Simple Interest

Find the simple interest and amount on each of the following :

(i) ₹ 8400 for 4 years at 8% per annum.

(ii) ₹ 50000 for 3 years $12\dfrac{1}{2}$% per annum.

(iii) ₹ 9275 for 2 years at $7\dfrac{1}{2}$% per annum.

Answer

(i)

Given:

P = ₹ 8400, R = 8%, T = 4 years

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

Amount = Principal + S.I.

Amount = ₹ 8400 + ₹ 2688

Amount = ₹ 11088

S.I. = ₹ 2688, Amount = ₹ 11088

(ii)

Given:

P = ₹ 50000, T = 3 years

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

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

Amount = Principal + S.I.

Amount = ₹ 50000 + ₹ 18750

Amount = ₹ 68750

S.I. = ₹ 18750, Amount = ₹ 68750

(iii)

Given:

P = ₹ 9275, T = 2 years

R = $7\dfrac{1}{2}$

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

Amount = Principal + S.I.

Amount = ₹ 9275 + ₹ 1391.25

Amount = ₹ 10666.25

S.I. = ₹ 1391.25, Amount = ₹ 10666.25

In what time will ₹ 7500 amount to ₹ 8625, if simple interest is reckoned at $7\dfrac{1}{2}$% per annum.

Answer

Given:

Principal = ₹ 7500, Amount = ₹ 8625

R = $7\dfrac{1}{2}$

S.I. = Amount - Principal

S.I. = ₹ 8625 - ₹ 7500 = ₹ 1125

Then

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

∴ Time = 2 years

In what time will ₹ 3600 amount to ₹ 4320 at 4% per annum simple interest?

Answer

Given:

P = ₹ 3600, A = ₹ 4320, R = 4%

S.I. = ₹ 4320 - ₹ 3600 = ₹ 720

Then

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

∴ Time = 5 years

At what rate per cent per annum will ₹ 6300 yield an interest of ₹ 2100 in 4 years?

Answer

Given:

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

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow R = \dfrac{S.I. \times 100}{P \times T}\% \\[1em] = \Big(\dfrac{2100 \times 100}{6300 \times 4}\Big)\% \\[1em] = \Big(\dfrac{2100 \times 1}{63 \times 4}\Big)\% \quad \text{[Dividing 100 and 6300 by 100]} \\[1em] = \Big(\dfrac{525 \times 1}{63 \times 1}\Big)\% \quad \text{[Dividing 2100 and 4 by 4]} \\[1em] = \Big(\dfrac{525}{63}\Big)\% \\[1em] = \dfrac{25}{3}\% \quad \text{[Dividing 525 and 63 by 21]} \\[1em] = 8\dfrac{1}{3}\%$

∴ Rate = $\bold {8\dfrac{1}{3}}\%$

At what rate per cent per annum simple interest will ₹ 66000 amount to ₹ 72720 in 2 years?

Answer

Given:

P = ₹ 66000, A = ₹ 72720, T = 2 years

S.I. = ₹ 72720 - ₹ 66000 = ₹ 6720

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow R = \dfrac{S.I. \times 100}{P \times T}\% \\[1em] = \Big(\dfrac{6720 \times 100}{66000 \times 2}\Big)\% \\[1em] = \Big(\dfrac{6720 \times 1}{660 \times 2}\Big)\% \quad \text{[Dividing 100 and 66000 by 100]} \\[1em] = \Big(\dfrac{3360 \times 1}{660 \times 1}\Big)\% \quad \text{[Dividing 6720 and 2 by 2]} \\[1em] = \dfrac{56}{11}\% \quad \text{[Dividing 3360 and 660 by 60]} \\[1em] = 5\dfrac{1}{11}\%$

∴ Rate = $\bold{5\dfrac{1}{11}\%}$ per annum

The simple interest on a sum of money for 5 years is $\dfrac{3}{5}$ of the sum. Find the rate per cent per annum.

Answer

Given:

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

R = ?

Let the sum (P) be x.

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

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow R = \dfrac{S.I. \times 100}{P \times T}\% \\[1em] = \left(\dfrac{(\dfrac{3}{5}x) \times 100}{x \times 5}\right)\% \\[1em] = \Big(\dfrac{3x \times 100}{5 \times x \times 5}\Big)\% \\[1em] = \dfrac{300x}{25x}\% \\[1em] = 12\%$

∴ Rate = 12% per annum

The simple interest on a certain sum for 3 years at 10% per annum is ₹ 829.50. Find the sum.

