Simple Interest and Compound Interest

Here,

P = ₹ 4,500, T = $2\frac{1}{2}$ years = $\frac{5}{2}$ years and R = $7\frac{2}{3}$% p.a. = $\frac{23}{3}$% p.a.

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

$= ₹ \left(4500 \times \frac{23}{3} \times \frac{5}{2} \times \frac{1}{100}\right)$ = ₹ 862.50

Amount = Principal + S.I. = ₹ (4,500 + 862.50) = ₹ 5,362.50

Hence, simple interest = ₹ 862.50 and amount = ₹ 5,362.50.

Here,

P = ₹ 6,360, T = 6 years 3 months = $6\frac{3}{12}$ years = $\frac{25}{4}$ years and R = 8% p.a.

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

$= ₹ \left(6360 \times 8 \times \frac{25}{4} \times \frac{1}{100}\right)$ = ₹ 3,180

Amount = Principal + S.I. = ₹ (6,360 + 3,180) = ₹ 9,540

Hence, simple interest = ₹ 3,180 and amount = ₹ 9,540.

Here,

P = ₹ 19,200, T = 11 months = $\frac{11}{12}$ years and R = $9\frac{3}{4}$% p.a. = $\frac{39}{4}$% p.a.

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

$= ₹ \left(19200 \times \frac{39}{4} \times \frac{11}{12} \times \frac{1}{100}\right)$ = ₹ 1,716

Amount = Principal + S.I. = ₹ (19,200 + 1,716) = ₹ 20,916

Hence, simple interest = ₹ 1,716 and amount = ₹ 20,916.

Here,

P = ₹ 58,400, T = 75 days = $\frac{75}{365}$ years = $\frac{15}{73}$ years and R = $6\frac{1}{2}$% p.a. = $\frac{13}{2}$% p.a.

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

$= ₹ \left(58400 \times \frac{13}{2} \times \frac{15}{73} \times \frac{1}{100}\right)$ = ₹ 780

Amount = Principal + S.I. = ₹ (58,400 + 780) = ₹ 59,180

Hence, simple interest = ₹ 780 and amount = ₹ 59,180.

The day on which the money is borrowed is not counted while the day on which it is returned is counted.

Month	Days
Oct.	13
Nov.	30
Dec.	31
Jan.	31
Feb.	28
Mar.	13

Time = (13 + 30 + 31 + 31 + 28 + 13) days = 146 days

= $\frac{146}{365}$ years = $\frac{2}{5}$ years

Also, P = ₹ 8,600 and R = 8% p.a.

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

$= ₹ \left(8600 \times 8 \times \frac{2}{5} \times \frac{1}{100}\right)$ = ₹ 275.20

Amount = Principal + S.I. = ₹ (8,600 + 275.20) = ₹ 8,875.20

Hence, simple interest = ₹ 275.20 and amount = ₹ 8,875.20.

Here,

P = ₹ 10,500, R = 7% p.a. and T = 5 years.

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

$= ₹ \left(\frac{10500 \times 7 \times 5}{100}\right)$ = ₹ 3,675

Amount = Principal + S.I. = ₹ (10,500 + 3,675) = ₹ 14,175

So, Sunidhi has to pay ₹ 14,175 in all.

Value of the watch = Amount − Cash paid = ₹ (14,175 − 13,000) = ₹ 1,175

Hence, the value of the watch is ₹ 1,175.

Here,

P = ₹ 7,560, S.I. = ₹ 1,102.50 and R = $6\frac{1}{4}$% p.a. = $\frac{25}{4}$% p.a.

$$\begin{aligned}\text{T} = \frac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \frac{100 \times 1102.50}{7560 \times \frac{25}{4}}\\[1em] = 100 \times 1102.50 \times \frac{1}{7560} \times \frac{4}{25}\\[1em] = \frac{441000}{189000}\\[1em] = \frac{7}{3}\end{aligned}$$

So, T = $\frac{7}{3}$ years = $2\frac{1}{3}$ years = 2 years 4 months.

Hence, the required time is 2 years 4 months.

Here,

P = ₹ 25,600, A = ₹ 35,664 and R = $9\frac{1}{4}$% p.a. = $\frac{37}{4}$% p.a.

S.I. = A − P = ₹ (35,664 − 25,600) = ₹ 10,064

$$\begin{aligned}\text{T} = \frac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \frac{100 \times 10064}{25600 \times \frac{37}{4}}\\[1em] = 100 \times 10064 \times \frac{1}{25600} \times \frac{4}{37}\\[1em] = \frac{4025600}{947200}\\[1em] = \frac{17}{4}\end{aligned}$$

So, T = $\frac{17}{4}$ years = $4\frac{1}{4}$ years = 4 years 3 months.

Hence, the required time is 4 years 3 months.

Here,

P = ₹ 1,625, A = ₹ 2,080 and T = $3\frac{1}{2}$ years = $\frac{7}{2}$ years.

S.I. = A − P = ₹ (2,080 − 1,625) = ₹ 455

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

$= \frac{100 \times 455}{1625 \times \frac{7}{2}}$%

$= \left(100 \times 455 \times \frac{1}{1625} \times \frac{2}{7}\right)$%

$= \frac{91000}{11375}$%

= 8%

Hence, the required rate is 8% p.a.

Here,

P = ₹ 6,720, S.I. = ₹ 1,911 and T = 3 years 3 months = $3\frac{3}{12}$ years = $\frac{13}{4}$ years.

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

$= \frac{100 \times 1911}{6720 \times \frac{13}{4}}$%

$= \left(100 \times 1911 \times \frac{1}{6720} \times \frac{4}{13}\right)$%

$= \frac{764400}{87360}$%

$= \frac{35}{4}$%

So, R = $\frac{35}{4}$% p.a. = $8\frac{3}{4}$% p.a.

Hence, the required rate is $\mathbf{8\frac{3}{4}}$% p.a.

Let the sum be ₹ $x$. Then,

P = ₹ $x$, A = ₹ $2x$, T = 12 years

S.I. = A − P = ₹ $(2x - x) = ₹ x$

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

$= \frac{100 \times x}{x \times 12}$%

$= \frac{100}{12}$%

$= \frac{25}{3}$%

= $8\frac{1}{3}$%

Hence, the required rate is $\mathbf{8\frac{1}{3}}$% p.a.

Let the sum be ₹ $x$. Then, S.I. = ₹ $\left(\frac{9}{16}x\right)$.

Let the rate be R% per annum and the time be R years.

$$\begin{aligned}\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \frac{9x}{16} = \frac{x \times \text{R} \times \text{R}}{100}\\[1em] \Rightarrow \text{R}^2 = \frac{9}{16} \times 100\\[1em] \Rightarrow \text{R}^2 = \frac{900}{16}\\[1em] \Rightarrow \text{R} = \sqrt{\frac{900}{16}}\\[1em] = \frac{\sqrt{900}}{\sqrt{16}}\\[1em] = \frac{30}{4}\\[1em] = \frac{15}{2}\end{aligned}$$

So, R = $\frac{15}{2} = 7\frac{1}{2}$

R% = $7\frac{1}{2}$%

Hence, rate = $\mathbf{7\frac{1}{2}}$% p.a. and time = $\mathbf{7\frac{1}{2}}$ years.

