Simple and Compound Interest

Given:

Principal (P) = ₹ 4,000

Rate of interest (R) = 7.5% p.a.

Time (T) = 3 years 3 months

= $3\frac{3}{12} \text{ years }= 3\frac{1}{4} \text{ years }= \frac{13}{4} \text{ years }$

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{4000 \times 7.5 \times \frac{13}{4}}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{4000 \times 7.5 \times 13}{100 \times 4} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{390000}{400} \\[1em] \Rightarrow \text{S.I.} = ₹ 975\end{aligned}$$

Amount = Principal + Interest

= ₹ 4,000 + ₹ 975

= ₹ 4,975

Hence, simple interest = ₹ 975 and amount = ₹ 4,975.

Given:

Simple interest (I) = ₹ 170.10

Rate of interest (R) = 6% p.a.

Time (T) = 2 years 3 months

= $2\frac{3}{12} \text{ years }= 2\frac{1}{4} \text{ years }= \frac{9}{4} \text{ years }$

As we know:

$\text{P} = \frac{\text{I} \times 100}{\text{R} \times \text{T}}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{P} = ₹ \frac{170.10 \times 100}{6 \times \frac{9}{4}} \\[1em] \Rightarrow \text{P} = ₹ \frac{170.10 \times 100 \times 4}{6 \times 9} \\[1em] \Rightarrow \text{P} = ₹ \frac{68040}{54} \\[1em] \Rightarrow \text{P} = ₹ 1{,}260\end{aligned}$$

Hence, the required sum of money = ₹ 1,260.

Given:

Principal (P) = ₹ 800

Simple interest (I) = ₹ 130

Time (T) = 2 years 6 months

= $2\frac{6}{12} \text{ years }= 2\frac{1}{2} \text{ years }= \frac{5}{2} \text{ years }$

As we know:

$\text{R} = \frac{\text{I} \times 100}{\text{P} \times \text{T}}$%

Substituting the values, we get

$\Rightarrow \text{R} = \frac{130 \times 100}{800 \times \frac{5}{2}}$%

$\Rightarrow \text{R} = \frac{130 \times 100 \times 2}{800 \times 5}$%

$\Rightarrow \text{R} = \frac{26000}{4000}$%

⇒ R = 6.5%

Hence, the rate of interest = 6.5% p.a.

Given:

Principal (P) = ₹ 3.3 lakhs = ₹ 3,30,000

Rate of interest (R) = 6.5% p.a.

Simple interest (I) = ₹ 75,075

As we know:

$\text{T} = \frac{\text{I} \times 100}{\text{P} \times \text{R}}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{T} = \frac{75075 \times 100}{330000 \times 6.5} \\[1em] \Rightarrow \text{T} = \frac{7507500}{2145000} \\[1em] \Rightarrow \text{T} = \frac{7}{2} \\[1em] \Rightarrow \text{T} = 3\frac{1}{2}\end{aligned}$$

Hence, the required time = $\mathbf{3\frac{1}{2}}$ years i.e. 3 years 6 months.

Given:

Simple interest (I) = ₹ 2,356.25

Rate of interest (R) = $7\frac{1}{4}$% p.a. = $\frac{29}{4}$% p.a.

Time (T) = $2\frac{1}{2}$ years

= $\frac{5}{2} \text{ years }$

As we know:

$\text{P} = \frac{\text{I} \times 100}{\text{R} \times \text{T}}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{P} = ₹ \frac{2356.25 \times 100}{\frac{29}{4} \times \frac{5}{2}} \\[1em] \Rightarrow \text{P} = ₹ \frac{2356.25 \times 100 \times 4 \times 2}{29 \times 5} \\[1em] \Rightarrow \text{P} = ₹ \frac{1885000}{145} \\[1em] \Rightarrow \text{P} = ₹ 13{,}000\end{aligned}$$

Hence, the required sum of money = ₹ 13,000.

Given:

Amount (A) = ₹ 11,300

Rate of interest (R) = 4% p.a.

Time (T) = 3 years 3 months

= $3\frac{3}{12} \text{ years }= 3\frac{1}{4} \text{ years }= \frac{13}{4} \text{ years }$

As we know:

$\text{A} = \Big(1 + \frac{\text{R} \times \text{T}}{100}\Big)\text{P}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 11300 = \Big(1 + \frac{4 \times \frac{13}{4}}{100}\Big)\text{P} \\[1em] \Rightarrow 11300 = \Big(1 + \frac{4 \times 13}{100 \times 4}\Big)\text{P} \\[1em] \Rightarrow 11300 = \Big(1 + \frac{13}{100}\Big)\text{P} \\[1em] \Rightarrow 11300 = \frac{113}{100}\text{P} \\[1em] \Rightarrow \text{P} = \frac{11300 \times 100}{113} \\[1em] \Rightarrow \text{P} = ₹ 10{,}000\end{aligned}$$

Hence, the required sum of money = ₹ 10,000.

