Algebra word problem -- Paying down a debt....

In summary, the person has an outstanding debt, and is trying to calculate the total amount of money he has spent on the debt and the amount of money he has left to spend. He solves for p and finds that he has spent 200 pounds and has left 110 pounds.
  • #1
paulmdrdo
89
2

Homework Statement


Q27hall_XXIV.jpg


Homework Equations


I can't start because I did not understand the statement in red underline. Please explain to me first. Then I may be able to solve it. Sorry english is not my first language. Hope you would understand. Thank you!

The Attempt at a Solution

 
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  • #2
paulmdrdo said:

Homework Statement


Q27hall_XXIV.jpg


Homework Equations


I can't start because I did not understand the statement in red underline. Please explain to me first. Then I may be able to solve it. Sorry english is not my first language. Hope you would understand. Thank you!

The Attempt at a Solution

The person has an old outstanding debt on which he pays 5% of his total income. This 5% of his income with which he pays is for interest on the outstanding debt which is calculated at the rate of 7.5%.
 
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  • #3
My understanding of this is that it is an exercise incorporating the fact that when you deal with the real world problems, you have to know what information is relevant and what is irrelevant. It is relevant that he pays 5% of his total income in restitution another loan, but I don't think it is it all relevant to this calculation what the rate of interest is.

(Of course in real life if he doesn't think he will need his savings for any purpose the person would be advised to use his savings to pay off that debt if he can).
 
  • #4
epenguin said:
My understanding of this is that it is an exercise incorporating the fact that when you deal with the real world problems, you have to know what information is relevant and what is irrelevant. It is relevant that he pays 5% of his total income in restitution another loan, but I don't think it is it all relevant to this calculation what the rate of interest is.

(Of course in real life if he doesn't think he will need his savings for any purpose the person would be advised to use his savings to pay off that debt if he can).
Since the problem asks for the amount of the outstanding debt, knowing the interest which accrues on that debt is essential to calculating that figure.

My interpretation of this situation is the borrower is paying the interest on the debt to keep it from falling in arrears, but alas, he does not appear to be paying down on the principal amount which was initially borrowed.
 
  • #5
Yes I really read the question and this time particularly the last few words and you are right, all the information given is necessary.

Which just shows I was right about something! :oldbiggrin:
 
  • #6
X-1/4x-1/3x-0.05x=110

Solving for x = 300 total money he had.

Solving for his debt

I=pr
15 = p0.075
Solving for p = 200 pounds ---> answer

Please check my work.
 
  • #7
I think your answer is right – but you don't need to be very uncertain. It is easier to start with the answer and see if it works than to go through the calculation again!

If he has £300 as you calculate, the third he spends is £100, the quarter he saves is £75, the 5% interest is £15. Adds up to £190, out of £300 so that leaves him £110 as you said, okay so far.
Now the interest @ 7.5%.? Well interest on £100 @ 7.5% is £7.5 , so £15 is the interest on £200. All OK.

Remember the practice of working backwards as well as forwards will help you your confidence in right answers, and catch out and help prevent you definitely committing wrong ones.
 
Last edited:

1. How do I set up an algebraic equation for paying down a debt?

To set up an algebraic equation for paying down a debt, you need to first identify the variables involved. Usually, this includes the initial amount owed, the interest rate, the monthly payment, and the number of months. Then, you can use the formula A = P(1 + r/n)^(nt) to calculate the total amount owed after t months, where A is the final amount, P is the initial amount, r is the interest rate, n is the number of times the interest is compounded per year, and t is the number of years.

2. How do I solve for the unknown variable in an algebraic word problem?

To solve for the unknown variable in an algebraic word problem, you can use algebraic manipulation and basic arithmetic operations to isolate the variable on one side of the equation. Remember to follow the order of operations and to perform the same operation on both sides of the equation to maintain balance.

3. Can I use substitution to solve an algebraic word problem?

Yes, substitution can be a useful method for solving an algebraic word problem. This involves replacing a variable with an equivalent expression and then solving the resulting equation. It is particularly helpful when there are multiple equations and variables involved.

4. How do I know if my solution to an algebraic word problem is correct?

To check if your solution to an algebraic word problem is correct, you can plug in the values of the variables into the original equation and see if it satisfies the equation. You can also use a graphing calculator to graph the equation and see if your solution lies on the graph.

5. Are there any common mistakes to avoid when solving algebraic word problems?

Yes, there are a few common mistakes to avoid when solving algebraic word problems. These include incorrectly setting up the equation, making errors in algebraic manipulation, and forgetting to consider all the given information in the problem. It is important to carefully read and understand the problem before attempting to solve it.

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