Compound Interest

Principal for first year = ₹8000.

Interest for the first year = ₹ $\frac{8000 \times 5 \times 1}{100}$ = ₹400.

Amount after one year = ₹8000 + ₹400 = ₹8400.

Principal for the second year = ₹8400.

Interest for the second year = ₹ $\frac{8400 \times 5 \times 1}{100}$ = ₹420.

Amount after 2 years = ₹8400 + ₹420 = ₹8820.

Compound interest for 2 years = Final amount - Principal = ₹8820 - ₹8000 = ₹820.

Hence, the amount and compound interest on ₹8000 at 5% per annum after 2 years is ₹8820 and ₹820 respectively.

(i) Principal for first year = ₹46875.

Interest for the first year = ₹ $\frac{46875 \times 4 \times 1}{100} = \frac{187500}{100}$ = ₹1875.

Hence, the interest for the first year = ₹1875.

(ii) Amount after one year = ₹46875 + ₹1875 = ₹48750.

Principal for the second year = ₹48750.

Interest for the second year = ₹ $\frac{48750 \times 4 \times 1}{100} = \frac{195000}{100}$ = ₹1950.

Amount after 2 years = ₹48750 + ₹1950 = ₹50700.

Hence, the amount standing to his credit at the end of the second year is ₹50700.

(iii) Principal for third year = ₹50700.

Interest for the third year = ₹ $\frac{50700 \times 4 \times 1}{100} = \frac{202800}{100}$ = ₹2028.

Hence, the interest for the third year = ₹2028.

Principal for first year = ₹8000.

Interest for the first year = ₹ $\frac{8000 \times 10 \times 1}{100}$ = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ $\frac{8800 \times 10 \times 1}{100}$ = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ $\frac{9680 \times 10 \times 1}{100}$ = ₹968.

Amount after 3 years = ₹9680 + ₹968 = ₹10648.

Hence, the compound interest for the second year on ₹8000 invested for 3 years at 10% p.a. is ₹880 and the sum due at the end of third year is ₹10648.

(i) Principal for first year = ₹12800.

Interest for the first year = ₹ $\frac{12800 \times 10 \times 1}{100} = \frac{128000}{100}$ = ₹1280.

Amount after first year = ₹12800 + ₹1280 = ₹14080

Hence, the sum due to Ramesh at the end of first year = ₹14080.

(ii) Principal for the second year = ₹14080.

Interest for the second year = ₹ $\frac{14080 \times 10 \times 1}{100} = \frac{140800}{100}$ = ₹1408.

Hence, the interest Ramesh earns for second year = ₹1408.

(iii) Amount after 2 years = ₹14080 + ₹1408 = ₹15488.

Interest for the third year = ₹ $\frac{15488 \times 10 \times 1}{100} = \frac{154880}{100}$ = ₹1548.80

Amount after 3 years = ₹15488 + ₹1548.80 = ₹17036.80

Hence, the total amount due to Ramesh at the end of third year is ₹17036.80

Given, S.I. = ₹1380, rate = 12% p.a. and time = 2 years.

(i) Let the sum of money be P, then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 1380 = \frac{P \times 12 \times 2}{100} \\[1em] \Rightarrow 138000 = 24P \\[1em] \Rightarrow P = \frac{138000}{24} \\[1em] \Rightarrow P = 5750.\end{aligned}$$

Hence, the sum of money is ₹5750.

(ii) Since, the rate of interest is 12% per annum, therefore, the rate of interest half-yearly = 6%.

Principal for first half-year = ₹5750.

Interest for first half-year = ₹ $\frac{5750 \times 6 \times 1}{100} = \frac{34500}{100}$ = ₹345.

∴ Amount after first half-year = ₹5750 + ₹345 = ₹6095.

Principal for the second half-year = ₹6095.

Interest for the 2nd half-year = ₹ $\frac{6095 \times 6 \times 1}{100} = \frac{36570}{100}$ = ₹365.70

∴ Compound interest on the above sum for one year payable half-yearly = ₹345 + ₹365.70 = ₹710.70

Hence, the compound interest on ₹5750 for one year payable half-yearly at the same rate is ₹710.70

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹11200.

Interest = Amount - Principal = ₹11200 - ₹10000 = ₹1200.

Interest = $\frac{P \times R \times T}{100}$

$$\begin{aligned}\Rightarrow 1200 = \frac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100R \\[1em] \Rightarrow R = \frac{1200}{100} \\[1em] \Rightarrow R = 12%.\end{aligned}$$

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹11200.

Interest for second year = $\frac{11200 \times 12 \times 1}{100} = \frac{134400}{100}$ = ₹1344.

Amount at the end of second year = ₹11200 + ₹1344 = ₹12544.

Hence, the amount at the end of second year = ₹12544.

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹5325.

Interest = Amount - Principal = ₹5325 - ₹5000 = ₹325.

Interest = $\frac{P \times R \times T}{100}$

$$\begin{aligned}\Rightarrow 325 = \frac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 325 = 50R \\[1em] \Rightarrow R = \frac{325}{50} \\[1em] \Rightarrow R = 6.5%.\end{aligned}$$

Hence, the rate of interest is 6.5% per annum.

(ii) Amount after first year = ₹5325.

Interest for second year = $\frac{5325 \times 6.5 \times 1}{100} = \frac{34612.5}{100}$ = ₹346.125.

Amount at the end of second year = ₹5325 + ₹346.125 = ₹5671.125.

Hence, the amount at the end of second year to the nearest rupee = ₹5671.

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹5600.

Interest = Amount - Principal = ₹5600 - ₹5000 = ₹600.

Interest = $\frac{P \times R \times T}{100}$

$$\begin{aligned}\Rightarrow 600 = \frac{5000 \times R \times 1}{100} \\[1em] \Rightarrow 600 = 50R \\[1em] \Rightarrow R = \frac{600}{50} \\[1em] \Rightarrow R = 12%.\end{aligned}$$

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹5600.

Principal for second year = ₹5600.

Interest for second year = $\frac{5600 \times 12 \times 1}{100} = \frac{67200}{100}$ = ₹672.

Hence, the interest accrued in second year = ₹672.

(iii) Amount after second year = ₹5600 + ₹672 = ₹6272

Principal for third year = ₹6272.

Interest for third year = $\frac{6272 \times 12 \times 1}{100} = \frac{75264}{100}$ = ₹752.64

Amount after third year = ₹6272 + ₹752.64 = ₹7024.64

Hence, the amount at the end of third year = ₹7024.64

Principal for first year = ₹2000.

Interest for the first year = ₹ $\frac{2000 \times 10 \times 1}{100}$ = ₹200.

Amount after one year = ₹2000 + ₹200 = ₹2200.

Principal for the second year = ₹2200.

Interest for the second year = ₹ $\frac{2200 \times 10 \times 1}{100}$ = ₹220

Amount after 2 years = ₹2200 + ₹220 = ₹2420.

Principal for next $\frac{1}{2}$ year = ₹2420.

Interest for the next $\frac{1}{2}$ year = ₹ $\frac{2420 \times 10 \times \frac{1}{2}}{100} = \frac{12100}{100}$ = ₹121.

Amount after $2\frac{1}{2}$ years = ₹2420 + ₹121 = ₹2541.

Compound interest for $2\frac{1}{2}$ years = Final amount - Principal = ₹2541 - ₹2000 = ₹541.

Hence, the amount and compound interest on ₹2000 at 10% per annum after $2\frac{1}{2}$ years is ₹2541 and ₹541 respectively.

Since, the rate of interest is 8% per annum, therefore, the rate of interest half-yearly = $\frac{1}{2}$ of 8% = 4%.

Principal for first half-year = ₹50000.

Interest for first half-year = $\frac{50000 \times 4 \times 1}{100} = \frac{200000}{100}$ = ₹2000.

Amount after first half-year = ₹50000 + ₹2000 = ₹52000.

Principal for second half-year = ₹52000.

Interest for the second half-year = $\frac{52000 \times 4 \times 1}{100} = \frac{208000}{100}$ = ₹2080.

Amount after one year = ₹52000 + ₹2080 = ₹54080.

Principal for third half-year = ₹54080.

Interest for the third half-year = $\frac{54080 \times 4 \times 1}{100} = \frac{216320}{100}$ = ₹2163.20

Amount after $1\frac{1}{2}$ year = ₹54080 + ₹2163.20 = ₹56243.20

Compound interest for $1\frac{1}{2}$ year = Final amount - principal = ₹56243.20 - ₹50000 = ₹6243.20

Hence, the amount and the compound interest on ₹50000 for $1\frac{1}{2}$ years at 8% per annum, the interest being compounded semi-annually are ₹56243.20 and ₹6243.20 respectively.

Principal for first year = ₹5000.

Interest for the first year = ₹ $\frac{5000 \times 6 \times 1}{100} = \frac{30000}{100}$ = ₹300.

Amount after one year = ₹5000 + ₹300 = ₹5300.

Principal for the second year = ₹5300.

Interest for the second year = ₹ $\frac{5300 \times 8 \times 1}{100} = \frac{42400}{100}$ = ₹424.

Amount after 2 years = ₹5300 + ₹424 = ₹5724.

Compound interest = Final amount - Principal = ₹5724 - ₹5000 = ₹724.

Hence, the amount and the compound interest are ₹5724 and ₹724 respectively.

(i) Principal for first year = ₹9600.

Interest for the first year = ₹ $\frac{9600 \times 10 \times 1}{100} = \frac{96000}{100}$ = ₹960.