Given:

Rate of interest (R) = $13\frac{1}{3}$% p.a. = $\frac{40}{3}$% p.a.

Let the sum of money (principal) be ₹ P.

As the sum of money triples itself,

Amount (A) = ₹ 3P

Interest = Amount - Principal

= ₹ 3P - ₹ P

= ₹ 2P

As we know:

$\text{T} = \frac{\text{I} \times 100}{\text{P} \times \text{R}}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{T} = \frac{2\text{P} \times 100}{\text{P} \times \frac{40}{3}} \\[1em] \Rightarrow \text{T} = \frac{2\text{P} \times 100 \times 3}{\text{P} \times 40} \\[1em] \Rightarrow \text{T} = \frac{600}{40} \\[1em] \Rightarrow \text{T} = 15\end{aligned}$$

Hence, the required time = 15 years.

Given:

Principal (P) = ₹ 4,050

Amount (A) = ₹ 4,576.50

Time (T) = 2 years

Interest = Amount - Principal

= ₹ 4,576.50 - ₹ 4,050

= ₹ 526.50

As we know:

$\text{R} = \frac{\text{I} \times 100}{\text{P} \times \text{T}}$%

Substituting the values, we get

$\Rightarrow \text{R} = \frac{526.50 \times 100}{4050 \times 2}$%

$\Rightarrow \text{R} = \frac{52650}{8100}$%

⇒ R = 6.5%

So, the rate of interest = 6.5% p.a.

Now, for the second case :

Principal (P) = ₹ 1 lakh = ₹ 1,00,000

Rate of interest (R) = 6.5% p.a.

Time (T) = 3 years

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{100000 \times 6.5 \times 3}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{1950000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 19{,}500\end{aligned}$$

Amount = Principal + Interest

= ₹ 1,00,000 + ₹ 19,500

= ₹ 1,19,500

Hence, ₹ 1 lakh will amount to ₹ 1,19,500 in 3 years.

Given:

For the second sum :

Principal (P) = ₹ 9,600

Rate of interest (R) = 10% p.a.

Time (T) = 3 years 6 months

= $3\frac{6}{12} \text{ years }= 3\frac{1}{2} \text{ years }= \frac{7}{2} \text{ years }$

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{9600 \times 10 \times \frac{7}{2}}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{9600 \times 10 \times 7}{100 \times 2} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{672000}{200} \\[1em] \Rightarrow \text{S.I.} = ₹ 3{,}360\end{aligned}$$

Twice of this interest = 2 × ₹ 3,360 = ₹ 6,720

Let the money invested at 7.5% p.a. for 2 years be ₹ $x$.

According to given information,

$$\begin{aligned}\Rightarrow \frac{x \times 7.5 \times 2}{100} = 6720 \\[1em] \Rightarrow \frac{15x}{100} = 6720 \\[1em] \Rightarrow x = \frac{6720 \times 100}{15} \\[1em] \Rightarrow x = \frac{672000}{15} \\[1em] \Rightarrow x = 44{,}800\end{aligned}$$

Hence, the required sum of money = ₹ 44,800.

Given:

Principal (P) = ₹ 6,000

Rate of interest (R) = 10% p.a.

Time (T) = 2 years

Principal for the first year = ₹ 6,000

$\Rightarrow \text{Interest for the first year} = ₹ \frac{6000 \times 10 \times 1}{100} = ₹ 600$

Amount at the end of first year = ₹ 6,000 + ₹ 600 = ₹ 6,600

Principal for the second year = ₹ 6,600

$\Rightarrow \text{Interest for the second year} = ₹ \frac{6600 \times 10 \times 1}{100} = ₹ 660$

Amount at the end of second year = ₹ 6,600 + ₹ 660 = ₹ 7,260

Compound interest = Final amount - Original principal

= ₹ 7,260 - ₹ 6,000

= ₹ 1,260

Hence, the compound interest on ₹ 6,000 for two years = ₹ 1,260.

Given:

Principal (P) = ₹ 9,375

Rate of interest (R) = 4% p.a.

Time (T) = 2 years

Principal for the first year = ₹ 9,375

$\Rightarrow \text{Interest for the first year} = ₹ \frac{9375 \times 4 \times 1}{100} = ₹ 375$

Amount at the end of first year = ₹ 9,375 + ₹ 375 = ₹ 9,750

Principal for the second year = ₹ 9,750

$\Rightarrow \text{Interest for the second year} = ₹ \frac{9750 \times 4 \times 1}{100} = ₹ 390$

Amount at the end of second year = ₹ 9,750 + ₹ 390 = ₹ 10,140

Compound interest = Final amount - Original principal

= ₹ 10,140 - ₹ 9,375

= ₹ 765

Hence, the compound interest that Salma has to pay after 2 years = ₹ 765.

Given:

Principal (P) = ₹ 12,000

Rate of interest (R) = 10% p.a.

