In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?

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Interest Word Problems
Simple and Compound Interest

The following table gives the Formulas for Simple Interest, Compound Interest, and Continuously Compounded Interest. Scroll down the page for more examples and solutions.

In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?

The Simple Interest Formula is given by

Simple Interest = Principal × Interest Rate × Time

I = Prt   where
The Principal (P) is the amount of money deposited or borrowed.
The Interest Rate (r) is a percent of the principal earned or paid.
The Time (t) is the length of time the money is deposited or borrowed.

Example:
Sarah deposits $4,000 at a bank at an interest rate of 4.5% per year. How much interest will she earn at the end of 3 years?

Solution:
Simple Interest = 4,000 × 4.5% × 3 = 540

She earns $540 at the end of 3 years.

Example:
Wanda borrowed $3,000 from a bank at an interest rate of 12% per year for a 2-year period. How much interest does she have to pay the bank at the end of 2 years?

Solution:
Simple Interest = 3,000 × 12% × 2 = 720

She has to pay the bank $720 at the end of 2 years.

Example:
Raymond bought a car for $40, 000. He took a $20,000 loan from a bank at an interest rate of 15% per year for a 3-year period. What is the total amount (interest and loan) that he would have to pay the bank at the end of 3 years?

Solution:
Simple Interest = 20,000 × 13% × 3 = 7,800

At the end of 3 years, he would have to pay

$20,000 + $7,800 = $27,800

Tutorial On Simple Interest

Examples:

  1. Ian is investing $4,000 for 2 years. The interest rate is 5.5%. How much interest will Ian earn after 2 years?
  2. Doug made a 3 year investment. The interest rate was 4.5%. After 3 years, he earned $675 in interest. How much was his original investment?
  3. Kim got a loan of $4700 to buy a used car. The interest rate is 7.5%. She paid $1057.50 in interest. How many years did it take her to pay off her loan?
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How To Solve Interest Problems Using The Simple Interest Formula?

Interest represents a change in money.

If you have a savings account, the interest will increase your balance based upon the interest rate paid by the bank.

If you have a loan, the interest will increase the amount you owe based upon the interest rate charged by the bank.

Examples:

  1. If you invest $3,500 in savings account that pays 4% simple interest, how much interest will you earn after 3 years? What ill the new balance be?
  2. You borrow $6000 from a loan shark. If you will owe $7200 in 18 months, what would be the simple interest rate?
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How To Use The Formula For Simple Interest To Find The Principal, The Rate Or The Time?

Examples:

  1. An investment earned $11.25 interest after 9 months. The rate was 5%. What was the principal?
  2. $2000 was invested for 3 years. It earned $204 in interest. What was the rate?
  3. A loan of $1200 had $36 in interest. The rate was 6%. What was the length of the loan?
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How To Solve Simple Interest Problems, Compound Interest Problems, Continuously Compounded Interest Problems, And Determining The Effective Rate Of Return?

Examples Of Simple Interest Problems:

  1. Joseph buys a new home using an interest only loan where he pays only the interest on the value of the home each month. The home is valued at $200,000 and Joesph pays 5% interest per year on the home. How much is his monthly interest payment?
  2. Anthony puts $10000 dollars into a savings account that pays interest every month at a rate of 1.8% per year. How much money does Anthony have after one month? If he leaves his original investment and the first month of interest in the account, how much will he have after the second month?

Examples Of Compound Interest Problems:

  1. Matt is saving for a new car. He invests $5000 into an account that pays 3% interest a year and is compound monthly. How much will he have after 5 years?
  2. Matt is planning to buy a car in three years. He wants to invest $5000 now and hopes to have $6000 to spend on the car when he buys it. What kind of interest rate would he need if his investment is compounded monthly?

Examples Of Continuously Compound Interest Problems:

  1. Lindsey invests $1000 into an account with 4% per year continuously compounded interest. How much will she have after 10 years? How long will it take for her investment to double?
  2. Tony and Matt both incest $5000 in an account that receives 3% interest annually for 10 years. Tony invests in an account that is compounded monthly. Matt invests in an account that is compounded continuously. Who made the better investment?

Example Of Effective Rate Of Return:

  1. If $2500 is invested at 5% compounded monthly, what is the effective rate of return. What is the effective rate of return if this investment is compounded semiannually?
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Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?



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Joe’s brother has just had a new baby, Emily. Joe decides that he would like to set up a savings vehicle in Emily’s name, to provide a nest egg for her when she is older. He knows how hard it is to save up money for a deposit on a mortgage, and wants to make it easier for Emily when she gets to that time on her life.

Joe finds a long term savings account offering a rate of 4.2% effective annual interest rate (eAPR). He decides to put $1,000 into the account to open it, and to set up an automatic deposit of $50 per month from his regular bank account.

Wondering how much this will amount to when Emily is 30 he enters $1000 into the Principal field, $50 into the Monthly Deposit field, 4.2 into the % Rate field, and 30 into the Years field.

Results

Hitting the Calculate button brings up the results of the savings calculator. After 30 years the final balance will be $39,484. Of this, $18,000 is from the $50 monthly deposits he made. The remaining $20,484 is the interest which accrued in the account over the years.

Joe plays around with the principal and monthly deposits to get a feel for how different inputs will affect the outcome. He notices that over such a long period, it is the size of the regular monthly payments that has the biggest effect on the final balance.

For instance, increasing his monthly contribution from $50 to $60 results in a final balance $7,000 higher than the smaller $50 contribution. He decides he can afford the extra $10 a month and resolves to increase his monthly deposit.

He also notes that the interest portion outweighs the amount he contributes. Looking at the ratio between interest and deposits (the green part of the chart compared to the yellow part), he sees that initially interest plays very little part in the growth. After 6 years, his deposits total $4,320, and the interest paid only $869. But as time goes on, the interest mounts up. By the 30th year, the interest totals $24,000, moving ahead of the deposits of $21,600 for the first time.

In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?

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In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?

In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?
Shortcut Trick \(12.5\% = \frac{1}{8}\)

Let P be (8)3 = 512

In what time will the interest on 3600 at the rate of 10% be equal to the interest on 1800 at the rate of 12.5% in 4 years?

CI of 3rd year = 64 + 8 + 8 + 1 = 81

81 = 1944

\(⇒ 512 = \frac{{512 \times 1944}}{{81}}\)

⇒ P = 12288

\(SI = \frac{{PRT}}{{100}}\)

\(⇒ \frac{{12288 \times 15 \times 5}}{{100}}\)

⇒ 9216

∴ The required simple interest is Rs. 9216.

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