Answer

Given:

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

Then

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

∴ Sum = ₹ 2765

What sum will yield an interest of ₹ 7840 in 2 years at $6\dfrac{1}{4}$% per annum?

Answer

Given:

S.I. = ₹ 7840, T = 2 years

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

Then

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

∴ Sum = ₹ 62720

At what rate per cent per annum simple interest will a sum treble itself in 16 years?

Answer

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

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow R = \dfrac{S.I. \times 100}{P \times T}\% \\[1em] = \Big(\dfrac{2x \times 100}{x \times 16}\Big)\% \\[1em] = \Big(\dfrac{200x}{16x}\Big)\% \\[1em] = 12\dfrac{1}{2}\%$

∴ Rate = $\bold{12\dfrac{1}{2}\%}$

A sum when reckoned at 6% per annum simple interest amounts to ₹ 4130 in 3 years. Find the sum.

Answer

Given:

A = ₹ 4130, T = 3 years, R = 6%

We have the formula:

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

∴ Sum = ₹ 3500

A sum of money put at 11% per annum simple interest amounts to ₹ 10370 in 2 years. What will it amount to in 3 years at the same rate?

Answer

Let the sum be ₹ 100.

Given:

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

Then

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

Amount = S.I. + Principal

Amount = ₹ 22 + ₹ 100

Amount = ₹ 122

If Amount is ₹ 122, then Sum = ₹ 100.

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

If Amount is ₹ 10370, then Sum =

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

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

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

Required Amount = S.I. + Principal

Required Amount = ₹ 2805 + ₹ 8500 = ₹ 11305

∴ Amount = ₹ 11305

A sum of money put at 9% per annum simple interest amounts to ₹ 10160 in 3 years. What will it amount to in 2 years at 8% per annum?

Answer

Let the sum be ₹ 100.

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

Then

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

Amount = S.I. + Principal

Amount = ₹ 27 + ₹ 100

Amount = ₹ 127

If Amount is ₹ 127, then Sum = ₹ 100.

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

If Amount is ₹ 10160, then Sum =

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

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

And

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

Required Amount = S.I. + Principal

Required Amount = ₹ 1280 + ₹ 8000 = ₹ 9280

∴ Amount = ₹ 9280

Which of the following statements is incorrect?

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

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

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

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

Answer

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

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

Hence, option 2 is the correct option.

If the principal is ₹ 600, then the amount to be paid at the end of 3 years at $7\dfrac{1}{2}$% p.a. simple interest will be

1. ₹ 735
2. ₹ 750
3. ₹ 775
4. ₹ 785

Answer

Given:

P = ₹ 600, T = 3 years

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

Then

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

Amount = S.I. + Principal

Amount = ₹ 135 + ₹ 600 = ₹ 735

Hence, option 1 is the correct option.

At what rate, ₹ 800 gives ₹ 208 as simple interest in 2 years?

1. 11%
2. 12%
3. 13%
4. 14%

Answer

Given:

Principal (P) = ₹ 800

Simple Interest (S.I.) = ₹ 208

Time (T) = 2 years

Then

$R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \Big(\dfrac{208 \times 100}{800 \times 2}\Big)\% \\[1em] = \dfrac{20800}{1600}\% \\[1em] = \dfrac{208}{16}\% \\[1em] = 13\%$

Hence, option 3 is the correct option.

In how many years will ₹ 900 give ₹ 351 as simple interest at 13% p.a.?

1. $1\dfrac{1}{2}$years

2. 2 years

3. $2\dfrac{1}{2}$

4. 3 years

Answer

Given:

Principal (P) = ₹ 900

Simple Interest (S.I.) = ₹ 351

Rate (R) = 13% p.a.

Then

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

Hence, option 4 is the correct option.

At what rate per cent per annum simple interest will a sum be double of itself in 8 years?

1. 10%

2. $12\dfrac{1}{2}$%

3. 15%

4. $17\dfrac{1}{2}$%

Answer

Given:

Let Principal (P) = x

Amount (A) = 2x

Time (T) = 8 years

S.I. = Amount - Principal

S.I. = 2x - x = x

Then

$R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \Big(\dfrac{x \times 100}{x \times 8}\Big)\% \\[1em] = \dfrac{100}{8}\% \\[1em] = \dfrac{25}{2}\% \\[1em] = 12\dfrac{1}{2}\%$

Hence, option 2 is the correct option.

