# Thread: A debt for a motorbike

1. ## A debt for a motorbike

Sorry - another question.

For the following, I got \$97,800 as an answer. Is this correct? It doesn't seem to make sense. I know I'm doing something wrong here, but what could it be?

Tula repays her father \$4890 to cover a debt for a motorbike which she incurred 15 months ago, agreeing to pay 4% p.a. How much did she borrow?

Many thanks.

Sorry - another question.

For the following, I got \$97,800 as an answer. Is this correct? It doesn't seem to make sense. I know I'm doing something wrong here, but what could it be?

Tula repays her father \$4890 to cover a debt for a motorbike which she incurred 15 months ago, agreeing to pay 4% p.a. How much did she borrow?

Many thanks.
Let the amount borrowed be $A_0$, for continuously compounded interest the amount owed is:

$A(t)=A_0 e^{rt}$

where $r$ is the interest rate.

Here we are told that at $t=1.25$ years, $A(t)=4890$ for $r=0.04$, now you need to find $A(1.25)$

CB

3. $A(t)=A_0 e^{rt}$
$4890 = A_0 e^{0.05}$ What significance does e have? I have not yet learned about it.

Is there an easier way?

Is there an easier way?
Depends on definition of easier?
Tula repays ... \$4890 to cover a debt ... incurred 15 months ago, agreeing to pay 4% p.a.

How much did she borrow?
At the end of the first year the interest (at 4%) amounted to:
InterestFor1Year = BORROWED x 0.04

Since this is "per annum" it is assumed that the compounding is on a yearly basis, that is the interest in added only once per year.

The amount owed at the end of the first year is
The original amount BORROWED plus the interest on that amount for 1 year:
Debt1Year = BORROWED + ( 0.04 x BORROWED )

The borrower needs to pay interest on the interest.
3 months is 1/4 year or 0.25
InterestFor3Months = Debt1Year x 0.04 x 0.25

DEBT or Total Amout repaid is \$4890.
That consists of:
The original amount BORROWED
The interest on that amount for 1 year
plus the interest on that interest for 1/4 year
plus the interest on the original amount BORROWED for an additional 1/4 year.

4890 = BORROWED + (0.04 x BORROWED) + ( (1.04 x BORROWED) x 0.04 x 0.25 )
1 equation 1 unknown

4890 = BORROWED + 0.04 x BORROWED + 0.0104 x BORROWED

4890 = BORROWED ( 1 + 0.04 + 0.0104 )
4890 = BORROWED x 1.0504

BORROWED = $\dfrac{4890}{1.0504}$

You should ALWAYS check your work.
From that number BORROWED
multiply it by 1.04 (& round to the nearest penny)
then
multiply that amount by (1 + 0.04/4)

The final amount should be \$4890.

There is really nothing simplier than using the exponentials for such problems.
.

$A(t)=A_0 e^{rt}$
$4890 = A_0 e^{0.05}$ What significance does e have? I have not yet learned about it.

Is there an easier way?
Or is this supposed to be a simple interest problem - like the other?

CB