Simple Interest

(i)

Given:

P = ₹ 8400, R = 8%, T = 4 years

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{ 8400 \times 8 \times 4}{100}\Big) \\[1em] = ₹ \Big(\frac{ 84 \times 8 \times 4}{1}\Big) \quad \text{[Dividing 8400 and 100 by 100]} \\[1em] = ₹ 84 \times 32 \\[1em] = ₹ 2688\end{aligned}$$

Amount = Principal + S.I.

Amount = ₹ 8400 + ₹ 2688

Amount = ₹ 11088

S.I. = ₹ 2688, Amount = ₹ 11088

(ii)

Given:

P = ₹ 50000, T = 3 years

R = $12\frac{1}{2}$% = $\frac{25}{2}$%

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \left(\frac{ 50000 \times \frac{25}{2} \times 3}{100}\right) \\[1em] = ₹ \Big(\frac{ 50000 \times 25 \times 3}{100 \times 2}\Big) \\[1em] = ₹ \Big(\frac{ 500 \times 25 \times 3}{1 \times 2}\Big) \quad \text{[Dividing 50000 and 100 by 100]} \\[1em] = ₹ \Big(\frac{ 250 \times 25 \times 3}{1 \times 1}\Big) \quad \text{[Dividing 500 and 2 by 2]} \\[1em] = ₹ 250 \times 25 \times 3 \\[1em] = ₹ 250 \times 75 \\[1em] = ₹ 18750\end{aligned}$$

Amount = Principal + S.I.

Amount = ₹ 50000 + ₹ 18750

Amount = ₹ 68750

S.I. = ₹ 18750, Amount = ₹ 68750

(iii)

Given:

P = ₹ 9275, T = 2 years

R = $7\frac{1}{2}$% = $\frac{15}{2}$%

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \left(\frac{ 9275 \times \frac{15}{2} \times 2}{100}\right) \\[1em] = ₹ \Big(\frac{ 9275 \times 15 \times 2}{100 \times 2}\Big) \\[1em] = ₹ \Big(\frac{ 371 \times 15 \times 2}{4 \times 2}\Big) \quad \text{[Dividing 9275 and 100 by 25]} \\[1em] = ₹ \Big(\frac{ 371 \times 15 \times 1}{4 \times 1}\Big) \\[1em] = ₹ \frac{5565}{4} \\[1em] = ₹ 1391.25\end{aligned}$$

Amount = Principal + S.I.

Amount = ₹ 9275 + ₹ 1391.25

Amount = ₹ 10666.25

S.I. = ₹ 1391.25, Amount = ₹ 10666.25

Given:

Principal = ₹ 7500, Amount = ₹ 8625

R = $7\frac{1}{2}$% = $\frac{15}{2}$%

S.I. = Amount - Principal

S.I. = ₹ 8625 - ₹ 7500 = ₹ 1125

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow T = \frac{S.I. \times 100}{P \times R} \\[1em] = \left(\frac{1125 \times 100}{7500 \times \frac{15}{2}}\right) \text{ years} \\[1em] = \Big(\frac{1125 \times 100 \times 2}{7500 \times 15}\Big) \text{ years} \\[1em] = \Big(\frac{1125 \times 1 \times 2}{75 \times 15}\Big) \text{ years} \quad \text{[Dividing 7500 and 100 by 100]} \\[1em] = \Big(\frac{15 \times 1 \times 2}{1 \times 15}\Big) \text{ years} \quad \text{[Dividing 1125 and 75 by 75]} \\[1em] = \Big(\frac{1 \times 1 \times 2}{1 \times 1}\Big) \text{ years} \\[1em] = 2 \text{ years}\end{aligned}$$

∴ Time = 2 years

Given:

P = ₹ 3600, A = ₹ 4320, R = 4%

S.I. = ₹ 4320 - ₹ 3600 = ₹ 720

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow T = \frac{S.I. \times 100}{P \times R} \\[1em] = \Big(\frac{720 \times 100}{3600 \times 4}\Big) \text{ years} \\[1em] = \Big(\frac{720 \times 1}{36 \times 4}\Big) \text{ years} \quad \text{[Dividing 100 and 3600 by 100]} \\[1em] = \Big(\frac{20 \times 1}{1 \times 4}\Big) \text{ years} \quad \text{[Dividing 720 and 36 by 36]} \\[1em] = \Big(\frac{5 \times 1}{1 \times 1}\Big) \text{ years} \quad \text{[Dividing 20 and 4 by 4]} \\[1em] = 5 \text{ years}\end{aligned}$$