Here,

S.I. = ₹ 406, T = 1 year 2 months = $1\frac{2}{12}$ years = $\frac{7}{6}$ years and R = $6\frac{1}{4}$% p.a. = $\frac{25}{4}$% p.a.

$$\begin{aligned}\text{P} = \frac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \frac{100 \times 406}{\frac{25}{4} \times \frac{7}{6}}\\[1em] = 100 \times 406 \times \frac{4}{25} \times \frac{6}{7}\\[1em] = \frac{974400}{175}\\[1em] = 5568\end{aligned}$$

Hence, the required sum is ₹ 5,568.

Let the sum be ₹ $x$.

Then, P = ₹ $x$, T = $2\frac{1}{2}$ years = $\frac{5}{2}$ years and R = $7\frac{1}{2}$% p.a. = $\frac{15}{2}$% p.a.

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

$= ₹ \left(x \times \frac{15}{2} \times \frac{5}{2} \times \frac{1}{100}\right) = ₹ \left(\frac{3x}{16}\right)$

Amount = P + S.I. = ₹ $\left(x + \frac{3x}{16}\right) = ₹ \left(\frac{19x}{16}\right)$

But, amount = ₹ 1,748.

$$\begin{aligned}\Rightarrow \frac{19x}{16} = 1748\\[1em] \Rightarrow x = 1748 \times \frac{16}{19}\\[1em] \Rightarrow x = 1472\end{aligned}$$

Hence, the required sum is ₹ 1,472.

Let the sum be ₹ P. Then, A = ₹ $\left(\frac{8}{5}\text{P}\right)$ and T = 5 years.

S.I. = A − P = ₹ $\left(\frac{8\text{P}}{5} - \text{P}\right) = ₹ \left(\frac{3\text{P}}{5}\right)$

$$\begin{aligned}\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \frac{3\text{P}}{5} = \frac{\text{P} \times \text{R} \times 5}{100}\\[1em] \Rightarrow \frac{3\text{P}}{5} = \frac{\text{PR}}{20}\\[1em] \Rightarrow \text{R} = \frac{3 \times 20}{5}\\[1em] \Rightarrow \text{R} = 12\end{aligned}$$

Hence, the rate of interest is 12% p.a.

For ₹ 8,560, we have P = ₹ 8,560, R = 12% p.a. and T = 5 years.

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

$= ₹ \left(\frac{8560 \times 12 \times 5}{100}\right)$ = ₹ 5,136

Now, S.I. = ₹ 5,136, R = $12\frac{1}{2}$% p.a. = $\frac{25}{2}$% p.a. and T = 4 years.

$$\begin{aligned}\text{P} = \frac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \frac{100 \times 5136}{\frac{25}{2} \times 4}\\[1em] = 100 \times 5136 \times \frac{2}{25} \times \frac{1}{4}\\[1em] = \frac{1027200}{100}\\[1em] = 10272\end{aligned}$$

Hence, the required sum is ₹ 10,272.

Here,

P = ₹ 1,250, A = ₹ 1,550 and T = 3 years.

S.I. = A − P = ₹ (1,550 − 1,250) = ₹ 300

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

$= \frac{100 \times 300}{1250 \times 3}$%

$= \frac{30000}{3750}$%

= 8%

So, the rate of interest is 8% p.a.

Now, P = ₹ 3,200, R = 8% p.a. and T = 4 years.

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

$= ₹ \left(\frac{3200 \times 8 \times 4}{100}\right)$ = ₹ 1,024

Amount = Principal + S.I. = ₹ (3,200 + 1,024) = ₹ 4,224

Hence, ₹ 3,200 will amount to ₹ 4,224 in 4 years.

S.I. for 3 years = (Amount after 5 years) − (Amount after 2 years)

= ₹ (4,160 − 3,224)

= ₹ 936

S.I. for 1 year = ₹ $\left(\frac{936}{3}\right)$ = ₹ 312

S.I. for 2 years = ₹ (312 × 2) = ₹ 624

∴ Sum = (Amount after 2 years) − (S.I. for 2 years)

= ₹ (3,224 − 624)

= ₹ 2,600

Now, P = ₹ 2,600, S.I. = ₹ 624 and T = 2 years.

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

= $\left(\frac{100 \times 624}{2600 \times 2}\right)$% p.a.

= 12% p.a.

Hence, the required sum is ₹ 2,600 and the rate of interest is 12% p.a.

Interest earned from B = ₹ $\left(\frac{2500 \times 7 \times 4}{100}\right)$ = ₹ 700

∴ Interest earned from C = ₹ (1,120 − 700) = ₹ 420

Now, for C, we have S.I. = ₹ 420, R = 7% p.a. and T = 4 years.

$$\begin{aligned}\text{P} = \frac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \frac{100 \times 420}{7 \times 4}\\[1em] = \frac{42000}{28}\\[1em] = 1500\end{aligned}$$

Hence, the sum lent to C is ₹ 1,500.

Let the sum be ₹ $x$. Then,

S.I. on ₹ $x$ for 3 years at 8% p.a. = ₹ $\left(\frac{x \times 8 \times 3}{100}\right) = ₹ \left(\frac{24x}{100}\right)$

S.I. on ₹ $x$ for 2 years at 9% p.a. = ₹ $\left(\frac{x \times 9 \times 2}{100}\right) = ₹ \left(\frac{18x}{100}\right)$

Since the simple interest for 3 years at 8% is ₹ 96 more than the simple interest for 2 years at 9%,

$$\begin{aligned}\Rightarrow \frac{24x}{100} - \frac{18x}{100} = 96\\[1em] \Rightarrow \frac{6x}{100} = 96\\[1em] \Rightarrow x = \frac{96 \times 100}{6}\\[1em] \Rightarrow x = 1600\end{aligned}$$

Hence, the required sum is ₹ 1,600.

Let each sum be ₹ P.

S.I. for $4\frac{1}{2}$ years = ₹ $\left(\frac{\text{P} \times 11 \times 9}{2 \times 100}\right)$

S.I. for $3\frac{1}{2}$ years = ₹ $\left(\frac{\text{P} \times 11 \times 7}{2 \times 100}\right)$

Since the difference between the two interests is ₹ 412.50,

$$\begin{aligned}\Rightarrow \frac{\text{P} \times 11 \times 9}{2 \times 100} - \frac{\text{P} \times 11 \times 7}{2 \times 100} = 412.50\\[1em] \Rightarrow \frac{11\text{P}}{200} \times (9 - 7) = 412.50\\[1em] \Rightarrow \frac{11\text{P}}{200} \times 2 = 412.50\\[1em] \Rightarrow \frac{11\text{P}}{100} = 412.50\\[1em] \Rightarrow \text{P} = \frac{412.50 \times 100}{11}\\[1em] \Rightarrow \text{P} = 3750\end{aligned}$$

Hence, each sum is ₹ 3,750.