At simple interest a sum becomes $\dfrac{7}{4}$ of itself in 5 years. The rate of interest is

1. 10% p.a.

2. 12% p.a.

3. $12\dfrac{1}{2}$% p.a.

4. 15% p.a.

Answer

Given:

Let the Principal (P) = ₹ 100

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

Amount = ₹ $\dfrac{7}{4} \times 100$

Amount = ₹ 7 x 25 = ₹ 175

Time (T) = 5 years

S.I. = Amount - Principal

S.I. = ₹ 175 - ₹ 100 = ₹ 75

Then

$R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \Big(\dfrac{75 \times 100}{100 \times 5}\Big)\% \\[1em] = \dfrac{75}{5}\% \\[1em] = 15\%$

Hence, option 4 is the correct option.

Fill in the blanks :

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

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

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

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

(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 ............... .

Answer

(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

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

S.I. = ₹ 312

(iv)

Given:

P = ₹ 2600, A = ₹ 3900, T = 5 years

S.I. = Amount - Principal

S.I. = ₹ 3900 - ₹ 2600 = ₹ 1300

Then

$R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \Big(\dfrac{1300 \times 100}{2600 \times 5}\Big)\% \\[1em] = \Big(\dfrac{1 \times 100}{2 \times 5}\Big)\% \\[1em] = \dfrac{100}{10}\% \\[1em] = 10\%$

R = 10%

(v) Given:

P = ₹ 3400, R = 8%, T = 5 years

Then

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

Amount = S.I. + Principal

Amount = ₹ 1360 + ₹ 3400

Amount = ₹ 4760

Write true (T) or false (F) :

(i) The total money to be paid back to the lender is called interest.

(ii) Amount = Principal + Interest

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

(iv) S.I. = $\dfrac{100 \times P}{R \times T}$

(v) If a man borrows ₹ 5200 at 6% p.a. simple interest, the amount he has to return at the end of 5 years is ₹ 6760.

Answer

(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. = $\dfrac{P \times R \times T}{100}$

(v) True
Reason —

P = ₹ 5200, R = 6%, T = 5 years

Then

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

Amount = S.I. + Principal

Amount = ₹ 1560 + ₹ 5200

Amount = ₹ 6760

Mr. Shah has two choices of investing his money. If he invests in a bank, he gets 8% simple interest. If he invests in his friend's company ADG, he gets 12% simple interest. Mr. Shah has ₹ 5,00,000 to invest.

(1) If Mr. Shah invests all the money with the bank, the interest received by him after 2 years will be:

1. ₹ 50,000
2. ₹ 60,000
3. ₹ 75,000
4. ₹ 80,000

(2) What sum invested in ADG company will amount to ₹ 4,80,000 in 5 years ?

1. ₹ 3,00,000
2. ₹ 3,50,000
3. ₹ 4,00,000
4. ₹ 4,20,000

(3) In what time will the money invested with the bank double itself ?

1. $7\dfrac{1}{2}$years

2. 9 years

3. $12\dfrac{1}{2}$years

4. $11\dfrac{3}{4}$years

(4) If Mr. Shah wishes to invest partly in the bank and partly in ADG company such that after 2 years he receives the same interest from both, then find the sum that he would invest in the bank.

1. ₹ 2,00,000
2. ₹ 2,50,000
3. ₹ 3,00,000
4. ₹ 3,50,000

Answer

(1)

Given:

P = ₹ 5,00,000, R = 8%, T = 2 years.

Then

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

Hence, option 4 is the correct option.

(2)

Given:

A = ₹ 4,80,000, R = 12%, T = 5 years.

Then

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

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

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

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

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

Hence, option 3 is the correct option.

Assertion: The interest on ₹ 700 at 5% p.a. for 12 months is ₹ 35.

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

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

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

Explanation

Given:

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

Then

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

The Assertion is True.

Reason:

S.I. = $\dfrac{\text{PRT}}{100}$, 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: Amount received after depositing ₹ 800 for a period of 3 years at the rate of 12% p.a. simple interest is ₹ 1096.

Reason: Amount = Principal + Interest.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

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

Explanation

Given:

P = ₹ 800, T = 3 years, R = 12%

Then

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

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.