∴ Time = 5 years

Given:

P = ₹ 6300, S.I. = ₹ 2100, T = 4 years

Then

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

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

$= \Big(\frac{2100 \times 100}{6300 \times 4}\Big)$%

$= \Big(\frac{2100 \times 1}{63 \times 4}\Big)$% $\quad$ [Dividing 100 and 6300 by 100]

$= \Big(\frac{525 \times 1}{63 \times 1}\Big)$% $\quad$ [Dividing 2100 and 4 by 4]

$= \Big(\frac{525}{63}\Big)$%

$= \frac{25}{3}$% $\quad$ [Dividing 525 and 63 by 21]

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

∴ Rate = $\bold {8\frac{1}{3}}$%

Given:

P = ₹ 66000, A = ₹ 72720, T = 2 years

S.I. = ₹ 72720 - ₹ 66000 = ₹ 6720

Then

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

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

$= \Big(\frac{6720 \times 100}{66000 \times 2}\Big)$%

$= \Big(\frac{6720 \times 1}{660 \times 2}\Big)$% $\quad$ [Dividing 100 and 66000 by 100]

$= \Big(\frac{3360 \times 1}{660 \times 1}\Big)$% $\quad$ [Dividing 6720 and 2 by 2]

$= \frac{56}{11}$% $\quad$ [Dividing 3360 and 660 by 60]

$= 5\frac{1}{11}$%

∴ Rate = $\mathbf{5\frac{1}{11}}$% per annum

Given:

S.I. for 5 years is $\frac{3}{5}$ of the sum.

R = ?

Let the sum (P) be x.

S.I. = $\frac{3}{5}x$ and T = 5 years.

Then

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

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

$= \left(\frac{(\frac{3}{5}x) \times 100}{x \times 5}\right)$%

$= \Big(\frac{3x \times 100}{5 \times x \times 5}\Big)$%

$= \frac{300x}{25x}$%

$= 12$%

∴ Rate = 12% per annum

Given:

S.I. = ₹ 829.50, R = 10%, T = 3 years.

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow P = \frac{S.I. \times 100}{R \times T} \\[1em] = ₹ \Big(\frac{829.50 \times 100}{10 \times 3}\Big) \\[1em] = ₹ \frac{82950}{30} \\[1em] = ₹ 2765\end{aligned}$$

∴ Sum = ₹ 2765

Given:

S.I. = ₹ 7840, T = 2 years

R = $6\frac{1}{4}$% = $\frac{25}{4}$%

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow P = \frac{S.I. \times 100}{R \times T} \\[1em] = ₹ \left(\frac{7840 \times 100}{\frac{25}{4} \times 2}\right) \\[1em] = ₹ \Big(\frac{7840 \times 100 \times 4}{25 \times 2}\Big) \\[1em] = ₹ \Big(\frac{7840 \times 100 \times 4}{50}\Big) \\[1em] = ₹ \Big(\frac{7840 \times 2 \times 4}{1}\Big) \\[1em] = ₹ 7840 \times 2 \times 4 \\[1em] = ₹ 62720\end{aligned}$$

∴ Sum = ₹ 62720

Given:

T = 16 years

Let the Principal (P) be x.

Treble means Amount (A) = 3x.

S.I. = Amount - Principal

S.I. = 3x - x = 2x

Then

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

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

$= \Big(\frac{2x \times 100}{x \times 16}\Big)$%

$= \Big(\frac{200x}{16x}\Big)$%

$= 12\frac{1}{2}$%

∴ Rate = $\bold{12\frac{1}{2}}$%

Given:

A = ₹ 4130, T = 3 years, R = 6%

We have the formula:

$$\begin{aligned}P = \frac{100 \times A}{100 + (R \times T)} \\[1em] = \frac{100 \times 4130}{100 + (6 \times 3)} \\[1em] = \frac{100 \times 4130}{100 + 18} \\[1em] = \frac{413000}{118} \\[1em] = ₹ 3500\end{aligned}$$

∴ Sum = ₹ 3500

Let the sum be ₹ 100.

Given:

R = 11% p.a., T = 2 years.

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em]= ₹ \Big(\frac{100 \times 11 \times 2}{100}\Big) \\[1em] = ₹ \Big(\frac{1 \times 11 \times 2}{1}\Big) \\[1em] = ₹ 11 \times 2 \\[1em] = ₹ 22\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 22 + ₹ 100

Amount = ₹ 122

If Amount is ₹ 122, then Sum = ₹ 100.

