Interest

Given:

P = ₹ 483

R = 5%

T = 2 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{483 \times 5 \times 2}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{4830}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ 48.3\end{aligned}$$

Hence, option 2 is the correct option.

Given:

P = ₹ 8,490

R = 5%

T = 73 days = $\frac{73}{365}$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{8,490 \times 5 \times 73}{100 \times 365}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{30,98,850}{36500}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ 84.9\end{aligned}$$

Hence, option 2 is the correct option.

Given:

A = 2P

T = 8 years

Let the rate be $r$.

$$\begin{aligned}\because \text{A = S.I. + P}\\[1em] \Rightarrow \text{S.I. = A - P}\\[1em] \Rightarrow \text{S.I. = 2P - P}\\[1em] \Rightarrow \text{S.I. = P}\end{aligned}$$

And we know,

$$\begin{aligned}\text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{P} = ₹ \Big(\frac{P \times r \times 8}{100}\Big)\\[1em] \Rightarrow \cancel{P} = ₹ \Big(\frac{\cancel{P} \times r \times 8}{100}\Big)\\[1em] \Rightarrow r = \frac{100}{8}%\\[1em] \Rightarrow r = \frac{25}{2}%\\[1em]\end{aligned}$$

When A becomes 3P, ⇒ S.I. = 2P

Let the time be $t$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{2P} = ₹ \Big(\frac{P \times 25 \times t}{2 \times 100}\Big)\\[1em] \Rightarrow 2\cancel{P} = ₹ \Big(\frac{\cancel{P} \times 25 \times t}{200}\Big)\\[1em] \Rightarrow t = ₹ \frac{400}{25}\\[1em] \Rightarrow t = 16\\[1em]\end{aligned}$$

Hence, option 1 is the correct option.

Given:

P = ₹ 5,000

T = 2 years

S.I. = ₹ 500

Let rate of interest be $r$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{500} = ₹ \Big(\frac{5,000 \times r \times 2}{100}\Big)\\[1em] \Rightarrow \text{500} = ₹ \frac{10,000r}{100}\\[1em] \Rightarrow \text{500} = ₹ 100r\\[1em] \Rightarrow r = \frac{500}{100}\\[1em] \Rightarrow r = 5\end{aligned}$$

Hence, option 2 is the correct option.

Given:

P = ₹ 7,000

R = 5%

S.I. = ₹ 1,400

Let time be $t$ years.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{1,400} = ₹ \Big(\frac{7,000 \times 5 \times t}{100}\Big)\\[1em] \Rightarrow \text{1,400} = ₹ \frac{35,000t}{100}\\[1em] \Rightarrow \text{1,400} = ₹ 350t\\[1em] \Rightarrow t = \frac{1,400}{350}\\[1em] \Rightarrow t = 4\end{aligned}$$

∴ Time = 4 years

Hence, option 4 is the correct option.

Given:

P = ₹ 750

R = 10%

T = 3 years 4 months

= $\Big(3 + \frac{4}{12}\Big)$ years

= $\Big(3 + \frac{1}{3}\Big)$ years

= $\frac{10}{3}$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{750 \times 10 \times 10}{3 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{75,000}{300}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ 250 \\[2em] \because \text{A = P + S.I.}\\[1em] \Rightarrow \text{A = ₹ 750 + ₹ 250}\\[1em] \Rightarrow \text{A = ₹ 1,000}\\[1em]\end{aligned}$$

Hence, S.I. = ₹ 250 and Amount = ₹ 1,000.

Given:

P = ₹ 5,000

R = 8%

To calculate time (T):

Dec = 8 days (31 - 23)

Jan = 31 days

Feb = 29 days

March = 31 days

April = 30 days

May = 31 days

June = 30 days

July = 29 days

Total = 219 days

T = 219 days

= $\Big(\frac{219}{365}\Big)$ years

= $\Big(\frac{3}{5}\Big)$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{5,000 \times 8 \times 3}{5 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{1,20,000}{500}\\[1em] \Rightarrow \text{S.I.} = ₹ 240 \\[2em] \because \text{A = P + S.I.}\\[1em] \Rightarrow \text{A = ₹ 5,000 + ₹ 240}\\[1em] \Rightarrow \text{A = ₹ 5,240}\\[1em]\end{aligned}$$

Hence, S.I. = ₹ 240 and Amount = ₹ 5,240.

Given:

P = ₹ 2,600

R = 1% per month = 12% per annum

T = 2 years 3 months

= $\Big(2 + \frac{3}{12}\Big)$ years

= $\Big(2 + \frac{1}{4}\Big)$ years

= $\Big(\frac{8 + 1}{4}\Big)$ years

= $\Big(\frac{9}{4}\Big)$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{2,600 \times 12 \times 9}{4 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{2,08,800}{400}\\[1em] \Rightarrow \text{S.I.} = ₹ 702 \\[2em] \because \text{A = P + S.I.}\\[1em] \Rightarrow \text{A = ₹ 2,600 + ₹ 702}\\[1em] \Rightarrow \text{A = ₹ 3,302}\\[1em]\end{aligned}$$

Hence, S.I. = ₹ 702 and Amount = ₹ 3,302.

Given:

P = ₹ 4,000

R = 2 paise per rupee per month

= $\frac{2}{100}$ per month

= $\frac{2}{100} \times 12$ per annum

= $\frac{24}{100}$ per annum

= 24% per annum

T = $\Big(1\frac{1}{3}\Big)$ years

= $\Big(\frac{4}{3}\Big)$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{4,000 \times 24 \times 4}{3 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{3,84,000}{300}\\[1em] \Rightarrow \text{S.I.} = ₹ 1,280 \\[2em] \because \text{A = P + S.I.}\\[1em] \Rightarrow \text{A = ₹ 4,000 + ₹ 1,280}\\[1em] \Rightarrow \text{A = ₹ 5,280}\\[1em]\end{aligned}$$

Hence, S.I. = ₹ 1,280 and Amount = ₹ 5,280.

Given:

P = ₹ 24,000

R = 7.5%

T = 4 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{24,000 \times 7.5 \times 4}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{7,20,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 7,200\end{aligned}$$ And $$\begin{aligned}\because \text{A = P + S.I.}\\[1em] \Rightarrow \text{A = ₹ 24,000 + ₹ 7,200}\\[1em] \Rightarrow \text{A = ₹ 31,200}\\[1em]\end{aligned}$$

Hence, Rohit will pay ₹ 31,200 at the end of 4 years.

Given:

S.I. = ₹ 7,008

R = 5%

T = 6 years 3 months

= $\Big(6 + \frac{3}{12}\Big)$ years

= $\Big(6 + \frac{1}{4}\Big)$ years

= $\frac{25}{4}$ years

Let the Principal amount be ₹ $P$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 7,008 = \Big(\frac{P \times 5 \times 25}{4 \times 100}\Big)\\[1em] \Rightarrow 7,008 = \Big(\frac{P \times 125}{400}\Big)\\[1em] \Rightarrow P = \Big(\frac{7,008 \times 400}{125}\Big)\\[1em] \Rightarrow P = \frac{28,03,200}{125}\\[1em] \Rightarrow P = 22,425.60\end{aligned}$$

Hence, the Principal amount be ₹ 22,425.60.

Given:

A = ₹ 4,000

R = 6.25%

T = 4 years

Let the Principal amount be ₹ $P$.

As we know, $$\begin{aligned}\text{A = S.I. + P}\\[1em] \Rightarrow 4,000 = \text{S.I. + P}\\[1em] \Rightarrow \text{S.I.} = 4,000 - P \\[1em]\end{aligned}$$

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 4,000 - P = \Big(\frac{P \times 6.25 \times 4}{100}\Big)\\[1em] \Rightarrow 4,000 - P = \frac{P \times 25}{100}\\[1em] \Rightarrow 4,000 - P = \frac{P}{4}\\[1em] \Rightarrow 4,000 = \frac{P}{4} + P\\[1em] \Rightarrow 4,000 = \frac{P}{4} + \frac{4P}{4}\\[1em] \Rightarrow 4,000 = \frac{(P + 4P)}{4}\\[1em] \Rightarrow 4,000 = \frac{5P}{4}\\[1em] \Rightarrow P = \frac{4,000 \times 4}{5}\\[1em] \Rightarrow P = \frac{16,000}{5}\\[1em] \Rightarrow P = 3,200\end{aligned}$$

Hence, the Principal amount be ₹ 3,200.

Given:

P = ₹ 630

T = 4 years

S.I. = ₹ 126

Let rate of interest be $r$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{126} = ₹ \Big(\frac{630 \times r \times 4}{100}\Big)\\[1em] \Rightarrow \text{126} = ₹ \frac{2,520r}{100}\\[1em] \Rightarrow \text{126} = ₹ \frac{252r}{10}\\[1em] \Rightarrow r = \frac{126 \times 10}{252}\\[1em] \Rightarrow r = 5\end{aligned}$$

Hence, the rate of interest = 5%.

Given:

T = $6\frac{1}{4}$ years

= $\frac{25}{4}$ years

A = 2P

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow \text{2P = P + S.I.}\\[1em] \Rightarrow \text{2P - P = S.I.}\\[1em] \Rightarrow \text{P = S.I.}\\[1em]\end{aligned}$$

Let rate of interest be $r$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{P} = ₹ \Big(\frac{P \times r \times 25}{4 \times 100}\Big)\\[1em] \Rightarrow \cancel {P} = ₹ \frac{\cancel {P} \times r \times 25}{400}\\[1em] \Rightarrow 1 = ₹ \frac{25r}{400}\\[1em] \Rightarrow \text{400} = 25r\\[1em] \Rightarrow r = \frac{400}{25}\\[1em] \Rightarrow r = 16\end{aligned}$$

Hence, the rate of interest = 16%.

