Compound Interest objective type question set 1

Answer: b. 10%

Explanation:
Let \( P \) be the principal amount, \( A \) be the total amount, and \( n \) be the number of years.
According to the formula for Compound Interest:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ \frac{3993}{3000} = \left(1 + \frac{r}{100}\right)^3 \]
\[ \frac{1331}{1000} = \left(1 + \frac{r}{100}\right)^3 \]
\[ \left(\frac{11}{10}\right)^3 = \left(1 + \frac{r}{100}\right)^3 \]
\[ \frac{11}{10} = 1 + \frac{r}{100} \]
\[ \frac{r}{100} = \frac{11}{10} – 1 = \frac{1}{10} \]
Rate (r) = 10%

Answer : b. Rs. 27040

Explanation:
According to the question:
Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ A = 25000 \left(1 + \frac{4}{100}\right)^2 \]
\[ A = 25000 \left(\frac{104}{100}\right)^2 \]
\[ A = 25000 \left(\frac{104 \times 104}{100 \times 100}\right) \]
\[ A = \frac{250 \times 104 \times 104}{100} \]
\[ A = 2.5 \times 10816 \]
\[ A = 27040 \]
Total Amount = 27,040

Answer : b. Rs. 3972

Solution:
According to the question:
Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ A = 12000 \left(1 + \frac{10}{100}\right)^3 \]
\[ A = 12000 \left(\frac{110}{100}\right)^3 \]
\[ A = 12000 \times \frac{11 \times 11 \times 11}{10 \times 10 \times 10} \]
\[ A = 12 \times 1331 \]
\[ A = 15972 \]

Compound Interest (C.I.) = Amount – Principal
\[ C.I. = 15972 – 12000 = 3972 \]
Compound Interest = 3,972

Answer : c. Rs. 12.5

Explanation:
Formula used for difference (D) between C.I. and S.I. for 2 years:
\[ \text{Difference} = P \left(\frac{R}{100}\right)^2 \]
Where \( P \) is the principal amount, \( R \) is the rate of interest, and \( n = 2 \) years.

Calculation:
\[ \text{Difference} = 5000 \left(\frac{5}{100}\right)^2 \]
\[ \text{Difference} = \frac{5000 \times 5 \times 5}{100 \times 100} \]
\[ \text{Difference} = \frac{5 \times 25}{10} = \frac{125}{10} \]
\[ \text{Difference} = 12.5 \]
The difference between C.I. and S.I. is 12.5.

Answer : b. 2 years

Explanation:
Let \( n \) be the number of years.
Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
Given: Amount (\( A \)) = 2420, Principal (\( P \)) = 2000, Rate (\( r \)) = 10%

\[ 2420 = 2000 \left(1 + \frac{10}{100}\right)^n \]
\[ \frac{2420}{2000} = \left(\frac{110}{100}\right)^n \]
\[ \frac{121}{100} = \left(\frac{11}{10}\right)^n \]
\[ \left(\frac{11}{10}\right)^2 = \left(\frac{11}{10}\right)^n \]
Comparing the powers:
n = 2 years

Answer : b. 232

Solution:
1. For Bank A (Simple Interest):
Principal (\( P_1 \)) = 6000, Rate (\( R_1 \)) = 5%, Time (\( T \)) = 2 years
\[ S.I. = \frac{6000 \times 5 \times 2}{100} = 600 \]

2. For Bank B (Compound Interest):
Principal (\( P_2 \)) = 5000, Rate (\( R_2 \)) = 8%, Time (\( n \)) = 2 years
Formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ A = 5000 \left(1 + \frac{8}{100}\right)^2 \]
\[ A = 5000 \left(\frac{108}{100}\right) \times \left(\frac{108}{100}\right) = 5832 \]
\[ C.I. = 5832 – 5000 = 832 \]

3. Final Difference:
Difference = \( C.I. – S.I. \)
\[ \text{Difference} = 832 – 600 = 232 \]
The difference between the interests is Rs. 232.

Answer : c. 40 years

Explanation:
Short Trick Method:
If a sum of money becomes \( x \) times in \( T \) years, then it will become \( x^n \) times in \( n \times T \) years.

According to the question:
Money becomes 2 times (\( x = 2 \)) in 10 years (\( T = 10 \)).
To become 16 times, we can write 16 as \( 2^4 \) (where \( n = 4 \)).

Time required = \( n \times T \)
Time required = \( 4 \times 10 = 40 \) years.
The money will become 16 times in 40 years.

