "then", for example?
I would look for relationships in the problem and use them to identify variables and set up equations. Here we have 2 relationships:
number of coins is ?
total value is ?
You can use them to either set up a system of 2 equations and 2 variables or express one variable through another and set up one equation with 1 variable. Give it a try.
 
cbarker12 said:
Dear Everyone,

Disclaimer: This question is not a homework question. I am using a outdate book (meaning that the book, College Algebra by Richard Heineman, is not taught out of) to use this technique (as you will read below).

I was given a document about problem solving technique by my previous tutoring supervisor and Intermediate Algebra instructor at my undergraduate university. The technique by using some transitional words (let, then, but, so, therefore) in the solution in order for the students not to be overwhelm and confused on the story problems. For instead,
"Suppose in designing a house, the living room is to be thrice as long as it is wide. The total area of the room is 507 square feet. What should be the length of the room?"

Solution to this example:
Let [MATH]w[/MATH] be the width. Then the length is [MATH] 3w[/MATH], and the area will be [MATH] (3w)(w)=3w^2[/MATH]. But we know the total area is 507. So [MATH]3w^2=507[/MATH]. After we divide 3 and take the positive root on both side of the equation, we discovered [MATH]w=13[/MATH]. Therefore (or any synonym), the length of the room is 39 feet because [MATH] 3 \cdot 13=39[/MATH].

Here is my problem, Exercise 15 problem 5: A vending machine contains $4.30 consisting of nickels and dimes. How many of each if the total number of coins is 60?
I have notes some facts in the problem: There is 4.30 dollars. We know that 2 nickels is equivalent to one dime. We have something with 60 coins.
But I am struck on how to set it up.

Here is what I think the solution should look like:

Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then the machine have a total of $4.30. But the machine has 60 coins all together. So....
Click to expand...
Let

the number of Nickels = N

the number of Dimes = D

But

Total number of coins is 60. Therefore

N + D = 60 ...........................................................................(1)
and
Total value of collection = 430 cents. Therefore

5N + 10D = 430 ...............................................................(2)

Now you have two equations and two unknowns ........................... solve for N and D.

Personally I think this is a bit convoluted - and it did NOT work for you. Maybe need more practice
 
lev888 said:
I am not a big fan of this method. It makes total sense after someone else solves the problem. But is it obvious to you what comes after "then", for example?
I would look for relationships in the problem and use them to identify variables and set up equations. Here we have 2 relationships:
number of coins is ?
total value is ?
You can use them to either set up a system of 2 equations and 2 variables or express one variable through another and set up one equation with 1 variable. Give it a try.
Click to expand...
In the textbook, the problem is under single variable linear equations. I thought about the system of equation first. But I am pretending to be a student learning this for the first time.

Subhotosh Khan said:
Let

the number of Nickels = N

the number of Dimes = D

But

Total number of coins is 60. Therefore

N + D = 60 ...........................................................................(1)
and
Total value of collection = 430 cents. Therefore

5N + 10D = 430 ...............................................................(2)

Now you have two equations and two unknowns ........................... solve for N and D.

Personally I think this is a bit convoluted - and it did NOT work for you. Maybe need more practice
Click to expand...

I agree, I need to practice more. I avoid word problems like the plague, but I will be teaching them soon enough so. I want to get more comfortable with them. I also want to use this method in physics and other higher applied math courses.
 
cbarker12 said:
In the textbook, the problem is under single variable linear equations. I thought about the system of equation first. But I am pretending to be a student learning this for the first time.
Click to expand...
You can "cheat" and use 2 variables, then use one of the relationships to reduce their number. E.g. in your first example instead of "Let w be the width. Then the length is 3w" you can say:
Let w be the width and l the length. Then l = 3w.
Area is 507, so lw = 507.
Use the first equation to get rid of l in the second: 3w*w = 507 - one equation with 1 variable.
When explaining the problem to students you just omit the explicit length variable and say "length is 3w".
 
Beer soaked recall follows.
cbarker12 said:
...
Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then the machine have a total of $4.30. But the machine has 60 coins all together. So....
Click to expand...
I've seen an old friend (Denis) do this kind of problem a lot with the following layout:
`.05x + .10(60-x) = 4.30`
 
cbarker12 said:
I was given a document about problem solving technique by my previous tutoring supervisor and Intermediate Algebra instructor at my undergraduate university. The technique by using some transitional words (let, then, but, so, therefore) in the solution in order for the students not to be overwhelm and confused on the story problems.
Click to expand...
I would not call this a problem-solving technique, but a writing style. It's a good style for writing solutions in a textbook, keeping it readable; but it is not a help for finding a solution. Generally, if you focus too much on the final form of anything you write (from a speech to a proof to a solution), you will miss details. So you start with a rough draft; but in problem solving, some orderly format for that is helpful.

What you want to do to solve a problem like this is to first understand what the problem says; then translate the givens into symbolic form, and then solve it in that form.

cbarker12 said:
Here is my problem, Exercise 15 problem 5: A vending machine contains $4.30 consisting of nickels and dimes. How many of each if the total number of coins is 60?
I have notes some facts in the problem: There is 4.30 dollars. We know that 2 nickels is equivalent to one dime. We have something with 60 coins.
But I am struck on how to set it up.

Here is what I think the solution should look like:

Let n be the number of nickels in the vending machine. Then the machine have a total of $4.30. But the machine has 60 coins all together. So....
Click to expand...