Given:

T = 8 months

= $\frac{8}{12}$ years

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

Let principal be P

S.I. = 0.06 P

Let rate of interest be $r$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{0.06P} = ₹ \Big(\frac{P \times r \times 2}{3 \times 100}\Big)\\[1em] \Rightarrow \frac{6P}{100} = ₹ \Big(\frac{P \times r \times 2}{300}\Big)\\[1em] \Rightarrow \frac{6}{100}\cancel {P} = ₹ \Big(\frac{\cancel {P} \times r \times 2}{300}\Big)\\[1em] \Rightarrow \frac{6}{100} = ₹ \frac{2r}{300}\\[1em] \Rightarrow r = \frac{6 \times 300}{100 \times 2}\\[1em] \Rightarrow r = \frac{1800}{200}\\[1em] \Rightarrow r = 9 %\end{aligned}$$

Hence, the rate of interest = 9%.

Given:

P = ₹ 950

R = 7%

S.I. = ₹ 399

Let time be $t$ years. $$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{399} = ₹ \Big(\frac{950 \times 7 \times t}{100}\Big)\\[1em] \Rightarrow \text{399} = ₹ \Big(\frac{6,650t}{100}\Big)\\[1em] \Rightarrow \text{399} = ₹ \Big(\frac{133t}{2}\Big)\\[1em] \Rightarrow t = \frac{399 \times 2}{133}\\[1em] \Rightarrow t = \frac{798}{133}\\[1em] \Rightarrow t = 6\end{aligned}$$

Hence, the time is 6 years.

Given:

P = ₹ 1,200

R = 3.5%

A = ₹ 1,536

As we know,

∵ A = P + S.I.

⇒ ₹ 1,536 = ₹ 1,200 + S.I.

⇒ S.I. = ₹ 1,536 - ₹ 1,200

⇒ S.I. = ₹ 336

Let time be $t$ years.

$$\begin{aligned}\because \text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{336} = \Big(\frac{1,200 \times 3.5 \times t}{100}\Big)\\[1em] \Rightarrow \text{336} = \Big(\frac{4,200t}{100}\Big)\\[1em] \Rightarrow \text{336} = 42t\\[1em] \Rightarrow t = \frac{336}{42}\\[1em] \Rightarrow t = 8\end{aligned}$$

Hence, the time is 8 years.

Given:

T = $6\frac{1}{4}$ years

= $\frac{25}{4}$ years

Let principal be P

S.I. = $\frac{3}{8}$ P

Let rate of interest be $r$.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \frac{3}{8}P = ₹ \Big(\frac{P \times r \times 25}{4 \times 100}\Big)\\[1em] \Rightarrow \frac{3P}{8} = ₹ \Big(\frac{P \times r \times 25}{400}\Big)\\[1em] \Rightarrow \frac{3}{8}\cancel {P} = ₹ \frac{\cancel {P} \times 25r}{400}\\[1em] \Rightarrow \frac{3}{8} = ₹ \frac{25r}{400}\\[1em] \Rightarrow r = \frac{3 \times 400}{8 \times 25}\\[1em] \Rightarrow r = \frac{1200}{200}\\[1em] \Rightarrow r = 6 %\end{aligned}$$

Hence, the rate of interest = 6%.

A = ₹ 10,210.20

R = 5%

To calculate time (T):

May = 7 days (31 -24)

Jun = 30 days

July = 31 days

August = 31 days

Sept = 30 days

Oct = 17 days

Total = 146 days

T = 146 days

= $\frac{146}{365}$ years

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

Let the Principal amount be ₹ $P$.

As we know,

$$\begin{aligned}\text{A = S.I. + P}\\[1em] \Rightarrow 10,210.20 = \text{S.I. + P}\\[1em] \Rightarrow \text{S.I.} = 10,210.20 - P \\[1em]\end{aligned}$$

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 10,210.20 - P = \Big(\frac{P \times 5 \times 2}{5 \times 100}\Big)\\[1em] \Rightarrow 10,210.20 - P = \Big(\frac{P \times 2}{100}\Big)\\[1em] \Rightarrow 10,210.20 - P = \frac{P}{50}\\[1em] \Rightarrow 10,210.20 = \frac{P}{50} + P\\[1em] \Rightarrow 10,210.20 = \frac{P}{50} + \frac{50P}{50}\\[1em] \Rightarrow 10,210.20 = \frac{(P + 50P)}{50}\\[1em] \Rightarrow 10,210.20 = \frac{51P}{50}\\[1em] \Rightarrow P = \frac{10,210.20 \times 50}{51}\\[1em] \Rightarrow P = \frac{510510}{51}\\[1em] \Rightarrow P = 10,010\end{aligned}$$

Hence, the Principal amount be ₹ 10,010.

Let the Principal amount be ₹ P.

S.I. = $\frac{5}{8}$ of P

R = 6%

Let time be $t$ years.

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \frac{5}{8} \times \text{P} = ₹ \Big(\frac{P \times 6 \times t}{100}\Big)\\[1em] \Rightarrow \frac{5}{8} \times \cancel{P} = ₹ \Big(\frac{\cancel{P} \times 3t}{50}\Big)\\[1em] \Rightarrow \frac{5}{8} = ₹ \Big(\frac{3t}{50}\Big)\\[1em] \Rightarrow t = \frac{5 \times 50}{8 \times 3}\\[1em] \Rightarrow t = \frac{250}{24}\\[1em] \Rightarrow t = 10\frac{10}{24}\\[1em] \Rightarrow t = 10\frac{5}{12}\\[1em]\end{aligned}$$

Hence, the time is 10 years and 5 months.

Given:

P = ₹ 7,000

R = 6%

T = 3 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{7,000 \times 6 \times 3}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{1,26,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 1,260\end{aligned}$$

P = ₹ 9,500

R = 5%

T = 3 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{9,500 \times 5 \times 3}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{1,42,500}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 1,425\end{aligned}$$

Total income from the interest = ₹ 1,260 + ₹ 1,425 = ₹ 2,685

Hence, total amount from the interest in 3 years = ₹ 2,685.

Given:

Total amount Raj borrows = ₹ 8,000

P = ₹ 4,500

R = 5%

T = 4 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{4,500 \times 5 \times 4}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{90,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 900\end{aligned}$$

Remaining Principal = ₹ (8,000 - 4500) = ₹ 3,500

R = 6%

T = 4 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{3,500 \times 6 \times 4}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{84,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 840\end{aligned}$$

Total interest paid in 4 years = ₹(900 + 840) = ₹ 1,740

Hence, total interest paid in 4 years = ₹ 1,740.

Given:

P = ₹ 5,000

R = 5%

T = 2 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{5,000 \times 5 \times 2}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{50,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 500\end{aligned}$$

And,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow\text{A} = ₹ 5,000 + ₹ 500\\[1em] \Rightarrow\text{A} = ₹ 5,500\end{aligned}$$

Hence, option 3 is the correct option.

Given:

A = 3P

T = 20 years

Let the rate be $r$.

$$\begin{aligned}\because \text{A = S.I. + P}\\[1em] \Rightarrow \text{S.I. = A - P}\\[1em] \Rightarrow \text{S.I. = 3P - P}\\[1em] \Rightarrow \text{S.I. = 2P}\end{aligned}$$

And we know,

$$\begin{aligned}\text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{2P} = ₹ \Big(\frac{P \times r \times 20}{100}\Big)\\[1em] \Rightarrow 2 \cancel{P} = ₹ \frac{\cancel{P} \times r \times 20}{100}\\[1em] \Rightarrow 2 = ₹ \frac{r}{5}\\[1em] \Rightarrow r = 2 \times 5\\[1em] \Rightarrow r = 10\end{aligned}$$

Hence, option 2 is the correct option.

Given:

S.I. = 0.5P

= $\frac{5}{10}$ P

T = 10 years

Let the rate be $r$.

We know,

$$\begin{aligned}\text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \frac{5}{10}P = ₹ \Big(\frac{P \times r \times 10}{100}\Big)\\[1em] \Rightarrow \frac{1}{2} \cancel{P} = ₹ \frac{\cancel{P} \times R \times 10}{100}\\[1em] \Rightarrow \frac{1}{2} = ₹ \frac{R}{10}\\[1em] \Rightarrow R = \frac{10}{2}\\[1em] \Rightarrow R = 5\end{aligned}$$

Hence, option 3 is the correct option.

Given:

P = ₹ 600

R = 5%

A = ₹ 720

Let the time be $t$.

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 720 = ₹ 600 + \text{S.I.}\\[1em] \Rightarrow \text{S.I.} = ₹ 720 - ₹ 600\\[1em] \Rightarrow \text{S.I.} = ₹ 120\\[1em]\end{aligned}$$

And

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 120 = \Big(\frac{600 \times 5 \times t}{100}\Big)\\[1em] \Rightarrow 120 = \frac{3,000 \times t}{100}\\[1em] \Rightarrow 120 = 30 \times t\\[1em] \Rightarrow t = \frac{120}{30}\\[1em] \Rightarrow t = 4\end{aligned}$$

Hence, option 2 is the correct option.

Given:

P = ₹ 8,000

R = 10%

T = 2 years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{8,000 \times 10 \times 2}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{1,60,000}{100}\\[1em] \Rightarrow \text{S.I.} = ₹ 1,600\end{aligned}$$

And,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow\text{A} = ₹ 8,000 + ₹ 1,600\\[1em] \Rightarrow\text{A} = ₹ 9,600\end{aligned}$$

Hence, option 1 is the correct option.

(i) Given:

P = ₹ 3,750

A = ₹ 4,620

T = 3 years

Let the rate of interest be $r$.

