(1) Principal = ₹ 2000, Rate = 5 p.c.p.a., Duration = 2 years

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 2000 × \((1+\frac{5}{100})^2\)

= 2000 × \((1+\frac{1}{20})^2\)

= 2000 × \((\frac{21}{20})^2\)

= 2000 x \(\frac{441}{400}\)

∴ A = 5 x 441 = ₹ 2205

Compound Interest (I) = A - P

∴ I = 2205 - 2000 = ₹ 205

(2) Principal = ₹ 5000, Rate = 8 p.c.p.a., Duration = 3 years

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 5000 × \((1+\frac{8}{100})^3\)

= 5000 × \((\frac{108}{100})^3\)

= 5000 × \((\frac{108×108×108}{100×100×100})\)

∴ A = ₹ 6298.56

Compound Interest (I) = A - P

∴ I = 6298.56 - 5000 = ₹ 1298.56

(3) Principal = ₹ 4000, Rate = 7.5 p.c.p.a., Duration = 2 years

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 4000 × \((1+\frac{7.5}{100})^2\)

= 4000 × \((\frac{107.5}{100})^2\)

= 4000 × \((\frac{107.5×107.5}{100×100})\)

∴ A = ₹ 4622.5

Compound Interest (I) = A - P

∴ I = 4622.5 - 4000 = ₹ 622.5

Given Data:

• Principal (P) = ₹ 12,500
• Rate of interest (R) = 12 p.c.p.a.
• Duration (N) = 3 years

To find the amount (A) compounded annually, we use the formula:

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 12500 × \((1+\frac{12}{100})^3\)

= 12500 × \((\frac{112}{100})^3\)

= 12500 × \((\frac{112×112×112}{100×100×100})\)

= \(\frac{125×112×112×112}{100×100}\)

∴ A = ₹ 17561.60

∴ Sameerrao should pay ₹ 17,561.60 to clear his loan.

Given Data:

• Principal (P): ₹ 8,000
• Rate of interest (R): 10 ​c.p.a., which is 10.5%
• Duration (N): 2 years

To find the amount (A) compounded annually, we use the formula:

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 8000 × \((1+\frac{10.5}{100})^2\)

= 8000 × \((\frac{110.5}{100})^2\)

= 8000 × \((\frac{110.5×110.5}{100×100})\)

= \(\frac{8×110.5×110.5}{10}\)

∴ A = ₹ 9768.20

Compound Interest (I) = A - P

∴ I = 9768.20 - 8000 = ₹ 1768.20

Shalaka will have to pay ₹ 1,768.20 in compound interest after two years

Given Data:

• Initial number of workers (P): 320
• Rate of increase per year (R): 25%
• Duration in years (N): 2 years

A = Number of workers after 2 years

A = P\((1+\frac{R}{100})^N\)

∴ A = 320 × \((1+\frac{25}{100})^2\)

= 320 × \((\frac{125}{100})^2\)

= 320 × \(\frac{125×125}{100×100}\)

= \(\frac{500000}{1000}\)

∴ A = 500

The number of workers after 2 years will be 500

Given Data:

• Initial number of sheep (P): 200
• Rate of increase per year (R): 8%
• Duration in years (N): 3 years

A = Number of sheeps after 3 years

A = P\((1+\frac{R}{100})^N\)

∴ A = 200\((1+\frac{8}{100})^3\)

= 200 × \((\frac{108}{100})^3\)

= 200 × \(\frac{108×108×108}{100×100×100}\)

= \(\frac{2×108×108×108}{100×100}\)

∴ A = 251.94 ≈ 252

The shepherd will have approximately 252 sheep after 3 years.

[Note : The answer in the textbook is 242 sheep whereas the correct answer is 252 sheep.]

Given Data:

• Initial number of trees (P): 40,000
• Rate of increase per year (R): 5%
• Duration in years (N): 3 years

A = Number of tress after 3 years

A = P\((1+\frac{R}{100})^N\)

∴ A = 40000\((1+\frac{5}{100})^3\)

= 40000 × \((\frac{105}{100})^3\)

= 40000 × \(\frac{105×105×105}{100×100×100}\)

= \(\frac{4×105×105×105}{100}\)

∴ A = 46305

The expected number of trees after 3 years is 46,305 trees

Given Data:

• Initial cost price (P): ₹ 2,50,000
• Rate of depreciation (R): 10% per year
• Duration (N): 2 years

When finding a reduced price (depreciation) over a period of time, the compound interest formula is used, but the rate of depreciation (R) is taken as negative because the value is decreasing

A = Price of the machine after two years.

