Compound Interest — Multiple Choice Questions

The compound interest on ₹ 3,750 for 2 years at 8% p.a., compounded annually is:

1. ₹ 604

2. ₹ 614

3. ₹ 624

4. ₹ 642

Answer

Given,

P = ₹ 3,750

n = 2 years

r = 8%

By formula,

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

Substituting values we get :

$\Rightarrow A = 3750 \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\dfrac{100 + 8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750\times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \dfrac{729}{625} \\[1em] \Rightarrow A = ₹ 4,374$

Compound interest = Final amount - Initial principal

= ₹ 4374 - ₹ 3750 = ₹ 624.

Hence, option 3 is correct option.

A man invests ₹ 46,875 at 4% p.a. compound interest for 3 years. The interest for the 1st year will be:

1. ₹ 1,785

2. ₹ 1,587

3. ₹ 1,875

4. ₹ 1,758

Answer

Given,

P = ₹ 46,875

n = 1 year

r = 4%

By formula,

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

Substituting values we get :

$\Rightarrow A = 46875 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] \Rightarrow A = 46875 \times \Big(\dfrac{100 + 4}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \Big(\dfrac{104}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \dfrac{26}{25} \\[1em] \Rightarrow A = ₹ 48,750.$

Compound interest = Final amount - Initial principal

= ₹ 48,750 - ₹ 46,875 = ₹ 1,875.

Hence, option 3 is correct option.

A man deposits ₹ 10,000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is:

1. ₹ 12,950.29

2. ₹ 12,905.29

3. ₹ 12,059.29

4. ₹ 12,095.29

Answer

Given,

P = ₹ 10,000

n = 3 years

r = 9%

By formula,

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

Substituting values we get :

$\Rightarrow A = 10000 \times \Big(1 + \dfrac{9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \Big(\dfrac{100 + 9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000\times \Big(\dfrac{109}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \dfrac{1295029}{1000000} \\[1em] \Rightarrow A = ₹ 12,950.29$

Hence, option 1 is correct option.

₹ 16,000 is deposited in a bank for three years. The rates of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes ₹ 21,384. The rate of interest for the third year will be:

1. 7%

2. 10%

3. 11%

4. 12%

Answer

Given,

P = ₹ 16,000

r₁ = 8%

r₂ = 12%

A = ₹ 21,384

Let the interest for third year be r₃.

By formula,

Amount = $P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big)$

Substituting the values in formula,

$\Rightarrow 21384 = 16000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{100 + 8}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{108}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{27}{25}\Big) \times \Big(\dfrac{28}{25}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = \dfrac{16000 \times 27 \times 28}{25 \times 25} \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 \times 25 \times 25= 16000 \times 27 \times 28 \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 25 \times 25}{16000 \times 27 \times 28} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 625}{16000 \times 756} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{13365000}{12096000} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 1 + \dfrac{r_3}{100} = 1.10 \\[1em] \Rightarrow \dfrac{r_3}{100} = 1.10 - 1 \\[1em] \Rightarrow r_3 = 0.10 \times 100\\[1em] \Rightarrow r_3 = 10$

Hence, option 2 is correct option.

A man borrows ₹ 5,000 at 12% compound interest p.a., interest payable every six months. He pays back ₹ 1,800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be:

1. ₹ 2,024.60

2. ₹ 2,204.60

3. ₹ 2,240.60

4. ₹ 2,402.60

Answer

For first six moths :

P = ₹ 5,000

T = 6 months = 0.5 year

R = 12%

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

$= \dfrac{5000 \times 12 \times 0.5}{100}$ = ₹ 300.

Amount = P + I = ₹ 5,000 + ₹ 300 = ₹ 5,300.

Amount payed at end of six months = ₹ 1,800.

Amount left at beginning of second six months = ₹ 5,300 - ₹ 1,800 = ₹ 3,500.

For next six months :

P = ₹ 3,500

R = 12%

T = 6 months = 0.5 year

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

$= \dfrac{3500 \times 12 \times 0.5}{100}$ = ₹ 210.

Amount = P + I = ₹ 3,500 + ₹ 210 = ₹ 3,710.

Amount payed at end of second six months = ₹ 1,800.