Let the first part be ₹ $x$.

Then, second part = ₹ $(6000 - x)$.

S.I. on ₹ $x$ at 12% p.a. for 9 months, i.e. $\frac{3}{4}$ years

= ₹ $\left(x \times 12 \times \frac{3}{4} \times \frac{1}{100}\right) = ₹ \left(\frac{9x}{100}\right)$

S.I. on ₹ $(6000 - x)$ at 10% p.a. for $1\frac{1}{2}$ years, i.e. $\frac{3}{2}$ years

= ₹ $\bigg\lbrace (6000 - x) \times 10 \times \frac{3}{2} \times \frac{1}{100} \bigg\rbrace = ₹ \bigg\lbrace \frac{15(6000 - x)}{100}\bigg\rbrace$

Since both the interests are equal,

$$\begin{aligned}\Rightarrow \frac{9x}{100} = \frac{15(6000 - x)}{100}\\[1em] \Rightarrow 9x = 90000 - 15x\\[1em] \Rightarrow 24x = 90000\\[1em] \Rightarrow x = 3750\end{aligned}$$

Second part = ₹ (6,000 − 3,750) = ₹ 2,250

Hence, the two parts are ₹ 3,750 and ₹ 2,250.

Let the first part be ₹ $x$.

Then, second part = ₹ $(13500 - x)$.

2 years 9 months = $2\frac{9}{12}$ years = $\frac{11}{4}$ years

1 year 8 months = $1\frac{8}{12}$ years = $\frac{5}{3}$ years

S.I. on ₹ $x$ at $\frac{25}{3}$% p.a. for $\frac{11}{4}$ years

S.I. = ₹ $\left(x \times \frac{25}{3} \times \frac{11}{4} \times \frac{1}{100}\right) = ₹ \left(\frac{11x}{48}\right)$

S.I. on ₹ $(13500 - x)$ at $\frac{15}{2}$% p.a. for $\frac{5}{3}$ years

S.I. = ₹ $\bigg\lbrace (13500 - x) \times \frac{15}{2} \times \frac{5}{3} \times \frac{1}{100}\bigg\rbrace = ₹ \bigg\lbrace \frac{13500 - x}{8}\bigg\rbrace$

Since the total interest received is ₹ 2,375,

$$\begin{aligned}\Rightarrow \frac{11x}{48} + \frac{13500 - x}{8} = 2375\\[1em] \Rightarrow 11x + 6(13500 - x) = 2375 \times 48\\[1em] \Rightarrow 11x + 81000 - 6x = 114000\\[1em] \Rightarrow 5x = 33000\\[1em] \Rightarrow x = 6600\end{aligned}$$

Second part = ₹ (13,500 − 6,600) = ₹ 6,900

Hence, the two parts are ₹ 6,600 and ₹ 6,900.

Let the sum be ₹ $x$. Then,

P = ₹ $x$, A = ₹ $4x$, R = $8\frac{1}{3}$% p.a. = $\frac{25}{3}$% p.a.

S.I. = A − P = ₹ $(4x - x) = ₹ 3x$

$$\begin{aligned}\text{T} = \frac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \frac{100 \times 3x}{x \times \frac{25}{3}}\\[1em] = 100 \times 3x \times \frac{1}{x} \times \frac{3}{25}\\[1em] = \frac{900}{25}\\[1em] = 36\end{aligned}$$

Hence, the sum will become 4 times of itself in 36 years.

Let the required sum be ₹ P.

S.I. on ₹ P at $\frac{20}{3}$% p.a. for 6 years = ₹ $\left(\frac{\text{P} \times 20 \times 6}{3 \times 100}\right)$

S.I. on ₹ 3,200 at $\frac{42}{5}$% p.a. for 7 years = ₹ $\left(\frac{3200 \times 42 \times 7}{5 \times 100}\right)$

Since the two simple interests are equal,

$$\begin{aligned}\Rightarrow \frac{\text{P} \times 20 \times 6}{3 \times 100} = \frac{3200 \times 42 \times 7}{5 \times 100}\\[1em] \Rightarrow \frac{2\text{P}}{5} = \frac{940800}{500}\\[1em] \Rightarrow \frac{2\text{P}}{5} = \frac{9408}{5}\\[1em] \Rightarrow 2\text{P} = 9408\\[1em] \Rightarrow \text{P} = 4704\end{aligned}$$

Hence, the required sum is ₹ 4,704.

Let the rate of interest be R% p.a. and T = $3\frac{1}{2}$ years = $\frac{7}{2}$ years.

Interest paid by Naveen = ₹ $\left(42000 \times \text{R} \times \frac{7}{2} \times \frac{1}{100}\right)$ = ₹ 1,470R

Interest paid by Praveen = ₹ $\left(55000 \times \text{R} \times \frac{7}{2} \times \frac{1}{100}\right)$ = ₹ 1,925R

Since Praveen has to pay ₹ 3,640 more interest than Naveen,

$$\begin{aligned}\Rightarrow 1925\text{R} - 1470\text{R} = 3640\\[1em] \Rightarrow 455\text{R} = 3640\\[1em] \Rightarrow \text{R} = \frac{3640}{455}\\[1em] \Rightarrow \text{R} = 8\end{aligned}$$

Hence, the rate of interest is 8% p.a.

Let sum of money be ₹ P and rate of interest be R%.

Case I : T = 2 years

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

$= \frac{\text{P} \times \text{R} \times 2}{100}$

Case II : T = 2 years, new rate = (R + 3)%

∴ S.I. = $\frac{\text{P} \times (\text{R} + 3) \times 2}{100}$

But the difference in these two S.I.'s is ₹ 720.

$$\begin{aligned}\therefore \frac{\text{P} \times (\text{R} + 3) \times 2}{100} - \frac{\text{P} \times \text{R} \times 2}{100} = 720\\[1em] \Rightarrow 2\text{PR} + 6\text{P} - 2\text{PR} = 720 \times 100\\[1em] \Rightarrow 6\text{P} = 72000\\[1em] \Rightarrow \text{P} = \frac{72000}{6}\\[1em] \Rightarrow \text{P} = 12000\end{aligned}$$

Hence, the required sum is ₹ 12,000.

Let the sum be ₹ P and the time be T years.

$$\begin{aligned}\frac{\text{P} \times 6 \times \text{T}}{100} = 960\\[1em] \Rightarrow 6\text{PT} = 96000\\[1em] \Rightarrow \text{PT} = 16000 \qquad \text{.....(i)}\end{aligned}$$

Also, when the time is (T + 5) years,

$$\begin{aligned}\frac{\text{P} \times 6 \times (\text{T} + 5)}{100} = 2160\\[1em] \Rightarrow 6\text{PT} + 30\text{P} = 216000\\[1em] \Rightarrow \text{PT} + 5\text{P} = 36000 \qquad \text{.....(ii)}\end{aligned}$$

Substituting PT = 16000 from equation (i) in equation (ii),

$$\begin{aligned}16000 + 5\text{P} = 36000\\[1em] \Rightarrow 5\text{P} = 36000 - 16000\\[1em] \Rightarrow 5\text{P} = 20000\\[1em] \Rightarrow \text{P} = 4000\end{aligned}$$

Hence, the required sum is ₹ 4,000.