Time (T) = 3 years

Principal for the first year = ₹ 12,000

$\Rightarrow \text{Interest for the first year} = ₹ \frac{12000 \times 10 \times 1}{100} = ₹ 1{,}200$

Amount at the end of first year = ₹ 12,000 + ₹ 1,200 = ₹ 13,200

Principal for the second year = ₹ 13,200

$\Rightarrow \text{Interest for the second year} = ₹ \frac{13200 \times 10 \times 1}{100} = ₹ 1{,}320$

Amount at the end of second year = ₹ 13,200 + ₹ 1,320 = ₹ 14,520

Principal for the third year = ₹ 14,520

$\Rightarrow \text{Interest for the third year} = ₹ \frac{14520 \times 10 \times 1}{100} = ₹ 1{,}452$

Amount at the end of third year = ₹ 14,520 + ₹ 1,452 = ₹ 15,972

Compound interest = Final amount - Original principal

= ₹ 15,972 - ₹ 12,000

= ₹ 3,972

Hence, the amount = ₹ 15,972 and the compound interest = ₹ 3,972.

Given:

Principal (P) = ₹ 46,875

Rate of interest (R) = 4% p.a.

Time (T) = 3 years

(i) Principal for the first year = ₹ 46,875

$\Rightarrow \text{Interest for the first year} = ₹ \frac{46875 \times 4 \times 1}{100} = ₹ 1{,}875$

Hence, the interest for the first year = ₹ 1,875.

(ii) Amount at the end of first year = ₹ 46,875 + ₹ 1,875 = ₹ 48,750

Principal for the second year = ₹ 48,750

$\Rightarrow \text{Interest for the second year} = ₹ \frac{48750 \times 4 \times 1}{100} = ₹ 1{,}950$

Amount at the end of second year = ₹ 48,750 + ₹ 1,950 = ₹ 50,700

Hence, the amount standing to his credit at the end of second year = ₹ 50,700.

(iii) Principal for the third year = ₹ 50,700

$\Rightarrow \text{Interest for the third year} = ₹ \frac{50700 \times 4 \times 1}{100} = ₹ 2{,}028$

Hence, the interest for the third year = ₹ 2,028.

Given:

Principal (P) = ₹ 6,000

Rate of interest (R) = 10% p.a.

Time (T) = 3 years

Principal for the first year = ₹ 6,000

$\Rightarrow \text{Interest for the first year} = ₹ \frac{6000 \times 10 \times 1}{100} = ₹ 600$

Amount at the end of first year = ₹ 6,000 + ₹ 600 = ₹ 6,600

Principal for the second year = ₹ 6,600

$\Rightarrow \text{Interest for the second year} = ₹ \frac{6600 \times 10 \times 1}{100} = ₹ 660$

Amount at the end of second year = ₹ 6,600 + ₹ 660 = ₹ 7,260

Principal for the third year = ₹ 7,260

$\Rightarrow \text{Interest for the third year} = ₹ \frac{7260 \times 10 \times 1}{100} = ₹ 726$

Amount at the end of third year = ₹ 7,260 + ₹ 726 = ₹ 7,986

Hence, the compound interest for the second year = ₹ 660 and the sum due at the end of third year = ₹ 7,986.

Given:

Principal (P) = ₹ 5,000

Rate of interest for the first year = 6% p.a.

Rate of interest for the second year = 8% p.a.

Principal for the first year = ₹ 5,000

$\Rightarrow \text{Interest for the first year} = ₹ \frac{5000 \times 6 \times 1}{100} = ₹ 300$

Amount at the end of first year = ₹ 5,000 + ₹ 300 = ₹ 5,300

Principal for the second year = ₹ 5,300

$\Rightarrow \text{Interest for the second year} = ₹ \frac{5300 \times 8 \times 1}{100} = ₹ 424$

Amount at the end of second year = ₹ 5,300 + ₹ 424 = ₹ 5,724

Compound interest = Final amount - Original principal

= ₹ 5,724 - ₹ 5,000

= ₹ 724

Hence, the amount = ₹ 5,724 and the compound interest = ₹ 724.

Given:

Principal (P) = ₹ 20,000

Rate of interest (R) = 8% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{20000 \times 8 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{320000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 3{,}200\end{aligned}$$

Compound interest :

Principal for the first year = ₹ 20,000

$\Rightarrow \text{Interest for the first year} = ₹ \frac{20000 \times 8 \times 1}{100} = ₹ 1{,}600$

Amount at the end of first year = ₹ 20,000 + ₹ 1,600 = ₹ 21,600

Principal for the second year = ₹ 21,600

$\Rightarrow \text{Interest for the second year} = ₹ \frac{21600 \times 8 \times 1}{100} = ₹ 1{,}728$

Compound interest for 2 years = ₹ 1,600 + ₹ 1,728 = ₹ 3,328

Difference between compound interest and simple interest = ₹ 3,328 - ₹ 3,200 = ₹ 128

Hence, the required difference = ₹ 128.

