Question Number 203222 by Tawa11 last updated on 12/Jan/24
XYZ Ltd wants to borrow money to finance a project.
They can pay N2500 monthly in repayments. If the
interest rate is 5% and the load will be paid back
fully in 30 years, how much can the company afford
to borrow?
They can pay N2500 monthly in repayments. If the
interest rate is 5% and the load will be paid back
fully in 30 years, how much can the company afford
to borrow?
Commented by mr W last updated on 13/Jan/24
$${P}=\frac{\mathrm{2500}}{\frac{\mathrm{0}.\mathrm{05}}{\mathrm{12}}}\left(\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{0}.\mathrm{05}}{\mathrm{12}}\right)^{\mathrm{30}×\mathrm{12}} }\right)\approx\mathrm{465700} \\ $$$${the}\:{derivation}\:{of}\:{formula}\:{see}\:{below}. \\ $$
Answered by mr W last updated on 13/Jan/24
![say the loan you take from the bank is P. the (yearly) interest rate is r. the mouthly payment is M. the mouthly interest rate is i=(r/(12)). at the very beginning: the amount of money you own the bank is P_0 =P. after one month: the amount of money you own the bank is P_1 =P(1+i)−M. after 2 months: the amount of money you own the bank is P_2 =P_1 (1+i)−M =P(1+i)^2 −M(1+i)−M after 3 months: the amount of money you own the bank is P_3 =P_2 (1+i)−M =P(1+i)^3 −M(1+i)^2 −M(1+i)−M ....... after n months: the amount of money you own the bank is P_n =P_(n−1) (1+i)−M =P(1+i)^n −M(1+i)^(n−1) −M(1+i)^(n−2) −...−M(1+i)−M =P(1+i)^n −M[(1+i)^(n−1) +(1+i)^(n−2) +...+(1+i)+1] =P(1+i)^n −M×(((1+i)^n −1)/((1+i)−1)) =(P−(M/i))(1+i)^n +(M/i) if you want to have paid back fully after n months, then P_n =0, i.e. (P−(M/i))(1+i)^n +(M/i)=0 ⇒P=(M/i)[1−(1/((1+i)^n ))] in current example: r=5% ⇒i=(r/(12))=((0.05)/(12)) M=2500 $ n=30×12=360 months P=((2500)/((0.05)/(12)))[1−(1/((1+((0.05)/(12)))^(30×12) ))]=465704 $](https://www.tinkutara.com/question/Q203227.png)
$${say}\:{the}\:{loan}\:{you}\:{take}\:{from}\:{the}\:{bank} \\ $$$${is}\:{P}.\:{the}\:\left({yearly}\right)\:{interest}\:{rate}\:{is}\:{r}. \\ $$$${the}\:{mouthly}\:{payment}\:{is}\:{M}. \\ $$$${the}\:{mouthly}\:{interest}\:{rate}\:{is}\:{i}=\frac{{r}}{\mathrm{12}}. \\ $$$${at}\:{the}\:{very}\:{beginning}: \\ $$$${the}\:{amount}\:{of}\:{money}\:{you}\:{own} \\ $$$${the}\:{bank}\:{is}\:{P}_{\mathrm{0}} ={P}. \\ $$$$ \\ $$$${after}\:{one}\:{month}: \\ $$$${the}\:{amount}\:{of}\:{money}\:{you}\:{own} \\ $$$${the}\:{bank}\:{is}\: \\ $$$${P}_{\mathrm{1}} ={P}\left(\mathrm{1}+{i}\right)−{M}. \\ $$$$ \\ $$$${after}\:\mathrm{2}\:{months}: \\ $$$${the}\:{amount}\:{of}\:{money}\:{you}\:{own} \\ $$$${the}\:{bank}\:{is}\: \\ $$$${P}_{\mathrm{2}} ={P}_{\mathrm{1}} \left(\mathrm{1}+{i}\right)−{M} \\ $$$$\:\:\:\:\:={P}\left(\mathrm{1}+{i}\right)^{\mathrm{2}} −{M}\left(\mathrm{1}+{i}\right)−{M} \\ $$$$ \\ $$$${after}\:\mathrm{3}\:{months}: \\ $$$${the}\:{amount}\:{of}\:{money}\:{you}\:{own} \\ $$$${the}\:{bank}\:{is}\: \\ $$$${P}_{\mathrm{3}} ={P}_{\mathrm{2}} \left(\mathrm{1}+{i}\right)−{M} \\ $$$$\:\:\:\:\:={P}\left(\mathrm{1}+{i}\right)^{\mathrm{3}} −{M}\left(\mathrm{1}+{i}\right)^{\mathrm{2}} −{M}\left(\mathrm{1}+{i}\right)−{M} \\ $$$$……. \\ $$$${after}\:{n}\:{months}: \\ $$$${the}\:{amount}\:{of}\:{money}\:{you}\:{own} \\ $$$${the}\:{bank}\:{is}\: \\ $$$${P}_{{n}} ={P}_{{n}−\mathrm{1}} \left(\mathrm{1}+{i}\right)−{M} \\ $$$$\:\:\:\:\:={P}\left(\mathrm{1}+{i}\right)^{{n}} −{M}\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} −{M}\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{2}} −…−{M}\left(\mathrm{1}+{i}\right)−{M} \\ $$$$\:\:\:\:\:={P}\left(\mathrm{1}+{i}\right)^{{n}} −{M}\left[\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} +\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{2}} +…+\left(\mathrm{1}+{i}\right)+\mathrm{1}\right] \\ $$$$\:\:\:\:\:={P}\left(\mathrm{1}+{i}\right)^{{n}} −{M}×\frac{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}{\left(\mathrm{1}+{i}\right)−\mathrm{1}} \\ $$$$\:\:\:\:\:=\left({P}−\frac{{M}}{{i}}\right)\left(\mathrm{1}+{i}\right)^{{n}} +\frac{{M}}{{i}} \\ $$$$ \\ $$$${if}\:{you}\:{want}\:{to}\:{have}\:{paid}\:{back}\:{fully} \\ $$$${after}\:{n}\:{months},\:{then}\:{P}_{{n}} =\mathrm{0},\:{i}.{e}. \\ $$$$\left({P}−\frac{{M}}{{i}}\right)\left(\mathrm{1}+{i}\right)^{{n}} +\frac{{M}}{{i}}=\mathrm{0} \\ $$$$\Rightarrow{P}=\frac{{M}}{{i}}\left[\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{{n}} }\right] \\ $$$$ \\ $$$${in}\:{current}\:{example}: \\ $$$${r}=\mathrm{5\%}\:\Rightarrow{i}=\frac{{r}}{\mathrm{12}}=\frac{\mathrm{0}.\mathrm{05}}{\mathrm{12}} \\ $$$${M}=\mathrm{2500}\:\$ \\ $$$${n}=\mathrm{30}×\mathrm{12}=\mathrm{360}\:{months} \\ $$$${P}=\frac{\mathrm{2500}}{\frac{\mathrm{0}.\mathrm{05}}{\mathrm{12}}}\left[\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{0}.\mathrm{05}}{\mathrm{12}}\right)^{\mathrm{30}×\mathrm{12}} }\right]=\mathrm{465704}\:\$ \\ $$
Commented by Tawa11 last updated on 13/Jan/24
$$\mathrm{Wow}\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 13/Jan/24
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$