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 4,620 = ₹ 3,750 + \text{S.I.}\\[1em] \Rightarrow \text{S.I.} = ₹ 4,620 - ₹ 3,750 \\[1em] \Rightarrow \text{S.I.} = ₹ 870\end{aligned}$$

And,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow ₹ 870 = \Big(\frac{₹ 3,750 \times r \times 3}{100}\Big)\\[1em] \Rightarrow ₹ 870 = \frac{11,250r}{100}\\[1em] \Rightarrow ₹ 870 = \frac{225r}{2}\\[1em] \Rightarrow r = \frac{2 \times 870}{225}\\[1em] \Rightarrow r = \frac{1740}{225}\\[1em] \Rightarrow r = \frac{116}{15}\\[1em] \Rightarrow r = 7\frac{11}{15}\end{aligned}$$

Hence, the rate of interest = $7\frac{11}{15}%$.

(ii) Given:

P = ₹ 7,500

R = $\frac{116}{15}%$

T = $5\frac{1}{2}$ years

= $\frac{11}{2}$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \Big(\frac{7,500 \times 116 \times 11}{15 \times 2 \times 100}\Big)\\[1em] \Rightarrow \text{S.I.} = ₹ \frac{95,70,000}{3000}\\[1em] \Rightarrow \text{S.I.} = ₹ 3,190\end{aligned}$$

And,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow\text{A} = ₹ 7,500 + ₹ 3,190\\[1em] \Rightarrow\text{A} = ₹ 10,690\end{aligned}$$

Hence, the amount = ₹ 10,690.

(i) Let the Principal amount be ₹ P.

A = 2P

T = 8 years

Let the rate be $r$.

$$\begin{aligned}\because \text{A = S.I. + P}\\[1em] \Rightarrow \text{S.I. = A - P}\\[1em] \Rightarrow \text{S.I. = 2P - P}\\[1em] \Rightarrow \text{S.I. = P}\end{aligned}$$

And we know,

$$\begin{aligned}\text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{P} = ₹ \Big(\frac{P \times r \times 8}{100}\Big)\\[1em] \Rightarrow \cancel{P} = ₹ \frac{\cancel{P} \times r \times 8}{100}\\[1em] \Rightarrow 1 = ₹ \frac{8r}{100}\\[1em] \Rightarrow r = \frac{100}{8}%\\[1em] \Rightarrow r = \frac{25}{2}%\\[1em] \Rightarrow r = 12.5%\end{aligned}$$

Hence, rate of interest = $12.5%$.

(ii) When A becomes 3P, ⇒ S.I. = 2P

Let the time be $t$ years

$$\begin{aligned}\because \text{S.I.} = ₹ \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{2P} = ₹ \Big(\frac{P \times 25 \times t}{2 \times 100}\Big)\\[1em] \Rightarrow 2\cancel{P} = ₹ \frac{\cancel{P} \times 25 \times t}{200}\\[1em] \Rightarrow t = \frac{400}{25}\\[1em] \Rightarrow t = 16\\[1em]\end{aligned}$$

Hence,the time is 16 years.

Given:

P = ₹ 4,000

A = ₹ 5,000

T = 8 years

Let the rate be $r$.

As we know, $$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 5,000 = ₹ 4,000 + S.I.\\[1em] \Rightarrow S.I. = ₹ 5,000 - ₹ 4,000 \\[1em] \Rightarrow S.I. = ₹ 1,000\end{aligned}$$

And

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow ₹ 1,000 = \Big(\frac{₹ 4,000 \times r \times 8}{100}\Big)\\[1em] \Rightarrow ₹ 1,000 = 40 \times r \times 8\\[1em] \Rightarrow ₹ 1,000 = 320 \times r\\[1em] \Rightarrow r = \frac{1,000}{320}%\\[1em] \Rightarrow r = \frac{25}{8}%\end{aligned}$$

When P = $₹ 2,100$

A = $₹ 2,800$

R = $\frac{25}{8}%$

Let the time be $t$.

As we know,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 2,800 = ₹ 2,100 + S.I.\\[1em] \Rightarrow S.I. = ₹ 2,800 - ₹ 2,100 \\[1em] \Rightarrow S.I. = ₹ 700\end{aligned}$$

And

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 700 = \Big(\frac{2,100 \times 25 \times t}{8 \times 100}\Big)\\[1em] \Rightarrow 700 = \frac{21 \times 25 \times t}{8}\\[1em] \Rightarrow 700 = \frac{525 \times t}{8}\\[1em] \Rightarrow t = \frac{700 \times 8}{525}\\[1em] \Rightarrow t = \frac{32}{3}\\[1em] \Rightarrow t = 10\frac{2}{3}\\[1em] \Rightarrow t = 10\frac{8}{12}\end{aligned}$$

Hence, the time = 10 years and 8 months.

Given:

P₁ = ₹ 7,500

R₁ = 5%

T₁ = 6 years

So, S.I.₁ =

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{7,500 \times 5 \times 6}{100}\Big)\\[1em] = \frac{2,25,000}{100}\\[1em] = 2,250\end{aligned}$$

R₂ = 6.5%

T₂ = 4 years

Let P₂ be $P$.

And S.I.₂ =

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{P \times 6.5 \times 4}{100}\Big)\\[1em] = \frac{26P}{100}\\[1em]\end{aligned}$$

As interest on both are same. Hence,

S.I.₁ = S.I.₂

$$\begin{aligned}\therefore 2,250 = \frac{26P}{100}\\[1em] \Rightarrow P = \frac{2,250 \times 100}{26}\\[1em] \Rightarrow P = \frac{2,25,000}{26}\\[1em] \Rightarrow P = 8,653.85\end{aligned}$$

Hence, the principal amount = ₹ 8,653.85.

Given:

Amount in 4 years = ₹ 3,825

⇒ P + S.I. of 4 years = = ₹ 3,825 ..........(1)

Amount in 6 years = ₹ 4,050

⇒ P + S.I. of 6 years = ₹ 4,050 ..........(2)

Subtracting equation (1) from (2), we get

S.I. of 2 years = $₹ 4,050 - ₹ 3,825$

= $₹ 225$

S.I. of 1 year = ₹ $\frac{225}{2}$

= $₹ 112.5$

S.I. of 4 years = $₹ 112.5 \times 4$

= $₹ 450$

From equation (1), we get:

P + ₹ 450 = ₹ 3,825

P = ₹ 3,825 - ₹ 450

P = ₹ 3,375

Now when P = ₹ 3,375

S.I. = ₹ 112.5

T = 1 year

Let the rate be $r$.

As we know

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 112.5 = \Big(\frac{3,375 \times r \times 1}{100}\Big)\\[1em] \Rightarrow 112.5 = \frac{3,375r}{100}\\[1em] \Rightarrow r = \frac{112.5 \times 100}{3375}\\[1em] \Rightarrow r = \frac{11250}{3375}\\[1em] \Rightarrow r = \frac{10}{3}\\[1em] \Rightarrow r = 3\frac{1}{3}\end{aligned}$$

Hence, principal = $₹ 3,375$ and rate = $3\frac{1}{3}%$

Given:

P = ₹ 3,750

T = 4 years

S.I. = one-fifth of P

= $\frac{1}{5} \times ₹ 3,750$

= $\frac{₹ 3,750}{5}$

= $₹ 750$

Let the rate be $r$.

As we know,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 750 = \Big(\frac{3,750 \times r \times 4}{100}\Big)\\[1em] \Rightarrow 750 = 75 \times r \times 2\\[1em] \Rightarrow 750 = 150r\\[1em] \Rightarrow r = \frac{750}{150}\\[1em] \Rightarrow r = 5\end{aligned}$$

When P = ₹ 3,750

R = 5%

T = 15 years

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{3,750 \times 5 \times 15}{100}\Big)\\[1em] = \Big(\frac{2,81,250}{100}\Big)\\[1em] = 2,812.50\end{aligned}$$

And,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] = 3,750 + 2,812.5 \\[1em] = 6,562.50\end{aligned}$$

Hence, rate = $5%$ and the amount = ₹ 6,562.50.

Given:

P = ₹ 1,950

A = ₹ 2,125.50

R = 5%

Let the time be $t$

As we know,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow ₹ 2,125.50 = ₹ 1,950 + S.I.\\[1em] \Rightarrow S.I. = ₹ 2,125.50 - ₹ 1,950 \\[1em] \Rightarrow S.I. = ₹ 175.50\end{aligned}$$

And,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 175.50 = \Big(\frac{1,950 \times 5 \times t}{100}\Big)\\[1em] \Rightarrow 175.50 = \frac{9,750t}{100}\\[1em] \Rightarrow t = \frac{175.50 \times 100}{9750}\\[1em] \Rightarrow t = \frac{9}{5}\\[1em] \Rightarrow t = 1\frac{4}{5}\end{aligned}$$

$1\frac{4}{5}$ years means 1 years and 292 days (∵ $\frac{4}{5}$ x 365 = 292).

As the money was lent on 5th January, 2011,

1 year from 5th January, 2011 = 5th January, 2012

And 292 days = Jan (26 days) + Feb (29 days) + March (31 days) + April (30 days) + May (31 days) + June (30 days) + July (31 days) + Aug (31 days) + Sept (30 days) + Oct (23 days)

Hence, the required date is 23^rd October, 2012.

Given:

P₁ = ₹ 2,400

T₁ = 3 years

Let the rate be $r$.

Hence,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text {S.I} = \Big(\frac{2,400 \times r \times 3}{100}\Big)\\[1em] \Rightarrow \text {S.I} = \Big(\frac{7,200r}{100}\Big)\\[1em] \Rightarrow \text {S.I} = 72r\\[1em]\end{aligned}$$

P₂ = ₹ 2,000

T₂ = 3 years

R = r%

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big(\frac{2,000 \times r \times 3}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big(\frac{6,000r}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = 60r\end{aligned}$$

According to the question,

Difference between the two S.I. = ₹ 60

$$\begin{aligned}72r - 60r = ₹ 60\\[1em] \Rightarrow 12r = ₹ 60\\[1em] \Rightarrow r = \frac{60}{12}\\[1em] \Rightarrow r = 5\end{aligned}$$

Hence, the rate is 5%.