If Amount is ₹ 1, then Sum = ₹ $\frac{100}{122}$.

If Amount is ₹ 10370, then Sum =

$$\begin{aligned}\phantom{=} ₹ \Big(\frac{100}{122} \times 10370\Big) \\[1em] = ₹ \Big(\frac{50}{61} \times 10370\Big) \\[1em] = ₹ \Big(\frac{518500}{61}\Big) \\[1em] = ₹ 8500\end{aligned}$$

Now, sum = ₹ 8500, R = 11% p.a., T = 3 years.

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{8500 \times 11 \times 3}{100}\Big) \\[1em] = ₹ \Big(\frac{85 \times 11 \times 3}{1}\Big) \\[1em] = ₹ 85 \times 11 \times 3 \\[1em] = ₹ 2805\end{aligned}$$

Required Amount = S.I. + Principal

Required Amount = ₹ 2805 + ₹ 8500 = ₹ 11305

∴ Amount = ₹ 11305

Let the sum be ₹ 100.

Given: R = 9% p.a., T = 3 years.

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{100 \times 9 \times 3}{100}\Big) \\[1em] = ₹ \Big(\frac{1 \times 9 \times 3}{1}\Big) \\[1em] = ₹ 9 \times 3 \\[1em] = ₹ 27\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 27 + ₹ 100

Amount = ₹ 127

If Amount is ₹ 127, then Sum = ₹ 100.

If Amount is ₹ 1, then Sum = ₹ $\frac{100}{127}$.

If Amount is ₹ 10160, then Sum =

$$\begin{aligned}\phantom{=} ₹ \Big(\frac{100}{127} \times 10160\Big) \\[1em] = ₹ \Big(\frac{1016000}{127}\Big) \\[1em] = ₹ 8000\end{aligned}$$

Now, sum = ₹ 8000, R = 8% p.a., T = 2 years.

And

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{8000 \times 8 \times 2}{100}\Big) \\[1em] = ₹ \Big(\frac{80 \times 8 \times 2}{1}\Big) \\[1em] = ₹ 80 \times 8 \times 2 \\[1em] = ₹ 1280\end{aligned}$$

Required Amount = S.I. + Principal

Required Amount = ₹ 1280 + ₹ 8000 = ₹ 9280

∴ Amount = ₹ 9280

The incorrect statement is R = $\frac{P \times T}{100 \times S.I.}$

The correct formula for Rate is R = $\frac{100 \times S.I.}{P \times T}$

Hence, option 2 is the correct option.

Given:

P = ₹ 600, T = 3 years

R = $7\frac{1}{2}$% = $\frac{15}{2}$%

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \left(\frac{600 \times \frac{15}{2} \times 3}{100}\right) \\[1em] = ₹ \Big(\frac{600 \times 15 \times 3}{100 \times 2}\Big) \\[1em] = ₹ \Big(\frac{6 \times 15 \times 3}{1 \times 2}\Big) \\[1em] = ₹ \Big(\frac{3 \times 15 \times 3}{1 \times 1}\Big) \\[1em] = ₹ 3 \times 15 \times 3 \\[1em] = ₹ 135\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 135 + ₹ 600 = ₹ 735

Hence, option 1 is the correct option.

Given:

Principal (P) = ₹ 800

Simple Interest (S.I.) = ₹ 208

Time (T) = 2 years

Then

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

$= \Big(\frac{208 \times 100}{800 \times 2}\Big)$%

$= \frac{20800}{1600}$%

$= \frac{208}{16}$%

$= 13$%

Hence, option 3 is the correct option.

Given:

Principal (P) = ₹ 900

Simple Interest (S.I.) = ₹ 351

Rate (R) = 13% p.a.

Then

$$\begin{aligned}T = \frac{S.I. \times 100}{P \times R} \\[1em] = \Big(\frac{351 \times 100}{900 \times 13}\Big) \\[1em] = \frac{35100}{11700} \\[1em] = \frac{351}{117} \\[1em] = 3 \text{ years}\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Let Principal (P) = x

Amount (A) = 2x

Time (T) = 8 years

S.I. = Amount - Principal

S.I. = 2x - x = x

Then

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

$= \Big(\frac{x \times 100}{x \times 8}\Big)$%

$= \frac{100}{8}$%

$= \frac{25}{2}$%

$= 12\frac{1}{2}$%

Hence, option 2 is the correct option.