Principal for the first year = ₹ 5,000

Interest for the first year = ₹ $\left(\frac{5000 \times 8 \times 1}{100}\right)$ = ₹ 400

Amount at the end of the first year = ₹ (5,000 + 400) = ₹ 5,400

Principal for the second year = ₹ 5,400

Interest for the second year = ₹ $\left(\frac{5400 \times 8 \times 1}{100}\right)$ = ₹ 432

Amount at the end of the second year = ₹ (5,400 + 432) = ₹ 5,832

∴ Amount = ₹ 5,832

Compound Interest = (Amount) − (Principal) = ₹ (5,832 − 5,000) = ₹ 832

Hence, amount = ₹ 5,832 and compound interest = ₹ 832.

Principal for the first year = ₹ 8,000

Interest for the first year = ₹ $\left(\frac{8000 \times 6 \times 1}{100}\right)$ = ₹ 480

Amount at the end of the first year = ₹ (8,000 + 480) = ₹ 8,480

Principal for the second year = ₹ 8,480

Interest for the second year = ₹ $\left(\frac{8480 \times 6 \times 1}{100}\right)$ = ₹ 508.80

Amount at the end of the second year = ₹ (8,480 + 508.80) = ₹ 8,988.80

∴ Amount = ₹ 8,988.80

Compound Interest = (Amount) − (Principal) = ₹ (8,988.80 − 8,000) = ₹ 988.80

Hence, amount = ₹ 8,988.80 and compound interest = ₹ 988.80.

Principal for the first year = ₹ 2,500

Interest rate during first year = 6% p.a.

Interest for the first year = ₹ $\left(\frac{2500 \times 6 \times 1}{100}\right)$ = ₹ 150

Amount at the end of the first year = ₹ (2,500 + 150) = ₹ 2,650

Principal for the second year = ₹ 2,650

Interest rate during second year = 8% p.a.

Interest for the second year = ₹ $\left(\frac{2650 \times 8 \times 1}{100}\right)$ = ₹ 212

Amount at the end of the second year = ₹ (2,650 + 212) = ₹ 2,862

∴ Amount = ₹ 2,862

Compound Interest = (Amount) − (Principal) = ₹ (2,862 − 2,500) = ₹ 362

Hence, amount = ₹ 2,862 and compound interest = ₹ 362.

Principal for the first year = ₹ 25,000

Interest for the first year = ₹ $\left(\frac{25000 \times 6 \times 1}{100}\right)$ = ₹ 1,500

Amount at the end of the first year = ₹ (25,000 + 1,500) = ₹ 26,500

Principal for the second year = ₹ 26,500

Interest for the second year = ₹ $\left(\frac{26500 \times 6 \times 1}{100}\right)$ = ₹ 1,590

Amount at the end of the second year = ₹ (26,500 + 1,590) = ₹ 28,090

Principal for the third year = ₹ 28,090

Interest for the third year = ₹ $\left(\frac{28090 \times 6 \times 1}{100}\right)$ = ₹ 1,685.40

Amount at the end of the third year = ₹ (28,090 + 1,685.40) = ₹ 29,775.40

∴ Amount = ₹ 29,775.40

Compound Interest = (Amount) − (Principal) = ₹ (29,775.40 − 25,000) = ₹ 4,775.40

Hence, amount = ₹ 29,775.40 and compound interest = ₹ 4,775.40.

Principal for the first year = ₹ 10,000

Interest for the first year = ₹ $\left(\frac{10000 \times 10 \times 1}{100}\right)$ = ₹ 1,000

Amount at the end of the first year = ₹ (10,000 + 1,000) = ₹ 11,000

Principal for the second year = ₹ 11,000

Interest for the second year = ₹ $\left(\frac{11000 \times 10 \times 1}{100}\right)$ = ₹ 1,100

Amount at the end of the second year = ₹ (11,000 + 1,100) = ₹ 12,100

Principal for the third year = ₹ 12,100

Interest for the third year = ₹ $\left(\frac{12100 \times 10 \times 1}{100}\right)$ = ₹ 1,210

Amount at the end of the third year = ₹ (12,100 + 1,210) = ₹ 13,310

∴ Amount = ₹ 13,310

Compound Interest = (Amount) − (Principal) = ₹ (13,310 − 10,000) = ₹ 3,310

Hence, amount = ₹ 13,310 and compound interest = ₹ 3,310.

Principal for the first year = ₹ 25,000

Interest for the first year = ₹ $\left(\frac{25000 \times 12 \times 1}{100}\right)$ = ₹ 3,000

Amount at the end of the first year = ₹ (25,000 + 3,000) = ₹ 28,000

Principal for the second year = ₹ 28,000

Interest for the second year = ₹ $\left(\frac{28000 \times 12 \times 1}{100}\right)$ = ₹ 3,360

Amount at the end of the second year = ₹ (28,000 + 3,360) = ₹ 31,360

Principal for the third year = ₹ 31,360

Interest for the third year = ₹ $\left(\frac{31360 \times 12 \times 1}{100}\right)$ = ₹ 3,763.20

Amount at the end of the third year = ₹ (31,360 + 3,763.20) = ₹ 35,123.20

Hence, Karim will have to pay ₹ 35,123.20 at the end of 3 years.

Principal for the first year = ₹ 15,625

Interest for the first year = ₹ $\left(\frac{15625 \times 8 \times 1}{100}\right)$ = ₹ 1,250

Amount at the end of the first year = ₹ (15,625 + 1,250) = ₹ 16,875

Principal for the second year = ₹ 16,875

Interest for the second year = ₹ $\left(\frac{16875 \times 8 \times 1}{100}\right)$ = ₹ 1,350

Amount at the end of the second year = ₹ (16,875 + 1,350) = ₹ 18,225

Principal for the third year = ₹ 18,225

Interest for the third year = ₹ $\left(\frac{18225 \times 8 \times 1}{100}\right)$ = ₹ 1,458

Amount at the end of the third year = ₹ (18,225 + 1,458) = ₹ 19,683

Hence, Manoj will get ₹ 19,683 after 3 years.

Here,

P = ₹ 16,000, R = $12\frac{1}{2}$% p.a. = $\frac{25}{2}$% p.a. and T = 2 years.