Let the first part be ₹ $x$ and the second part be ₹ $(15,600 - x)$.

Hence

P₁ = ₹ $x$

R₁ = 5%

T₁ = 5 years

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big(\frac{x \times 5 \times 5}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \frac{25x}{100}\\[1em] \Rightarrow \text{S.I.} = \frac{x}{4}\\[1em]\end{aligned}$$

P₂ = ₹ $15,600 - x$

R₂ = $4\frac{1}{2}%$

= $\frac{9}{2}%$

T₂ = 6 years

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big[\frac{(15,600 - x) \times 9 \times 6}{2 \times 100}\Big]\\[1em] \Rightarrow \text{S.I.} = \frac{54(15,600 - x)}{200}\\[1em] \Rightarrow \text{S.I.} = \frac{27(15,600 - x)}{100}\\[1em]\end{aligned}$$

Both interest are equal.

$$\begin{aligned}\frac{x}{4} = \frac{27(15,600 - x)}{100}\\[1em] \Rightarrow\frac{x}{4} = \frac{4,21,200 - 27x}{100}\\[1em] \Rightarrow\frac{x}{4} + \frac{27x}{100} = \frac{4,21,200}{100}\\[1em] \Rightarrow\frac{25x}{100} + \frac{27x}{100} = 4,212\\[1em] \Rightarrow\frac{(25x + 27x)}{100} = 4,212\\[1em] \Rightarrow\frac{52x}{100} = 4,212\\[1em] \Rightarrow x = \frac{4,212\times 100}{52}\\[1em] \Rightarrow x = \frac{4,21,200}{52}\\[1em] \Rightarrow x = 8,100\end{aligned}$$

Other amount will be ₹ (15,600 - 8,100) = ₹ 7,500.

Hence, ₹ 8,100 and ₹ 7,500 are the two parts.

Given:

S.I. = $\frac{16}{25}$ of P

Rate of interest = Time = $x$

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \frac{16}{25}P = \Big(\frac{P \times x \times x}{100}\Big)\\[1em] \Rightarrow \frac{16}{25}\cancel{P} = \Big(\frac{\cancel{P} \times x \times x}{100}\Big)\\[1em] \Rightarrow x^2 = \frac{16\times100}{25}\\[1em] \Rightarrow x^2 = 64\\[1em] \Rightarrow x = \sqrt{64}\\[1em] \Rightarrow x = 8\\[1em]\end{aligned}$$

Hence, rate = 8% and time = 8 years.

Let the first part be ₹ $x$ and the second part be ₹ $(9,000 - x)$.

Hence,

P₁ = ₹ $x$

R₁ = 16%

T₁ = 2 years

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big(\frac{x \times 16 \times 2}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \frac{32x}{100}\\[1em] \Rightarrow \text{S.I.} = \frac{8x}{25}\\[1em]\end{aligned}$$

P₂ = ₹ $9,000 - x$

R₂ = $6%$

T₂ = 3 years

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{S.I.} = \Big[\frac{(9,000 - x) \times 6 \times 3}{100}\Big]\\[1em] \Rightarrow \text{S.I.} = \frac{18(9,000 - x)}{100}\\[1em] \Rightarrow \text{S.I.} = \frac{9(9,000 - x)}{50}\\[1em]\end{aligned}$$

Both interest are equal.

$$\begin{aligned}\frac{8x}{25} = \frac{9(9,000 - x)}{50}\\[1em] \Rightarrow\frac{8x}{25} = \frac{81,000 - 9x}{50}\\[1em] \Rightarrow\frac{8x}{25} + \frac{9x}{50} = \frac{81,000}{50}\\[1em] \Rightarrow\frac{16x}{50} + \frac{9x}{50} = 1,620\\[1em] \Rightarrow\frac{(16x + 9x)}{50} = 1,620\\[1em] \Rightarrow\frac{25x}{50} = 1,620\\[1em] \Rightarrow x = \frac{1,620\times 50}{25}\\[1em] \Rightarrow x = \frac{81,000}{25}\\[1em] \Rightarrow x = 3,240\end{aligned}$$

Other amount will be ₹ (9,000 - 3,240) = ₹ 5,760.

Hence, ₹ 3,240 and ₹ 5,760 are the two parts.

For 1st year:

P = ₹ 1,000

R = 20%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{1,000 \times 20 \times 1}{100}\Big)\\[1em] = ₹ \frac{20,000}{100}\\[1em] = ₹ 200\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 1,000 + 200\\[1em] = ₹ 1,200\end{aligned}$$

For 2nd year:

P = ₹ 1,200

R = 20%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{1,200 \times 20 \times 1}{100}\Big)\\[1em] = ₹ \frac{24,000}{100}\\[1em] = ₹ 240\end{aligned}$$

And

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 1,200 + 240\\[1em] = ₹ 1,440\end{aligned}$$

And

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 1,440 - ₹ 1,000\\[1em] = ₹ 440\end{aligned}$$

Hence, option 3 is the correct option.

For 1st year:

P = ₹ 2,000

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{2,000 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{20,000}{100}\\[1em] = ₹ 200\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 2,000 + 200\\[1em] = ₹ 2,200\end{aligned}$$

For 2nd year:

P = ₹ 2,200

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{2,200 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{22,000}{100}\\[1em] = ₹ 220\end{aligned}$$

And

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 2,200 + 220\\[1em] = ₹ 2,420\end{aligned}$$

Hence, option 3 is the correct option.

P = ₹ 6,000

R = 8%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{6,000 \times 8 \times 1}{100}\Big)\\[1em] = ₹ \frac{48,000}{100}\\[1em] = ₹ 480\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 6,000 + 480\\[1em] = ₹ 6,480\end{aligned}$$

And

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 6,480 - ₹ 6,000\\[1em] = ₹ 480\end{aligned}$$

Difference between S.I. and C.I. = ₹ 480 - ₹ 480 = 0.

Hence, option 2 is the correct option.

For 1st year:

P = ₹ 4,000

R = 5%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{4,000 \times 5 \times 1}{100}\Big)\\[1em] = ₹ \frac{20,000}{100}\\[1em] = ₹ 200\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 4,000 + 200\\[1em] = ₹ 4,200\end{aligned}$$

For 2nd year:

P = ₹ 4,200

R = 5%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{4,200 \times 5 \times 1}{100}\Big)\\[1em] = ₹ \frac{21,000}{100}\\[1em] = ₹ 210\end{aligned}$$

And

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 4,200 + 210\\[1em] = ₹ 4,410\end{aligned}$$

And

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 4,410 - ₹ 4,000\\[1em] = ₹ 410\end{aligned}$$

Difference between the C.I. in 1 year and compound in interest in 2 years = ₹ 410 - ₹ 200 = ₹ 210

Hence, option 2 is the correct option.

(i) For 1st year:

P = ₹ 8,000

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{8,000 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{80,000}{100}\\[1em] = ₹ 800\end{aligned}$$

Hence, interest for first year = ₹ 800.

(ii)

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 8,000 + 800\\[1em] = ₹ 8,800\end{aligned}$$

So, principal for the second year = ₹ 8,800.

(iii) For 2nd year:

P = ₹ 8,800

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{8,800 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{88,000}{100}\\[1em] = ₹ 880\end{aligned}$$

Hence, interest for the second year = ₹ 880.

(iv)

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 8,800 + 880\\[1em] = ₹ 9,680\end{aligned}$$

So, final amount at the end of the second year = ₹ 9,680

(v)

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 9,680 - ₹ 8,000\\[1em] = ₹ 1,680\end{aligned}$$

Hence, compound interest earned in 2 years = ₹ 1,680.

(i) For 1st year:

P = ₹ 20,000

R = 8%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{20,000 \times 8 \times 1}{100}\Big)\\[1em] = ₹ \frac{1,60,000}{100}\\[1em] = ₹ 1,600\end{aligned}$$

Hence, interest for first year = ₹ 1,600.

(ii)

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 20,000 + 1,600\\[1em] = ₹ 21,600\end{aligned}$$

For 2nd year:

P = ₹ 21,600

R = 8%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{21,600 \times 8 \times 1}{100}\Big)\\[1em] = ₹ \frac{1,72,800}{100}\\[1em] = ₹ 1,728\end{aligned}$$

Hence, interest for the second year = ₹ 1,728.

(iii)

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 21,600 + 1,728\\[1em] = ₹ 23,328\end{aligned}$$

So, final amount at the end of the second year = ₹ 23,328.

(iv)

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 23,328 - ₹ 20,000\\[1em] = ₹ 3,328\end{aligned}$$

Hence, compound interest for 2 years = ₹ 3,328.

Given:

P = ₹ 12,000

R = 10%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 12,000\Big[1 + \frac{10}{100}\Big]^2\\[1em] = 12,000\Big[1 + \frac{1}{10}\Big]^2\\[1em] = 12,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^2\\[1em] = 12,000\Big[\frac{(10 + 1)}{10}\Big]^2\\[1em] = 12,000\Big[\frac{11}{10}\Big]^2\\[1em] = 12,000\Big[\frac{121}{100}\Big]\\[1em] = \Big[\frac{1,45,200}{1000}\Big]\\[1em] = 14,520\end{aligned}$$

Also

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 14,520 - ₹ 12,000\\[1em] = ₹ 2,520\end{aligned}$$

Hence, amount = ₹ 14,520 compound interest = ₹ 2,520.

Given:

P = ₹ 10,000

R = 8%

n = 3 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 10,000\Big[1 + \frac{8}{100}\Big]^3\\[1em] = 10,000\Big[1 + \frac{2}{25}\Big]^3\\[1em] = 10,000\Big[\frac{25}{25} + \frac{2}{25}\Big]^3\\[1em] = 10,000\Big[\frac{(25 + 2)}{25}\Big]^3\\[1em] = 10,000\Big[\frac{27}{25}\Big]^3\\[1em] = 10,000\Big[\frac{19683}{15625}\Big]\\[1em] = \Big[\frac{19,68,30,000}{15625}\Big]\\[1em] = 12,597.12\end{aligned}$$

Also

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 12,597.12 - ₹ 10,000\\[1em] = ₹ 2,597.12\end{aligned}$$

Hence, amount = ₹ 12,597.12 compound interest = ₹ 2,597.12.

