Question

10,000,000 of money is invested for 2 years at 10% interest. Determine the Final Money if the money is compounded each:
A. Every 3 Months
B. Every Month
C. Every day

± 50 points....please help, I'm very confused, Compound interest mathematics​

Answer 1

If the principal does not remain the same for the entire time period due to the addition of interest to the principal after certain time period to form a new principal, then the interest obtained in this way is called Compound interest.

Amount obtained in Compound interest is given by :

$\bigstar\;\;\boxed{\mathsf{Amount = Principal\bigg(1 + \dfrac{Rate\;of \;interest}{100}\bigg)^{Conversion\;periods}}}$

Note : Conversion period is the time from one interest period to the next interest period. If the interest is compounded annually then there is one conversion period in an year. If the interest is compounded semi-annually then there are two conversion periods in an year. if the interest is compounded quarterly then there are four conversion periods in an year.

Problem :

Given : 10000000 is invested for two years at 10% interest

$\implies\boxed{\begin{minipage}{4 cm}\bigstar\;\;\textsf{Principal = 10000000}\\\\\bigstar\;\;\textsf{Time period = 2 years}\\\\\bigstar\;\;\textsf{Rate of interest = 10\%}\end{minipage}}$

A. If the money is compounded each : Every 3 Months

If the money is compounded for every three months, then there are 4 conversion periods in an year. As the invested time period is of 2 years, there will be 8 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{4}}{100}\bigg)^8}$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{40}\bigg)^8}$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{41}{40}\bigg)^8}$

$\mathsf{\implies Amount = 10000000 \times \dfrac{41^8}{40^8}}$

$\implies \large{\boxed{\mathsf{Amount = \dfrac{41^8}{4^8 \times 10}}}}$

B. If the money is compounded each : Every Month

If the money is compounded for every month, then there are 12 conversion periods in an year. As the invested time period is of 2 years, there will be 24 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{12}}{100}\bigg)^{24}}$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{120}\bigg)^{24}}$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{121}{120}\bigg)^{24}}$

$\mathsf{\implies Amount = 10000000 \times \dfrac{121^{24}}{120^{24}}}$

$\implies \large{\boxed{\mathsf{Amount = \dfrac{121^{24}}{12^{24} \times 10^{17}}}}}$

C. If the money is compounded each : Every Day

If the money is compounded each and every day, As there are 365 days in an year, there will be 365 conversion periods in an year. As the invested time period is of 2 years, there will be 730 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{365}}{100}\bigg)^{730}}$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{3650}\bigg)^{730}}$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{3651}{3650}\bigg)^{730}}$

$\mathsf{\implies Amount = 10000000 \times \dfrac{3651^{730}}{3650^{730}}}$

$\implies \large\boxed{\mathsf{Amount = \dfrac{3651^{730}}{365^{730} \times 10^{723}}}}$

Answer 2

Step-by-step explanation:

If the principal does not remain the same for the entire time period due to the addition of interest to the principal after certain time period to form a new principal, then the interest obtained in this way is called Compound interest.

Amount obtained in Compound interest is given by :

$\bigstar\;\;\boxed{\mathsf{Amount = Principal\bigg(1 + \dfrac{Rate\;of \;interest}{100}\bigg)^{Conversion\;periods}}}★ </p><p>Amount=Principal(1+ </p><p>100</p><p>Rate \\ of \\ interest</p><p> </p><p> ) </p><p>Conversionperiods$

Note : Conversion period is the time from one interest period to the next interest period. If the interest is compounded annually then there is one conversion period in an year. If the interest is compounded semi-annually then there are two conversion periods in an year. if the interest is compounded quarterly then there are four conversion periods in an year.