Given:

Principal (P) = ₹ 15,000

Rate of interest (R) = 10% p.a. compounded annually

Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 15{,}000\Big(1 + \frac{10}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 15{,}000\Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 15{,}000 \times \frac{11}{10} \times \frac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 18{,}150\end{aligned}$$

C.I. = A - P

= ₹ 18,150 - ₹ 15,000

= ₹ 3,150

Hence, the amount = ₹ 18,150 and the compound interest = ₹ 3,150.

Given:

Principal (P) = ₹ 1,56,250

Rate of interest = 8% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\frac{8}{2}$% = 4%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 1{,}56{,}250\Big(1 + \frac{4}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}56{,}250\Big(\frac{26}{25}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}56{,}250 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} \\[1em] \Rightarrow \text{A} = ₹ 1{,}75{,}760\end{aligned}$$

C.I. = A - P

= ₹ 1,75,760 - ₹ 1,56,250

= ₹ 19,510

Hence, the amount = ₹ 1,75,760 and the compound interest = ₹ 19,510.

Given:

Principal (P) = ₹ 1,00,000

Rate of interest = 4% p.a. compounded quarterly

As the interest is compounded quarterly,

Rate per conversion period (R) = $\frac{4}{4}$% = 1%

Number of conversion periods (n) = $\frac{9}{3}$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 1{,}00{,}000\Big(1 + \frac{1}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}00{,}000\Big(\frac{101}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}00{,}000 \times \frac{101}{100} \times \frac{101}{100} \times \frac{101}{100} \\[1em] \Rightarrow \text{A} = ₹ 1{,}03{,}030.10\end{aligned}$$

C.I. = A - P

= ₹ 1,03,030.10 - ₹ 1,00,000

= ₹ 3,030.10

Hence, the amount = ₹ 1,03,030.10 and the compound interest = ₹ 3,030.10.

Given:

Principal (P) = ₹ 4,800

Rate of interest (R) = 5% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{4800 \times 5 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{48000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 480\end{aligned}$$

Compound interest :

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 4{,}800\Big(1 + \frac{5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 4{,}800\Big(\frac{21}{20}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 4{,}800 \times \frac{21}{20} \times \frac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 5{,}292\end{aligned}$$

C.I. = A - P = ₹ 5,292 - ₹ 4,800 = ₹ 492

Difference between compound interest and simple interest = ₹ 492 - ₹ 480 = ₹ 12

Hence, the required difference = ₹ 12.

Given:

Principal (P) = ₹ 3,125

Rate of interest for the first year (R₁) = 4% p.a.

Rate of interest for the second year (R₂) = 5% p.a.

Rate of interest for the third year (R₃) = 6% p.a.

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}_1}{100}\Big)\Big(1 + \frac{\text{R}_2}{100}\Big)\Big(1 + \frac{\text{R}_3}{100}\Big)$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 3{,}125\Big(1 + \frac{4}{100}\Big)\Big(1 + \frac{5}{100}\Big)\Big(1 + \frac{6}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 3{,}125 \times \frac{26}{25} \times \frac{21}{20} \times \frac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 3{,}617.25\end{aligned}$$

C.I. = A - P

= ₹ 3,617.25 - ₹ 3,125

= ₹ 492.25

Hence, the compound interest = ₹ 492.25.

Given:

Principal (P) = ₹ 1,32,000

Rate of interest (R) = 15% p.a. compounded yearly

Time = 2 years 4 months

First we find the amount at the end of 2 years.

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 1{,}32{,}000\Big(1 + \frac{15}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}32{,}000\Big(\frac{23}{20}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}32{,}000 \times \frac{23}{20} \times \frac{23}{20} \\[1em] \Rightarrow \text{A} = ₹ 1{,}74{,}570\end{aligned}$$

As the remaining time is not a complete conversion period, we consider simple interest for the last partial period.

Principal for the next 4 months = ₹ 1,74,570

Time = 4 months

= $\frac{4}{12} \text{ year }= \frac{1}{3} \text{ year }$

$$\begin{aligned}\Rightarrow \text{Interest} = ₹ \frac{174570 \times 15 \times \frac{1}{3}}{100} \\[1em] \Rightarrow \text{Interest} = ₹ \frac{174570 \times 15 \times 1}{100 \times 3} \\[1em] \Rightarrow \text{Interest} = ₹ \frac{2618550}{300} \\[1em] \Rightarrow \text{Interest} = ₹ 8{,}728.50\end{aligned}$$

Amount at the end of 2 years and 4 months = ₹ 1,74,570 + ₹ 8,728.50 = ₹ 1,83,298.50

Hence, Khushi will pay ₹ 1,83,298.50 to clear the loan.

Given:

Principal (P) = ₹ 18,000

Rate of interest (R) = 8% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{18000 \times 8 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{288000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 2{,}880\end{aligned}$$

Compound interest :

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 18{,}000\Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 18{,}000\Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 18{,}000 \times \frac{27}{25} \times \frac{27}{25} \\[1em] \Rightarrow \text{A} = ₹ 20{,}995.20\end{aligned}$$

C.I. = A - P = ₹ 20,995.20 - ₹ 18,000 = ₹ 2,995.20

Extra amount to be paid = ₹ 2,995.20 - ₹ 2,880 = ₹ 115.20

Hence, Kabir would have to pay ₹ 115.20 extra.

