# Algebraic World Problem

1. Apr 25, 2012

### S.R

1. The problem statement, all variables and given/known data
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? 2. Relevant equations 3. The attempt at a solution Any help would be appreciated. Let x represent the original cost: 2. Apr 25, 2012 ### mtayab1994 I would say work in reverse start from the$5 and work your way back.

3. Apr 25, 2012

### Staff: Mentor

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. Apr 25, 2012

Or just do the inverse of what he spent starting out from $5. 5. Apr 25, 2012 ### cesaruelas 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. Apr 25, 2012 ### S.R Thanks for the advice. I know -35 is incorrect, that's why I posted the problem. 7. Apr 25, 2012 ### cesaruelas 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. Apr 25, 2012 ### S.R Is S2: [((x-(x/2+10))/2]-10? Last edited: Apr 25, 2012 9. Apr 25, 2012 ### cesaruelas I would only change the last element and make it a + 10, the rest is okay according to me. 10. Apr 25, 2012 ### S.R By "plus$10" does this mean Frank spends an additional $10 or that he spends one half plus$10?

11. Apr 25, 2012

### Staff: Mentor

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. Apr 25, 2012

### S.R

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. Apr 25, 2012

### Staff: Mentor

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. Apr 25, 2012

### S.R

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. Apr 25, 2012

### Staff: Mentor

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. Apr 25, 2012