Answer : d. 20%

Solution:
Let \( P \) be the principal and \( r \) be the rate of interest.
Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
According to the question:
\[ 1440 = P \left(1 + \frac{r}{100}\right)^2 \quad \text{—(eq 1)} \]
\[ 1728 = P \left(1 + \frac{r}{100}\right)^3 \quad \text{—(eq 2)} \]

Dividing equation (2) by equation (1):
\[ \frac{1728}{1440} = \frac{P \left(1 + \frac{r}{100}\right)^3}{P \left(1 + \frac{r}{100}\right)^2} \]
\[ \frac{1728}{1440} = 1 + \frac{r}{100} \]
\[ \frac{r}{100} = \frac{1728}{1440} – 1 \]
\[ \frac{r}{100} = \frac{1728 – 1440}{1440} \]
\[ r = \frac{288}{1440} \times 100 \]
Rate (r) = 20%

Answer : a. 5 %

Explanation:
Let \( P \) be the principal, \( A \) be the amount, and \( r \) be the rate of interest.
Given: \( P = 16000 \), \( A = 18522 \), and \( n = 3 \) years.

Formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ 18522 = 16000 \left(1 + \frac{r}{100}\right)^3 \]
\[ \frac{18522}{16000} = \left(1 + \frac{r}{100}\right)^3 \]
\[ \frac{9261}{8000} = \left(1 + \frac{r}{100}\right)^3 \]
\[ \left(\frac{21}{20}\right)^3 = \left(1 + \frac{r}{100}\right)^3 \]

Taking the cube root on both sides:
\[ \frac{21}{20} = 1 + \frac{r}{100} \]
\[ \frac{r}{100} = \frac{21}{20} – 1 \]
\[ \frac{r}{100} = \frac{1}{20} \]
\[ r = \frac{100}{20} = 5 \]
Rate (r) = 5% per annum

Answer : c. Rs. 1331

Explanation:
To find the interest for the 4th year specifically, we find the difference between the Amount after 4 years and the Amount after 3 years.

1. Amount after 4 years (\( A_4 \)):
\[ A_4 = 10000 \left(1 + \frac{10}{100}\right)^4 \]
\[ A_4 = 10000 \left(\frac{11}{10}\right)^4 \]
\[ A_4 = 10000 \times \frac{14641}{10000} = 14641 \]

2. Amount after 3 years (\( A_3 \)):
\[ A_3 = 10000 \left(1 + \frac{10}{100}\right)^3 \]
\[ A_3 = 10000 \left(\frac{11}{10}\right)^3 \]
\[ A_3 = 10000 \times \frac{1331}{1000} = 13310 \]

3. Interest for the 4th year:
\[ \text{Interest} = A_4 – A_3 \]
\[ \text{Interest} = 14641 – 13310 = 1331 \]
The compound interest for the 4th year is 1,331.

Answer : c. Rs. 1250

Explanation:
Let \( P \) be the principal amount.
Given: Amount (\( A \)) = 1352, Rate (\( r \)) = 4%, Time (\( n \)) = 2 years.

Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ 1352 = P \left(1 + \frac{4}{100}\right)^2 \]
\[ 1352 = P \left(\frac{104}{100}\right)^2 \]
\[ 1352 = P \left(\frac{26}{25}\right)^2 \]

Solving for \( P \):
\[ P = 1352 \times \frac{25}{26} \times \frac{25}{26} \]
\[ P = 2 \times 25 \times 25 \]
\[ P = 1250 \]
The principal amount is Rs. 1,250.

Answer : c. 3 years

Explanation:
Let \( P \) be the principal, \( A \) be the amount, and \( n \) be the number of years.
Given: \( P = 1000 \), \( A = 1331 \), and \( r = 10\% \).

Formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ 1331 = 1000 \left(1 + \frac{10}{100}\right)^n \]
\[ \frac{1331}{1000} = \left(\frac{110}{100}\right)^n \]
\[ \frac{1331}{1000} = \left(\frac{11}{10}\right)^n \]

Since \( 1331 = 11^3 \) and \( 1000 = 10^3 \):
\[ \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n \]

Comparing the powers on both sides:
n = 3 years

Answer : d. Rs. 50,000

Explanation:
Let \( P \) be the principal amount and \( R \) be the rate of interest.
Given: Difference (\( CI – SI \)) = 80, Rate (\( R \)) = 4%, Time (\( n \)) = 2 years.

Formula used for 2 years: \[ CI – SI = P \left(\frac{R}{100}\right)^2 \]
\[ 80 = P \left(\frac{4}{100}\right)^2 \]
\[ 80 = P \left(\frac{1}{25}\right)^2 \]

Solving for \( P \):
\[ P = 80 \times 25 \times 25 \]
\[ P = 80 \times 625 \]
\[ P = 50000 \]
The principal amount is Rs. 50,000.

Answer : d. Rs. 50,000

Explanation:
When interest is compounded half-yearly, the rate is halved and the time is doubled.
Given: \( P = 12000 \), \( r = 5\% \) per annum, \( n = 1 \) year.
New Rate (\( R \)) = \( \frac{5}{2} = 2.5\% \)
New Time (\( T \)) = \( 1 \times 2 = 2 \) half-years.