Given:

Principal (P) = ₹ 75,000

Rate of interest = 12% p.a.

Time = $1\frac{1}{2}$ years

(i) Compounded annually :

First we find the amount at the end of 1 year.

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 75{,}000\Big(1 + \frac{12}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 75{,}000 \times \frac{28}{25} \\[1em] \Rightarrow \text{A} = ₹ 84{,}000\end{aligned}$$

As the remaining time is not a complete conversion period, we consider simple interest for the last partial period.

Principal for the next $\frac{1}{2}$ year = ₹ 84,000

$$\begin{aligned}\Rightarrow \text{Interest} = ₹ \frac{84000 \times 12 \times \frac{1}{2}}{100} \\[1em] \Rightarrow \text{Interest} = ₹ \frac{84000 \times 12 \times 1}{100 \times 2} \\[1em] \Rightarrow \text{Interest} = ₹ \frac{1008000}{200} \\[1em] \Rightarrow \text{Interest} = ₹ 5{,}040\end{aligned}$$

Amount at the end of $1\frac{1}{2}$ years = ₹ 84,000 + ₹ 5,040 = ₹ 89,040

Hence, the amount when compounded annually = ₹ 89,040.

(ii) Compounded half-yearly :

As the interest is compounded half-yearly,

Rate per conversion period (R) = $\frac{12}{2}$% = 6%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 75{,}000\Big(1 + \frac{6}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 75{,}000\Big(\frac{53}{50}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 75{,}000 \times \frac{53}{50} \times \frac{53}{50} \times \frac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 89{,}326.20\end{aligned}$$

Hence, the amount when compounded half-yearly = ₹ 89,326.20.

Given:

Principal (P) = ₹ 10,000

Rate of interest (R) = 7% p.a. compounded annually

(i) Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \frac{7}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000\Big(\frac{107}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \frac{107}{100} \times \frac{107}{100} \\[1em] \Rightarrow \text{A} = ₹ 11{,}449\end{aligned}$$

Hence, the amount received by him at the end of 2 years = ₹ 11,449.

(ii) Principal for the third year = ₹ 11,449

$$\begin{aligned}\Rightarrow \text{Interest for the third year} = ₹ \frac{11449 \times 7 \times 1}{100} \\[1em] = ₹ \frac{80143}{100} \\[1em] = ₹ 801.43\end{aligned}$$

Hence, the interest for the 3rd year = ₹ 801.43.

Given:

Amount (A) = ₹ 9,261

Rate of interest (R) = 5% p.a. compound interest

Number of conversion periods (n) = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 9261 = \text{P}\Big(1 + \frac{5}{100}\Big)^3 \\[1em] \Rightarrow 9261 = \text{P}\Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow 9261 = \text{P} \times \frac{9261}{8000} \\[1em] \Rightarrow \text{P} = \frac{9261 \times 8000}{9261} \\[1em] \Rightarrow \text{P} = ₹ 8{,}000\end{aligned}$$

Hence, the required sum of money = ₹ 8,000.

Given:

Amount (A) = ₹ 1,40,608

Rate of interest = 8% p.a. compounded half-yearly

As the interest is compounded half-yearly,

Rate per conversion period (R) = $\frac{8}{2}$% = 4%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 140608 = \text{P}\Big(1 + \frac{4}{100}\Big)^3 \\[1em] \Rightarrow 140608 = \text{P}\Big(\frac{26}{25}\Big)^3 \\[1em] \Rightarrow 140608 = \text{P} \times \frac{17576}{15625} \\[1em] \Rightarrow \text{P} = \frac{140608 \times 15625}{17576} \\[1em] \Rightarrow \text{P} = ₹ 1{,}25{,}000\end{aligned}$$

Hence, the required sum invested = ₹ 1,25,000.

Given:

Principal (P) = ₹ 2,000

Amount (A) = ₹ 2,315.25

Number of conversion periods (n) = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 2315.25 = 2000\Big(1 + \frac{\text{R}}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \frac{2315.25}{2000} \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \frac{9261}{8000} \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow 1 + \frac{\text{R}}{100} = \frac{21}{20} \\[1em] \Rightarrow \frac{\text{R}}{100} = \frac{21}{20} - 1 \\[1em] \Rightarrow \frac{\text{R}}{100} = \frac{1}{20} \\[1em] \Rightarrow \text{R} = \frac{100}{20} \\[1em] \Rightarrow \text{R} = 5\end{aligned}$$

Hence, the required rate = 5% per annum.

Given:

Principal (P) = ₹ 40,000

Amount (A) = ₹ 46,305

Time = $1\frac{1}{2}$ years, compound interest payable half yearly

As the interest is payable half yearly,

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

Let the rate per conversion period be R%.