For simple interest :

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

$= ₹ \left(16000 \times \frac{25}{2} \times 2 \times \frac{1}{100}\right)$ = ₹ 4,000

For compound interest :

Principal for the first year = ₹ 16,000

Interest for the first year = ₹ $\left(16000 \times \frac{25}{2} \times \frac{1}{100}\right)$ = ₹ 2,000

Amount at the end of the first year = ₹ (16,000 + 2,000) = ₹ 18,000

Principal for the second year = ₹ 18,000

Interest for the second year = ₹ $\left(18000 \times \frac{25}{2} \times \frac{1}{100}\right)$ = ₹ 2,250

Amount at the end of the second year = ₹ (18,000 + 2,250) = ₹ 20,250

C.I. = (Amount) − (Principal) = ₹ (20,250 − 16,000) = ₹ 4,250

∴ S.I. : C.I. = 4000 : 4250 = 16 : 17

Hence, the required ratio is 16 : 17.

Clearly, we have

P = ₹ 1,00,000, R = 10% p.a. and $n$ = 3 years.

Using the formula, $\text{A} = \text{P}\left(1 + \frac{\text{R}}{100}\right)^n$, we get :

$$\begin{aligned}\text{Amount after 3 years} = 100000 \times \left(1 + \frac{10}{100}\right)^3\\[1em] = 100000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\\[1em] = 133100\end{aligned}$$

Hence, Mr Dubey will have to pay ₹ 1,33,100 to the bank after 3 years.

Clearly, we have

P = ₹ 10,240, R = $12\frac{1}{2}$% p.a. = $\frac{25}{2}$% p.a. and n = 3 years.

$$\begin{aligned}\text{Amount after 3 years} = 10240 \times \left(1 + \frac{25}{2 \times 100}\right)^3\\[1em] = 10240 \times \frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\\[1em] = 14580\end{aligned}$$

∴ C.I. after 3 years = Amount − Principal = ₹ (14,580 − 10,240) = ₹ 4,340

Hence, amount = ₹ 14,580 and compound interest = ₹ 4,340.

Present population = 176400 and rate of increase = 5% p.a.

$$\begin{aligned}\text{Population after 2 years} = 176400 \times \left(1 + \frac{5}{100}\right)^2\\[1em] = 176400 \times \frac{21}{20} \times \frac{21}{20}\\[1em] = 194481\end{aligned}$$

Hence, the population after 2 years will be 194481.

Here,

P = ₹ 50,000, r₁ = 4% p.a., r₂ = 5% p.a. and r₃ = 10% p.a.

$$\begin{aligned}\text{Capital amount after 3 years} = 50000 \times \left(1 + \frac{4}{100}\right) \times \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10}{100}\right)\\[1em] = 50000 \times \frac{26}{25} \times \frac{21}{20} \times \frac{11}{10}\\[1em] = 60060\end{aligned}$$

Hence, Amit's capital amount after 3 years will be ₹ 60,060.

Here,

height 3 years ago = 125 cm and rate of increase = 20% p.a.

$$\begin{aligned}\text{Present height} = 125 \times \left(1 + \frac{20}{100}\right)^3\\[1em] = 125 \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\\[1em] = 216\end{aligned}$$

Hence, the present height of the mango tree is 216 cm.

Here,

population 2 years ago = 10000, r₁ = 5% p.a. and r₂ = 6% p.a.

$$\begin{aligned}\text{Present population} = 10000 \times \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{6}{100}\right)\\[1em] = 10000 \times \frac{21}{20} \times \frac{53}{50}\\[1em] = 11130\end{aligned}$$

Hence, the present population of the town is 11130.

Here,

P = ₹ 16,000, R = $7\frac{1}{2}$% p.a. = $\frac{15}{2}$% p.a. and T = 2 years.

Interest paid by Mahesh (simple interest) :

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

$= ₹ \left(16000 \times \frac{15}{2} \times 2 \times \frac{1}{100}\right)$ = ₹ 2,400

Interest received by Mahesh (compound interest) :

$$\begin{aligned}\text{Amount after 2 years} = 16000 \times \left(1 + \frac{15}{2 \times 100}\right)^2\\[1em] = 16000 \times \frac{43}{40} \times \frac{43}{40}\\[1em] = 18490\end{aligned}$$

C.I. = Amount − Principal = ₹ (18,490 − 16,000) = ₹ 2,490

∴ Gain = C.I. − S.I. = ₹ (2,490 − 2,400) = ₹ 90

Hence, Mahesh gains ₹ 90 at the end of 2 years.

Here,

present value = ₹ 6,25,000 and rate of depreciation = 8% p.a.

$$\begin{aligned}\text{Value of the machine after 2 years} = 625000 \times \left(1 - \frac{8}{100}\right)^2\\[1em] = 625000 \times \frac{23}{25} \times \frac{23}{25}\\[1em] = 529000\end{aligned}$$

Hence, the value of the machine after 2 years will be ₹ 5,29,000.

Here,

present value = ₹ 3,48,000, r₁ = 10% p.a. and r₂ = 20% p.a.

$$\begin{aligned}\text{Value of the car after 2 years} = 348000 \times \left(1 - \frac{10}{100}\right) \times \left(1 - \frac{20}{100}\right)\\[1em] = 348000 \times \frac{9}{10} \times \frac{4}{5}\\[1em] = 250560\end{aligned}$$

Hence, the value of the car after 2 years will be ₹ 2,50,560.

Here,

P = ₹ 1,20,000, R = 8% per annum = 4% per half year

and n = 1 year = 2 half years.

$$\begin{aligned}\text{Amount after 1 year} = 120000 \times \left(1 + \frac{4}{100}\right)^2\\[1em] = 120000 \times \frac{26}{25} \times \frac{26}{25}\\[1em] = 129792\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (1,29,792 − 1,20,000) = ₹ 9,792

Hence, amount = ₹ 1,29,792 and compound interest = ₹ 9,792.

Here,

P = ₹ 32,500, R = 12% per annum = 6% per half year

and n = 1 year = 2 half years.

$$\begin{aligned}\text{Amount after 1 year} = 32500 \times \left(1 + \frac{6}{100}\right)^2\\[1em] = 32500 \times \frac{53}{50} \times \frac{53}{50}\\[1em] = 36517\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (36,517 − 32,500) = ₹ 4,017

Hence, amount = ₹ 36,517 and compound interest = ₹ 4,017.

Here,

P = ₹ 24,000, R = 10% per annum = 5% per half year

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

$$\begin{aligned}\text{Amount after } 1\frac{1}{2} \text{ years} = 24000 \times \left(1 + \frac{5}{100}\right)^3\\[1em] = 24000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\\[1em] = 27783\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (27,783 − 24,000) = ₹ 3,783

Hence, amount = ₹ 27,783 and compound interest = ₹ 3,783.

Here,

P = ₹ 10,000, R = $\frac{12}{4}$% per quarter = 3% per quarter

and n = 6 months = 2 quarters.

$$\begin{aligned}\text{Amount after 6 months} = 10000 \times \left(1 + \frac{3}{100}\right)^2\\[1em] = 10000 \times \frac{103}{100} \times \frac{103}{100}\\[1em] = 10609\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (10,609 − 10,000) = ₹ 609

Hence, amount = ₹ 10,609 and compound interest = ₹ 609.