Amount after one year = ₹9600 + ₹960 = ₹10560.

Hence, the amount due at the end of first year = ₹10560.

(ii) Principal for second year = ₹10560.

Interest for the second year = ₹ $\frac{10560 \times 10 \times 1}{100} = \frac{105600}{100}$ = ₹1056.

Amount after 2 years = ₹10560 + ₹1056 = ₹11616.

Hence, the amount due at the end of second year = ₹11616.

(iii) Compound interest = Final amount - Principal = ₹11616 - ₹9600 = ₹2016.

Hence, the compound interest earned in 2 years = ₹2016.

(iv) Difference between (ii) and (i) = ₹11616 - ₹10560 = ₹1056.

Interest on the above sum for 1 year = $\frac{1056 \times 10 \times 1}{100} = \frac{10560}{100}$ = ₹105.60

Hence, the difference = ₹1056 and interest earned on it = ₹105.60

(v) Principal for third year = ₹11616.

Interest for the third year = ₹ $\frac{11616 \times 10 \times 1}{100} = \frac{116160}{100}$ = ₹1161.60

Hence, the compound interest for third year = ₹1161.60

Let the sum of money be ₹x.

Given, simple interest = ₹1600.

$$\begin{aligned}\therefore 1600 = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 1600 = \frac{x \times 10 \times 2}{100} \\[1em] \Rightarrow 1600 = \frac{20x}{100} \\[1em] \Rightarrow 160000 = 20x \\[1em] \Rightarrow x = \frac{160000}{20} \\[1em] \Rightarrow x = ₹8000.\end{aligned}$$

Principal for first year = ₹8000.

Interest for the first year = ₹ $\frac{8000 \times 10 \times 1}{100} = \frac{80000}{100}$ = ₹800.

Amount after one year = ₹8000 + ₹800 = ₹8800.

Principal for the second year = ₹8800.

Interest for the second year = ₹ $\frac{8800 \times 10 \times 1}{100} = \frac{88000}{100}$ = ₹880.

Amount after 2 years = ₹8800 + ₹880 = ₹9680.

Principal for the third year = ₹9680.

Interest for the third year = ₹ $\frac{9680 \times 10 \times 1}{100} = \frac{96800}{100}$ = ₹968

Amount after 3 years = ₹9680 + ₹968 = ₹10648

Compound interest = Final amount - Principal = ₹10648 - ₹8000 = ₹2648.

Hence, the amount and compound interest due = ₹10648 and ₹2648 respectively.

Simple interest = $\frac{P \times R \times T}{100}.$

∴ S.I. = $\frac{20000 \times 10 \times 2.5}{100} = \frac{500000}{100}$ = ₹5000.

Calculating, compound interest :

Principal for first year = ₹20000.

Interest for first year = $\frac{20000 \times 10 \times 1}{100} = \frac{200000}{100}$ = ₹2000.

Amount after first year = ₹20000 + ₹2000 = ₹22000.

Principal for second year = ₹22000.

Interest for second year = $\frac{22000 \times 10 \times 1}{100} = \frac{220000}{100}$ = ₹2200.

Amount after second year = ₹22000 + ₹2200 = ₹24200.

Principal for next $\frac{1}{2}$ year = ₹24200.

Interest for next $\frac{1}{2}$ year = $\frac{24200 \times 10 \times \frac{1}{2}}{100} = \frac{242000}{200}$ = ₹1210.

Amount after $2\frac{1}{2}$ year = ₹24200 + ₹1210 = ₹25410.

Compound interest = Final amount - principal = ₹25410 - ₹20000 = ₹5410.

Difference between C.I. and S.I. = ₹5410 - ₹5000 = ₹410.

Hence, Venkat gained ₹410 after $2\frac{1}{2}$ years.

Principal for first year = ₹6000, rate = 5%.

Interest for first year = $\frac{6000 \times 5 \times 1}{100} = \frac{30000}{100}$ = ₹300.

Amount after first year = ₹6000 + ₹300 = ₹6300.

Money refunded at the end of first year = ₹1200.

Principal for second year = ₹6300 - ₹1200 = ₹5100.

Interest for second year = $\frac{5100 \times 5 \times 1}{100} = \frac{25500}{100}$ = ₹255.

Amount after second year = ₹5100 + ₹255 = ₹5355.

Money refunded at the end of second year = ₹1200.

Principal for third year = ₹5355 - ₹1200 = ₹4155.

Hence, the amount outstanding at the beginning of third year = ₹4155.

For 1st year,

P = ₹ 1,600

R = 5%

T = 1 year

Using formula,

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

Substituting the values, we get

$$\begin{aligned}\Rightarrow I = \frac{1,600 \times 5 \times 1}{100}\\[1em] = 16 \times 5 \\[1em] = 80.\end{aligned}$$

A = P + I = ₹ 1,600 + ₹ 80 = ₹ 1,680

For 2nd year,

P = ₹ 1,680 + ₹ 1,600 (Amount deposited at beginning of every year) = ₹3,280

R = 5%

T = 1 year

$$\begin{aligned}\Rightarrow I = \frac{3,280 \times 5 \times 1}{100}\\[1em] = \frac{16,400}{100} \\[1em] = 164.\end{aligned}$$

A = P + I = ₹ 3,280 + ₹ 164 = ₹ 3,444.

Gain = Total balance - money invested

= ₹ 3,444 - (₹ 1,600 x 2)

= ₹ 3,444 - ₹ 3,200

= ₹ 244.

Hence, at the end of 2 years amount = ₹ 3,444 and the gain = ₹ 244.

Principal for first year = ₹100000, rate = 11%.

Interest for first year = $\frac{100000 \times 11 \times 1}{100} = \frac{1100000}{100}$ = ₹11000.

Amount after first year = ₹100000 + ₹11000 = ₹111000.

Money refunded at the end of first year = ₹41000.

Principal for second year = ₹111000 - ₹41000 = ₹70000.

Interest for second year = $\frac{70000 \times 11 \times 1}{100} = \frac{770000}{100}$ = ₹7700.

Amount after second year = ₹70000 + ₹7700 = ₹77700.

Money refunded at the end of second year = ₹47700.

Principal for third year = ₹77700 - ₹47700 = ₹30000.

Hence, the amount outstanding at the beginning of third year = ₹30000.

Principal for first year = ₹50000, rate = 12%.

Interest for first year = $\frac{50000 \times 12 \times 1}{100} = \frac{600000}{100}$ = ₹6000.

Amount after first year = ₹50000 + ₹6000 = ₹56000.

Money refunded at the end of first year = ₹33000.

Principal for second year = ₹56000 - ₹33000 = ₹23000, rate = 15%.

Interest for second year = $\frac{23000 \times 15 \times 1}{100} = \frac{345000}{100}$ = ₹3450.

Amount after second year = ₹23000 + ₹3450 = ₹26450.

Hence, Jaya must pay ₹26450 at the end of second year to clear her debt.

Formula for calculating amount,

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

Using formula we get,

$$\begin{aligned}A = ₹5000\Big(1 + \frac{6}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\frac{106}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\frac{53}{50}\Big)^2 \\[1em] = ₹5000 \times \frac{53}{50} \times \frac{53}{50} \\[1em] = ₹\frac{14045000}{2500} \\[1em] = ₹5618.\end{aligned}$$

Compound interest = Final amount - Principal = ₹5618 - ₹5000 = ₹618.

Hence, the amount and the compound interest on ₹5000 for 2 years at 6% per annum is ₹5618 and ₹618 respectively.

Formula for calculating amount,

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

Using formula we get,

$$\begin{aligned}A = ₹8000\Big(1 + \frac{10}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\frac{110}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\frac{11}{10}\Big)^4 \\[1em] = ₹8000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\[1em] = ₹\frac{117128000}{10000} \\[1em] = ₹11712.80\end{aligned}$$

Compound interest = Final amount - Principal = ₹11712.80 - ₹8000 = ₹3712.80

Hence, the amount and the compound interest on ₹8000 for 4 years at 10% per annum is ₹11712.80 and ₹3712.80 respectively.

Since rate of interest is 5% per annum, therefore rate of interest per conversion period (half-yearly) = 2.5%.

As the money is invested for one year, therefore,

n (the number of conversion periods) = 2.

$$\begin{aligned}\therefore A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹7400 \times \Big(1 + \frac{2.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\frac{102.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\frac{1025}{1000}\Big)^2 \\[1em] = ₹7400 \times \Big(\frac{41}{40}\Big)^2 \\[1em] = ₹7400 \times \frac{41}{40} \times \frac{41}{40} \\[1em] = ₹ \frac{12439400}{1600} \\[1em] = ₹7774.625\end{aligned}$$

Hence, amount = ₹7774.625

Since, compound interest is reckoned semi-annually,

rate = $\frac{10}{2}$ % = 5%.

n (no. of conversion periods) = 3 half-years.

$$\begin{aligned}\therefore A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹5000 \times \Big(1 + \frac{5}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\frac{105}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\frac{21}{20}\Big)^2 \\[1em] = ₹5000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\[1em] = ₹5000 \times \frac{9261}{8000} \\[1em] = ₹5788.125\end{aligned}$$

C.I. = Amount - Principal = ₹5788.125 - ₹5000 = ₹788.125

Hence, amount = ₹5788.125 and compound interest = ₹788.125

Since rate of interest is 4% per annum, therefore rate of interest per conversion period (quarterly) = $\frac{1}{4} \times 4$ = 1%.