In my initial work, I would not try to write out sentences. I would make a table something like this:

Total value: $4.30​
Total number of coins: 60​
Value of nickel: 0.05​
Value of dime: 0.10​

(At this point, I might decide to write all values in cents rather than dollars to make it easier. I won't, but it's a good idea. That's one reason this is a rough draft, so we can change course as needed.)

Now I have to define a variable. (If you had stated the context, allowing only a single variable, you would have had more appropriate help quicker; that's worth knowing when you tutor, that you need to know the student's context.)

Let's define, as you did,

Let n = number of nickels.​

Now we have another unknown quantity, the number of dimes. So we want

Then _____ = number of dimes.​

What goes in the blank?

Notice that what drives this is not a grammatical description of our final write-up, but seeing what we need next at each phase.

Next, you can write an equation.
 
Dr.Peterson said:
I would not call this a problem-solving technique, but a writing style. It's a good style for writing solutions in a textbook, keeping it readable; but it is not a help for finding a solution. Generally, if you focus too much on the final form of anything you write (from a speech to a proof to a solution), you will miss details. So you start with a rough draft; but in problem solving, some orderly format for that is helpful.

What you want to do to solve a problem like this is to first understand what the problem says; then translate the givens into symbolic form, and then solve it in that form.



In my initial work, I would not try to write out sentences. I would make a table something like this:

Total value: $4.30​
Total number of coins: 60​
Value of nickel: 0.05​
Value of dime: 0.10​

(At this point, I might decide to write all values in cents rather than dollars to make it easier. I won't, but it's a good idea. That's one reason this is a rough draft, so we can change course as needed.)

Now I have to define a variable. (If you had stated the context, allowing only a single variable, you would have had more appropriate help quicker; that's worth knowing when you tutor, that you need to know the student's context.)

Let's define, as you did,

Let n = number of nickels.​

Now we have another unknown quantity, the number of dimes. So we want

Then _____ = number of dimes.​

What goes in the blank?

Notice that what drives this is not a grammatical description of our final write-up, but seeing what we need next at each phase.

Next, you can write an equation.
Click to expand...
The answer to the blank is 2n? Because the number of nickels in dimes are double.
 
cbarker12 said:
The answer to the blank is 2n? Because the number of nickels in dimes are double.
Click to expand...
Where does it say that? Are you confusing the values of dimes and nickels with their numbers? I could have any combination of nickels and dimes, regardless of their relative values.

What you do know is that there are 60 coins in all, which are either nickels or dimes. So what expression tells you how many dimes there are?
 
I just checked my pockets to see if you were correct or not. I had three dimes and seven nickels. I did not have twice as many nickels as dimes!
 
Subhotosh Khan said:
No - it should be 60 - n

Where did you get that information?

The value of 1 dime is is double that of value of 1 nickel - not in number.

Please study response #6 - write back if you miss something.
Click to expand...
Sorry for the mistake. I kinda don't want the answer given to me. I want to figure it out. I want to struggle with math. And not just find the answer, haha, there is. I did it too many times in my undergraduate career. I do appreciate the answer in #6, but that would be my last resort.
 
You have 60 coins consisting of nickels and dimes.
If I have 5 nickels, then I have 60-5 dimes.
If I have 15 nickels, then I have 60-15 dimes.
If I have 28 nickels, then I have 60-28 dimes.
If I have 33 nickels, then I have 60-33 dimes.
If I have 42 nickels, then I have 60-42 dimes.
If I have 51 nickels, then I have 60-51 dimes.
If I have 55 nickels, then I have 60-55 dimes.
If I have n nickels, then I have __ dimes.
 
For completeness, Here is the solution to the problem.

Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then [MATH]60-n [/MATH] be the number of dimes in the same machine, and the following expression for this problem is [MATH].05n+.10(60-n)[/MATH]. But we know the coins inside of the machine add up to $4.30. So [MATH].05n+.10(60-n)=4.30[/MATH]. Using basic algebraic techniques, we find out that [MATH]n=34[/MATH]. Therefore, there are 34 nickels, and 26 dimes (60-34) in the machine.
 
cbarker12 said:
the following expression for this problem is .05n+.10(60−n)\displaystyle .05n+.10(60-n).
Click to expand...
I don't understand the above. Where did it come from? An expression is a value, it should correspond to some quantity in the problem, not to the problem itself. If you wrote "The following expression is the total value of coins" it would make sense.
 
cbarker12 said:
For completeness, Here is the solution to the problem.

Let [MATH]n[/MATH] be the number of nickels in the vending machine. Then [MATH]60-n [/MATH] be the number of dimes in the same machine, and the following expression for this problem is [MATH].05n+.10(60-n)[/MATH]. But we know the coins inside of the machine add up to $4.30. So [MATH].05n+.10(60-n)=4.30[/MATH]. Using basic algebraic techniques, we find out that [MATH]n=34[/MATH]. Therefore, there are 34 nickels, and 26 dimes (60-34) in the machine.
Click to expand...
The last and very important thing to do is to CHECK that your answer satisfy the given conditions:

cbarker12 said:
the total number of coins is 60
Click to expand...

34 + 26 = 60 ............................ ♥

cbarker12 said:
machine contains $4.30
Click to expand...

34 * 0.05 + 26 * 0.10 = 1.70 + 2.60 = 4.30 ................................... ♥
 
lev888 said:
I don't understand the above. Where did it come from? An expression is a value, it should correspond to some quantity in the problem, not to the problem itself. If you wrote "The following expression is the total value of coins" it would make sense.
Click to expand...
I thought about it and I type quicker than I think, sometimes.
 
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