Compound Interest
- The compound interest on ₹ 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly, is :
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We find interest on compounded half-yearly,
Here , P = ₹ 10000, r = 4 / 2 = 2%, n = 2 × 2 = 4 years
Using the given formula ,A = P 
1 + r 
n 100 A = 10,000 1 + 2 4 100 A = 10,000 51 4 = 10824.3216 50
Correct Option: B
We find interest on compounded half-yearly,
Here , P = ₹ 10000, r = 4 / 2 = 2%, n = 2 × 2 = 4 years
Using the given formula ,A = P 1 + r n 100 A = 10,000 1 + 2 4 100 A = 10,000 51 4 = 10824.3216 50
∴ Compound Interest = Amount - Principal
Compound Interest = 10,824.3216 – 10,000 = ₹ 824.32
- At what percent per annum will ₹ 3000 amounts to ₹ 3993 in 3 years if the interest is compounded annually?
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Here , P = ₹ 3000, A = ₹ 3993, n = 3 years
Using the given formula ,A = P 1 + r n 100 ∴ 1 + r n = A 100 P 1 + r 3 = 3993 = 1331 100 3000 1000 1 + r 3 = 11 3 100 10
Correct Option: B
Here , P = ₹ 3000, A = ₹ 3993, n = 3 years
Using the given formula ,A = P 1 + r n 100 ∴ 1 + r n = A 100 P 1 + r 3 = 3993 = 1331 100 3000 1000 1 + r 3 = 11 3 100 10 ⇒ 1 + r = 11 100 10 ⇒ r = 11 − 1 100 10 ⇒ r = 1 ⇒ r = 100 100 10 10
∴ r = 10%
- A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in
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Suppose the Principal be P and rate of interest be R%.
According to question ,2P = P 
1 + R 15 100 2 = 1 + R 15 100
Cubing on both sides, we have8 = 1 1 + R 45 100
On Multiplying by P both sides ,8P = P 1 + R 45 100
Correct Option: A
Suppose the Principal be P and rate of interest be R%.
According to question ,2P = P 1 + R 15 100 2 = 1 + R 15 100
Cubing on both sides, we have8 = 1 1 + R 45 100
On Multiplying by P both sides ,8P = P 1 + R 45 100
Required time = 45 years
We can find required answer with the help of given formula :
Here, m = 2, t = 15 years
It becomes 8 times = 23 times
in t × n years= 15 × 3 = 45 years
- If a sum of money compounded annually becomes 1.44 times of itself in 2 years, then the rate of interest per annum is
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Given in question , Time = 2 years , Rate = R%
Suppose Principal be P and Amount = 1.44P
Using the given formula ,A = P 1 + R T 100 ⇒ 1.44P = P 1 + R 2 100 ⇒ (1.2)2 = 1 + R 2 100 ⇒ 1 + R = 1.2 100
⇒ R = 0.2 × 100 = 20%
We can find required answer with the help of given formula :
Here, n = 1.44 , t = 2 years
R% = (n1 / t − 1) × 100%
Correct Option: D
Given in question , Time = 2 years , Rate = R%
Suppose Principal be P and Amount = 1.44P
Using the given formula ,A = P 1 + R T 100 ⇒ 1.44P = P 1 + R 2 100 ⇒ (1.2)2 = 1 + R 2 100 ⇒ 1 + R = 1.2 100
⇒ R = 0.2 × 100 = 20%
We can find required answer with the help of given formula :
Here, n = 1.44 , t = 2 years
R% = (n1 / t − 1) × 100%
R% = [(1.44)1/2 − 1] × 100%
R% = [(1.2) − 1] × 100%
R% = 0.2 × 100% = 20%
- A sum of money becomes 1.331 times in 3 years as compound interest. The rate of interest is
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As per the given in question ,
Time = 3 years , Rate = R%
Suppose principal = ₹ 1000 , then amount = ₹ 1331∴ A = P 1 + R T 100 ⇒ 1331 = 1000 1 + R 3 100 ⇒ 1331 = 1 + R 3 1000 100 ⇒ 11 3 = 1 + R 3 10 100 ⇒ 1 + R = 11 100 10 ⇒ R = 1 100 10 ⇒ R = 1 × 100 = 10% 10
Second Method to solve this question :
Here, n = 1.331, t = 3 years
R% = (n1/t − 1) × 100%
Correct Option: C
As per the given in question ,
Time = 3 years , Rate = R%
Suppose principal = ₹ 1000 , then amount = ₹ 1331∴ A = P 1 + R T 100 ⇒ 1331 = 1000 1 + R 3 100 ⇒ 1331 = 1 + R 3 1000 100 ⇒ 11 3 = 1 + R 3 10 100 ⇒ 1 + R = 11 100 10 ⇒ R = 1 100 10 ⇒ R = 1 × 100 = 10% 10
Second Method to solve this question :
Here, n = 1.331, t = 3 years
R% = (n1/t − 1) × 100%
R% = [(1.331)1/3 − 1] × 100%
R% =[1.1 − 1] × 100%
R% = 0.1 × 100% = 10%