Here,

P = ₹ 15,625, R = $\frac{16}{4}$% per quarter = 4% per quarter

and n = 9 months = 3 quarters.

$$\begin{aligned}\text{Amount after 9 months} = 15625 \times \left(1 + \frac{4}{100}\right)^3\\[1em] = 15625 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\\[1em] = 17576\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (17,576 − 15,625) = ₹ 1,951

Hence, amount = ₹ 17,576 and compound interest = ₹ 1,951.

Here,

P = ₹ 25,60,000, R = $\frac{10}{4}$% per quarter = $\frac{5}{2}$% per quarter

and n = 1 year = 4 quarters.

$$\begin{aligned}\text{Amount after 1 year} = 2560000 \times \left(1 + \frac{5}{2 \times 100}\right)^4\\[1em] = 2560000 \times \frac{41}{40} \times \frac{41}{40} \times \frac{41}{40} \times \frac{41}{40}\\[1em] = 2825761\end{aligned}$$

∴ C.I. = (Amount) − (Principal) = ₹ (28,25,761 − 25,60,000) = ₹ 2,65,761

Hence, amount = ₹ 28,25,761 and compound interest = ₹ 2,65,761.

Let the sum be ₹ P.

Here,

A = ₹ 840, R = 8% p.a. and T = 5 years.

$$\begin{aligned}\text{A} = \text{P} + \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow 840 = \text{P} + \frac{\text{P} \times 8 \times 5}{100}\\[1em] \Rightarrow 840 = \frac{\text{P}(100 + 5 \times 8)}{100}\\[1em] \Rightarrow \text{P} = \frac{100 \times 840}{100 + 5 \times 8}\end{aligned}$$

Hence, Option 4 is the correct option.

Here,

P = ₹ 64, A = ₹ 83.20 and T = 2 years.

S.I. = A − P = ₹ (83.20 − 64) = ₹ 19.20

R = $\frac{100 \times \text{S.I.}}{\text{P} \times \text{T}}$% p.a. = $\left(\frac{100 \times 19.20}{64 \times 2}\right)$% p.a. = 15% p.a.

Now, P = ₹ 86, R = 15% p.a. and T = 4 years.

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

$= ₹ \left(\frac{86 \times 15 \times 4}{100}\right)$ = ₹ 51.60

Amount = ₹ (86 + 51.60) = ₹ 137.60

Hence, Option 4 is the correct option.

Let the sum be ₹ P.

Here,

S.I. = ₹ $x$, R = $x$% p.a. and T = $x$ years.

$$\begin{aligned}\text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow x = \frac{\text{P} \times x \times x}{100}\\[1em] \Rightarrow 100x = \text{P}x^2\\[1em] \Rightarrow \text{P} = \frac{100}{x}\end{aligned}$$

Hence, Option 3 is the correct option.

Let the capital be ₹ P.

Fall in the rate of interest = $\left(13 - 12\frac{1}{2}\right)$% p.a. = $\frac{1}{2}$% p.a.

$$\begin{aligned}\Rightarrow \frac{\text{P} \times \frac{1}{2} \times 1}{100} = 104\\[1em] \Rightarrow \frac{\text{P}}{200} = 104\\[1em] \Rightarrow \text{P} = 104 \times 200\\[1em] \Rightarrow \text{P} = 20800\end{aligned}$$

Hence, Option 1 is the correct option.

Let the sum lent by A be ₹ P. Then, the sum lent by B = ₹ 2P.

Interest received by A = ₹ $\left(\frac{\text{P} \times 5 \times 10}{100}\right) = ₹ \left(\frac{\text{P}}{2}\right)$

Interest received by B = ₹ $\left(\frac{2\text{P} \times 5 \times 5}{100}\right) = ₹ \left(\frac{\text{P}}{2}\right)$

So, A and B will get the same amount as interest.

Hence, Option 3 is the correct option.

Interest paid by B to A = ₹ $\left(\frac{720 \times 8 \times 3}{100}\right)$ = ₹ 172.80

Interest received by B from C = ₹ $\left(720 \times \frac{21}{2} \times 2 \times \frac{1}{100}\right)$ = ₹ 151.20

As ₹ 151.20 < ₹ 172.80, so B has a loss in the whole transaction.

Loss = ₹ (172.80 − 151.20) = ₹ 21.60

Hence, Option 1 is the correct option.

Here,

P = ₹ 800, A = ₹ 920 and T = 3 years.

S.I. = A − P = ₹ (920 − 800) = ₹ 120

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

= $\left(\frac{100 \times 120}{800 \times 3}\right)$% p.a. = 5% p.a.

New rate = (5 + 3)% p.a. = 8% p.a.

New S.I. = ₹ $\left(\frac{800 \times 8 \times 3}{100}\right)$ = ₹ 192

New amount = ₹ (800 + 192) = ₹ 992

Hence, Option 1 is the correct option.

Let the rate of interest be R% p.a.

Since the total simple interest received is ₹ 900,

$$\begin{aligned}\Rightarrow \frac{6000 \times \text{R} \times 2}{100} + \frac{1500 \times \text{R} \times 4}{100} = 900\\[1em] \Rightarrow 120\text{R} + 60\text{R} = 900\\[1em] \Rightarrow 180\text{R} = 900\\[1em] \Rightarrow \text{R} = 5\end{aligned}$$

Hence, Option 2 is the correct option.

Let the sum be ₹ P. If the money gets doubled in 5 years, then S.I. = ₹ P and T = 5 years.

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

= $\left(\frac{100 \times \text{P}}{\text{P} \times 5}\right)$% p.a.

= 20% p.a.

So, Statement (i) is false and Statement (ii) is true.

If the money gets doubled in 5 years, then S.I. for 5 years = ₹ P.

∴ S.I. for 10 years = ₹ 2P

Amount after 10 years = ₹ (P + 2P) = ₹ 3P

So, the money becomes three times and not four times in 10 years. Thus, Statement (iii) is false.

Therefore, (ii) alone is correct.

Hence, Option 2 is the correct option.

Let the sum be ₹ P. Then, A = ₹ 2P and S.I. = ₹ (2P − P) = ₹ P.

Also, R = $6\frac{1}{4}$% p.a. = $\frac{25}{4}$% p.a.

$$\begin{aligned}\text{T} = \frac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \frac{100 \times \text{P}}{\text{P} \times \frac{25}{4}}\\[1em] = 100 \times \text{P} \times \frac{1}{\text{P}} \times \frac{4}{25}\\[1em] = \frac{400}{25}\\[1em] = 16\end{aligned}$$

Hence, Option 4 is the correct option.

Let the sum be ₹ P. Then, A = ₹ $\left(\frac{8}{5}\text{P}\right)$ and T = 5 years.

S.I. = A − P = ₹ $\left(\frac{8\text{P}}{5} - \text{P}\right) = ₹ \left(\frac{3\text{P}}{5}\right)$

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

$= \left(100 \times \frac{3\text{P}}{5} \times \frac{1}{\text{P} \times 5}\right)$%

$= \frac{60}{5}$%

= 12%

Hence, Option 4 is the correct option.