As the money is invested for 9 months, therefore,

n (the number of conversion periods) = $\frac{9}{3}$ = 3.

$$\begin{aligned}\therefore A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹100000 \times \Big(1 + \frac{1}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\frac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\frac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \frac{101}{100} \times \frac{101}{100} \times \frac{101}{100} \\[1em] = ₹\frac{100000 \times 101 \times 101 \times 101}{1000000} \\[1em] = ₹\frac{1030301}{10} \\[1em] = ₹103030.10\end{aligned}$$

Compound interest = Final amount - Principal = ₹103030.10 - ₹100000 = ₹3030.10

Hence, amount = ₹103030.10 and compound interest = ₹3030.10.

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

Putting values in formula we get,

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

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(1 + \frac{5}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\frac{105}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\frac{21}{20}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\frac{441}{400} - 1\Big] \\[1em] = ₹4800 \times \Big[\frac{441 - 400}{400}\Big] \\[1em] = ₹4800 \times \Big[\frac{41}{400}\Big] \\[1em] = ₹\frac{196800}{400} \\[1em] = ₹492.\end{aligned}$$

C.I. - S.I. = ₹492 - ₹480 = ₹12.

Hence, the difference between C.I. and S.I. = ₹12.

Since interest is calculated half yearly, hence rate = $\frac{4}{2}$ % = 2%

Time = 2 years or 4 half-years.

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

Putting values in formula we get,

$$\begin{aligned}S.I. = \frac{₹2500 \times 2 \times 4}{100} \\[1em] = ₹\frac{20000}{100} \\[1em] = ₹200.\end{aligned}$$

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(1 + \frac{2}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\frac{102}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\frac{51}{50}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\frac{6765201}{6250000} - 1\Big] \\[1em] = ₹2500 \times \Big[\frac{6765201 - 6250000}{6250000}\Big] \\[1em] = ₹2500 \times \Big[\frac{515201}{6250000}\Big] \\[1em] = ₹\frac{515201}{2500} \\[1em] = ₹206.084\end{aligned}$$

C.I. - S.I. = ₹206.084 - ₹200 = ₹6.084

Hence, the difference between C.I. and S.I. = ₹6.084

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

For first year, P = ₹2000 and rate = 4%.

Using formula,

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹2000 \times \Big(1 + \frac{4}{100}\Big)^1 \\[1em] = ₹2000 \times \Big(\frac{104}{100}\Big) \\[1em] = ₹2000 \times \frac{26}{25} \\[1em] = ₹80 \times 26 \\[1em] = ₹2080.\end{aligned}$$

For second year, P = ₹2080 and rate = 3%.

Using formula,

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹2080 \times \Big(1 + \frac{3}{100}\Big)^1 \\[1em] = ₹2080 \times \Big(\frac{103}{100}\Big) \\[1em] = ₹\frac{214240}{100} \\[1em] = ₹2142.40\end{aligned}$$

C.I. = Final Amount - Principal = ₹2142.40 - ₹2000 = 142.40

Hence, the amount = ₹2142.40 and compound interest = ₹142.40

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

For first year, P = ₹3125 and rate = 4%.

Using formula,

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = ₹3125 \times \Big(1 + \frac{4}{100}\Big)^1 \\[1em] = ₹3125 \times \Big(\frac{104}{100}\Big) \\[1em] = ₹3125 \times \frac{26}{25} \\[1em] = ₹\frac{81250}{25} \\[1em] = ₹3250.\end{aligned}$$

For second year, P = ₹3250 and rate = 5%.

Using formula,

$$\begin{aligned}A = ₹3250 \times \Big(1 + \frac{5}{100}\Big)^1 \\[1em] = ₹3250 \times \Big(\frac{105}{100}\Big) \\[1em] = ₹3250 \times \frac{21}{20} \\[1em] = ₹\frac{68250}{20} \\[1em] = ₹3412.50\end{aligned}$$

For third year, P = ₹3412.50 and rate = 6%.

Using formula,

$$\begin{aligned}A = ₹3412.50 \times \Big(1 + \frac{6}{100}\Big)^1 \\[1em] = ₹3412.50 \times \Big(\frac{106}{100}\Big) \\[1em] = ₹3412.50 \times \frac{53}{50} \\[1em] = ₹\frac{180862.5}{50} \\[1em] = ₹3617.25\end{aligned}$$

C.I. = Final Amount - Principal = ₹3617.25 - ₹3125 = ₹492.25

Hence, compound interest = ₹492.25

Let principal = P.

Given, A = ₹9261, rate = 5%, n = 3.

We know,

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

Putting values in formula we get,

$$\begin{aligned}9261 = P\Big(1 + \frac{5}{100}\Big)^3 \\[1em] 9261 = P\Big(\frac{105}{100}\Big)^3 \\[1em] 9261 = P\Big(\frac{21}{20}\Big)^3 \\[1em] 9261 = P \times \frac{9261}{8000} \\[1em] P = \frac{9261 \times 8000}{9261} \\[1em] P = ₹8000.\end{aligned}$$

Hence, ₹8000 will amount to ₹9261 in 3 years at 5% per annum compound interest.

Let principal = P.

Given, A = ₹7803.

Since interest is compounded semi-annually,

rate = $\frac{4}{2}$% = 2%.

n (the number of conversion periods) = 2.

We know,

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

Putting values in formula we get,

$$\begin{aligned}7803 = P\Big(1 + \frac{2}{100}\Big)^2 \\[1em] 7803 = P\Big(\frac{102}{100}\Big)^2 \\[1em] 7803 = P\Big(\frac{51}{50}\Big)^2 \\[1em] 7803 = P \times \frac{2601}{2500} \\[1em] P = \frac{7803 \times 2500}{2601} \\[1em] P = ₹7500.\end{aligned}$$

Hence, ₹7500 will amount to ₹7803 in 1 year at 4% per annum compounded semi-annually.

Let principal = P.

Given, A = ₹132651.

Since interest is compounded half-yearly,

rate = $\frac{4}{2}$% = 2%.

n (the number of conversion periods) = 3.

We know,

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

Putting values in formula we get,

$$\begin{aligned}132651 = P\Big(1 + \frac{2}{100}\Big)^3 \\[1em] 132651 = P\Big(\frac{102}{100}\Big)^3 \\[1em] 132651 = P\Big(\frac{51}{50}\Big)^3 \\[1em] 132651 = P \times \frac{132651}{125000} \\[1em] P = \frac{132651 \times 125000}{132651} \\[1em] P = ₹125000.\end{aligned}$$

Hence, ₹125000 will amount to ₹132651 in $1\frac{1}{2}$ year at 4% per annum compounded semi-annually.

Let principal = P,

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 5712 = P\Big[\Big(1 + \frac{4}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\frac{104}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\frac{26}{25}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\frac{676}{625} - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\frac{676 - 625}{625}\Big] \\[1em] \Rightarrow 5712 = P \times \frac{51}{625} \\[1em] \Rightarrow P = \frac{5712 \times 625}{51} \\[1em] \Rightarrow P = 112 \times 625 \\[1em] \Rightarrow P = ₹70000.\end{aligned}$$

Hence, principal = ₹70000.

Let rate of interest be r% per annum.

Given, ₹1200 amounts to ₹1275 after one year.

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 1275 = 1200\Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow \frac{1275}{1200} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{1275}{1200} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{1275 - 1200}{1200} = \frac{r}{100} \\[1em] \Rightarrow \frac{75}{1200} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{7500}{1200} \\[1em] \Rightarrow r = \frac{25}{4} \\[1em] \Rightarrow r = 6\frac{1}{4}%.\end{aligned}$$

Principal for second year = ₹1275.

Interest for second year = $\frac{P \times R \times T}{100}$.

Substituting value we get,

$$\begin{aligned}\text{Interest } = \frac{₹1275 \times \frac{25}{4} \times 1}{100} \\[1em] = \frac{₹1275 \times 25}{400} \\[1em] = ₹\frac{31875}{400} \\[1em] = ₹79.6875 \approx ₹80.\end{aligned}$$

Hence, the interest for the second year = ₹80.

Let rate of interest = r.

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

Given, A = ₹2500, P = ₹2304, n = 2.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 2500 = 2304\Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{2500}{2304} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{50}{48}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{50}{48} = \Big(1 + \frac{r}{100}\Big) \\[1em] \Rightarrow \frac{50}{48} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{50 - 48}{48} = \frac{r}{100} \\[1em] \Rightarrow \frac{2}{48} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{2}{48} \times 100 \\[1em] \Rightarrow r = \frac{200}{48} = 4\frac{1}{6}%.\end{aligned}$$

Hence, rate of interest = $4\frac{1}{6}$%.

Let rate of interest = r.

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

Let Principal = P.

Given, sum becomes $\frac{25}{16}$ times of itself in two years,

∴ A = $\frac{25}{16}$P.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow \frac{25}{16}P = P\Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{25}{16} = \Big(1 + \frac{r}{100}\Big)^2\\[1em] \Rightarrow \Big(\frac{5}{4}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2\\[1em]\end{aligned}$$

Taking square root on both sides,

$$\begin{aligned}\Rightarrow \frac{5}{4} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{5}{4} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{5 - 4}{4} = \frac{r}{100} \\[1em] \Rightarrow \frac{1}{4} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{100}{4} \\[1em] \Rightarrow r = 25%.\end{aligned}$$

Hence, rate of interest = 25%.

Let rate of interest = r.