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 46305 = 40000\Big(1 + \frac{\text{R}}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \frac{46305}{40000} \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \frac{9261}{8000} \\[1em] \Rightarrow \Big(1 + \frac{\text{R}}{100}\Big)^3 = \Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow 1 + \frac{\text{R}}{100} = \frac{21}{20} \\[1em] \Rightarrow \frac{\text{R}}{100} = \frac{1}{20} \\[1em] \Rightarrow \text{R} = 5\end{aligned}$$

So, the half yearly rate of interest = 5%.

Rate of interest per annum = 2 × 5% = 10%

Hence, the rate of interest = 10% per annum.

Given:

Principal (P) = ₹ 15,625

Amount (A) = ₹ 17,576

Rate of interest (R) = 4% p.a. compound interest

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 17576 = 15625\Big(1 + \frac{4}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{26}{25}\Big)^n = \frac{17576}{15625} \\[1em] \Rightarrow \Big(\frac{26}{25}\Big)^n = \Big(\frac{26}{25}\Big)^3 \\[1em] \Rightarrow n = 3\end{aligned}$$

Hence, the required time = 3 years.

Given:

Principal (P) = ₹ 16,000

Amount (A) = ₹ 18,522

Rate of interest = 10% p.a. compounded semi-annually

As the interest is compounded semi-annually,

Rate per conversion period (R) = $\frac{10}{2}$% = 5%

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 18522 = 16000\Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^n = \frac{18522}{16000} \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^n = \frac{9261}{8000} \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^n = \Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow n = 3\end{aligned}$$

So, the number of conversion periods = 3 half years.

Time period = $\frac{3}{2}$ years

= $1\frac{1}{2} \text{ years }$

Hence, the time period of the investment = $\mathbf{1\frac{1}{2}}$ years.

(i) The money borrowed (lent or invested) is called principal.

(ii) The additional money paid by the borrower to the moneylender in lieu of the money used is called interest.

(iii) In simple interest, the principal remains constant for the whole loan period.

(iv) In compound interest the principal goes on changing every conversion period.

(v) The time after which the interest is added each time to form a new principal is called conversion period.

(vi) If the interest is compounded semi-annually, then semi-annually rate is half of annual rate.

(i) False

Reason — The banks, post offices and insurance companies calculate the interest for one year and then add it to the principal to get the principal for the next year. So, the interest paid by them is compound interest and not simple interest.

(ii) True

Reason — In compound interest, the amount at the end of a year becomes the principal for the next year. So, the interest of a year is calculated on the amount of the previous year.

(iii) False

Reason — In compound interest, the interest of each conversion period is added to the principal to form the new principal. So, the principal goes on changing and does not remain constant.

(iv) True

Reason — The time from one specified interest period to the next period, i.e., the period after which the interest is added to the principal, is called conversion period.

(v) False

Reason — If the interest is compounded quarterly, the interest is added after every three months. So, there are 4 conversion periods in a year and not 2.

Given:

Principal (P) = ₹ 1,000

Rate of interest (R) = 10% p.a.

Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 1{,}000\Big(1 + \frac{10}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}000 \times \frac{11}{10} \times \frac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 1{,}210\end{aligned}$$

C.I. = A - P = ₹ 1,210 - ₹ 1,000 = ₹ 210

Hence, option 2 is the correct option.

Given:

Principal (P) = ₹ 5,000

Rate of interest = 20% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\frac{20}{2}$% = 10%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \frac{10}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 6{,}655\end{aligned}$$

C.I. = A - P = ₹ 6,655 - ₹ 5,000 = ₹ 1,655

Hence, option 2 is the correct option.

Given:

Principal (P) = ₹ 10,000

Rate of interest = 8% p.a. compounded quarterly

As the interest is compounded quarterly,

Rate per conversion period (R) = $\frac{8}{4}$% = 2%

Number of conversion periods (n) = $\frac{6}{3}$ = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \frac{2}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \frac{51}{50} \times \frac{51}{50} \\[1em] \Rightarrow \text{A} = ₹ 10{,}404\end{aligned}$$

C.I. = A - P = ₹ 10,404 - ₹ 10,000 = ₹ 404

Hence, option 3 is the correct option.

As the interest is compounded half yearly, the half yearly rate is half of the annual rate and there are two conversion periods in a year.

Rate per conversion period (R) = $\frac{8}{2}$% = 4%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

Hence, option 1 is the correct option.

As the interest is compounded quarterly, the quarterly rate is one fourth of the annual rate and there are four conversion periods in a year.

Rate per conversion period (R) = $\frac{4}{4}$% = 1%

Number of conversion periods (n) = 2 × 4 = 8

Hence, option 3 is the correct option.

In simple interest, the interest of every year is calculated on the original principal, whereas in compound interest, the interest of every conversion period is calculated on the amount of the previous conversion period.

