L.C. MATHS

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## Financial Maths

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Compound Interest

Has not appeared

Depreciation

Has not appeared

Instalment Savings & Pensions

Loans & Mortgages

## 2020 Paper 1 Question 5

(a) A couple agree to take out a €$$250{,}000$$ mortgage in order to purchase a new home.
The loan is to be paid back monthly over $$25$$ years with the repayments due at the end of each month. The bank charges an annual percentage rate (APR) which is equivalent to a monthly rate of $$0.287\%$$.
Using the amortisation formula, or otherwise, find the couples’ monthly repayment on the mortgage. Give your answer in euro correct to the nearest cent.

$$1{,}244.06\mbox{ euro}$$

Solution

\begin{align}A&=P\frac{i(1+i)^t}{(1+i)^t-1}\\&=(250{,}000)\frac{(0.00287)(1+0.00287)^{300}}{(1+0.00287)^{300}-1}\\&\approx1{,}244.06\mbox{ euro}\end{align}

Video Walkthrough
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(b) Another couple agree to take out a mortgage of €$$350{,}000$$, at a rate of $$0.3\%$$ per month, in order to purchase a new home. This loan is also to be paid back monthly over $$25$$ years with the repayments due at the end of each month. The amount of each repayment is €$$1771$$.

After exactly $$11$$ years of repayments, the couple receive a financial windfall.
They decide to repay the remaining balance on the mortgage.

Write down a series (including the first two and last two terms) which shows the total of the present values of all the remaining monthly repayments due over the remaining $$14$$ years of the mortgage (after the last monthly repayment at the end of year $$11$$).

Hence, find how much the couple will need to repay in order to clear their mortgage entirely.

Series: $$\dfrac{1771}{1.003}+\dfrac{1771}{1.003^2}+…+\dfrac{1771}{1.003^{167}}+\dfrac{1771}{1.003^{168}}$$

Repayment: $$233{,}438.25\mbox{ euro}$$

Solution

Series

\begin{align}\frac{1771}{1.003}+\frac{1771}{1.003^2}+…+\frac{1771}{1.003^{167}}+\frac{1771}{1.003^{168}}\end{align}

$\,$

Repayment

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{168}&=\frac{\frac{1771}{1.003}\left[1-\left(\frac{1}{1.003}\right)^{168}\right]}{1-\frac{1}{1.003}}\\&\approx233{,}438.25\mbox{ euro}\end{align}

Video Walkthrough

## 2018 Paper 2 Question 8(c)

(c) Acme Confectionery has an employee pension plan. For an employee who qualifies for the full pension, Acme Confectionery will pay a sum of €$$20{,}000$$ on the day of retirement. It will then pay a sum on the same date each subsequent year for the next $$25$$ years. Each year the employee is paid a sum that is $$1\%$$ more than the amount paid in the previous year.
What sum of money must the company have set aside on the day of retirement in order to fund this pension? Assume an annual interest rate (AER) of $$2.4\%$$.

$$440{,}132.40\mbox{ euro}$$

Solution

\begin{align}\mbox{Amount}=20{,}000S_{26}\end{align}

\begin{align}a=1&&r=\frac{1.01}{1.024}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Amount}&=20{,}000S_{26}\\&=20{,}000\left[\frac{a\left(1-\left(\frac{1.01}{1.024}\right)^{26}\right)}{1-\frac{1.01}{1.024}}\right]\\&\approx440{,}132.40\mbox{ euro}\end{align}

Video Walkthrough

## 2017 Paper 1 Question 8

(a) When a loan of €$$P$$ is repaid in equal repayments of amount €$$A$$ ,at the end of each of $$t$$ equal periods of time, where $$i$$ is the periodic compound interest rate (expressed as a decimal), the formula below can be used to find the amount of each repayment.

\begin{align}A=P\frac{i(1+i)^t}{((1+i)^t-1)}\end{align}

Show how this formula is derived. You may use the formula for the sum of a finite geometric series.

Solution

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P=\frac{a(1-r^t)}{1-r}\end{align}

\begin{align}a=\frac{A}{1+i}&&r=\frac{1}{1+i}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P&=\frac{\frac{A}{1+i}\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1-\frac{1}{1+i}}\\&=\frac{A\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1+i-1}\\&=\frac{A[(1+i)^t-1]}{i(1+i)^t}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=P\frac{i(1+i)^t}{(1+i)^t-1}\end{align}

as required.

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(b) Alex has a credit card debt of €$$5000$$. One method of clearing this debt is to make a fixed repayment at the end of each month. The amount of this repayment is $$2.5\%$$ of the original debt.

(i) What is the fixed monthly repayment, €$$A$$, required to pay the debt of €$$5000$$?