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

Given, A = ₹2315.25, P = ₹2000, n = 3.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 2315.25 = 2000\Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{2315.25}{2000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{231525}{200000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{9261}{8000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^3 = \Big(1 + \frac{r}{100}\Big)^3 \\[1em]\end{aligned}$$

Taking cube root on both sides we get,

$$\begin{aligned}\Rightarrow \frac{21}{20} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{21}{20} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{21 - 20}{20} = \frac{r}{100} \\[1em] \Rightarrow \frac{1}{20} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{100}{20} \\[1em] \Rightarrow r = 5%.\end{aligned}$$

Hence, rate of interest = 5%.

Let rate of interest per annum be r% per annum, i.e. $\frac{r}{2}$% half-yearly.

n = 2 years or 4 half-years.

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

Given, A = ₹48620.25 and P = ₹40000.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 48620.25 = 40000\Big(1 + \frac{\frac{r}{2}}{100}\Big)^4 \\[1em] \Rightarrow \frac{48620.25}{40000} = \Big(1 + \frac{r}{200}\Big)^4 \\[1em] \Rightarrow \frac{4862025}{4000000} = \Big(1 + \frac{r}{200}\Big)^4 \\[1em] \Rightarrow \frac{194481}{160000} = \Big(1 + \frac{r}{200}\Big)^4 \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^4 = \Big(1 + \frac{r}{200}\Big)^4 \\[1em] \Rightarrow \frac{21}{20} = 1 + \frac{r}{200} \\[1em] \Rightarrow \frac{21}{20} - 1 = \frac{r}{200} \\[1em] \Rightarrow \frac{21 - 20}{20} = \frac{r}{200} \\[1em] \Rightarrow \frac{1}{20} = \frac{r}{200} \\[1em] \Rightarrow r = \frac{200}{20} \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Hence, rate of interest = 10% per annum.

Let rate of interest per annum be r% per annum, i.e. $\frac{r}{2}$% half-yearly.

n = $1\frac{1}{2}$ years or 3 half-years.

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

Let principal be P,

∴ A = $\frac{216}{125}P$.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow \frac{216}{125}P = P\Big(1 + \frac{\frac{r}{2}}{100}\Big)^3 \\[1em] \Rightarrow \frac{216}{125} = \Big(1 + \frac{r}{200}\Big)^3\\[1em] \Rightarrow \Big(\frac{6}{5}\Big)^3 = \Big(1 + \frac{r}{200}\Big)^3\\[1em]\end{aligned}$$

Taking cube root on both sides,

$$\begin{aligned}\frac{6}{5} = 1 + \frac{r}{200} \\[1em] \frac{6}{5} - 1 = \frac{r}{200} \\[1em] \frac{6 - 5}{5} = \frac{r}{200} \\[1em] \frac{1}{5} = \frac{r}{200} \\[1em] r = \frac{200}{5} \\[1em] r = 40%.\end{aligned}$$

Hence, rate of interest = 40% per annum.

Let the rate of interest be r% per annum.

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

Substituting value we get,

$$\begin{aligned}\Rightarrow 88200 = 80000\Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{88200}{80000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{441}{400} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{21}{20} \\[1em] \Rightarrow \frac{r}{100} = \frac{21}{20} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{21 - 20}{20} \\[1em] \Rightarrow \frac{r}{100} = \frac{1}{20} \\[1em] \Rightarrow r = \frac{1}{20} \times 100 \\[1em] \Rightarrow r = 5%.\end{aligned}$$

After 3 years,

$$\begin{aligned}A = ₹80000\Big(1 + \frac{5}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\frac{105}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\frac{21}{20}\Big)^3 \\[1em] = ₹80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\[1em] = ₹80000 \times \frac{9261}{8000} \\[1em] = ₹92610.\end{aligned}$$

Hence, the rate of interest = 5% per annum and amount after 3 years = ₹92610.

Let the rate of interest be r% per annum.

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

As sum amounts to ₹5292 in 2 years, substituting value we get,

$\therefore 5292 = P\Big(1 + \frac{r}{100}\Big)^2$ .....(Eq. 1)

As sum amounts to ₹5556.60 in 3 years, substituting value we get,

$\therefore 5556.60 = P\Big(1 + \frac{r}{100}\Big)^3$ .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$$\begin{aligned}\Rightarrow 1 + \frac{r}{100} = \frac{5556.60}{5292} \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{555660}{529200} \\[1em] \Rightarrow \frac{r}{100} = \frac{555660}{529200} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{26460}{529200} \\[1em] \Rightarrow r = \frac{26460}{529200} \times 100 \\[1em] \Rightarrow r = \frac{26460}{5292} \\[1em] \Rightarrow r = 5%.\end{aligned}$$

Substituting value of r in Eq. 1 we get,

$$\begin{aligned}\Rightarrow 5292 = P\Big(1 + \frac{5}{100}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(1 + \frac{1}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(\frac{21}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P \times \frac{441}{400} \\[1em] \Rightarrow P = 5292 \times \frac{400}{441} \\[1em] \Rightarrow P = 12 \times 400 \\[1em] \Rightarrow P = ₹4800.\end{aligned}$$

Hence, sum = ₹4800 and rate = 5%.

Let the rate of interest be r% per annum.

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

As sum amounts to ₹798.60 in 3 years substituting value we get,

$\therefore 798.60 = P\Big(1 + \frac{r}{100}\Big)^3$ .....(Eq. 1)

As sum amounts to ₹878.46 in 4 years substituting value we get,

$\therefore 878.46 = P\Big(1 + \frac{r}{100}\Big)^4$ .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$$\begin{aligned}\Rightarrow 1 + \frac{r}{100} = \frac{878.46}{798.60} \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{87846}{79860} \\[1em] \Rightarrow \frac{r}{100} = \frac{87846}{79860} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{87846 - 79860}{79860} \\[1em] \Rightarrow \frac{r}{100} = \frac{7986}{79860} \\[1em] \Rightarrow \frac{r}{100} = \frac{1}{10} \\[1em] \Rightarrow r = \frac{100}{10} \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Substituting value of r in Eq. 1 we get,

$$\begin{aligned}\Rightarrow 798.60 = P\Big(1 + \frac{10}{100}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(1 + \frac{1}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(\frac{11}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P \times \frac{1331}{1000} \\[1em] \Rightarrow P = 798.60 \times \frac{1000}{1331} \\[1em] \Rightarrow P = \frac{798600}{1331} \\[1em] \Rightarrow P = ₹600.\end{aligned}$$

Hence, sum = ₹600 and rate = 10%.

Let the time be n years.

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 17576 = 15625\Big(1 + \frac{4}{100}\Big)^n \\[1em] \Rightarrow 17576 = 15625\Big(\frac{104}{100}\Big)^n \\[1em] \Rightarrow \frac{17576}{15625} = \Big(\frac{26}{25}\Big)^n \\[1em] \Rightarrow \Big(\frac{26}{25}\Big)^3 = \Big(\frac{26}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.\end{aligned}$$

Hence, ₹15625 will amount to ₹17576 in 3 years at 4% per annum compound interest.

Let the time be n years.

$C.I. = P \Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big].$

Substituting values we get,

$$\begin{aligned}\Rightarrow 496.50 = 1500\Big[\Big(1 + \frac{10}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 496.50 = 1500\Big(\frac{110}{100}\Big)^n - 1500 \\[1em] \Rightarrow 496.50 + 1500 = 1500\Big(\frac{110}{100}\Big)^n \\[1em] \Rightarrow \frac{1996.50}{1500} = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow \frac{199650}{150000} = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow \frac{1331}{1000} = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.\end{aligned}$$

Hence, in 3 years ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually.

Let the time be n years.

$C.I. = P \Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big].$

Substituting values we get,

$$\begin{aligned}\Rightarrow 3246.40 = 12500\Big[\Big(1 + \frac{8}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 3246.40 = 12500\Big(\frac{108}{100}\Big)^n - 12500 \\[1em] \Rightarrow 3246.40 + 12500 = 12500\Big(\frac{108}{100}\Big)^n \\[1em] \Rightarrow \frac{15746.40}{12500} = \Big(\frac{27}{25}\Big)^n \\[1em] \Rightarrow \frac{1574640}{1250000} = \Big(\frac{27}{25}\Big)^n \\[1em] \Rightarrow \frac{19683}{15625} = \Big(\frac{27}{25}\Big)^n \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^3 = \Big(\frac{27}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.\end{aligned}$$

Hence, in 3 years ₹12500 yield ₹3246.40 as compound interest at 8% per annum compounded annually.

Rate = 10% p.a. i.e. \dfrac{10%}{2} = 5% compounded semi-annually.

Let time be n half-years.

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 18522 = 16000\Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow 18522 = 16000\Big(\frac{105}{100}\Big)^n \\[1em] \Rightarrow \frac{18522}{16000} = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow \frac{9261}{8000} = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^3 = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3 \text{ half-years}.\end{aligned}$$

n = 3 half-years i.e. $1\frac{1}{2}$ years.

Hence, time = $1\frac{1}{2}$ years.

If the rates are different then formula is,

$A = P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Substituting values we get,

$$\begin{aligned}A = P\Big(1 + \frac{5}{100}\Big)\Big(1 + \frac{6}{100}\Big) \\[1em] \Rightarrow 2782.50 = P\Big(1 + \frac{1}{20}\Big)\Big(1 + \frac{3}{50}\Big) \\[1em] \Rightarrow 2782.50 = P \times \frac{21}{20} \times \frac{53}{50} \\[1em] \Rightarrow P = \frac{2782.50 \times 20 \times 50}{21 \times 53} \\[1em] \Rightarrow P = ₹2500.\end{aligned}$$

Hence, sum = ₹2500.

