How Much Money Did Frank Initially Have for School Supplies?

  • Thread starter S.R
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In summary, an algebraic world problem is a real-world situation that can be solved using mathematical equations and expressions. To set up and solve these problems, one must identify unknown quantities and their relationships, use variables and equations, and apply algebraic operations. These problems can be used in everyday life to solve practical problems and require a good understanding of basic algebraic concepts and critical thinking skills.
  • #1
S.R
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Homework Statement


Frank bought supplies for school. In the first store, he spent half his money plus $10. In the second store, he spent half of what he had left plus $10. In the third store, he spent 80% of what he had left. He came home with $5. How much did Frank start out with?

Homework Equations


The Attempt at a Solution


Any help would be appreciated.

Let x represent the original cost:
cYcdD.png
 
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  • #2
I would say work in reverse start from the $5 and work your way back.
 
  • #3
S.R said:

Homework Statement


Frank bought supplies for school. In the first store, he spent half his money plus $10. In the second store, he spent half of what he had left plus $10. In the third store, he spent 80% of what he had left. He came home with $5. How much did Frank start out with?


Homework Equations





The Attempt at a Solution


Any help would be appreciated.

Let x represent the original cost:
cYcdD.png

Obviously, x = -35 is not the right answer.

Write a single equation that has on one side the amounts he spent in the three stores + what he had left, and on the other side, the total amount he started with (x).
 
  • #4
Mark44 said:
Obviously, x = -35 is not the right answer.

Write a single equation that has on one side the amounts he spent in the three stores + what he had left, and on the other side, the total amount he started with (x).

Or just do the inverse of what he spent starting out from $5.
 
  • #5
The problem is your second equation... the problem reads, he spent half of what he had left and what you wrote would mean he spent half of what he spent in the first store. You made the same mistake when writing the last one, the problem reads 80% of what he had left and you wrote 80% of what he spent. Try rewriting those and work the solution again.
 
  • #6
Mark44 said:
Obviously, x = -35 is not the right answer.

Write a single equation that has on one side the amounts he spent in the three stores + what he had left, and on the other side, the total amount he started with (x).

Thanks for the advice. I know -35 is incorrect, that's why I posted the problem.
 
  • #7
Also, the assumption that S3 is equal to 5 is wrong, that's what was left, not wht he spent in the third store (which is what you should write in S3.
 
  • #8
cesaruelas said:
Also, the assumption that S3 is equal to 5 is wrong, that's what was left, not wht he spent in the third store (which is what you should write in S3.

Is S2: [((x-(x/2+10))/2]-10?
 
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  • #9
S.R said:
Is S2: [((x-(x/2+10))/2]-10?

I would only change the last element and make it a + 10, the rest is okay according to me.
 
  • #10
cesaruelas said:
I would only change the last element and make it a + 10, the rest is okay according to me.

By "plus $10" does this mean Frank spends an additional $10 or that he spends one half plus $10?
 
  • #11
S.R said:
In the first store, he spent half his money plus $10
If he starts off with x dollars, the above can be translated to x/2 + 10, the amount he spent in the first store.
 
  • #12
Mark44 said:
If he starts off with x dollars, the above can be translated to x/2 + 10, the amount he spent in the first store.

Yes. In the second store he spends (x-a)/2+10, where a=x/2+10. Correct? I'm still unclear on how to solve this problem.
 
  • #13
Looks good to me.

In store 1 he spent x/2 + 10, so he has left x - (x/2 + 10), or x/2 - 10.

In store 2 he spent half of what remained, plus $10.

Maybe it would help to make a table, listing how much he had when he went into each store, and how much when he came out.
 
  • #14
Mark44 said:
Looks good to me.

In store 1 he spent x/2 + 10, so he has left x - (x/2 + 10), or x/2 - 10.

In store 2 he spent half of what remained, plus $10.

Maybe it would help to make a table, listing how much he had when he went into each store, and how much when he came out.

Store 1: he spent x/2+10, so he has left x/2-10.

Store 2: he spent [(x/2-10)/2]+10 = x/4+5, so he has left (x/2-10)-(x/4+5) = x/4-15.

Store 3: he spent 4/5*[x/4-15] = x/5-12, so he has left (x/4-15)-(x/5-12) = x/20-3.

Is this correct?
 
  • #15
S.R said:
Store 1: he spent x/2+10, so he has left x/2-10.

Store 2: he spent [(x/2-10)/2]+10 = x/4+5, so he has left (x/2-10)-(x/4+5) = x/4-15.

Store 3: he spent 4/5*[x/4-15] = x/5-12, so he has left (x/4-15)-(x/5-12) = x/20-3.

Is this correct?

Continue doing what you're doing until you get an equation that you can solve for x. Once you have a value, check it.
 
  • #16
Mark44 said:
Continue doing what you're doing until you get an equation that you can solve for x. Once you have a value, check it.

Nevermind: answer is 160.
 
Last edited:

1. What is an algebraic world problem?

An algebraic world problem is a real-world situation that can be solved using mathematical equations and expressions. It involves using variables and operations to represent the unknown quantities and relationships between them.

2. How do you set up and solve an algebraic world problem?

To set up and solve an algebraic world problem, you first need to identify the unknown quantities and the relationships between them. Then, you can use variables to represent these quantities and write equations that describe the relationships. Finally, you can use algebraic operations to solve for the unknown quantities.

3. What are some common types of algebraic world problems?

Some common types of algebraic world problems include problems involving linear equations, quadratic equations, systems of equations, and inequalities. These problems can involve various real-world situations, such as distance, time, money, and rates.

4. How can algebraic world problems be used in everyday life?

Algebraic world problems can be used in everyday life to solve practical problems, make decisions, and understand relationships between quantities. For example, they can be used to calculate the cost of a purchase with a discount, plan a road trip based on distance and speed, or determine the optimal pricing strategy for a business.

5. What skills are needed to solve algebraic world problems?

To solve algebraic world problems, one needs to have a good understanding of basic algebraic concepts such as variables, equations, and operations. They also need critical thinking skills, problem-solving skills, and the ability to apply mathematical knowledge to real-world situations.

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