Let the required time be T years.

S.I. on ₹ 780 at 5% p.a. = ₹ $\left(\frac{780 \times 5 \times \text{T}}{100}\right)$ = ₹ 39T

S.I. on ₹ 600 at $\frac{13}{2}$% p.a. = ₹ $\left(600 \times \frac{13}{2} \times \text{T} \times \frac{1}{100}\right)$ = ₹ 39T

As the two interests are equal to 39T for every value of T, so the two simple interests are always equal.

Hence, Option 4 is the correct option.

Let each sum be ₹ P.

S.I. for 5 years = ₹ $\left(\frac{\text{P} \times 15 \times 5}{100}\right) = ₹ \left(\frac{75\text{P}}{100}\right)$

S.I. for $3\frac{1}{2}$ years = ₹ $\left(\text{P} \times 15 \times \frac{7}{2} \times \frac{1}{100}\right) = ₹ \left(\frac{105\text{P}}{200}\right)$

Since the difference between the two interests is ₹ 144,

$$\begin{aligned}\Rightarrow \frac{75\text{P}}{100} - \frac{105\text{P}}{200} = 144\\[1em] \Rightarrow \frac{150\text{P} - 105\text{P}}{200} = 144\\[1em] \Rightarrow \frac{45\text{P}}{200} = 144\\[1em] \Rightarrow \text{P} = \frac{144 \times 200}{45}\\[1em] \Rightarrow \text{P} = 640\end{aligned}$$

Hence, Option 3 is the correct option.

Let the money lent at 12% be ₹ $x$. Then, the money lent at $12\frac{1}{2}$% = ₹ $(2500 - x)$.

Since the total annual interest is ₹ 306,

$$\begin{aligned}\Rightarrow \frac{x \times 12 \times 1}{100} + (2500 - x) \times \frac{25}{2} \times 1 \times \frac{1}{100} = 306\\[1em] \Rightarrow \frac{12x}{100} + \frac{25(2500 - x)}{200} = 306\\[1em] \Rightarrow 24x + 25(2500 - x) = 306 \times 200\\[1em] \Rightarrow 24x + 62500 - 25x = 61200\\[1em] \Rightarrow -x = -1300\\[1em] \Rightarrow x = 1300\end{aligned}$$

Hence, Option 3 is the correct option.

Let the first part be ₹ $x$. Then, the second part = ₹ $(625 - x)$.

Since the interest on the first part after 2 years equals the interest on the second part after 4 years,

$$\begin{aligned}\Rightarrow \frac{x \times 5 \times 2}{100} = \frac{(625 - x) \times 10 \times 4}{100}\\[1em] \Rightarrow 10x = 40(625 - x)\\[1em] \Rightarrow 10x = 25000 - 40x\\[1em] \Rightarrow 50x = 25000\\[1em] \Rightarrow x = 500\end{aligned}$$

Second part = ₹ (625 − 500) = ₹ 125

Hence, Option 1 is the correct option.

Here,

P = ₹ 540, R = $16\frac{2}{3}$% p.a. = $\frac{50}{3}$% p.a. and n = 2 years.

$$\begin{aligned}\text{Amount after 2 years} = 540 \times \left(1 + \frac{50}{3 \times 100}\right)^2\\[1em] = 540 \times \frac{7}{6} \times \frac{7}{6}\\[1em] = 735\end{aligned}$$

∴ C.I. = Amount − Principal = ₹ (735 − 540) = ₹ 195

Hence, Option 3 is the correct option.

Here,

P = ₹ 600, R = 10% p.a. and T = 1 year.

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

$= ₹ \left(\frac{600 \times 10 \times 1}{100}\right)$ = ₹ 60

For compound interest reckoned half-yearly, rate = 5% per half year and $n$ = 2 half years.

$$\begin{aligned}\text{Amount after 1 year} = 600 \times \left(1 + \frac{5}{100}\right)^2\\[1em] = 600 \times \frac{21}{20} \times \frac{21}{20}\\[1em] = 661.50\end{aligned}$$

C.I. = ₹ (661.50 − 600) = ₹ 61.50

∴ Difference = C.I. − S.I. = ₹ (61.50 − 60) = ₹ 1.50

Hence, Option 2 is the correct option.

Let the sum be ₹ P.

Here,

S.I. = ₹ 256, R = $\frac{25}{2}$% p.a. and T = 2 years.

$$\begin{aligned}\text{P} = \frac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \frac{100 \times 256}{\frac{25}{2} \times 2}\\[1em] = 100 \times 256 \times \frac{2}{25} \times \frac{1}{2}\\[1em] = 1024\end{aligned}$$

Now, for compound interest,

$$\begin{aligned}\text{Amount after 2 years} = 1024 \times \left(1 + \frac{25}{2 \times 100}\right)^2\\[1em] = 1024 \times \frac{9}{8} \times \frac{9}{8}\\[1em] = 1296\end{aligned}$$

∴ C.I. = ₹ (1,296 − 1,024) = ₹ 272

Hence, Option 4 is the correct option.

(i) If S.I. and C.I. denote simple interest and compound interest respectively on the same sum at the same rate of interest for the same period, then the relation between S.I. and C.I. is C.I. ≥ S.I.

(ii) The basic difference between simple interest and compound interest is that in simple interest, the principal remains same throughout the loan period while in compound interest it varies.

(iii) If the compound interest (compounded annually) on a sum of money at a certain rate of interest for $n$ years is equal to the simple interest on the same sum at the same rate for the same period, then $n$ = 1.

(iv) Amount = Principal + Interest

In simple interest, the interest is calculated on the original principal throughout the loan period, whereas in compound interest the interest of each period is added to the principal for the next period.

For a period of 1 year, both are equal and for a period of more than 1 year, the compound interest is greater than the simple interest.

So, S.I. ≤ C.I. and hence Assertion (A) is true.

The formula S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ is the correct formula for simple interest, so Reason (R) is true.

However, the formula S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ alone does not explain why the compound interest is greater than or equal to the simple interest. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, Option 2 is the correct option.

The amount on a sum of ₹ P for $n$ years at $r$% per annum, compounded annually, is given by

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

and the compound interest is given by

$\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \frac{r}{100}\right)^n - 1 \bigg\rbrace$

The given formula $\text{P}\left(1 + \frac{rn}{100}\right)$ is the amount under simple interest and not the compound interest. So, Assertion (A) is false.

Compound Interest = Final amount − Original amount is the correct relation, so Reason (R) is true.

Hence, Option 4 is the correct option.

Let the principal be ₹ P. Then, S.I. = ₹ $\left(\frac{9}{16}\text{P}\right)$.