Given,

Interest for first year = ₹225

Interest for second year = ₹240.

Difference = ₹15.

Here, ₹15 is the interest on ₹225 for 1 year.

(i) We know that,

$$\begin{aligned}\text{Rate} = \frac{S.I. \times 100}{P \times T} \\[1em] = \frac{15 \times 100}{225 \times 1} \\[1em] = \frac{20}{3} \\[1em] = 6\frac{2}{3}%.\end{aligned}$$

Hence, rate of interest = $6\frac{2}{3}%.$

(ii) We know that,

$$\begin{aligned}\text{P} = \frac{S.I. \times 100}{R \times T} \\[1em] = \frac{225 \times 100}{\frac{20}{3} \times 1} \\[1em] = \frac{22500}{\frac{20}{3}} \\[1em] = \frac{67500}{20} \\[1em] = ₹3375.\end{aligned}$$

Hence, the sum = ₹3375.

(iii) Here,

Amount after second year = ₹3375 + ₹225 + ₹240 = ₹3840.

Interest earned in third year = $\frac{3840 \times \frac{20}{3} \times 1}{100} = \frac{3840 \times 20 \times 1}{300}$ = ₹256.

Hence, interest earned in third year = ₹256.

Given,

Let Sum (P) = ₹x,

Rate (R) = 5% p.a.

Period (n) = 2 years.

We know that,

$\text{S.I.} = \frac{\text{PRT}}{100}$

Substituting values we get,

$$\begin{aligned}\text{S.I.} = \frac{x \times 5 \times 2}{100} \\[1em] = ₹\frac{x}{10}.\end{aligned}$$

When interest is compounded annually,

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

Substituting values,

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

C.I. = A - P = $₹\frac{441x}{400} - ₹x = ₹\frac{41x}{400}.$

$$\begin{aligned}\text{C.I.} - \text{S.I.} = ₹\frac{41x}{400} - ₹\frac{x}{10} \\[1em] = ₹\frac{41x - 40x}{400} \\[1em] = ₹\frac{x}{400}\end{aligned}$$

Given,

Difference = ₹25.

$$\begin{aligned}\therefore \frac{x}{400} = 25 \\[1em] \Rightarrow x = 25 \times 400 \\[1em] \Rightarrow x = ₹10000.\end{aligned}$$

Hence, sum of money = ₹10000.

Let Sum (P) = ₹x.

Given,

Rate = 10% p.a. or 5% half-yearly.

Period = 1 year or 2 half-years.

We know that,

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

Substituting values,

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

C.I. = A - P = $₹\frac{441x}{400} - ₹x = ₹\frac{41x}{400}.$

We know that,

$\text{S.I.} = \frac{\text{PRT}}{100}$

Substituting values we get,

$$\begin{aligned}\text{S.I.} = \frac{x \times 10 \times 1}{100} \\[1em] = ₹\frac{x}{10}.\end{aligned}$$

$$\begin{aligned}\text{C.I.} - \text{S.I.} = ₹\frac{41x}{400} - ₹\frac{x}{10} \\[1em] = ₹\frac{41x - 40x}{400} \\[1em] = ₹\frac{x}{400}\end{aligned}$$

Given,

Difference = ₹15,

$$\begin{aligned}\therefore \frac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹6000.\end{aligned}$$

Hence, sum of money = ₹6000.

Given,

Amount after one year = ₹1250,

Amount after second year = ₹1375.

Difference = ₹1375 - ₹1250 = ₹125.

So, ₹125 is the interest on ₹1250 for 1 year.

We know that,

$$\begin{aligned}R = \frac{S.I. \times 100}{P \times T} \\[1em] = \frac{125 \times 100}{1250 \times 1} \\[1em] = 10%.\end{aligned}$$

Hence, rate of interest = 10%.

Let principal be ₹P and rate of interest be R% p.a.

According to the first condition of the question,

S.I. on ₹P for 3 years = ₹225.

$$\begin{aligned}\therefore \frac{P \times R \times T}{100} = 225 \\[1em] \Rightarrow \frac{P \times R \times 3}{100} = 225 \\[1em] \Rightarrow P \times R = \frac{225 \times 100}{3} \\[1em] \Rightarrow P \times R = 7500 \\[1em] \Rightarrow P = \frac{7500}{R} .......(i)\end{aligned}$$

According to the second condition of the question,

C.I. on ₹P for 2 years at R% p.a. = ₹153.

$$\begin{aligned}\therefore P\Big[\Big(1 + \frac{R}{100}\Big)^n - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(1 + \frac{R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(\frac{100 + R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\frac{(100 + R)^2}{100^2} - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\frac{(100 + R)^2 - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\frac{100^2 + R^2 + 200R - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\frac{R^2 + 200R}{100^2}\Big] = 153 \\[1em]\end{aligned}$$

Using value of P from Eq 1 above:

$$\begin{aligned}\Rightarrow \frac{7500}{R} \times \frac{R(R + 200)}{100^2} = 153 \\[1em] \Rightarrow \frac{7500(R + 200)}{100^2} = 153 \\[1em] \Rightarrow R + 200 = \frac{153 \times 100 \times 100}{7500} \\[1em] \Rightarrow R + 200 = \frac{15300}{75} \\[1em] \Rightarrow R + 200 = 204 \\[1em] \Rightarrow R = 204 - 200 = 4%.\end{aligned}$$

Substituting value of R in (i) we get,

$P = \frac{7500}{4} = ₹1875.$

Hence, rate of interest = 4% and sum = ₹1875.

Case 1:

When compounded annually,

rate for first year = 10%, rate for next $\frac{1}{2}$ year = 5%.

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

On substituting values,

$$\begin{aligned}A = ₹8000\Big(1 + \frac{10}{100}\Big)\Big(1 + \frac{5}{100}\Big) \\[1em] = ₹8000\Big(\frac{110}{100}\Big)\Big(\frac{105}{100}\Big) \\[1em] = ₹8000 \times \frac{11}{10} \times \frac{21}{20} \\[1em] = ₹(40 \times 11 \times 21) \\[1em] = ₹9240.\end{aligned}$$

C.I. = A - P = ₹9240 - ₹8000 = ₹1240.

Case 2:

When compounded semi-annually,

rate = 5%, n = 3.

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

On substituting values,

$$\begin{aligned}A = ₹8000\Big(1 + \frac{5}{100}\Big)^3 \\[1em] = ₹8000\Big(\frac{105}{100}\Big)^3 \\[1em] = ₹8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\[1em] = ₹(21 \times 21 \times 21) \\[1em] = ₹9261.\end{aligned}$$

C.I. = A - P = ₹9261 - ₹8000 = ₹1261.

Difference between two C.I. = ₹1261 - ₹1240 = ₹21.

Let the sum lent out be ₹x.

When C.I. is reckoned yearly,

Rate = 20%

$$\begin{aligned}\therefore A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = x\Big(1 + \frac{20}{100}\Big)^2 \\[1em] = x\Big(1 + \frac{1}{5}\Big)^2 \\[1em] = x\Big(\frac{6}{5}\Big)^2 \\[1em] = \frac{36x}{25}\end{aligned}$$

C.I. = A - P = $\frac{36x}{25} - x = \frac{11x}{25}.$

When C.I. is reckoned half-yearly,

Rate = $\frac{20}{2}$ % = 10%.

n (no. of conversion periods) = 4 half-years.

$$\begin{aligned}\therefore A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = x\Big(1 + \frac{10}{100}\Big)^4 \\[1em] = x\Big(1 + \frac{1}{10}\Big)^4 \\[1em] = x\Big(\frac{11}{10}\Big)^4 \\[1em] = \frac{14641x}{10000}\end{aligned}$$

C.I. = A - P = $\frac{14641x}{10000} - x = \frac{4641x}{10000}.$

Given, difference between C.I. in both cases = ₹482.

$$\begin{aligned}\therefore \frac{4641x}{10000} - \frac{11x}{25} = 482 \\[1em] \Rightarrow \frac{4641x - 4400x}{10000} = 482 \\[1em] \Rightarrow \frac{241x}{10000} = 482 \\[1em] \Rightarrow x = \frac{482 \times 10000}{241} \\[1em] \Rightarrow x = ₹20000.\end{aligned}$$

Hence, sum lent out = ₹20000.

Let sum be ₹x and rate be r%.

Since C.I. is reckoned half-yearly, rate = $\frac{r}{2}$%

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

In one year,

n = 2

A = ₹13230

rate = $\frac{r}{2}$% as interest is calculated half-yearly.

Substituting values in formula,

$$\begin{aligned}13230 = x\Big(1 + \frac{\frac{r}{2}}{100}\Big)^2 \\[1em] 13230 = x\Big(1 + \frac{r}{200}\Big)^2 ........(i)\end{aligned}$$

In one and half year,

n = 3

A = ₹13891.50

rate = $\frac{r}{2}$% as interest is calculated half-yearly.