Given:

Let the Principal (P) = ₹ 100

Amount (A): The problem says the sum becomes $\frac{7}{4}$ of itself.

Amount = ₹ $\frac{7}{4} \times 100$

Amount = ₹ 7 x 25 = ₹ 175

Time (T) = 5 years

S.I. = Amount - Principal

S.I. = ₹ 175 - ₹ 100 = ₹ 75

Then

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

$= \Big(\frac{75 \times 100}{100 \times 5}\Big)$%

$= \frac{75}{5}$%

$= 15$%

Hence, option 4 is the correct option.

(i) The money borrowed for a certain period is called principal.

(ii) The rate per cent per annum is the interest on ₹ 100 for 1 year.

(iii) The simple interest on ₹ 800 invested at 13% per annum for 3 years is ₹ 312.

(iv) If ₹ 2600 becomes ₹ 3900 in 5 years, then the rate of interest is 10%.

(v) If the principal is ₹ 3400, then the amount to be paid at the end of 5 years at 8% p.a. simple interest will be ₹ 4760.

Explanation

(i) In financial math, the initial sum of money you borrow or invest before interest is added is always called the Principal.

(ii) The word "percent" literally means "per hundred." So, a rate of 8% means you pay ₹ 8 for every ₹ 100 borrowed over one year.

(iii)

Given:

P = ₹ 800, R = 13%, T = 3 years

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{800 \times 13 \times 3}{100}\Big) \\[1em] = ₹ \Big(\frac{8 \times 13 \times 3}{1}\Big) \\[1em] = ₹ 8 \times 13 \times 3 \\[1em] = ₹ 312\end{aligned}$$

S.I. = ₹ 312

(iv)

Given:

P = ₹ 2600, A = ₹ 3900, T = 5 years

S.I. = Amount - Principal

S.I. = ₹ 3900 - ₹ 2600 = ₹ 1300

Then

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

$= \Big(\frac{1300 \times 100}{2600 \times 5}\Big)$%

$= \Big(\frac{1 \times 100}{2 \times 5}\Big)$%

$= \frac{100}{10}$%

$= 10$%

R = 10%

(v) Given:

P = ₹ 3400, R = 8%, T = 5 years

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{3400 \times 8 \times 5}{100}\Big) \\[1em] = ₹ \Big(\frac{34 \times 8 \times 5}{1}\Big) \\[1em] = ₹ 34 \times 40 \\[1em] = ₹ 1360\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 1360 + ₹ 3400

Amount = ₹ 4760

(i) False
Reason — The total money (Principal + Interest) is called the Amount. Interest is only the "extra" fee charged for borrowing the money.

(ii) True
Reason — This is the fundamental formula for calculating the total value of a loan or investment at the end of a time period.

(iii) False
Reason — The rate per cent is the interest on ₹ 100 for 1 year. (Per cent = per hundred).

(iv) False
Reason — This formula is inverted. The correct formula for Simple Interest is:

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

(v) True
Reason —

P = ₹ 5200, R = 6%, T = 5 years

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{5200 \times 6 \times 5}{100}\Big) \\[1em] = ₹ \Big(\frac{52 \times 6 \times 5}{1}\Big) \\[1em] = ₹ 52 \times 6 \times 5 \\[1em] = ₹ 1560\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 1560 + ₹ 5200

Amount = ₹ 6760

(1)

Given:

P = ₹ 5,00,000, R = 8%, T = 2 years.

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{5,00,000 \times 8 \times 2}{100}\Big) \\[1em] = ₹ 5,000 \times 8 \times 2 \\[1em] = ₹ 80,000\end{aligned}$$

Hence, option 4 is the correct option.

(2)

Given:

A = ₹ 4,80,000, R = 12%, T = 5 years.

Then

$$\begin{aligned}P = \frac{100 \times A}{100 + (R \times T)} \\[1em] = ₹ \Big(\frac{100 \times 4,80,000}{100 + (12 \times 5)}\Big) \\[1em] = ₹ \Big(\frac{100 \times 4,80,000}{100 + 60}\Big) \\[1em] = ₹ \Big(\frac{4,80,00,000}{160}\Big) \\[1em] = ₹ 3,00,000\end{aligned}$$

Hence, option 1 is the correct option.