Amount left at beginning of third six months = ₹ 3,710 - ₹ 1,800 = ₹ 1,910

For next six months :

P = ₹ 1,910

R = 12%

T = 6 months = 0.5 year

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

$= \dfrac{1910 \times 12 \times 0.5}{100}$ = ₹ 114.6

Amount due at the end of third year = P + I = ₹ 1,910 + ₹ 114.6 = ₹ 2,024.60

Hence, option 1 is correct option.

The compound interest for the second year on ₹ 8,000 invested for 3 years at 10% p.a. is:

1. ₹ 780

2. ₹ 880

3. ₹ 890

4. ₹ 1,080

Answer

Given,

P = ₹ 8,000

T = 1 year

r = 10%

For the first year,

By formula,

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

$= \dfrac{8000 \times 10 \times 1}{100}$ = ₹ 800.

Amount = P + I = ₹ 8,000 + ₹ 800 = ₹ 8,800.

Amount at beginning of second year = ₹ 8,800.

P = ₹ 8,800

T = 1 year

r = 10%

For the first year,

By formula,

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

$= \dfrac{8800 \times 10 \times 1}{100}$ = ₹ 880.

The compound interest for the second year is ₹ 880

Hence, option 2 is correct option.

A person took a loan of ₹ 6,000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12%, the amount he had to pay after 2 years will be:

1. ₹ 7,329

2. ₹ 7,932

3. ₹ 7,292

4. ₹ 7,392

Answer

Given,

P = ₹ 6,000

r₁ = 10%

r₂ = 12%

n = 2 years

By formula,

Amount = $P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big)$

Substituting the values in formula,

$\Rightarrow A = 6000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] \Rightarrow A = \dfrac{6000 \times 11 \times 28}{10 \times 25} \\[1em] \Rightarrow A = ₹ 7,392.$

Hence, option 4 is correct option.

Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to ₹ 6,720. The rate of interest p.a. is:

1. 8%

2. 10%

3. 12%

4. 14%

Answer

Given,

P = ₹ 6,000

n = 1 year

A = ₹ 6,720

Let the rate of interest be r,

By formula,

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

Substituting values we get :

$\Rightarrow 6720 = 6000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{6720}{6000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = 1.12 \\[1em] \Rightarrow \dfrac{r}{100} = 1.12 - 1 \\[1em] \Rightarrow r = 0.12 \times 100 \\[1em] \Rightarrow r = 12$

Hence, option 3 is correct option.

The compound interest on ₹ 8,640 for 3 years at 8% p.a. is:

1. ₹ 2,345

2. ₹ 3,245

3. ₹ 3,425

4. ₹ 3,452

Answer

Given,

P = ₹ 8,640

n = 3 year

r = 8%

By formula,

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

Substituting values we get :

$\Rightarrow A = 8640 \times \Big(1 + \dfrac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{100 + 8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \dfrac{19683}{15625} \\[1em] \Rightarrow A = ₹ 10,883.9$

Compound interest = Final amount - Initial principal

= ₹ 10,883.9 - ₹ 8,640

= ₹ 2,243.9

If the interest is compounded half-yearly, then, C.I. when the principal is ₹ 7,400, the rate of interest is 5% p.a. and the duration is one year, is:

1. ₹ 373.63

2. ₹ 374.63

3. ₹ 373.36

4. ₹ 373

Answer

Given,

P = ₹ 7,400

r = 5%

n = 1 year

Given,

When interest is compounded half-yearly.

By formula,

$A = P \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}$

Substituting values we get :

$\Rightarrow A = 7400 \times \Big(1 + \dfrac{5}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{200 + 5}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{205}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{41}{40}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \dfrac{1681}{1600} \\[1em] \Rightarrow A = 7774.625$

Compound interest = Final amount - Initial principal

= ₹ 7774.625 - ₹ 7400

= ₹ 374.625 ≈ ₹ 374.63

Hence, option 2 is correct option.

The simple interest on a sum of money for 2 years at 4% per annum is ₹ 340. The compound interest on this sum for one year payable half-yearly at the same rate is:

1. ₹ 170.70

2. ₹ 107.70

3. ₹ 171.70

4. ₹ 270.70

Answer

Given,

I = ₹ 340

R = 4%

T = 2 years

Let sum of money be ₹ P.