For the first conversion period, the interest is the same in both the cases. From the second conversion period onwards, the principal in compound interest becomes greater than the original principal.

So, if the number of conversion periods ≥ 2, then the compound interest is greater than the simple interest.

Hence, option 4 is the correct option.

Given:

Principal (P) = ₹ 6,000

Amount (A) = ₹ 7,986

Rate of interest (R) = 10% p.a. compounded annually

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 7986 = 6000\Big(1 + \frac{10}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^n = \frac{7986}{6000} \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^n = \frac{1331}{1000} \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^n = \Big(\frac{11}{10}\Big)^3 \\[1em] \Rightarrow n = 3\end{aligned}$$

Hence, option 2 is the correct option.

In compound interest, the interest of each conversion period is calculated on the amount of the previous period, which keeps on increasing, while in simple interest it is always calculated on the original principal. So, for the same principal, rate of interest and time period, the compound interest is greater than the simple interest when the number of conversion periods ≥ 2.

Statement I is true.

The amount at the end of a year becomes the principal for the next year, so compound interest is calculated on the amount of the previous year.

Statement II is true.

Hence, option 3 is the correct option.

Principal (P) = ₹ 15,000, Rate of interest (R) = 10% p.a., n = 3 years

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 15{,}000\Big(1 + \frac{10}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 15{,}000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 19{,}965\end{aligned}$$

C.I. = A - P = ₹ 19,965 - ₹ 15,000 = ₹ 4,965

As the interest earned is ₹ 4,965 and not ₹ 4,500, Statement I is false.

Statement II is true, as the formula for simple interest is $\frac{\text{PRT}}{100}$.

Hence, option 2 is the correct option.

Given:

Principal (P) = ₹ 5,000

Rate of interest (R) = 6% p.a., interest payable yearly

Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \frac{6}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000\Big(\frac{53}{50}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \frac{53}{50} \times \frac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 5{,}618\end{aligned}$$

C.I. = A - P

= ₹ 5,618 - ₹ 5,000

= ₹ 618

Hence, the amount = ₹ 5,618 and the compound interest = ₹ 618.

Given:

Principal (P) = ₹ 7,400

Rate of interest = 5% p.a., interest payable half yearly

As the interest is payable half yearly,

Rate per conversion period (R) = $\frac{5}{2}$% = 2.5%

Number of conversion periods (n) = 1 × 2 = 2

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 7{,}400\Big(1 + \frac{2.5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 7{,}400\Big(\frac{41}{40}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 7{,}400 \times \frac{41}{40} \times \frac{41}{40} \\[1em] \Rightarrow \text{A} = ₹ 7{,}774.625\end{aligned}$$

C.I. = A - P

= ₹ 7,774.625 - ₹ 7,400

= ₹ 374.625

Hence, the amount = ₹ 7,774.625 and the compound interest = ₹ 374.625.

Given:

Principal (P) = ₹ 5,000

Rate of interest (R) = 8% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{S.I.} = ₹ \frac{5000 \times 8 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \frac{80000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 800\end{aligned}$$

Compound interest :

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000\Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \frac{27}{25} \times \frac{27}{25} \\[1em] \Rightarrow \text{A} = ₹ 5{,}832\end{aligned}$$

C.I. = A - P = ₹ 5,832 - ₹ 5,000 = ₹ 832

Difference between C.I. and S.I. = ₹ 832 - ₹ 800 = ₹ 32

Hence, the required difference = ₹ 32.

Given:

Principal (P) = ₹ 10,000

Rate of interest = 10% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\frac{10}{2}$% = 5%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \frac{5}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000\Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 11{,}576.25\end{aligned}$$

C.I. = A - P = ₹ 11,576.25 - ₹ 10,000 = ₹ 1,576.25

When compounded annually :

Amount at the end of 1 year

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \frac{10}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \frac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 11{,}000\end{aligned}$$

As the remaining time is not a complete conversion period, we consider simple interest for the last partial period.

$$\begin{aligned}\Rightarrow \text{Interest for the next } \frac{1}{2} \text{ year} = ₹ \frac{11000 \times 10 \times \frac{1}{2}}{100} \\[1em] = ₹ \frac{11000 \times 10 \times 1}{100 \times 2} \\[1em] = ₹ 550\end{aligned}$$

Amount at the end of $1\frac{1}{2}$ years = ₹ 11,000 + ₹ 550 = ₹ 11,550

C.I. = ₹ 11,550 - ₹ 10,000 = ₹ 1,550

As ₹ 1,576.25 is greater than ₹ 1,550, the interest received on compounding half yearly is more.

Hence, the amount = ₹ 11,576.25 and the compound interest = ₹ 1,576.25. Yes, this interest is more than the interest he would get if it was compounded annually.