Given:

P = ₹ 5,000

T = 2 years

R₁ = 10%

R₂ = 12%

As we know,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R_1}{100}\Big]\Big[1 + \frac{R_2}{100}\Big]\\[1em] = 5,000\Big[1 + \frac{10}{100}\Big]\Big[1 + \frac{12}{100}\Big]\\[1em] = 5,000\Big[1 + \frac{1}{10}\Big]\Big[1 + \frac{3}{25}\Big]\\[1em] = 5,000\Big[\frac{10}{10} + \frac{1}{10}\Big]\Big[\frac{25}{25} + \frac{3}{25}\Big]\\[1em] = 5,000\Big[\frac{(10 + 1)}{10}\Big]\Big[\frac{(25 + 3)}{25}\Big]\\[1em] = 5,000\Big[\frac{11}{10}\Big]\Big[\frac{28}{25}\Big]\\[1em] = \Big[\frac{15,40,000}{250}\Big]\\[1em] = ₹ 6,160\end{aligned}$$

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ 6,160 - ₹ 5,000\\[1em] = ₹ 1,160\end{aligned}$$

Hence, compound interest = ₹ 1,160

Given:

P = ₹ 15,000

T = 3 years

R₁ = 6%

R₂ = 8%

R₃ = 10%

As we know,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R_1}{100}\Big]\Big[1 + \frac{R_2}{100}\Big]\Big[1 + \frac{R_3}{100}\Big]\\[1em] = 15,000\Big[1 + \frac{6}{100}\Big]\Big[1 + \frac{8}{100}\Big]\Big[1 + \frac{10}{100}\Big]\\[1em] = 15,000\Big[1 + \frac{3}{50}\Big]\Big[1 + \frac{2}{25}\Big]\Big[1 + \frac{1}{10}\Big]\\[1em] = 15,000\Big[\frac{50}{50} + \frac{3}{50}\Big]\Big[\frac{25}{25} + \frac{2}{25}\Big]\Big[\frac{10}{10} + \frac{1}{10}\Big]\\[1em] = 15,000\Big[\frac{(50 + 3)}{50}\Big]\Big[\frac{(25 + 2)}{25}\Big]\Big[\frac{(10 + 1)}{10}\Big]\\[1em] = 15,000\Big[\frac{53}{50}\Big]\Big[\frac{27}{25}\Big]\Big[\frac{11}{10}\Big]\\[1em] = \Big[\frac{23,61,15,000}{12500}\Big]\\[1em] = 18,889.20\end{aligned}$$

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ 18,889.20 - ₹ 15,000\\[1em] = ₹ 3,889.20\end{aligned}$$

Hence, compound interest = ₹ 3,889.20

Given:

P = ₹ 16,000

n = 3 years

R = 5%

As we know,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 16,000\Big[1 + \frac{5}{100}\Big]^3\\[1em] = 16,000\Big[1 + \frac{1}{20}\Big]^3\\[1em] = 16,000\Big[\frac{20}{20} + \frac{1}{20}\Big]^3\\[1em] = 16,000\Big[\frac{(20 + 1)}{20}\Big]^3\\[1em] = 16,000\Big[\frac{21}{20}\Big]^3\\[1em] = 16,000\Big[\frac{9261}{8000}\Big]\\[1em] = \Big[\frac{14,81,76,000}{8000}\Big]\\[1em] = 18,522\end{aligned}$$

Hence, amount = ₹ 18,522

For 1st year:

P = ₹ 40,000

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{40,000 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{4,00,000}{100}\\[1em] = ₹ 4,000\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 40,000 + 4,000\\[1em] = ₹ 44,000\end{aligned}$$

For 2nd year:

P = ₹ 44,000

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{44,000 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{4,40,000}{100}\\[1em] = ₹ 4,400\end{aligned}$$

Hence, compound interest = ₹ 4,400.

For 1st year:

P = ₹ 15,000

R = 6%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{15,000 \times 6 \times 1}{100}\Big)\\[1em] = ₹ \frac{90,000}{100}\\[1em] = ₹ 900\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 15,000 + 900\\[1em] = ₹ 15,900\end{aligned}$$

For 2nd year:

P = ₹ 15,900

R = 6%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{15,900 \times 6 \times 1}{100}\Big)\\[1em] = ₹ \frac{954,000}{100}\\[1em] = ₹ 954\end{aligned}$$

Hence, compound interest = ₹ 954.

(i) For 1st year:

P = ₹ 9,600

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{9,600 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{96,000}{100}\\[1em] = ₹ 960\end{aligned}$$

Hence, interest for first year = ₹ 960.

(ii)

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 9,600 + 960\\[1em] = ₹ 10,560\end{aligned}$$

So, amount at the end of the first year = ₹ 10,560.

(iii) For 2nd year:

P = ₹ 10,560

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{10,560 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{1,05,600}{100}\\[1em] = ₹ 1,056\end{aligned}$$

Hence, interest for the second year = ₹ 1,056.

(iv)

$$\begin{aligned}\text{Amount at end of 2nd year = P + Interest}\\[1em] = ₹ 10,560 + 1,056\\[1em] = ₹ 11,616\end{aligned}$$

For 3rd year:

P = ₹ 11,616

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{11,616 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{1,16,160}{100}\\[1em] = ₹ 1,161.60\end{aligned}$$

Hence, interest for the third year = ₹ 1,161.60.

(i) For 1st year:

P = ₹ 5,000

T = 1 year

A = ₹ 5,600

As we know,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow 5,600 = 5,000 + S.I.\\[1em] \Rightarrow S.I. = 5,600 - 5,000\\[1em] \Rightarrow S.I. = 600\end{aligned}$$

Let the rate be $r$

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 600 = ₹ \Big(\frac{5,000 \times r \times 1}{100}\Big)\\[1em] \Rightarrow 600 = ₹ \Big(\frac{5,000r}{100}\Big)\\[1em] \Rightarrow 600 = 50r\\[1em] \Rightarrow r = \frac{600}{50}\\[1em] \Rightarrow r = 12\end{aligned}$$

Hence, rate of interest = 12%.

(ii) For 2nd year:

P = ₹ 5,600

R = 12%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{5,600 \times 12 \times 1}{100}\Big)\\[1em] = ₹ \frac{67,200}{100}\\[1em] = ₹ 672\end{aligned}$$

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ (5,600 + 672) \\[1em] = ₹ 6,272\end{aligned}$$

P = ₹ 7,500

R = 8%

T = 2 years

For simple interest

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{7,500 \times 8 \times 2}{100}\Big)\\[1em] = \Big(\frac{1,20,000}{100}\Big)\\[1em] = ₹ 1,200\end{aligned}$$

For compound interest

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 7,500\Big[1 + \frac{8}{100}\Big]^2\\[1em] = 7,500\Big[1 + \frac{2}{25}\Big]^2\\[1em] = 7,500\Big[\frac{25}{25} + \frac{2}{25}\Big]^2\\[1em] = 7,500\Big[\frac{(25 + 2)}{25}\Big]^2\\[1em] = 7,500\Big[\frac{27}{25}\Big]^2\\[1em] = 7,500\Big[\frac{729}{625}\Big]^2\\[1em] = \Big[\frac{54,67,500}{625}\Big]^2\\[1em] = ₹ 8,748\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ (8,748 - 7,500)\\[1em] = ₹ 1,248\end{aligned}$$

Difference between C.I. and S.I.

= ₹ (1,248 - 1,200)

= ₹ 48

The difference between C.I. and S.I. = ₹ 48.

P = ₹ 8,000

R = 10%

T = 3 years

For simple interest

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{8,000 \times 10 \times 3}{100}\Big)\\[1em] = \Big(\frac{2,40,000}{100}\Big)\\[1em] = ₹ 2,400\end{aligned}$$

For compound interest

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 8,000\Big[1 + \frac{10}{100}\Big]^3\\[1em] = 8,000\Big[1 + \frac{1}{10}\Big]^3\\[1em] = 8,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^3\\[1em] = 8,000\Big[\frac{(10 + 1)}{10}\Big]^3\\[1em] = 8,000\Big[\frac{11}{10}\Big]^3\\[1em] = 8,000\Big[\frac{1,331}{1000}\Big]\\[1em] = \Big[\frac{10,64,80,000}{1000}\Big]\\[1em] = ₹ 10,648\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P} \\[1em] = ₹ (10,648 - 8,000) \\[1em] = ₹ 2,648\end{aligned}$$

Difference between C.I. and S.I.

= ₹ (2,648 - 2,400)

= ₹ 248

The difference between C.I. and S.I. = ₹ 248.

For simple interest

P = ₹ 40,000

R = 10.5%

T = 2 years

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{40,000 \times 10.5 \times 2}{100}\Big)\\[1em] = \Big(\frac{8,40,000}{100}\Big)\\[1em] = ₹ 8,400\end{aligned}$$

For compound interest

P = ₹ 40,000

R = 10%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 40,000\Big[1 + \frac{10}{100}\Big]^2\\[1em] = 40,000\Big[1 + \frac{1}{10}\Big]^2\\[1em] = 40,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^2\\[1em] = 40,000\Big[\frac{(10 + 1)}{10}\Big]^2\\[1em] = 40,000\Big[\frac{11}{10}\Big]^2\\[1em] = 40,000\Big[\frac{121}{100}\Big]\\[1em] = \Big[\frac{48,40,000}{100}\Big]\\[1em] = ₹ 48,400\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ (48,400 - 40,000)\\[1em] = ₹ 8,400\end{aligned}$$

∴ C.I. is equals to S.I.