Let the rate be R% per annum and the time be R years.

$$\begin{aligned}\Rightarrow \text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \frac{9\text{P}}{16} = \frac{\text{P} \times \text{R} \times \text{R}}{100}\\[1em] \Rightarrow \text{R}^2 = \frac{9}{16} \times 100\\[1em] \Rightarrow \text{R}^2 = \frac{900}{16}\\[1em] \Rightarrow \text{R} = \sqrt{\frac{900}{16}}\\[1em] = \frac{\sqrt{900}}{\sqrt{16}}\\[1em] = \frac{30}{4}\\[1em] = \frac{15}{2}\end{aligned}$$

So, time = R years = $\frac{15}{2}$ years = $7\frac{1}{2}$ years.

Hence, Option 4 is the correct option.

Let the principal be ₹ P.

Here,

R = 5% p.a. and n = 2 years.

$$\begin{aligned}\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \frac{5}{100}\right)^2 - 1 \bigg\rbrace\\[1em] \Rightarrow 1025 = \text{P}\left(\frac{21}{20} \times \frac{21}{20} - 1\right)\\[1em] \Rightarrow 1025 = \text{P}\left(\frac{441 - 400}{400}\right)\\[1em] \Rightarrow 1025 = \frac{41\text{P}}{400}\\[1em] \Rightarrow \text{P} = \frac{1025 \times 400}{41}\\[1em] \Rightarrow \text{P} = 10000\end{aligned}$$

Amount = Principal + C.I. = ₹ (10,000 + 1,025) = ₹ 11,025

Hence, Option 1 is the correct option.

Option 1 : P = ₹ 1,00,000, R = 20% p.a. and T = 2 years.

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

$= ₹ \left(\frac{100000 \times 20 \times 2}{100}\right)$ = ₹ 40,000

Option 2 : P = ₹ 1,00,000, R = 15% p.a. and n = 2 years.

$$\begin{aligned}\text{Amount after 2 years} = 100000 \times \left(1 + \frac{15}{100}\right)^2\\[1em] = 100000 \times \frac{23}{20} \times \frac{23}{20}\\[1em] = 132250\end{aligned}$$

C.I. = ₹ (1,32,250 − 1,00,000) = ₹ 32,250

As ₹ 32,250 < ₹ 40,000, so borrowing at 15% p.a. compounded annually costs less.

Hence, Option 2 is the correct option.

Interest rate = 10% per 2 years = $\frac{10}{2}$% per annum = 5% per annum

As the interest is compounded half-yearly,

Rate = $\frac{5}{2}$% per half year = 2.5% per half year

Time = 4 years = (4 × 2) half years = 8 half years

$\text{Amount after 4 years} = 5000\left(1 + \frac{2.5}{100}\right)^8$

Hence, Option 3 is the correct option.

Assuming that the schemes use simple interest,

Scheme 1 : P = ₹ 1,00,000, R = 3% p.a. and T = 2 years.

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

$= ₹ \left(\frac{100000 \times 3 \times 2}{100}\right)$ = ₹ 6,000

Scheme 2 : P = ₹ 1,00,000, R = 2% p.a. and T = 3 years.

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

$= ₹ \left(\frac{100000 \times 2 \times 3}{100}\right)$ = ₹ 6,000

As the interest earned is the same in both the schemes, both will give the same amount.

Hence, Option 3 is the correct option.

For Reena : As the interest is compounded annually and the period is only 6 months, no compounding takes place within the period.

Interest = ₹ $\left(5000 \times 10 \times \frac{1}{2} \times \frac{1}{100}\right)$ = ₹ 250

Amount = ₹ (5,000 + 250) = ₹ 5,250

For Vineeta : Rate = 5% per half year and $n$ = 6 months = 1 half year.

Amount = ₹ $\bigg\lbrace 5000 \times \left(1 + \frac{5}{100}\right) \bigg\rbrace = ₹ \left(5000 \times \frac{21}{20}\right)$ = ₹ 5,250

As both the amounts are equal, both will get the same amount at the time of maturity.

Hence, Option 3 is the correct option.

Let the sum be ₹ P.

When the sum doubles itself, A = ₹ 2P and S.I. = ₹ (2P − P) = ₹ P in 10 years.

So, S.I. for 10 years = ₹ P.

When the sum trebles itself, A = ₹ 3P and S.I. = ₹ (3P − P) = ₹ 2P.

$\text{Time taken} = \frac{2\text{P}}{\text{P}} \times 10$

= 2 x 10

= 20

Hence, Option 3 is the correct option.

Let the sum be ₹ P.

Here,

R = 10% p.a. and $n$ = 3 years.

$$\begin{aligned}\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \frac{10}{100}\right)^3 - 1 \bigg\rbrace\\[1em] \Rightarrow 993 = \text{P}\left(\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} - 1\right)\\[1em] \Rightarrow 993 = \text{P}\left(\frac{1331 - 1000}{1000}\right)\\[1em] \Rightarrow 993 = \frac{331\text{P}}{1000}\\[1em] \Rightarrow \text{P} = \frac{993 \times 1000}{331}\\[1em] \Rightarrow \text{P} = 3000\end{aligned}$$

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

$= ₹ \left(\frac{3000 \times 10 \times 3}{100}\right)$ = ₹ 900

Hence, Option 2 is the correct option.

Let the sum be ₹ P.

Here,

R = 5% p.a. and T = 2 years.

S.I. = $\frac{\text{P} \times 5 \times 2}{100} = ₹ \left(\frac{\text{P}}{10}\right)$

$$\begin{aligned}\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \frac{5}{100}\right)^2 - 1 \bigg\rbrace\\[1em] = \text{P}\left(\frac{21}{20} \times \frac{21}{20} - 1\right)\\[1em] = \text{P}\left(\frac{441 - 400}{400}\right)\\[1em] = \frac{41\text{P}}{400}\end{aligned}$$

Given,

The difference between compound interest and simple interest = ₹ 150.

$$\begin{aligned}\Rightarrow \frac{41\text{P}}{400} - \frac{\text{P}}{10} = 150\\[1em] \Rightarrow \frac{41\text{P} - 40\text{P}}{400} = 150\\[1em] \Rightarrow \frac{\text{P}}{400} = 150\\[1em] \Rightarrow \text{P} = 60000\end{aligned}$$

Hence, Option 4 is the correct option.

Here, the price increases at the rate of 10% every 6 months, i.e. the rate is 10% per half year.

Time = 2 years = (2 × 2) half years = 4 half years

Price of 1 kg spice after 2 years = $12000\left(1 + \frac{10}{100}\right)^4$

Hence, Option 2 is the correct option.

Let the rate be R% per annum and the time be T years.

As $x$ is the simple interest on $y$,

$x = \frac{y \times \text{R} \times \text{T}}{100} \qquad \text{.....(i)}$

As $y$ is the simple interest on $z$,

$y = \frac{z \times \text{R} \times \text{T}}{100} \qquad \text{.....(ii)}$

Dividing equation (i) by equation (ii),

$$\begin{aligned}\frac{x}{y} = \frac{y}{z}\\[1em] \Rightarrow y^2 = xz\end{aligned}$$

Hence, Option 3 is the correct option.