By formula,

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

$\Rightarrow 340 = \dfrac{P \times 4 \times 2}{100} \\[1em] \Rightarrow 340 = \dfrac{P \times 8}{100} \\[1em] \Rightarrow P = \dfrac{340 \times 100}{8} \\[1em] \Rightarrow P = ₹ 4,250.$

Given, interest is compounded half-yearly.

By formula,

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}$

Substituting values we get :

$\Rightarrow A = 4250 \Big(1 + \dfrac{4}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 4250 \Big(\dfrac{200 + 4}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(\dfrac{204}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(1.02\Big)^2 \\[1em] \Rightarrow A = 4250 \times 1.0404 \\[1em] \Rightarrow A = ₹ 4,421.7$

Compound interest = Amount - Principal = ₹ 4,421.7 - ₹ 4,250 = ₹ 171.70.

Hence, option 3 is correct option.

The compound interest on a certain sum of money at 5% p.a. for two years is ₹ 246. The simple interest on the same sum for three years at 6% p.a. will be:

1. ₹ 432

2. ₹ 430.50

3. ₹ 432.75

4. ₹ 431.75

Answer

Given,

I = ₹ 246

R = 5%

n = 2 years

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

Compound interest = Amount - Principal

$CI = P\Big(1 + \dfrac{r}{100}\Big)^{n} - P$

$\Rightarrow 246 = P \Big(1 + \dfrac{5}{100}\Big)^{2} - P \\[1em] \Rightarrow 246 = P \Big(\dfrac{100 + 5}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P \Big(\dfrac{105}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P [(1.05)^2 - 1] \\[1em] \Rightarrow 246 = P [1.1025 - 1] \\[1em] \Rightarrow 246 = 0.1025P \\[1em] \Rightarrow \dfrac{246}{0.1025} = P\\[1em] \Rightarrow P = ₹ 2,400.$

For calculating Simple interest,

P = ₹ 2,400

R = 6%

T = 3 years

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

$\Rightarrow I = \dfrac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow I = \dfrac{43200}{100} \\[1em] \Rightarrow I = ₹ 432$

Hence, option 1 is correct option.

Ramesh wants to get ₹ 6,050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is:

1. ₹ 5,500

2. ₹ 5,000

3. ₹ 5,600

4. ₹ 5,800

Answer

Given,

A = ₹ 6,050

R = 10%

n = 2 years

Let amount he needs to keep be ₹ P.

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

Substituting values we get :

$\Rightarrow 6050 = P \Big(1 + \dfrac{10}{100}\Big)^{2} \\[1em] \Rightarrow 6050 = P \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P (1.1)^2 \\[1em] \Rightarrow 6050 = P \times 1.21 \\[1em] \Rightarrow \dfrac{6050}{1.21} = P \\[1em] \Rightarrow P = ₹ 5,000. \\[1em]$

Hence, option 2 is correct option.

The difference between the compound and simple interest on a certain sum deposited for 2 years at 5% p.a. is ₹ 12. The sum will be :

1. ₹ 4,500

2. ₹ 4,600

3. ₹ 4,800

4. ₹ 5,000

Answer

By formula,

Given,

T = 2 years

r = 5%

Let sum of money be ₹ P.

By formula,

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 5 \times 2}{100} \\[1em] = \dfrac{P}{10}.$

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{5}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{105}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{21}{20}\Big)^2 - P\\[1em] = P \times \dfrac{441}{400} - P \\[1em] = \dfrac{441P}{400} - P \\[1em] = \dfrac{441P - 400P}{400} \\[1em] = \dfrac{41P}{400}.$

Given,

Difference between S.I. and C.I. = ₹ 12

$\Rightarrow \dfrac{41P}{400} - \dfrac{P}{10} = 12 \\[1em] \Rightarrow \dfrac{41P - 40P}{400} = 12 \\[1em] \Rightarrow \dfrac{P}{400} = 12 \\[1em] \Rightarrow P = 400 \times 12 \\[1em] \Rightarrow P = ₹ 4,800.$

Hence, option 3 is correct option.

At what rate of compound interest p.a. will ₹ 20,000 amount to ₹ 26,620 in 3 years?

1. 4%

2. 6%

3. 8%

4. 10%

Answer

Given,

A = ₹ 26,620

P = ₹ 20,000

n = 3 years

Let rate of interest be r.