Formula used: \[ A = P \left(1 + \frac{r}{200}\right)^{2n} \]
\[ A = 12000 \left(1 + \frac{5}{200}\right)^2 \]
\[ A = 12000 \left(1 + \frac{1}{40}\right)^2 \]
\[ A = 12000 \left(\frac{41}{40}\right)^2 \]
\[ A = 12000 \times \frac{41 \times 41}{40 \times 40} \]
\[ A = 12000 \times \frac{1681}{1600} \]
\[ A = 7.5 \times 1681 \]
Amount = 12607.5

Answer : c. Rs. 7260

Solution:
Let \( P \) be the principal, \( r \) be the rate, and \( n \) be the time in years.
Given: \( P = 6000 \), \( r = 10\% \), \( n = 2 \).

Formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ A = 6000 \left(1 + \frac{10}{100}\right)^2 \]
\[ A = 6000 \left(\frac{11}{10}\right)^2 \]
\[ A = 6000 \times \frac{11 \times 11}{10 \times 10} \]
\[ A = 60 \times 121 \]
Amount (A) = 7260

Answer : c. Rs. 250

Explanation:
Let \( P \) be the principal amount.
Given: Amount (\( A \)) = 270.40, Rate (\( r \)) = 4%, Time (\( n \)) = 2 years.

Formula used: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
\[ 270.40 = P \left(1 + \frac{4}{100}\right)^2 \]
\[ 270.40 = P \left(\frac{104}{100}\right)^2 \]

Solving for \( P \):
\[ P = 270.40 \times \frac{100}{104} \times \frac{100}{104} \]
\[ P = \frac{27040}{100} \times \frac{100}{104} \times \frac{100}{104} \]
\[ P = \frac{27040 \times 100}{104 \times 104} \]
\[ P = 250 \]
The principal amount is Rs. 250.

Answer: a. 1.5 Years

Explanation:
When interest is compounded half-yearly, the rate is halved and the time is doubled (\( 2n \)).
Given: Principal (\( P \)) = 800, Amount (\( A \)) = 926.1, Rate (\( r \)) = 10%.

Formula used: \[ A = P \left(1 + \frac{r}{200}\right)^{2n} \]
\[ 926.1 = 800 \left(1 + \frac{10}{200}\right)^{2n} \]
\[ \frac{926.1}{800} = \left(\frac{21}{20}\right)^{2n} \]
\[ \frac{9261}{8000} = \left(\frac{21}{20}\right)^{2n} \]

Recognizing the cubes (\( 21^3 = 9261 \) and \( 20^3 = 8000 \)):
\[ \left(\frac{21}{20}\right)^3 = \left(\frac{21}{20}\right)^{2n} \]

Comparing the exponents:
\[ 2n = 3 \]
\[ n = \frac{3}{2} = 1.5 \]
Time = 1.5 years (or 1 year 6 months)

Answer: c. Rs. 27.27

Explanation:
: Simple interest at 10,000 for 3 years will be 900
Compound interest will be 927.27
Difference between both of them will be 927.27-900 = 27.27

Answer: c. 18 years

Explanation:
Short Trick Method:
If a sum of money becomes \( x \) times in \( T \) years, then it will become \( x^n \) times in \( n \times T \) years.

According to the question:
The money becomes 2 times (\( x = 2 \)) in 3 years (\( T = 3 \)).
To become 64 times, we can write 64 as \( 2^6 \) (where \( n = 6 \)).

Time required = \( n \times T \)
Time required = \( 6 \times 3 = 18 \) years.
The money will become 64 times in 18 years.

Answer: b. 1.5 year

Explanation:
When interest is compounded half-yearly, the rate is halved and the time is doubled (\( 2n \)).
Given: Principal (\( P \)) = 64000, Amount (\( A \)) = 68921, Rate (\( r \)) = 5%.

Formula: \[ A = P \left(1 + \frac{r}{200}\right)^{2n} \]
\[ 68921 = 64000 \left(1 + \frac{5}{200}\right)^{2n} \]
\[ \frac{68921}{64000} = \left(1 + \frac{1}{40}\right)^{2n} \]
\[ \frac{68921}{64000} = \left(\frac{41}{40}\right)^{2n} \]

Recognizing the cubes (\( 41^3 = 68921 \) and \( 40^3 = 64000 \)):
\[ \left(\frac{41}{40}\right)^3 = \left(\frac{41}{40}\right)^{2n} \]

Comparing the exponents:
\[ 2n = 3 \]
\[ n = \frac{3}{2} = 1.5 \]
Time = 1.5 years (or 1 year 6 months)