Hence, Both give equal interest.

For simple interest

P = ₹ 24,000

R = 13%

T = 2 years

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{24,000 \times 13 \times 2}{100}\Big)\\[1em] = \Big(\frac{6,24,000}{100}\Big)\\[1em] = ₹ 6,240\end{aligned}$$

For compound interest

P = ₹ 24,000

R = 12%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 24,000\Big[1 + \frac{12}{100}\Big]^2\\[1em] = 24,000\Big[1 + \frac{3}{25}\Big]^2\\[1em] = 24,000\Big[\frac{25}{25} + \frac{3}{25}\Big]^2\\[1em] = 24,000\Big[\frac{(25 + 3)}{25}\Big]^2\\[1em] = 24,000\Big[\frac{28}{25}\Big]^2\\[1em] = 24,000\Big[\frac{784}{625}\Big]\\[1em] = \Big[\frac{1,88,16,000}{625}\Big]\\[1em] = ₹ 30,105.60\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ (30,105.6 - 24,000)\\[1em] = ₹ 6,105.60\end{aligned}$$

Total interest = ₹ 6,240 + ₹ 6,105.60 = ₹ 12,345.60

Hence, total interest earned by Mr. Sharma in 2 years = ₹ 12,345.60

For 1st year:

P = ₹ 12,000

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{12,000 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{1,20,000}{100}\\[1em] = ₹ 1,200\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 12,000 + 1,200\\[1em] = ₹ 13,200\end{aligned}$$

Hence, the amount at the end of first year, before making the repayment = ₹ 13,200

(ii) The amount at the end of first year, after making the repayment = ₹ 13,200 - ₹ 8,000

= ₹ 5,200

Hence, the amount at the end of first year, after making the repayment = ₹ 5,200

(iii) So, the principal amount for second year = ₹ 5,200

(iv) For 2nd year:

P = ₹ 5,200

R = 10%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{5,200 \times 10 \times 1}{100}\Big)\\[1em] = ₹ \frac{52,000}{100}\\[1em] = ₹ 520\end{aligned}$$

And

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 5,200 + 520\\[1em] = ₹ 5,720\end{aligned}$$

Hence, the amount to be paid at the end of the second year to clear the account = ₹ 5,720.

For 1st year:

P = ₹ 16,000

R = 15%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{16,000 \times 15 \times 1}{100}\Big)\\[1em] = ₹ \frac{2,40,000}{100}\\[1em] = ₹ 2,400\end{aligned}$$

And

$$\begin{aligned}\text{Amount = P + Interest}\\[1em] = ₹ 16,000 + 2,400\\[1em] = ₹ 18,400\end{aligned}$$

The amount at the end of first year, after making the repayment = ₹ 18,400 - ₹ 9,000

= ₹ 9,400

Hence, the amount at the end of first year, after making the repayment = ₹ 9,400

For 2nd year:

P = ₹ 9,400

R = 15%

T = 1 year

$$\begin{aligned}\text{Interest} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = ₹ \Big(\frac{9,400 \times 15 \times 1}{100}\Big)\\[1em] = ₹ \frac{141,000}{100}\\[1em] = ₹ 1,410\end{aligned}$$

And

$$\begin{aligned}\text{Final amount = P + Interest}\\[1em] = ₹ 9,400 + 1,410\\[1em] = ₹ 10,810\end{aligned}$$

Hence, the amount to be paid at the end of the second year to clear the account = ₹ 10,810.

(i) For simple interest

T = 5 years

R = 8%

S.I. = ₹ 12,000

Let the principal amount be $P$.

As we know,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R\times T}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\frac{P\times 8 \times 5}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\frac{40P}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\frac{2P}{5}\Big)\\[1em] \Rightarrow 12,000 = \frac{2P}{5}\\[1em] \Rightarrow P = \frac{5 \times 12,000}{2} \\[1em] \Rightarrow P = \frac{60,000}{2} \\[1em] \Rightarrow P = 30,000\end{aligned}$$

Hence, the principal amount = ₹ 30,000.

(ii) For compound interest

P = ₹ 30,000

R = 10%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] \Rightarrow\text{A} = 30,000\Big[1 + \frac{10}{100}\Big]^2\\[1em] = 30,000\Big[1 + \frac{1}{10}\Big]^2\\[1em] = 30,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^2\\[1em] = 30,000\Big[\frac{(10 + 1)}{10}\Big]^2\\[1em] = 30,000\Big[\frac{11}{10}\Big]^2\\[1em] = 30,000\Big[\frac{121}{100}\Big]\\[1em] = \Big[\frac{36,30,000}{100}\Big]\\[1em] = 36,300\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] \Rightarrow C.I. = 36,300 - 30,000\\[1em] = 6,300\end{aligned}$$

Hence, the compound interest ₹ 6,300

Given:

P = ₹ 12,000

R = 10%

n = 1 years

When the interest is compounded half-yearly:

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{2 \times 100}\Big]^{2\times n}\\[1em] = 12,000\Big[1 + \frac{10}{200}\Big]^{2\times1}\\[1em] = 12,000\Big[1 + \frac{1}{20}\Big]^2\\[1em] = 12,000\Big[\frac{20}{20} + \frac{1}{20}\Big]^2\\[1em] = 12,000\Big[\frac{(20 + 1)}{20}\Big]^2\\[1em] = 12,000\Big[\frac{21}{20}\Big]^2\\[1em] = 12,000\Big[\frac{441}{400}\Big]\\[1em] = \Big[\frac{52,92,000}{400}\Big]\\[1em] = 13,230\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] \Rightarrow \text{C.I.} = 13,230 - 12,000\\[1em] = 1,230\end{aligned}$$

Hence, the amount = ₹ 13,230 and the compound interest ₹ 1,230

Given:

P = ₹ 8,000

R = 20%

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

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

When the interest is compounded half-yearly:

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{2 \times 100}\Big]^{2\times n}\\[1em] = 8,000\Big[1 + \frac{20}{200}\Big]^{2\times\frac{3}{2}}\\[1em] = 8,000\Big[1 + \frac{1}{10}\Big]^3\\[1em] = 8,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^3\\[1em] = 8,000\Big[\frac{(10 + 1)}{10}\Big]^3\\[1em] = 8,000\Big[\frac{11}{10}\Big]^3\\[1em] = 8,000\Big[\frac{1331}{1000}\Big]\\[1em] = \Big[\frac{1,06,48,000}{1000}\Big]\\[1em] = 10,648\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] \Rightarrow \text{C.I.} = 10,648 - 8,000\\[1em] = 2,648\end{aligned}$$

Hence, the amount = ₹ 10,648 and the compound interest = ₹ 2,648

Given:

P = ₹ 24,000

R = 10%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^{n}\\[1em] \Rightarrow\text{A} = 24,000\Big[1 + \frac{10}{100}\Big]^2\\[1em] = 24,000\Big[1 + \frac{1}{10}\Big]^2\\[1em] = 24,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^2\\[1em] = 24,000\Big[\frac{(10 + 1)}{10}\Big]^2\\[1em] = 24,000\Big[\frac{11}{10}\Big]^2\\[1em] = 24,000\Big[\frac{121}{100}\Big]\\[1em] = \Big[\frac{29,04,000}{100}\Big]\\[1em] = 29,040\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] \Rightarrow \text{C.I.} = 29,040 - 24,000\\[1em] = 5,040\end{aligned}$$

Hence, the amount = ₹ 29,040 and the compound interest = ₹ 5,040

Given:

T = 3 years

R = 8%

S.I. = ₹ 720

Let the principal be $P$.

As we know,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 720 = \Big(\frac{P \times 8 \times 3}{100}\Big)\\[1em] \Rightarrow 720 = \Big(\frac{24P}{100}\Big)\\[1em] \Rightarrow 720 = \Big(\frac{6P}{25}\Big)\\[1em] \Rightarrow P = \Big(\frac{25 \times 720}{6}\Big)\\[1em] \Rightarrow P = \Big(\frac{18,000}{6}\Big)\\[1em] \Rightarrow P = 3,000\end{aligned}$$

Hence, option 3 is the correct option.

Given:

T = 5 years

A = ₹ $\frac{5}{4}$ of P

Let the principal be $P$ and the rate be $r$.

As we know,

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow\frac{5}{4}P = P + S.I.\\[1em] \Rightarrow\text{S.I.} = \frac{5}{4}P - P\\[1em] \Rightarrow\text{S.I.} = \frac{5}{4}P - \frac{4}{4}P\\[1em] \Rightarrow\text{S.I.} = \frac{(5 - 4)}{4}P\\[1em] \Rightarrow\text{S.I.} = \frac{1}{4}P\end{aligned}$$

And

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \frac{1}{4} \times P = \Big(\frac{P \times r \times 5}{100}\Big)\\[1em] \Rightarrow \frac{1}{4}P = \Big(\frac{5Pr}{100}\Big)\\[1em] \Rightarrow \frac{1}{4}\cancel{P} = \Big(\frac{5r\cancel{P}}{100}\Big)\\[1em] \Rightarrow \frac{1}{4} = \Big(\frac{5r}{100}\Big)\\[1em] \Rightarrow \frac{1}{4} = \Big(\frac{r}{20}\Big)\\[1em] \Rightarrow r = \Big(\frac{1 \times 20}{4}\Big)\\[1em] \Rightarrow r = \Big(\frac{20}{4}\Big)\\[1em] \Rightarrow r = 5\end{aligned}$$

Hence, option 2 is the correct option.

Let the principal amount be ₹ x.

Let the first part be ₹ $\frac{3}{5}$ of $x$ and the second part be ₹ $x - \frac{3}{5}x$ = $\frac{5}{5}x - \frac{3}{5}x$ = $\frac{2}{5}x$.