(3)

Given:

R = 8%

Let P = x, Amount = 2x

S.I. = Amount - P

S.I. = 2x - x = x

Then

$$\begin{aligned}T = \frac{100 \times \text{S.I.}}{P \times R} \\[1em] = \Big(\frac{100 \times x}{x \times 8}\Big) \text{ years} \\[1em] = \frac{100}{8} \text{ years} \\[1em] = \frac{25}{2} \text{ years} \\[1em] = 12\frac{1}{2} \text{ years}\end{aligned}$$

Hence, option 3 is the correct option.

(4)

Given:

R (Bank) = 8%

R (ADG) = 12%

T = 2 years

Total Sum = ₹ 5,00,000.

Let Bank Investment = x, then ADG Investment = (5,00,000 - x)

Then

$$\begin{aligned}S.I. \text {(Bank)} = S.I. \text {(ADG)} \\[1em] \Rightarrow \frac{x \times 8 \times 2}{100} = ₹ \frac{(5,00,000 - x) \times 12 \times 2}{100} \quad \text{[Cancel 100 and 2 from both sides]} \\[1em] \Rightarrow 8x = ₹ (5,00,000 - x) \times 12 \\[1em] \Rightarrow \frac{8x}{12} = ₹ (5,00,000 - x) \\[1em] \Rightarrow \frac{2x}{3} = ₹ (5,00,000 - x) \\[1em] \Rightarrow \frac{2x}{3} + x = ₹ 5,00,000 \\[1em] \Rightarrow \frac{5x}{3} = ₹ 5,00,000 \\[1em] \Rightarrow x = ₹ \frac{5,00,000 \times 3}{5} \\[1em] \Rightarrow x = ₹ \frac{1,00,000 \times 3}{1} \\[1em] \Rightarrow x = ₹ 3,00,000\end{aligned}$$

Hence, option 3 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

Given:

P = ₹ 700, R = 5%, T = 12 months = 1 year

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{700 \times 5 \times 1}{100}\Big) \\[1em] = ₹ \Big(\frac{7 \times 5 \times 1}{1}\Big) \\[1em] = ₹ 7 \times 5 \\[1em] = ₹ 35\end{aligned}$$

The Assertion is True.

Reason:

S.I. = $\frac{\text{PRT}}{100}$, where P = principal, R = rate per cent per annum and T = time in months.

This is False.

In the standard formula, T must always be in years. If time is given in months, it must be converted (divided by 12) before using this specific formula.

Hence, option 3 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

Given:

P = ₹ 800, T = 3 years, R = 12%

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{800 \times 12 \times 3}{100}\Big) \\[1em] = ₹ \Big(\frac{8 \times 12 \times 3}{1}\Big) \\[1em] = ₹ 8 \times 12 \times 3 \\[1em] = ₹ 288\end{aligned}$$

Amount = S.I. + Principal

Amount = ₹ 288 + ₹ 800

Amount = ₹ 1088

Since the Assertion claims the amount is ₹ 1096, the Assertion is False.

Reason:

Amount = Principal + Interest.

This is True. This is the correct mathematical definition of Amount.

Hence, option 4 is the correct option.

Given:

S.I. = $\frac{4}{9}$ of the principal

Time and Rate are numerically equal.

Let the Principal (P) be x.

S.I. = $\frac{4}{9}x$

Let R = T = y

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow \frac{4}{9}x = \frac{x \times y \times y}{100} \\[1em] \Rightarrow \frac{4}{9} = \frac{y^2}{100} \\[1em] \Rightarrow y^2 = \frac{4 \times 100}{9} \\[1em] \Rightarrow y^2 = \frac{400}{9} \\[1em] \Rightarrow y = \sqrt{\frac{400}{9}} \\[1em] \Rightarrow y = \frac{20}{3} \\[1em] \Rightarrow y = 6\frac{2}{3}\end{aligned}$$

∴ Rate = $6\frac{2}{3}$%

Hence, option 3 is the correct option.

Given:

P = ₹ 5500, R = 8%, T = 5 years

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \Big(\frac{5500 \times 8 \times 5}{100}\Big) \\[1em] = ₹ \Big(\frac{55 \times 8 \times 5}{1}\Big) \\[1em] = ₹ 55 \times 40 \\[1em] = ₹ 2200\end{aligned}$$

Amount = Principal + S.I.

Amount = ₹ 5500 + ₹ 2200

Amount = ₹ 7700

The farmer cleared the account by giving ₹ 6000 and a cow.