Substituting values in formula,

$$\begin{aligned}13891.50 = x\Big(1 + \frac{\frac{r}{2}}{100}\Big)^3 \\[1em] 13891.50 = x\Big(1 + \frac{r}{200}\Big)^3 ........(ii)\end{aligned}$$

Dividing eqn. (ii) by (i),

$$\begin{aligned}\Rightarrow \frac{13891.50}{13230} = \frac{x\Big(1 + \frac{r}{200}\Big)^3}{x\Big(1 + \frac{r}{200}\Big)^2} \\[1em] \Rightarrow \frac{138915}{132300} = 1 + \frac{r}{200} \\[1em] \Rightarrow \frac{r}{200} = \frac{138915}{132300} - 1 \\[1em] \Rightarrow \frac{r}{200} = \frac{138915 - 132300}{132300} \\[1em] \Rightarrow \frac{r}{200} = \frac{6615}{132300} \\[1em] \Rightarrow r = \frac{6615 \times 200}{132300} \\[1em] \Rightarrow r = \frac{1323000}{132300} \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Putting value of r in eq. (i) we get,

$$\begin{aligned}\Rightarrow 13230 = x\Big(1 + \frac{10}{200}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(1 + \frac{1}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(\frac{21}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x \times \frac{441}{400} \\[1em] \Rightarrow x = \frac{13230 \times 400}{441} \\[1em] \Rightarrow x = ₹12000.\end{aligned}$$

Hence, sum = ₹12000 and rate = 10% per annum.

By formula,

V = $V_0\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Putting values in formula we get,

$$\begin{aligned}V = 200000\Big(1 + \frac{10}{100}\Big)\Big(1 + \frac{15}{100}\Big) \\[1em] = 200000 \times \frac{110}{100} \times \frac{115}{100} \\[1em] = 20 \times 110 \times 115 \\[1em] = 253000.\end{aligned}$$

Hence, the population of the town at the end of two years is 253000.

By formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = 15625\Big(1 + \frac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\frac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\frac{26}{25}\Big)^3 \\[1em] = 15625 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} \\[1em] = \frac{15625 \times 17576}{15625} \\[1em] = 17576.\end{aligned}$$

Increase in population = 17576 - 15625 = 1951.

Hence, the increase in population in next 3 years = 1951.

(i) By formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = 6760000\Big(1 + \frac{4}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\frac{104}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\frac{26}{25}\Big)^2 \\[1em] = 6760000 \times \frac{26}{25} \times \frac{26}{25} \\[1em] = \frac{6760000 \times 676}{625} \\[1em] = 10816 \times 676 \\[1em] = 7311616.\end{aligned}$$

Hence, the population after 2 years = 7311616.

(ii) Let population 2 years ago be P and present population will be the final population.

By formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 6760000 = P \times \Big(1 + \frac{4}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\frac{104}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\frac{26}{25}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \frac{26}{25} \times \frac{26}{25} \\[1em] \Rightarrow 6760000 = P \times \frac{676}{625} \\[1em] \Rightarrow P = \frac{6760000 \times 625}{676} \\[1em] \Rightarrow P = 625 \times 10000 \\[1em] \Rightarrow P = 6250000.\end{aligned}$$

Hence, the population 2 years ago was 6250000.

Depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = ₹9000\Big(1 - \frac{5}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\frac{95}{100}\Big)^2 \\[1em] = ₹9000 \times \Big(\frac{19}{20}\Big)^2 \\[1em] = ₹9000 \times \frac{19}{20} \times \frac{19}{20} \\[1em] = ₹\frac{9000 \times 361}{400} \\[1em] = ₹8122.5\end{aligned}$$

Depreciation = ₹9000 - ₹8122.5 = ₹877.5

Hence, the total depreciation in its value at the end of 2 years = ₹877.5

Depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = ₹24000\Big(1 - \frac{5}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\frac{95}{100}\Big)^3 \\[1em] = ₹24000 \times \Big(\frac{19}{20}\Big)^3 \\[1em] = ₹24000 \times \frac{19}{20} \times \frac{19}{20} \times \frac{19}{20} \\[1em] = ₹\frac{24000 \times 6859}{8000} \\[1em] = ₹(3 \times 6859) \\[1em] = ₹20577.\end{aligned}$$

Hence, the value of scooter after 3 years = ₹20577.

Let yearly produce 2 years ago be P. Then present production will be the final production.

By formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 2187 = P \times \Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\frac{108}{100}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow 2187 = P \times \frac{27}{25} \times \frac{27}{25} \\[1em] \Rightarrow 2187 = P \times \frac{729}{625} \\[1em] \Rightarrow P = \frac{2187 \times 625}{729} \\[1em] \Rightarrow P = 3 \times 625 \\[1em] \Rightarrow P = 1875.\end{aligned}$$

Hence, the yearly produce of wheat two years ago was 1875 quintals.

Depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Let value three years ago be V₀.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow ₹411540 = V_0\Big(1 - \frac{5}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\frac{95}{100}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0\Big(\frac{19}{20}\Big)^3 \\[1em] \Rightarrow ₹411540 = V_0 \times \frac{19}{20} \times \frac{19}{20} \times \frac{19}{20} \\[1em] \Rightarrow V_0 = ₹\frac{411540 \times 20 \times 20 \times 20}{19 \times 19 \times 19} \\[1em] \Rightarrow V_0 = ₹\frac{411540 \times 8000}{6859} \\[1em] \Rightarrow V_0 = ₹(60 \times 8000) \\[1em] \Rightarrow V_0 = ₹480000.\end{aligned}$$

Hence, the value of property 3 years ago was ₹480000.

Let rate of depreciation be r%.

Depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 14440 = 16000\Big(1 - \frac{r}{100}\Big)^2 \\[1em] \frac{14440}{16000} = \Big(1 - \frac{r}{100}\Big)^2 \\[1em] \frac{1444}{1600} = \Big(1 - \frac{r}{100}\Big)^2 \\[1em] \Big(\frac{38}{40}\Big)^2 = \Big(1 - \frac{r}{100}\Big)^2 \\[1em] 1 - \frac{r}{100} = \frac{38}{40} \\[1em] \frac{r}{100} = 1 - \frac{38}{40} \\[1em] \frac{r}{100} = \frac{2}{40} \\[1em] r = \frac{200}{40} \\[1em] r = 5%.\end{aligned}$$

Hence, the rate of depreciation = 5%.

Let rate of growth be r% per annum.

Growth formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 92610 = 80000\Big(1 + \frac{r}{100}\Big)^3 \\[1em] \frac{92610}{80000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \frac{9261}{8000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Big(\frac{21}{20}\Big)^3 = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] 1 + \frac{r}{100} = \frac{21}{20} \\[1em] \frac{r}{100} = \frac{21}{20} - 1 \\[1em] \frac{r}{100} = \frac{21 - 20}{20} \\[1em] \frac{r}{100} = \frac{1}{20} \\[1em] r = \frac{100}{20} \\[1em] r = 5%.\end{aligned}$$

Hence, the annual rate of growth of production of cars is 5%.

Let value be depreciated from ₹500000 to ₹364500 in n years.

Depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 364500 = 500000\Big(1 - \frac{10}{100}\Big)^n \\[1em] \Rightarrow \frac{364500}{500000} = \Big(\frac{100 - 10}{100}\Big)^n \\[1em] \Rightarrow \frac{729}{1000} = \Big(\frac{90}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{9}{10}\Big)^3 = \Big(\frac{9}{10}\Big)^n \\[1em] \therefore n = 3.\end{aligned}$$

Hence, the value of machine will be depreciated from ₹500000 to ₹364500 in 3 years.

When rate of interest are different then total amount is given by,

$A = P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Substituting values we get,

$$\begin{aligned}A = ₹2500000\Big(1 + \frac{5}{100}\Big)\Big(1 + \frac{10}{100}\Big) \\[1em] = ₹2500000\Big(1 + \frac{1}{20}\Big)\Big(1 + \frac{1}{10}\Big) \\[1em] = ₹2500000 \times \frac{21}{20} \times \frac{11}{10} \\[1em] = \frac{₹2500000 \times 21 \times 11}{20 \times 10} \\[1em] = ₹12500 \times 21 \times 11 \\[1em] = ₹2887500.\end{aligned}$$

Profit = Amount - Principal = ₹2887500 - ₹2500000 = ₹387500

Hence, profit = ₹387500.

Let initial value of property be V₀.

By growth formula,

$V = V_0\Big(1 + \frac{r}{100}\Big)^n.$

Substituting values we get,

$$\begin{aligned}V = V_0\Big(1 + \frac{25}{100}\Big)^3 \\[1em] = V_0\Big(1 + \frac{1}{4}\Big)^3 \\[1em] = V_0\Big(\frac{5}{4}\Big)^3 \\[1em] = V_0 \times \frac{125}{64} \\[1em] = \frac{125V_0}{64}.\end{aligned}$$

Change in value (C) = Final value - Initial value.

$$\begin{aligned}\therefore C = \frac{125V_0}{64} - V_0 \\[1em] = \frac{125V_0 - 64V_0}{64} \\[1em] = \frac{61V_0}{64}.\end{aligned}$$

Percentage increase = $\frac{\text{Change in value}}{\text{Original value}} \times 100.$

Substituting values we get,

$$\begin{aligned}\text{Percentage increase} = \frac{\frac{61V_0}{64}}{V_0} \times 100 \\[1em] = \frac{61V_0}{64V_0} \times 100 \\[1em] = \frac{6100}{64} \\[1em] = \frac{1525}{16} \\[1em] = 95\frac{5}{16}%.\end{aligned}$$

Hence, after 3 years the value of property will increase by $95\frac{5}{16}%.$

Since, value of land grows uniformly at the rate of 30% p.a. hence, by growth formula,

$V = V_0\Big(1 + \frac{r}{100}\Big)^n$.