Given:

Amount (A) = ₹ 1,32,651

Rate of interest = 4% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\frac{4}{2}$% = 2%

Number of conversion periods (n) = $1\frac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 132651 = \text{P}\Big(1 + \frac{2}{100}\Big)^3 \\[1em] \Rightarrow 132651 = \text{P}\Big(\frac{51}{50}\Big)^3 \\[1em] \Rightarrow 132651 = \text{P} \times \frac{132651}{125000} \\[1em] \Rightarrow \text{P} = \frac{132651 \times 125000}{132651} \\[1em] \Rightarrow \text{P} = ₹ 1{,}25{,}000\end{aligned}$$

Hence, the required sum invested = ₹ 1,25,000.

Given:

Principal (P) = ₹ 12,500

Compound interest (C.I.) = ₹ 3,246.40

Rate of interest (R) = 8% p.a. compounded annually

Amount = Principal + Compound interest

= ₹ 12,500 + ₹ 3,246.40

= ₹ 15,746.40

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 15746.40 = 12500\Big(1 + \frac{8}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^n = \frac{15746.40}{12500} \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^n = \frac{1574640}{1250000} \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^n = \frac{19683}{15625} \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^n = \Big(\frac{27}{25}\Big)^3 \\[1em] \Rightarrow n = 3\end{aligned}$$

Hence, the required time = 3 years.

Given:

Principal (P) = ₹ 2,500

Rate of interest for the first year (R₁) = 5% p.a.

Rate of interest for the second year (R₂) = 6% p.a.

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}_1}{100}\Big)\Big(1 + \frac{\text{R}_2}{100}\Big)$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 2{,}500\Big(1 + \frac{5}{100}\Big)\Big(1 + \frac{6}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 2{,}500 \times \frac{21}{20} \times \frac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 2{,}782.50\end{aligned}$$

C.I. = A - P

= ₹ 2,782.50 - ₹ 2,500

= ₹ 282.50

Hence, the amount = ₹ 2,782.50 and the compound interest = ₹ 282.50.

Given:

Rate of interest (R) = 5% p.a. compounded annually

Difference between the interests of third year and first year = ₹ 102.50

Let the sum of money (principal) be ₹ P.

$\Rightarrow \text{Interest for the first year} = ₹ \frac{\text{P} \times 5 \times 1}{100} = ₹ \frac{\text{P}}{20}$

Amount at the end of second year

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ \text{P}\Big(1 + \frac{5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ \text{P} \times \frac{21}{20} \times \frac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ \frac{441\text{P}}{400}\end{aligned}$$

This amount is the principal for the third year.

$\Rightarrow \text{Interest for the third year} = ₹ \frac{441\text{P}}{400} \times \frac{5}{100} = ₹ \frac{441\text{P}}{8000}$

According to given information,

$$\begin{aligned}\Rightarrow \frac{441\text{P}}{8000} - \frac{\text{P}}{20} = 102.50 \\[1em] \Rightarrow \frac{441\text{P}}{8000} - \frac{400\text{P}}{8000} = 102.50 \\[1em] \Rightarrow \frac{41\text{P}}{8000} = 102.50 \\[1em] \Rightarrow \text{P} = \frac{102.50 \times 8000}{41} \\[1em] \Rightarrow \text{P} = \frac{820000}{41} \\[1em] \Rightarrow \text{P} = ₹ 20{,}000\end{aligned}$$

Hence, the required sum = ₹ 20,000.

Given:

Principal (P) = ₹ 42,000

Time = 2 years

Difference between compound interest and simple interest = ₹ 105

(i) Let the rate of interest be R% per annum.

As we know:

$\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \frac{42000 \times \text{R} \times 2}{100} = ₹ 840\text{R}$

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \frac{\text{R}}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \frac{2\text{R}}{100} + \frac{\text{R}^2}{10000}\Big) \\[1em] \Rightarrow \text{A} = ₹ \Big(42000 + 840\text{R} + \frac{42\text{R}^2}{10}\Big)\end{aligned}$$

C.I. = A - P = ₹ $\Big(840\text{R} + \frac{42\text{R}^2}{10}\Big)$

According to given information,

$$\begin{aligned}\Rightarrow \Big(840\text{R} + \frac{42\text{R}^2}{10}\Big) - 840\text{R} = 105 \\[1em] \Rightarrow \frac{42\text{R}^2}{10} = 105 \\[1em] \Rightarrow \text{R}^2 = \frac{105 \times 10}{42} \\[1em] \Rightarrow \text{R}^2 = 25 \\[1em] \Rightarrow \text{R} = \pm 5\end{aligned}$$

As the rate of interest cannot be negative, we reject R = −5.

$\Rightarrow$ R = 5

Hence, the rate of interest = 5% p.a.

(ii) Amount at the end of first year

As we know:

$\text{A} = \text{P}\Big(1 + \frac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$$\begin{aligned}\Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \frac{5}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 42{,}000 \times \frac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 44{,}100\end{aligned}$$

Principal for the second year = ₹ 44,100

$\Rightarrow \text{Interest for the second year} = ₹ \frac{44100 \times 5 \times 1}{100} = ₹ 2{,}205$

Hence, the compound interest earned in second year = ₹ 2,205.