Cost of cow = Amount - ₹ 6000

Cost of cow = ₹ 7700 - ₹ 6000 = ₹ 1700

Hence, option 3 is the correct option.

Given:

P = ₹ 1200, S.I. = ₹ 432

Time and Rate are numerically equal.

Let R = T = y

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 432 = \frac{1200 \times y \times y}{100} \\[1em] \Rightarrow 432 = \frac{1200 \times y^2}{100} \\[1em] \Rightarrow 432 = 12 \times y^2 \\[1em] \Rightarrow y^2 = \frac{432}{12} \\[1em] \Rightarrow y^2 = 36 \\[1em] \Rightarrow y = \sqrt{36} \\[1em] \Rightarrow y = ±6\end{aligned}$$

Since time cannot be negative,

∴ Time = 6 years

Hence, option 2 is the correct option.

Given:

T = 5 years

Let the Principal (P) be x.

Money is doubled, so Amount (A) = 2x

S.I. = Amount - Principal

S.I. = 2x - x = x

Then

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

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

$= \Big(\frac{x \times 100}{x \times 5}\Big)$%

$= \frac{100}{5}$%

$= 20$%

∴ Rate = 20% per annum

Hence, option 1 is the correct option.

Given:

The sum doubles in 10 years.

Let the Principal (P) be x.

When sum is doubled, Amount = 2x

S.I. = Amount - Principal

S.I. = 2x - x = x and T = 10 years

Then

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

$= \Big(\frac{x \times 100}{x \times 10}\Big)$%

$= \frac{100}{10}$%

$= 10$%

Now, when sum is tripled, Amount = 3x

S.I. = 3x - x = 2x and R = 10%

Then

$$\begin{aligned}T = \frac{S.I. \times 100}{P \times R} \\[1em] = \Big(\frac{2x \times 100}{x \times 10}\Big) \text{ years} \\[1em] = \frac{200}{10} \text{ years} \\[1em] = 20 \text{ years}\end{aligned}$$

∴ Time = 20 years

Hence, option 3 is the correct option.

Given:

Total S.I. for 6 years = ₹ 29,250

For first 2 years, R = 7% p.a., T = 2 years

For next 4 years, R = 16% p.a., T = 4 years

Let the sum (P) be x.

S.I. for first 2 years :

$$\begin{aligned}S.I._1 = \frac{P \times R \times T}{100} \\[1em] = \frac{x \times 7 \times 2}{100} \\[1em] = \frac{14x}{100}\end{aligned}$$

S.I. for next 4 years :

$$\begin{aligned}S.I._2 = \frac{P \times R \times T}{100} \\[1em] = \frac{x \times 16 \times 4}{100} \\[1em] = \frac{64x}{100}\end{aligned}$$

Total S.I. = S.I.₁ + S.I.₂

$$\begin{aligned}\Rightarrow ₹ 29250 = \frac{14x}{100} + \frac{64x}{100} \\[1em] \Rightarrow ₹ 29250 = \frac{14x + 64x}{100} \\[1em] \Rightarrow ₹ 29250 = \frac{78x}{100} \\[1em] \Rightarrow x = ₹ \frac{29250 \times 100}{78} \\[1em] \Rightarrow x = ₹ \frac{2925000}{78} \\[1em] \Rightarrow x = ₹ 37500\end{aligned}$$

∴ Sum = ₹ 37,500

Hence, option 3 is the correct option.

Given:

P = ₹ 32,000, R = 8.5% p.a.

Time period is from 10th February, 2019 to 24th April, 2019.

Number of days (excluding the day of borrowing and including the day of repayment) :

February (from 11th to 28th) = 18 days

March = 31 days

April (from 1st to 24th) = 24 days

Total number of days = 18 + 31 + 24 = 73 days

T = $\frac{73}{365}$ year = $\frac{1}{5}$ year

Then

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = ₹ \left(\frac{32000 \times 8.5 \times \frac{1}{5}}{100}\right) \\[1em] = ₹ \Big(\frac{32000 \times 8.5}{100 \times 5}\Big) \\[1em] = ₹ \Big(\frac{32000 \times 8.5}{500}\Big) \\[1em] = ₹ \Big(\frac{320 \times 8.5}{5}\Big) \\[1em] = ₹ \Big(\frac{2720}{5}\Big) \\[1em] = ₹ 544\end{aligned}$$

∴ S.I. = ₹ 544

Hence, option 1 is the correct option.