A = P\((1+\frac{R}{100})^N\)

∴ A = 250000\((1+\frac{-10}{100})^2\)

= 250000\((\frac{90}{100})^2\)

= 250000 × \(\frac{90×90}{100×100}\)

= 25 x 90 x 90

∴ A = ₹ 202,500

Total Depreciation = Initial Price (P) - Final Value (A)

= 2,50,000 − 2,02,500

= ₹ 47,500

The total depreciation in the price of the machine after two years is ₹ 47,500

Given Data:

• Final Amount (A): ₹ 4036.80
• Rate of Interest (R): 16 p.c.p.a.
• Duration (N): 2 years

Total amount in compound interest is A = P\((1+\frac{R}{100})^N\)

∴ 4036.80 = P\((1+\frac{R}{100})^N\)

∴ 4036.80 = P\((1+\frac{16}{100})^2\)

∴ 4036.80 = P\((\frac{116}{100})^2\)

∴ 4036.80 = P\((\frac{116×116}{100×100}\)

∴ P = \(\frac{4036.80×100×100}{116×116}\) = 3000

∴ P =  ₹ 3000

Compound interest I = A − P

∴ I = 4036.80 − 3000

∴ I = ₹1036.80

The compound interest is ₹ 1036.80

Given Data:

• Principal (P): ₹ 15,000
• Rate of Interest (R): 12 p.c.p.a.
• Duration (N): 3 years

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ A = 15000\((1+\frac{12}{100})^3\)

= 15000\((\frac{112}{100})^3\)

= 15000 × \((\frac{112×112×112}{100×100×100})\)

= \(\frac{15×112×112×112}{1000}\)

∴ A = ₹ 21073.92

The amount required to settle the loan after 3 years is ₹ 21,073.92

Given Data:

• Final Amount (A): ₹ 13,924
• Duration (N): 2 years
• Rate of Interest (R): 18 p.c.p.a.

Amount by compound interest:

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ 13924 = P\((1+\frac{18}{100})^2\)

∴ 13924 = P\((\frac{118}{100})^2\)

∴ 13924 = P\((\frac{118×118}{100×100})\)

P = \(\frac{13924×100×100}{118×118}\)

∴ P = ₹ 10000

The principal was ₹ 10,000

Given Data:

• Initial population (P): 16,000
• Population after two years (A): 17,640
• Duration (N): 2 years

Increased values over time:  

A = P\((1+\frac{R}{100})^N\)

∴ 17640 = 16000\((1+\frac{R}{100})^2\)

\(\frac{17640}{16000}\) = \((1+\frac{R}{100})^2\)

\(\frac{441}{400}\) = \((1+\frac{R}{100})^2\)

Take the square root of both sides:

∴ \(\frac{21}{20}\) = \((1+\frac{R}{100})\)

\(\frac{R}{100}\) = \(\frac{21}{20}\) − 1 = \(\frac{21-20}{20}\)

\(\frac{R}{100}\) = \(\frac{1}{20}\)

∴  R = \(\frac{100}{20}\) = 5

The rate of increase in the population is 5 p.c.p.a.

Given Data:

Principal (P): ₹ 700

Final Amount (A): ₹ 847

Rate of Interest (R): 10 p.c.p.a.

N = ?

Amount by compound interest is:

Amount (A) = P\((1+\frac{R}{100})^N\)

∴ 847 = 700\((1+\frac{10}{100})^N\)

∴ 847 = 700 x \((\frac{110}{100})^N\)

\(\frac{847}{700}\) = \((\frac{11}{10})^N\)

\(\frac{121}{100}\) = \((\frac{11}{10})^N\)

\((\frac{11}{10})^2\) = \((\frac{11}{10})^N\)

∴ N = 2

The principal of ₹ 700 will amount to ₹ 847 in 2 years.

Given Data:

• Principal (P): ₹ 20,000
• Rate of Interest (R): 8 p.c.p.a.
• Duration (N): 2 years

Calculate Simple Interest (SI)

Using the formula for simple interest: I = ​\(\frac{PNR}{100}\)

SI = ​\(\frac{20000×2×8}{100}\)

SI = 200 × 16

SI = ₹3,200

Calculate Compound Interest (CI) :

Total amount (A) = P\((1+\frac{R}{100})^N\)

A = 20000\((1+\frac{8}{100})^2\)

= 20000\((\frac{108}{100})^2\)

= 20000 × \(\frac{108×108}{100×100}\)

= 2 x 108 x 108

∴ A = ₹23,328

Compound interest (I) = Amount − Principal.

CI = 23328 – 20000

CI = ₹3,328

Difference = Compound Interest – Simple Interest

Difference = 3328 − 3200

Difference = ₹ 128

The difference between the simple interest and compound interest on ₹ 20,000 at 8 p.c.p.a. for 2 years is ₹ 128