Hence

P₁ = ₹ $\frac{3x}{5}$

R₁ = 20%

T₁ = 1 year

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{3x \times 20 \times 1}{5 \times 100}\Big)\\[1em] = \frac{60x}{500}\\[1em] = \frac{3x}{25}\end{aligned}$$

P₂ = ₹ $\frac{2x}{5}$

R₂ = $20%$

T₂ = 1 year

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = \frac{2x}{5} \Big[1 + \frac{20}{100}\Big]^1\\[1em] = \frac{2x}{5} \Big[1 + \frac{1}{5}\Big]\\[1em] = \frac{2x}{5} \Big[\frac{5}{5} + \frac{1}{5}\Big]\\[1em] = \frac{2x}{5} \Big[\frac{(5 + 1)}{5}\Big]\\[1em] = \frac{2x}{5} \Big[\frac{6}{5}\Big]\\[1em] = \frac{12x}{25} \\[1em]\end{aligned}$$

And

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = \frac{12x}{25} - \frac{2x}{5} \\[1em] = \frac{12x}{25} - \frac{5\times 2x}{5 \times 5} \\[1em] = \frac{12x}{25} - \frac{10x}{25} \\[1em] = \frac{(12 - 10)x}{25} \\[1em] = \frac{2x}{25} \\[1em]\end{aligned}$$

Total amount = ₹ 1,000

$$\begin{aligned}\frac{3x}{25} + \frac{2x}{25} = 1,000\\[1em] \Rightarrow\frac{(3x + 2x)}{25} = 1,000\\[1em] \Rightarrow\frac{5x}{25} = 1,000\\[1em] \Rightarrow\frac{x}{5} = 1,000\\[1em] \Rightarrow x = 1,000 \times 5\\[1em] \Rightarrow x = 5,000\end{aligned}$$

Hence, option 4 is the correct option.

Given:

P = ₹ 1,000

R = 5%

n = 2 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 1,000\Big[1 + \frac{5}{100}\Big]^2\\[1em] = 1,000\Big[1 + \frac{1}{20}\Big]^2\\[1em] = 1,000\Big[\frac{20}{20} + \frac{1}{20}\Big]^2\\[1em] = 1,000\Big[\frac{(20 + 1)}{20}\Big]^2\\[1em] = 1,000\Big[\frac{21}{20}\Big]^3\\[1em] = 1,000\Big[\frac{441}{400}\Big]\\[1em] = \Big[\frac{4,41,000}{400}\Big]\\[1em] = 1,102.50\end{aligned}$$

Hence, option 2 is the correct option.

When the interest is compounded half-yearly:

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{2 \times 100}\Big]^{2\times n}\\[1em]\end{aligned}$$

2 x n means time is doubled.

Hence, option 2 is the correct option.

On a certain sum, at the same rate of interest and for the same time period,

For a time period of one year, compound interest (CI) = simple interest (SI).

But for more than one year, CI is always greater than SI, because interest is added to the principal after each compounding period, increasing the amount on which interest is calculated.

On a certain sum, at the same rate of interest and for the same time period, compound interest is always greater or equal to the simple interest.

So, statement 1 is false.

In compound interest, the principal does not remain constant — it increases after each compounding period because interest is added to it.

So, statement 2 is false.

Hence, option 2 is the correct option.

Let the principal be ₹ P.

S.I. = $\frac{9}{16}$ x P

Rate of interest = a%

Time = a years

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{9}{16}P = \frac{P \times a \times a}{100}\\[1em] \Rightarrow \frac{9}{16} \cancel{P} = \frac{\cancel{P} \times a \times a}{100}\\[1em] \Rightarrow \frac{9}{16} = \frac{a \times a}{100}\\[1em] \Rightarrow \frac{9}{16} = \frac{a^2}{100}\\[1em] \Rightarrow a^2 = \frac{9 \times 100}{16}\\[1em] \Rightarrow a^2 = \frac{900}{16}\\[1em] \Rightarrow a = \sqrt{\frac{900}{16}}\\[1em] \Rightarrow a = \frac{30}{4}\\[1em] \Rightarrow a = \frac{15}{2}\\[1em] \Rightarrow a = 7\frac{1}{2}\end{aligned}$$

Time = $7\frac{1}{2}$ years.

So, assertion (A) is true.

By formula;

$$\begin{aligned}\Rightarrow S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow T = \frac{\text{S.I.} \times 100}{P \times R} \\[1em] \Rightarrow \text{Time } = \frac{\text{Simple Interest} \times 100}{\text{Principal} \times \text{Rate}}\end{aligned}$$

So, reason (R) is true and reason (R) clearly explains assertion (A).

Hence, option 1 is the correct option.

We know that,

If time is greater than 1 year, then compound interest on the same sum at the same rate of interest for same time will always be greater than simple interest.

∴ If simple interest on ₹ 15,000 in 2 years at 6% p.a. is ₹ 1,800. Then compound interest on the same sum at the same rate of interest for 2 years will never be less than ₹ 1,800.

So, assertion (A) is true.

For a given principal, rate and time, both simple interest and compound interest are equal for the 1^st year.

So, reason (R) is true and reason (R) does not explains assertion (A).

Hence, option 2 is the correct option.

Given,

P = ₹ 8,000

R = 5%

T = 2 years

By formula,

$$\begin{aligned}\text{Amount }= P\Big(1 + \frac{R}{100}\Big)^n \\[1em] \text{Amount after one year }= 8000\Big(1 + \frac{5}{100}\Big)^1\\[1em] = 8000(1 + 0.05)\\[1em] = 8000 \times (1.05)\\[1em] = ₹8,400.\end{aligned}$$

Amount at the end of first year = ₹ 8,400

For 2nd year :

P = ₹ 8,400

R = 5%

T = 1 year

$$\begin{aligned}\text{Interest for 2nd year} = \frac{P \times R \times T}{100} \\[1em] = \frac{8,400 \times 5 \times 1}{100} \\[1em] = ₹ 420.\end{aligned}$$

∴ Compound interest for the 2^nd year on ₹ 8,000 at 5% p.a. is ₹ 420.

So, assertion (A) is false.

By formula,

Compound interest for 2 years = Amount at the end of 2^nd year - original sum.

So, reason (R) is true.

Hence, option 4 is the correct option.

In simple interest, the interest is calculated only on the original principal and remains the same every year.

Thus, SI for the 1st year is equal to SI for the 4th year—in fact, it's equal for every year.

So, assertion (A) is true.

Amount at the end of 4^th year - original sum = Total simple interest for 4 years

That formula gives the total simple interest for 4 years, not just for the 4th year.

So, reason (R) is false.

Hence, option 3 is the correct option.

For Mohan,

P = ₹ 4,800

T = $4\frac{1}{2}$ years

= $\frac{9}{2}$ years

Let the rate be $r$

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{4,800 \times r \times 9}{2 \times 100}\Big)\\[1em] = \Big(\frac{43,200r}{200}\Big)\\[1em] = 216r\end{aligned}$$

For Shyam,

P = ₹ 2,500

T = 6 years

Let the rate be $r$

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{2,500 \times r \times 6}{100}\Big)\\[1em] = \Big(\frac{15,000r}{100}\Big)\\[1em] = 150r\end{aligned}$$

Total interest = ₹ 2,196

₹ 216r + ₹ 150r = ₹ 2,196

₹ 366r = ₹ 2,196

r = $\frac{2196}{366}$

r = $6%$

Hence, the rate per cent per annum = 6%.

Given:

P = ₹ 2,550

R = 7.5%

T = 8 months = $\frac{8}{12}$ years

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

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{2,550 \times 7.5 \times 2}{3 \times 100}\Big)\\[1em] = \Big(\frac{38,250}{300}\Big)\\[1em] = 127.50\end{aligned}$$

And

$$\begin{aligned}\text{A = P + S.I.}\\[1em] \Rightarrow \text{A} = 2,550 + 127.50\\[1em] \Rightarrow \text{A} = 2,677.50\end{aligned}$$

Mohan paid in cash = ₹ 1,422.50

Price of the television = Amount - Paid in cash

= ₹ 2,677.50 - ₹ 1,422.50

= ₹ 1,255

Hence, the cost of television = ₹ 1,255

Let the first part be ₹ $x$ and the second part be ₹ $(10,800 - x)$.

Hence

P₁ = ₹ $x$

R₁ = 18%

T₁ = 1 year

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] = \Big(\frac{x \times 18 \times 1}{100}\Big)\\[1em] = \frac{18x}{100}\\[1em] = \frac{9x}{50}\\[1em]\end{aligned}$$

P₂ = ₹ $10,800 - x$

R₂ = $20%$

T₂ = 1 year

$$\begin{aligned}\text{S.I.} = \Big[\frac{P \times R \times T}{100}\Big]\\[1em] = \Big[\frac{(10,800 - x) \times 20 \times 1}{100}\Big]\\[1em] = \frac{20(10,800 - x)}{100}\\[1em] = \frac{(10,800 - x)}{5}\\[1em]\end{aligned}$$

Total amount = ₹ 2,060

$$\begin{aligned}\Rightarrow \frac{9x}{50} + \frac{(10,800 - x)}{5} = 2,060\\[1em] \Rightarrow \frac{9x}{50} + \frac{10(10,800 - x)}{50} = 2,060\\[1em] \Rightarrow \frac{9x}{50} + \frac{(1,08,000 - 10x)}{50} = 2,060\\[1em] \Rightarrow \frac{(9x + 1,08,000 - 10x)}{50} = 2,060\\[1em] \Rightarrow 1,08,000 - x = 2,060 \times 50\\[1em] \Rightarrow 1,08,000 - x = 1,03,000\\[1em] \Rightarrow x = 1,08,000 - 1,03,000\\[1em] \Rightarrow x = 5,000\\[1em]\end{aligned}$$

Other part = ₹ 10,800 - ₹ 5,000

= ₹ 5,800

Hence, the two parts are ₹ 5,000 at 18% and ₹ 5,800 at 20%.