(ii) The annual percentage rate (APR) charged on debt by the credit card company is $$21.75\%$$, fixed for the term of the debt. Find as a percentage, correct to $$3$$ significant figures, the monthly interest rate that is equivalent to an APR of $$21.75\%$$.

(iii) Assume Alex pays the fixed monthly repayment, €$$A$$,each month and does not have
any further transactions on that card. Complete the table below to show how the balance of the debt of €$$5000$$ is reducing each month for the first three months, assuming an APR of $$21.75\%$$, charged and compounded monthly.

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$3$$

(iv) Using the formula you derived on the previous page, or otherwise, find how long it would take to pay off a credit card debt of €$$5000$$, using the repayment method outlined at the beginning of part (b) above.

(v) Alex decides to borrow €$$5000$$ from the local Credit Union to pay off this credit card debt of €$$5000$$. The APR charge for the Credit Union loan is $$8.5\%$$ fixed for the term of the loan. The loan is to be repaid in equal weekly repayments, at the end of each week, for $$156$$ weeks. Find the amount of each weekly repayment.

(vi) How much will Alex save by paying off the credit card debt using the loan from the Credit Union instead of paying the fixed repayment from part (b)(i) each month to the credit card company?

(i) $$125\mbox{ euro}$$

(ii) $$1.65\%$$.

(iii)

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$125$$

$$82.50$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$125$$

$$81.80$$

$$43.20$$

$$4{,}914.30$$

$$3$$

$$125$$

$$81.09$$

$$43.91$$

$$4{,}870.39$$

(iv) $$66\mbox{ months}$$

(v) $$36.16\mbox{ euro}$$

(vi) $$2{,}609.04\mbox{ euro}$$

Solution

(i)

\begin{align}A&=5{,}000\times\left(\frac{2.5}{100}\right)\\&=125\mbox{ euro}\end{align}

(ii)

\begin{align}(1+i)^{1/12}&=(1.2175)^{1/12}\\&=1.01653…\end{align}

Therefore, the monthly interest rate is $$1.65\%$$.

(iii)

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$125$$

$$82.50$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$125$$

$$81.80$$

$$43.20$$

$$4{,}914.30$$

$$3$$

$$125$$

$$81.09$$

$$43.91$$

$$4{,}870.39$$

(iv)

\begin{align}A=P\frac{i(1+i)^t}{(1+i)^t-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A[(1+i)^t-1]=Pi(1+i)^t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A(1+i)^t-A=Pi(1+i)^t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(1+i)^t(A-Pi)=A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(1+i)^t=\frac{A}{A-Pi}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln(1+i)^t=\ln\left(\frac{A}{A-Pi}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t\ln(1+i)=\ln\left(\frac{A}{A-Pi}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{1}{\ln(1+i)}\times\ln\left(\frac{A}{A-Pi}\right)\\&=\frac{1}{\ln\left(1+\frac{1.65}{100}\right)}\times\ln\left(\frac{125}{125-5000\left(\frac{1.65}{100}\right)}\right)\\&\approx66\mbox{ months}\end{align}

(v)

\begin{align}A&=P\frac{i(1+i)^t}{(1+i)^t-1}\\&=(5000)\frac{(1.085^{1/52}-1)(1.085)^3}{(1.085)^3-1}\\&\approx36.16\mbox{ euro}\end{align}

(vi)

\begin{align}125\times66-(36.16)(156)=2{,}609.04\mbox{ euro}\end{align}

Video Walkthrough

## 2015 Paper 1 Question 6

(a) Donagh is arranging a loan and is examining two different repayment options.

(i) Bank A will charge him a monthly interest rate of $$0.35\%$$. Find, correct to three
significant figures, the annual percentage rate (APR) that is equivalent to a monthly
interest rate of $$0.35\%$$.

(ii) Bank B will charge him a rate that is equivalent to an APR of $$4.5\%$$. Find, correct to three significant figures, the monthly interest rate that is equivalent to an APR of $$4.5\%$$.

(i) $$4.28\%$$

(ii) $$0.367\%$$

Solution

(i)

\begin{align}(1+i)^t&=(1+0.0035)^{12}\\&=1.042818…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{APR}\approx4.28\%\end{align}

(ii)

\begin{align}(1+i)^{12}=1.045\end{align}

\begin{align}\downarrow\end{align}

\begin{align}i&=\sqrt[12]{1.045}-1\\&=0.0036748…\\&\approx0.367\%\end{align}

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(b) Donagh borrowed €$$80{,}000$$ at a monthly interest rate of $$0.35\%$$, fixed for the term of the loan, from Bank $$A$$. The loan is to be repaid in equal monthly repayments over ten years. The first repayment is due one month after the loan is issued. Calculate, correct to the nearest euro, the amount of each monthly repayment.

$$818\mbox{ euro}$$