Question

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Answer 1

$\large\underline{\sf{Solution-}}$

Case :-1 of simple interest.

We know,

Simple Interest on a certain sum of money Rs p at the rate of r % per annum for t years is given by

$\boxed{ \bf{SI \: = \dfrac{p \times r \times t}{100} }}$

where,

SI is simple interest

p is Principal

r is rate of interest per annum

t is time period.

Here,

$\rm :\longmapsto\:p = Rs \: 20000$

$\rm :\longmapsto\:r = 10 \: \% \: per \: annum$

$\rm :\longmapsto\:t = 2 \: \dfrac{1}{2} = \dfrac{5}{2} \: years$

So,

On substituting the values in above formula, we get

$\rm :\longmapsto\:SI = \dfrac{20000 \times 10 \times 5}{200}$

$\bf\implies \:SI \: = \: Rs \: 5000$

So,

Amount to be paid = Rs 20000 + Rs 5000 = Rs 25000

Case :- 2 of Compound interest

We know,

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded half yearly for n years, is

$\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}$

where,

• P is Principal

• r is rate of interest per annum

• n is time period.

So,

Here,

• P = Rs 25000

• R = 10 % per annum

• Time = 1.5 years

So, on substituting these values in above formula, we get

$\rm :\longmapsto\:\tt{ Amount=25000\bigg(1+\dfrac{10}{200}\bigg)^{2 \times 1.5}}$

$\rm :\longmapsto\:\tt{ Amount=25000\bigg(1+\dfrac{1}{20}\bigg)^{3}}$

$\rm :\longmapsto\:\tt{ Amount=25000\bigg(\dfrac{20 + 1}{20}\bigg)^{3}}$

$\rm :\longmapsto\:\tt{ Amount=25000\bigg(\dfrac{21}{20}\bigg)^{3}}$

$\bf\implies \:Amount = Rs \: 28940.625$

Additional Information :-

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded yearly for n years, is

$\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}$

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded quarterly for n years, is

$\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}$

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded monthly for n years, is

$\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{1200}\bigg)^{12n}}$

Answer 2

Answer:

Solution−

Case :-1 of simple interest.

We know,

Simple Interest on a certain sum of money Rs p at the rate of r % per annum for t years is given by

\boxed{ \bf{SI \: = \dfrac{p \times r \times t}{100} }}

SI=

100

p×r×t

where,

SI is simple interest

p is Principal

r is rate of interest per annum

t is time period.

Here,

\rm :\longmapsto\:p = Rs \: 20000:⟼p=Rs20000

\rm :\longmapsto\:r = 10 \: \% \: per \: annum:⟼r=10%perannum

\rm :\longmapsto\:t = 2 \: \dfrac{1}{2} = \dfrac{5}{2} \: years:⟼t=2

2

1

=

2

5

years

So,

On substituting the values in above formula, we get

\rm :\longmapsto\:SI = \dfrac{20000 \times 10 \times 5}{200}:⟼SI=

200

20000×10×5

\bf\implies \:SI \: = \: Rs \: 5000⟹SI=Rs5000

So,

Amount to be paid = Rs 20000 + Rs 5000 = Rs 25000

Case :- 2 of Compound interest

We know,

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded half yearly for n years, is

\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}:⟼Amount=P(1+

200

r

)

2n

where,

P is Principal

r is rate of interest per annum

n is time period.

So,

Here,

P = Rs 25000

R = 10 % per annum

Time = 1.5 years

So, on substituting these values in above formula, we get

\rm :\longmapsto\:\tt{ Amount=25000\bigg(1+\dfrac{10}{200}\bigg)^{2 \times 1.5}}:⟼Amount=25000(1+

200

10

)

2×1.5

\rm :\longmapsto\:\tt{ Amount=25000\bigg(1+\dfrac{1}{20}\bigg)^{3}}:⟼Amount=25000(1+

20

1

)

3

\rm :\longmapsto\:\tt{ Amount=25000\bigg(\dfrac{20 + 1}{20}\bigg)^{3}}:⟼Amount=25000(

20

20+1

)

3

\rm :\longmapsto\:\tt{ Amount=25000\bigg(\dfrac{21}{20}\bigg)^{3}}:⟼Amount=25000(

20

21

)

3

\bf\implies \:Amount = Rs \: 28940.625⟹Amount=Rs28940.625

Additional Information :-

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded yearly for n years, is

\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}:⟼Amount=P(1+

100

r

)

n

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded quarterly for n years, is

\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}:⟼Amount=P(1+

400

r

)

4n

Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded monthly for n years, is

\rm :\longmapsto\:\tt{ Amount=P\bigg(1+\dfrac{r}{1200}\bigg)^{12n}}:⟼Amount=P(1+

1200

r

)

12n