Substituting values we get value of land after 3 years,

$$\begin{aligned}V = ₹180000\Big(1 + \frac{30}{100}\Big)^3 \\[1em] = ₹180000\Big(\frac{130}{100}\Big)^3 \\[1em] = ₹180000\Big(\frac{13}{10}\Big)^3 \\[1em] = ₹180000 \times \frac{13}{10} \times \frac{13}{10} \times \frac{13}{10} \\[1em] = ₹180000 \times \frac{2197}{1000} \\[1em] = ₹180 \times 2197 \\[1em] = ₹395460.\end{aligned}$$

Given, the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter,

Hence, value of car after 3 years = $V = V_0\Big(1 - \frac{r_1}{100}\Big)\Big(1 - \frac{r_2}{100}\Big)\Big(1 - \frac{r_3}{100}\Big)$

Substituting values, we get value of car after 3 years,

$$\begin{aligned}V = ₹320000\Big(1 - \frac{20}{100}\Big)\Big(1 - \frac{15}{100}\Big)\Big(1 - \frac{15}{100}\Big) \\[1em] = ₹320000 \times \frac{80}{100} \times \frac{85}{100} \times \frac{85}{100} \\[1em] = ₹(32 \times 4 \times 17 \times 85) \\[1em] = ₹184960.\end{aligned}$$

Present value of land and car = ₹180000 + ₹320000 = ₹500000.

Total value of land and car after 3 years = ₹395460 + ₹184960 = ₹580420.

Profit = Amount - Initial value = ₹580420 - ₹500000 = ₹80420.

After 3 years profit of Mr. Durani would be ₹80420.

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = 1000 \times \Big[\Big(1 + \frac{10}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\frac{110}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\frac{11}{10}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\frac{121}{100} - 1\Big] \\[1em] = 1000 \times \frac{21}{100} \\[1em] = ₹210.\end{aligned}$$

Hence, Option 3 is the correct option.

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

Putting values in formula we get,

$$\begin{aligned}A = ₹8000 \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\frac{110}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\frac{11}{10}\Big)^2 \\[1em] = ₹8000 \times \frac{11}{10} \times \frac{11}{10} \\[1em] = ₹8000 \times \frac{121}{100} \\[1em] = ₹9680.\end{aligned}$$

Hence, Option 3 is the correct option.

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = ₹12000 \times \Big[\Big(1 + \frac{10}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\frac{110}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\frac{11}{10}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\frac{121}{100} - 1\Big] \\[1em] = ₹12000 \times \frac{21}{100} \\[1em] = ₹120 \times 21 \\[1em] = ₹2520.\end{aligned}$$

Hence, Option 2 is the correct option.

Rate = 8% i.e. $\frac{8}{2}$% = 4% when compounded semi-annually.

n (no. of conversion periods) = 2 half-years.

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = ₹10000 \times \Big[\Big(1 + \frac{4}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\frac{104}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\frac{52}{50}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\frac{2704}{2500} - 1\Big] \\[1em] = ₹10000 \times \frac{2704 - 2500}{2500} \\[1em] = ₹10000 \times \frac{204}{2500} \\[1em] = ₹\frac{2040000}{2500} \\[1em] = ₹816.\end{aligned}$$

Hence, Option 1 is the correct option.

Rate = 20% i.e. $\frac{20}{2}$% = 10% when compounded semi-annually.

n (no. of conversion periods) = 3 half-years.

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = ₹5000 \times \Big[\Big(1 + \frac{10}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\frac{110}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\frac{11}{10}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\frac{1331}{1000} - 1\Big] \\[1em] = ₹5000 \times \frac{1331 - 1000}{1000} \\[1em] = ₹5000 \times \frac{331}{1000} \\[1em] = ₹1655.\end{aligned}$$

Hence, Option 2 is the correct option.

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

As, in compound interest the interest is always calculated on the compounded principal whereas in simple interest, the interest is calculated on the initial principle so for conversion periods ≥ 2 compound interest will be greater than simple interest.

Hence, Option 4 is the correct option.

By growth formula,

$V = V_0\Big(1 + \frac{r}{100}\Big)^n$

Substituting values in formula,

$$\begin{aligned}V = 1200000\Big(1 + \frac{8}{100}\Big)^2 \\[1em] = 1200000\Big(\frac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\frac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\frac{27}{25}\Big)^2 \\[1em] = 1200000 \times \frac{27}{25} \times \frac{27}{25} \\[1em] = 1200000 \times \frac{729}{625} \\[1em] = 1920 \times 729 \\[1em] = 1399680.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

P = ₹ 1,000

R = 10%

T = 1 year

Using C.I. formula,

C.I. = $P\Big[\Big(1 + \frac{R}{100}\Big)^n - 1\Big]$

Substituting the values, we get

$$\begin{aligned}C.I. = 1,000 \times \Big[\Big(1 + \frac{10}{100}\Big)^1 - 1\Big]\\[1em] = 1,000 \times \Big[\Big(1 + \frac{1}{10}\Big) - 1\Big]\\[1em] = 1,000 \times \Big(1 + \frac{1}{10} - 1\Big)\\[1em] = 1,000 \times \Big(\frac{1}{10}\Big)\\[1em] = 100.\end{aligned}$$

Calculating S.I.,

P = ₹ 1,000

R = 10%

T = 1 year

Using formula,

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

Substituting the values, we get

$$\begin{aligned}\Rightarrow S.I. = \frac{1,000 \times 10 \times 1}{100}\\[1em] = 10 \times 10 \\[1em] = 100.\end{aligned}$$

Since,

C.I. = S.I. = ₹ 100.

Therefore, we can say that simple and compound interest is the same for first conversion period at the same rate of interest.

∴ Both statements are true.

Hence, option 1 is the correct option.

The formula, V = V_o $\Big(1 + \frac{r}{100}\Big)^n$ is indeed the correct formula used for calculating future value in problems involving compound growth (like population growth, compound interest, etc.), where:

V is the final value.

V_o is the initial value.

r is the rate of growth (or interest) per period.

n is the number of periods.

This formula accurately reflects how a quantity increases by a certain percentage over multiple periods, with the growth compounding on the new total each period.

∴ Reason (R) is true.

Given,

V_o = 10,000

r = 10%.

n = 5 years

Substituting values we get :

$$\begin{aligned}V = 10,000 \times \Big(1 + \frac{10}{100}\Big)^5\\[1em] = 10,000 \times \Big(1 + \frac{1}{10}\Big)^5\\[1em] = 10,000 \times \Big(\frac{10 + 1}{10}\Big)^5\\[1em] = 10,000 \times \Big(\frac{11}{10}\Big)^5\\[1em] = 10,000 \times \frac{1,61,051}{1,00,000}\\[1em] = 16,105.10\end{aligned}$$

Since population cannot be in fraction, the population in 2020 would be approximately 16,105.

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Simple interest is calculated only on the original principal amount. The interest earned each period remains constant.

Compound interest is calculated on the original principal amount + all accumulated interest from previous periods. This means the principal for calculating interest increases with each conversion period.

The person earning compound interest will always have a higher amount back at the end of the investment period, if the conversion period is greater than one.

∴ Assertion (A) is false.

The principal for each conversion period increases for the compound interest calculation. This is true and is the fundamental principle of compound interest.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

When interest is compounded yearly,

P = ₹10000, r = 10%, T = 1.

C.I. = $\frac{10000 \times 10 \times 1}{100}$ = ₹1000.

When interest is compounded half-yearly,

P = ₹10000, r = $\frac{10}{2}$% = 5%, T = 2.

For first half-year,

C.I. = $\frac{10000 \times 5 \times 1}{100}$ = ₹500.

Amount after first half-year = ₹1000 + ₹500 = ₹1500.

Principal for second half-year = ₹1500.

C.I. = $\frac{10500 \times 5 \times 1}{100}$ = ₹525.

Amount after second half-year = ₹1500 + ₹525 = ₹2025.

C.I. = Final amount - Principal = ₹2025 - ₹1000 = ₹1025.

Difference in C.I. in both cases = ₹1025 - ₹1000 = ₹25.

Hence, the interest would be ₹25 more, if the sum was lent at 10% per annum, interest being compounded half-yearly.

Let the rate of interest be r% per annum.

Given, ₹3072 amounts to ₹3264 in one year.

∴ Compound interest = Final amount - Principal = ₹3264 - ₹3072 = ₹192.

$$\begin{aligned}\therefore 192 = \frac{3072 \times r \times 1}{100} \\[1em] \Rightarrow r = \frac{19200}{3072} \\[1em] \Rightarrow r = 6.25%.\end{aligned}$$

Principal for second year = ₹3264.

C.I.= $\frac{3264 \times 6.25 \times 1}{100} = \frac{20400}{100}$ = ₹204.

Amount after second year = ₹3264 + ₹204 = ₹3468.

Hence, the rate of interest = 6.25% and the amount after second year = ₹3468.

Let principal be ₹P.

We know,

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

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 28090 = P\Big(1 + \frac{6}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\frac{106}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\frac{53}{50}\Big)^2 \\[1em] \Rightarrow 28090 = P \times \frac{53}{50} \times \frac{53}{50} \\[1em] \Rightarrow P = \frac{28090 \times 50 \times 50}{53 \times 53} \\[1em] \Rightarrow P = \frac{28090 \times 2500}{2809} \\[1em] \Rightarrow P = ₹25000.\end{aligned}$$

C.I. = Final amount - Principal = ₹28090 - ₹25000 = ₹3090.

Hence, the principal = ₹25000 and compound interest = ₹3090.

(i) Let the sum be ₹P.