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

$\Rightarrow 26620 = 20000 \Big(1 + \dfrac{R}{100}\Big)^{3} \\[1em] \Rightarrow \dfrac{26620}{20000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = \Big(1 + \dfrac{R}{100}\Big) \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{11-10}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{100}{10} = R \\[1em] \Rightarrow R = 10$

Hence, option 4 is correct option.

In what time will ₹ 5,000 amount to ₹ 5,832 at 8% rate of compound interest p.a.?

1. 2 years

2. 4 years

3. 6 years

4. 8 years

Answer

Given,

Amount = ₹ 5,832

Principal = ₹ 5,000

R = 8%

Let time be n years.

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

$\Rightarrow 5832 = 5000 \Big(1 + \dfrac{8}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{5832}{5000} = \Big(\dfrac{100 + 8}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = \Big(\dfrac{108}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = (1.08)^n \\[1em] \Rightarrow (1.08)^2 = (1.08)^n \\[1em] \Rightarrow n = 2$

Hence, option 1 is correct option.

A machine depreciates at the rate of 10% of its value at the beginning of a year. If the present value of a machine is ₹ 8,000, its value after 3 years will be:

1. ₹ 5,382

2. ₹ 5,832

3. ₹ 5,238

4. ₹ 5,638

Answer

Given,

V = ₹ 8,000

n = 3 years

R = 10%

Value of machine after n years = $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

$\text{Value of machine after 3 years }= 8000 \Big(1 - \dfrac{10}{100}\Big)^{3} \\[1em] = 8000 \Big(\dfrac{100 - 10}{100}\Big)^3 \\[1em] = 8000 \Big(\dfrac{90}{100}\Big)^3 \\[1em] = 8000 (0.9)^3 \\[1em] = 8000 \times 0.729 \\[1em] = ₹ 5,832.$

Value of machine after 3 years = ₹ 5,832

Hence, option 2 is correct option.

The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be:

1. 253000

2. 235000

3. 203500

4. 352000

Answer

Given,

P = 200000

r₁ = 10% p.a.

r₂ = 15% p.a.

By formula,

Population after two years = $P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big)$

Substituting the values in formula,

$\text{Population after two years }=200000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{15}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 15}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{115}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{23}{20}\Big) \\[1em] =200000 \times 1.10 \times 1.15\\[1em] =253000.$

Hence, option 1 is correct option.

A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for ₹ 10,000 and is sold for ₹ 7,744. The number of years, that the machine was used is:

1. 2

2. 4

3. 6

4. 8

Answer

Given,

Initial value (P) = ₹ 10,000

Depreciated value (A) = ₹ 7,744

R = 12%

Value of machine after n years = $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

$\Rightarrow 7744 = 10000 \Big(1 - \dfrac{12}{100}\Big)^{n} \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{100 - 12}{100}\Big)^n \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \dfrac{7744}{10000} = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{88}{100}\Big)^2 = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow n = 2.$

Hence, option 1 is correct option.

The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are ₹ 25,000 and ₹ 19,360 respectively, then the annual rate of depreciation is:

1. 8%

2. 10%

3. 12%

4. 14%

Answer

Given,

Initial value (P) = ₹ 25,000

Depreciated value (A) = ₹ 19,360

n = 2 years

Let rate of depreciation be R.

Value of machine after n years = $\Big[V \times \Big(1 - \dfrac{R}{100}\Big)^n \Big]$

$\Rightarrow 19360 = 25000 \Big(1 - \dfrac{R}{100}\Big)^{2} \\[1em] \Rightarrow 19360 = 25000 \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{19360}{25000} = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{484}{625} = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{22}{25}\Big)^2 = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{22}{25} = 1 - \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{R}{100} = 1 - \dfrac{22}{25} \\[1em] \Rightarrow \dfrac{R}{100} = \dfrac{25 - 22}{25} \\[1em] \Rightarrow \dfrac{R}{100} = \dfrac{3}{25} \\[1em] \Rightarrow R = \dfrac{3 \times 100}{25} \\[1em] \Rightarrow R = 3 \times 4\\[1em] \Rightarrow R = 12$

Hence, option 3 is correct option.