Given:

P = ₹ 16,000

R = 5%

n = 3 years

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{100}\Big]^n\\[1em] = 16,000\Big[1 + \frac{5}{100}\Big]^3\\[1em] = 16,000\Big[1 + \frac{1}{20}\Big]^3\\[1em] = 16,000\Big[\frac{20}{20} + \frac{1}{20}\Big]^3\\[1em] = 16,000\Big[\frac{(20 + 1)}{20}\Big]^3\\[1em] = 16,000\Big[\frac{21}{20}\Big]^3\\[1em] = 16,000\Big[\frac{9,261}{8,000}\Big]\\[1em] = \Big[\frac{14,81,76,000}{8,000}\Big]\\[1em] = 18,522\end{aligned}$$

Also

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 18,522 - ₹ 16,000\\[1em] = ₹ 2,522\end{aligned}$$

Hence, amount = ₹ 18,522 compound interest = ₹ 2,522.

Given:

P = ₹ 20,000

R = 10%

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

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

Since, interest is compounded half-yearly,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{2 \times 100}\Big]^{n \times 2}\\[1em] = 20,000\Big[1 + \frac{10}{200}\Big]^{2\times \frac{3}{2}}\\[1em] = 20,000\Big[1 + \frac{1}{20}\Big]^3\\[1em] = 20,000\Big[\frac{20}{20} + \frac{1}{20}\Big]^3\\[1em] = 20,000\Big[\frac{(20 + 1)}{20}\Big]^3\\[1em] = 20,000\Big[\frac{21}{20}\Big]^3\\[1em] = 20,000\Big[\frac{9,261}{8,000}\Big]\\[1em] = \Big[\frac{18,52,20,000}{8,000}\Big]\\[1em] = 23,152.50\end{aligned}$$

Also

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 23,152.50 - ₹ 20,000\\[1em] = ₹ 3,152.50\end{aligned}$$

Hence, amount = ₹ 23,152.50 and compound interest = ₹ 3,152.50.

Given:

P = ₹ 32,000

R = 20%

n = 1 year

Since, interest is compounded half-yearly,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R}{2 \times 100}\Big]^{n \times 2}\\[1em] = 32,000\Big[1 + \frac{20}{200}\Big]^{2\times 1}\\[1em] = 32,000\Big[1 + \frac{1}{10}\Big]^2\\[1em] = 32,000\Big[\frac{10}{10} + \frac{1}{10}\Big]^2\\[1em] = 32,000\Big[\frac{(10 + 1)}{10}\Big]^2\\[1em] = 32,000\Big[\frac{11}{10}\Big]^2\\[1em] = 32,000\Big[\frac{121}{100}\Big]\\[1em] = \Big[\frac{38,72,000}{100}\Big]\\[1em] = 38,720\end{aligned}$$

Also

$$\begin{aligned}\text{Compound Interest = Final amount - Original Principal}\\[1em] = ₹ 38,720 - ₹ 32,000\\[1em] = ₹ 6,720\end{aligned}$$

Hence, amount = ₹ 38,720 and compound interest = ₹ 6,720.

Given:

P = ₹ 4,000

T = 2 years

R₁ = 10%

R₂ = 15%

As we know, $$\begin{aligned}\text{A} = P\Big[1 + \frac{R_1}{100}\Big]\Big[1 + \frac{R_2}{100}\Big]\\[1em] = 4,000\Big[1 + \frac{10}{100}\Big]\Big[1 + \frac{15}{100}\Big]\\[1em] = 4,000\Big[1 + \frac{1}{10}\Big]\Big[1 + \frac{3}{20}\Big]\\[1em] = 4,000\Big[\frac{10}{10} + \frac{1}{10}\Big]\Big[\frac{20}{20} + \frac{3}{20}\Big]\\[1em] = 4,000\Big[\frac{(10 + 1)}{10}\Big]\Big[\frac{(20 + 3)}{20}\Big]\\[1em] = 4,000\Big[\frac{11}{10}\Big]\Big[\frac{23}{20}\Big]\\[1em] = \Big[\frac{10,12,000}{200}\Big]\\[1em] = ₹ 5,060\end{aligned}$$

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ 5,060 - ₹ 4,000\\[1em] = ₹ 1,060\end{aligned}$$

Hence, the amount = ₹ 5,060 and the compound interest = ₹ 1,060.

Given:

P = ₹ 10,000

T = 3 years

R₁ = 10%

R₂ = 15%

R₃ = 20%

As we know,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{R_1}{100}\Big]\Big[1 + \frac{R_2}{100}\Big]\Big[1 + \frac{R_3}{100}\Big]\\[1em] = 10,000\Big[1 + \frac{10}{100}\Big]\Big[1 + \frac{15}{100}\Big]\Big[1 + \frac{20}{100}\Big]\\[1em] = 10,000\Big[1 + \frac{1}{10}\Big]\Big[1 + \frac{3}{20}\Big]\Big[1 + \frac{1}{5}\Big]\\[1em] = 10,000\Big[\frac{10}{10} + \frac{1}{10}\Big]\Big[\frac{20}{20} + \frac{3}{20}\Big]\Big[\frac{5}{5} + \frac{1}{5}\Big]\\[1em] = 10,000\Big[\frac{(10 + 1)}{10}\Big]\Big[\frac{(20 + 3)}{20}\Big]\Big[\frac{(5 + 1)}{5}\Big]\\[1em] = 10,000\Big[\frac{11}{10}\Big]\Big[\frac{23}{20}\Big]\Big[\frac{6}{5}\Big]\\[1em] = \Big[\frac{1,51,80,000}{1000}\Big]\\[1em] = 15,180\end{aligned}$$

$$\begin{aligned}\text{C.I. = A - P}\\[1em] = ₹ 15,180 - ₹ 10,000\\[1em] = ₹ 5,180\end{aligned}$$

Hence, the amount = ₹ 15,180 and the compound interest = ₹ 5,180.

Given:

Amount in 2 years = ₹ 3,224

⇒ P + S.I. of 2 years = = ₹ 3,224 ..........(1)

Amount in 5 years = ₹ 4,160

⇒ P + S.I. of 5 years = ₹ 4,160 ..........(2)

Subtracting equation (1) from (2), we get

S.I. of 3 years = ₹ 4,160 - ₹ 3,224 = ₹ 936

S.I. of 1 year = ₹ $\frac{936}{3}$ = $₹ 312$

S.I. of 2 years = $₹ 312 \times 2$ = $₹ 624$

From equation (1), we get:

P + ₹ 624 = ₹ 3,224

P = ₹ 3,224 - ₹ 624

P = ₹ 2,600

Now when P = ₹ 2,600

S.I. = ₹ 312

T = 1 year

Let the rate be $r$.

As we know,

$$\begin{aligned}\text{S.I.} = \Big(\frac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow 312 = \Big(\frac{2,600 \times r \times 1}{100}\Big)\\[1em] \Rightarrow 312 = \frac{2,600r}{100}\\[1em] \Rightarrow 312 = 26r\\[1em] \Rightarrow r = \frac{312}{26}\\[1em] \Rightarrow r = 12\end{aligned}$$

Hence, principal = $₹ 2,600$ and rate = $12%$

Given:

P = ₹ 5,000

n = 2 years

A = ₹ 5,832

Let the rate be $r$.

As we know,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{r}{100}\Big]^n\\[1em] \Rightarrow 5,832 = 5,000\Big[1 + \frac{r}{100}\Big]^2\\[1em] \Rightarrow \frac{5,832}{5000} = \Big[1 + \frac{r}{100}\Big]^2\\[1em] \Rightarrow \frac{729}{625} = \Big[1 + \frac{r}{100}\Big]^2\\[1em] \Rightarrow \sqrt{\frac{729}{625}} = \Big[1 + \frac{r}{100}\Big]\\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{27}{25}\\[1em] \Rightarrow \frac{r}{100} = \frac{27}{25} - 1\\[1em] \Rightarrow \frac{r}{100} = \frac{27}{25} - \frac{25}{25}\\[1em] \Rightarrow \frac{r}{100} = \frac{(27 - 25)}{25}\\[1em] \Rightarrow \frac{r}{100} = \frac{2}{25}\\[1em] \Rightarrow r = \frac{2 \times 100}{25}\\[1em] \Rightarrow r = \frac{200}{25}\\[1em] \Rightarrow r = 8\end{aligned}$$

Hence, the rate of interest = $8%$.

Given:

P = ₹ 16,000

R = 10%

A = ₹ 18,522

Let the time be $n$.

Since the interest is compounded half-yearly,

$$\begin{aligned}\text{A} = P\Big[1 + \frac{r}{2 \times 100}\Big]^{2\times n}\\[1em] \Rightarrow 18,522 = 16,000\Big[1 + \frac{10}{200}\Big]^{2\times n}\\[1em] \Rightarrow \frac{18,522}{16,000} = \Big[1 + \frac{10}{200}\Big]^{2\times n}\\[1em] \Rightarrow \frac{18,522}{16,000} = \Big[1 + \frac{1}{20}\Big]^{2\times n}\\[1em] \Rightarrow \frac{18,522}{16,000} = \Big[\frac{20}{20} + \frac{1}{20}\Big]^{2\times n}\\[1em] \Rightarrow \frac{18,522}{16,000} = \Big[\frac{(20 + 1)}{20}\Big]^{2\times n}\\[1em] \Rightarrow \frac{18,522}{16,000} = \Big[\frac{21}{20}\Big]^{2\times n}\\[1em] \Rightarrow \frac{9,261}{8,000} = \Big[\frac{21}{20}\Big]^{2\times n}\\[1em] \Rightarrow \Big[\frac{21}{20}\Big]^3 = \Big[\frac{21}{20}\Big]^{2\times n}\\[1em]\end{aligned}$$

Hence, $2 \times n = 3$

$n = \frac{3}{2}$ years

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

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