Problem :

Given : 10000000 is invested for two years at 10% interest

$\begin{gathered}\implies\boxed{\begin{minipage}{4 cm}\bigstar\;\;\textsf{Principal = 10000000}\\\\\bigstar\;\;\textsf{Time period = 2 years}\\\\\bigstar\;\;\textsf{Rate of interest = 10\%}\end{minipage}}\end{gathered}$

A. If the money is compounded each : Every 3 Months

If the money is compounded for every three months, then there are 4 conversion periods in an year. As the invested time period is of 2 years, there will be 8 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{4}}{100}\bigg)^8}⟹Amount=10000000(1+ </p><p>100</p><p>4</p><p>10</p><p> </p><p> </p><p> </p><p> ) </p><p>8$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{40}\bigg)^8}⟹Amount=10000000(1+ </p><p>40</p><p>1</p><p> </p><p> ) </p><p>8$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{41}{40}\bigg)^8}⟹Amount=10000000( </p><p>40</p><p>41</p><p> </p><p> ) </p><p>8</p><p>$

$\mathsf{\implies Amount = 10000000 \times \dfrac{41^8}{40^8}}⟹Amount=10000000× </p><p>40 </p><p>8</p><p> </p><p>41 </p><p>8$

$\implies \large{\boxed{\mathsf{Amount = \dfrac{41^8}{4^8 \times 10}}}}⟹ </p><p>Amount= </p><p>4 </p><p>8</p><p> ×10</p><p>41 </p><p>8$

B. If the money is compounded each : Every Month

If the money is compounded for every month, then there are 12 conversion periods in an year. As the invested time period is of 2 years, there will be 24 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{12}}{100}\bigg)^{24}}⟹Amount=10000000(1+ </p><p>100</p><p>12</p><p>10</p><p> </p><p> </p><p> </p><p> ) </p><p>24$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{120}\bigg)^{24}}⟹Amount=10000000(1+ </p><p>120</p><p>1</p><p> </p><p> ) </p><p>24$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{121}{120}\bigg)^{24}}⟹Amount=10000000( </p><p>120</p><p>121</p><p> </p><p> ) </p><p>24</p><p>$

$\mathsf{\implies Amount = 10000000 \times \dfrac{121^{24}}{120^{24}}}⟹Amount=10000000× </p><p>120 </p><p>24</p><p> </p><p>121 </p><p>24</p><p>$

$\implies \large{\boxed{\mathsf{Amount = \dfrac{121^{24}}{12^{24} \times 10^{17}}}}}⟹ </p><p>Amount= </p><p>12 </p><p>24</p><p> ×10 </p><p>17</p><p> </p><p>121 </p><p>24</p><p>$

C. If the money is compounded each : Every Day

If the money is compounded each and every day, As there are 365 days in an year, there will be 365 conversion periods in an year. As the invested time period is of 2 years, there will be 730 conversion periods in 2 years

Substituting all the values in the Amount formula of C.I, We get :

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{\frac{10}{365}}{100}\bigg)^{730}}⟹Amount=10000000(1+ </p><p>100</p><p>365</p><p>10</p><p> </p><p> </p><p> </p><p> ) </p><p>730$

$\mathsf{\implies Amount = 10000000\bigg(1 + \dfrac{1}{3650}\bigg)^{730}}⟹Amount=10000000(1+ </p><p>3650</p><p>1</p><p> </p><p> ) </p><p>730$

$\mathsf{\implies Amount = 10000000\bigg(\dfrac{3651}{3650}\bigg)^{730}}⟹Amount=10000000( </p><p>3650</p><p>3651</p><p> </p><p> ) </p><p>730$

$\mathsf{\implies Amount = 10000000 \times \dfrac{3651^{730}}{3650^{730}}}⟹Amount=10000000× </p><p>3650 </p><p>730</p><p> </p><p>3651 </p><p>730</p><p>$

$\implies \large\boxed{\mathsf{Amount = \dfrac{3651^{730}}{365^{730} \times 10^{723}}}}⟹ </p><p>Amount= </p><p>365 </p><p>730</p><p> ×10 </p><p>723</p><p> </p><p>3651 </p><p>730</p><p> </p><p>$