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

C.I. when sum is lent at 5% for 2 years,

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \frac{5}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \frac{1}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\frac{21}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\frac{441}{400} - 1\Big] \\[1em] = P\Big[\frac{441 - 400}{400} \Big] \\[1em] = P \times \frac{41}{400} \\[1em] = \frac{41P}{400}.\end{aligned}$$

C.I. when sum is lent at 6% for 2 years,

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{6}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \frac{3}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\frac{53}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\frac{2809}{2500} - 1\Big] \\[1em] = P \times \frac{2809 - 2500}{2500} \\[1em] = \frac{309P}{2500}.\end{aligned}$$

Given, difference in C.I. = ₹422.

$$\begin{aligned}\therefore \frac{309P}{2500} - \frac{41P}{400} = 422 \\[1em] \Rightarrow \frac{309P \times 4 - 41P \times 25}{10000} = 422 \\[1em] \Rightarrow \frac{1236P - 1025P}{10000} = 422 \\[1em] \Rightarrow \frac{211P}{10000} = 422 \\[1em] \Rightarrow P = \frac{422 \times 10000}{211} \\[1em] \Rightarrow P = ₹20000.\end{aligned}$$

Hence, equal sum = ₹20000.

(ii) C.I. when sum is lent at 5% for 2 years = $\frac{41P}{400}.$

$$\begin{aligned}C.I. = \frac{41 \times 20000}{400} \\[1em] = 41 \times 50 \\[1em] = ₹2050.\end{aligned}$$

C.I. when sum is lent at 6% for 2 years = $\frac{309P}{2500}.$

$$\begin{aligned}C.I. = \frac{309 \times 20000}{2500} \\[1em] = 309 \times 8 \\[1em] = ₹2472.\end{aligned}$$

Hence, C.I. = ₹2050 when sum is lent at 5% for 2 years and C.I. = ₹2472 when sum is lent at 6% for 2 years.

Let the sum be ₹x and rate be r%.

Given, S.I. for 2 years = ₹1280.

$$\begin{aligned}\therefore \frac{P \times R \times T}{100} = 1280 \\[1em] \Rightarrow \frac{x \times r \times 2}{100} = 1280 \\[1em] \Rightarrow x \times r = \frac{1280 \times 100}{2} \\[1em] \Rightarrow x \times r = 64000 .......(i)\end{aligned}$$

Given, C.I. for 2 years = ₹1331.20

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(1 + \frac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(\frac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\frac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\frac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\frac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\frac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 1331.20 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 13312000 = 64000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \frac{13312000}{64000} \\[1em] \Rightarrow r + 200 = 208 \\[1em] \Rightarrow r = 8%.\end{aligned}$$

Putting value of r in Eq. (i) we get,

$$\begin{aligned}\Rightarrow x \times 8 = 64000 \\[1em] \Rightarrow x = \frac{64000}{8} \\[1em] \Rightarrow x = ₹8000.\end{aligned}$$

Hence, sum = ₹8000 and rate = 8%.

Let the sum be ₹P.

S.I. = $\frac{P \times 10 \times 3}{100} = \frac{3P}{10}.$

By formula,

C.I. = $P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \frac{10}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\frac{110}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\frac{11}{10}\Big)^3 - 1\Big] \\[1em] = P\Big[\frac{1331}{1000} - 1\Big] \\[1em] = P\Big[\frac{1331 - 1000}{1000} \Big] \\[1em] = P \times \frac{331}{1000} \\[1em] = \frac{331P}{1000}.\end{aligned}$$

Given, difference between C.I. and S.I. = ₹232.50

$$\begin{aligned}\therefore \frac{331P}{1000} - \frac{3P}{10} = 232.50 \\[1em] \Rightarrow \frac{331P- 300P}{1000} = 232.50 \\[1em] \Rightarrow \frac{31P}{1000} = 232.50 \\[1em] \Rightarrow P = \frac{232.50 \times 1000}{31} \\[1em] \Rightarrow P = \frac{232500}{31} \\[1em] \Rightarrow P = ₹7500.\end{aligned}$$

Hence, sum = ₹7500.

Let the sum be ₹x and rate be r%.

Given, S.I. for 3 years = ₹1080.

$$\begin{aligned}\therefore \frac{P \times R \times T}{100} = 1080 \\[1em] \Rightarrow \frac{x \times r \times 3}{100} = 1080 \\[1em] \Rightarrow x \times r = \frac{1080 \times 100}{3} \\[1em] \Rightarrow x \times r = 36000 .......(i)\end{aligned}$$

Given, C.I. for 2 years = ₹741.60

$$\begin{aligned}C.I. = P\Big[\Big(1 + \frac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(1 + \frac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(\frac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\frac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\frac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\frac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\frac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 741.60 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 7416000 = 36000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \frac{7416000}{36000} \\[1em] \Rightarrow r + 200 = 206 \\[1em] \Rightarrow r = 6%.\end{aligned}$$

Hence, the rate of interest = 6%.

(ii) Putting value of r in (i) we get,

$$\begin{aligned}\Rightarrow x \times r = 36000 \\[1em] \Rightarrow 6x = 36000 \\[1em] \Rightarrow x = ₹6000.\end{aligned}$$

Hence, principal = ₹6000.

Since, interest is compounded semi-annually, rate = \dfrac{10%}{2} = 5%.

Let time be n half-years,

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

Substituting values in formula we get,

$$\begin{aligned}\Rightarrow 2646 = 2400\Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow \frac{2646}{2400} = \Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow \frac{2646}{2400} = \Big(\frac{105}{100}\Big)^n \\[1em] \Rightarrow \frac{441}{400} = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^2 = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow n = 2.\end{aligned}$$

Time = 2 half-years or 1 year.

Hence, in 1 year ₹2400 amounts to ₹2646 at 10% p.a. compounded semi-annually.

Let rate of interest be r% p.a. i.e. $\frac{r}{2}$% if compounded half-yearly.

n = $1\frac{1}{2}$ years or 3 half-years.

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

Substituting values in formula we get,

$$\begin{aligned}\Rightarrow 79860 = 60000\Big(1 + \frac{r}{200}\Big)^3 \\[1em] \Rightarrow \frac{79860}{60000} = \Big(1 + \frac{r}{200}\Big)^3 \\[1em] \Rightarrow \frac{1331}{1000} = \Big(1 + \frac{r}{200}\Big)^3 \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(1 + \frac{r}{200}\Big)^3 \\[1em] \Rightarrow \frac{11}{10} = 1 + \frac{r}{200} \\[1em] \Rightarrow \frac{11}{10} - 1 = \frac{r}{200} \\[1em] \Rightarrow \frac{1}{10} = \frac{r}{200} \\[1em] \Rightarrow r = \frac{200}{10} = 20%.\end{aligned}$$

Hence, the rate of interest = 20% per annum.

Net growth rate = 9.2% - 1.7% = 7.5%.

By growth formula,

V = $V_0\Big(1 + \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = 320000\Big(1 + \frac{7.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\frac{107.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\frac{1075}{1000}\Big)^3 \\[1em] = 320000 \times \Big(\frac{43}{40}\Big)^3 \\[1em] = 320000 \times \frac{43}{40} \times \frac{43}{40} \times \frac{43}{40} \\[1em] \\[1em] = 320000 \times \frac{79507}{64000} \\[1em] = 5 \times 79507 \\[1em] = 397535.\end{aligned}$$

Hence, the population of the town after 3 years = 397535.

(i) By depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Let initial value of car be V₀.

Putting values in formula we get,

$$\begin{aligned}\Rightarrow 315600 = V_0\Big(1 - \frac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(1 - \frac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\frac{80}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\frac{4}{5}\Big)^2 \\[1em] \Rightarrow 315600 = V_0 \times \frac{4}{5} \times \frac{4}{5} \\[1em] \Rightarrow 315600 = V_0 \times \frac{16}{25} \\[1em] \Rightarrow V_0 = 315600 \times \frac{25}{16} \\[1em] \Rightarrow V_0 = ₹493125.\end{aligned}$$

Hence, the purchase price of car was ₹493125.

(ii) By depreciation formula,

V = $V_0\Big(1 - \frac{r}{100}\Big)^n$

Putting values in formula we get,

$$\begin{aligned}V = 315600\Big(1 - \frac{20}{100}\Big)^3 \\[1em] = 315600 \times \Big(\frac{80}{100}\Big)^3 \\[1em] = 315600 \times \Big(\frac{4}{5}\Big)^3 \\[1em] = 315600 \times \Big(\frac{64}{125}\Big) \\[1em] = \frac{20198400}{125} \\[1em] = ₹161587.20\end{aligned}$$

Hence, the value of car after 3 years = ₹161587.20

Given,

Investment = ₹400000

Loss in first year = 4%,

Profit in second year = 5%,

Profit in third year = 10%.

$$\begin{aligned}\therefore \text{Amt. after 3 yrs.} = ₹400000\Big(1 - \frac{4}{100}\Big)\Big(1 + \frac{5}{100}\Big)\Big(1 + \frac{10}{100}\Big) \\[1em] = ₹400000 \times \frac{96}{100} \times \frac{105}{100} \times \frac{110}{100} \\[1em] = ₹400000 \times \frac{24}{25} \times \frac{21}{20} \times \frac{11}{10} \\[1em] = ₹400000 \times \frac{5544}{5000} \\[1em] = ₹(80 \times 5544) \\[1em] = ₹443520.\end{aligned}$$

Net profit = Amount - Principal = ₹443520 - ₹400000 = ₹43520.

Hence, net profit after 3 years = ₹43520.