Compound Interest
- An amount of money appreciates to ₹ 7,000 after 4 years and to ₹ 10,000 after 8 years at a certain compound interest compounded annually. The initial amount of money was
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Here , A1 = ₹ 7,000 , T1 = 4 years and A2 = ₹ 10,000 , T2 = 8 years
Using the given formula ,A = P 
1 + R 
T 100 ⇒ 7000 = P 1 + R 4 ..... (i) 100
10000 = P1 + R 8 ..... (ii) 100
Dividing equation (ii) by (i)10000 = 1 + R 4 7000 100 ⇒ 10 = 1 + R 4 7 100
From equation (i),7000 = P × 10 7
⇒ P = ₹ 4900
We can find required answer with the help of given formula :
Here, b – a = 8 – 4 = 4 and B = Rs 10,000 , A = Rs,7000R% = 
B 1/n − 1 
× 100% A
Correct Option: B
Here , A1 = ₹ 7,000 , T1 = 4 years and A2 = ₹ 10,000 , T2 = 8 years
Using the given formula ,A = P 1 + R T 100 ⇒ 7000 = P 1 + R 4 ..... (i) 100
10000 = P1 + R 8 ..... (ii) 100
Dividing equation (ii) by (i)10000 = 1 + R 4 7000 100 ⇒ 10 = 1 + R 4 7 100
From equation (i),7000 = P × 10 7
⇒ P = ₹ 4900
We can find required answer with the help of given formula :
Here, b – a = 8 – 4 = 4 and B = Rs 10,000 , A = Rs,7000R% = B 1/n − 1 × 100% A R% = 10000 1/4 − 1 7000 R% = 10 1/4 − 1 7 ⇒ 1 + R = 10 1/4 100 7 ⇒ 1 + R 4 = 10 100 7
Using ,∵ Amount = P 1 + R 4 100 ⇒ 7000 = P × 10 7
∴ P = Rs. 4900
- The compound interest on a certain sum for two successive years are ₹ 225 and ₹ 238.50. The rate of interest per annum is :
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As per the given in question ,
Difference = 238.50 – 225 = ₹ 13.50 = S.I. on ₹ 225 for 1 year∴ Rate = S.I. × 100 Principal × Time Rate = 13.50 × 100 = 6% per annum 225 × 1
Second Method :
Here, b – a = 1 and B = Rs 238.50 , A = Rs,225R% = B − 1 × 100% A
Correct Option: D
As per the given in question ,
Difference = 238.50 – 225 = ₹ 13.50 = S.I. on ₹ 225 for 1 year∴ Rate = S.I. × 100 Principal × Time Rate = 13.50 × 100 = 6% per annum 225 × 1
Second Method :
Here, b – a = 1 and B = Rs 238.50 , A = Rs,225R% = B − 1 × 100% A R% = 238.50 − 1 × 100% 225 R% = 238.50 − 225 × 100% 225 R% = 13.5 × 100% = 6% 225
- A certain amount of money at r%, compounded annually after two and three years becomes ₹ 1440 and ₹ 1728 respectively is
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Given in question , A1 = ₹ 1440 , T1 = 2 years
and A2 = ₹ 1728 , T2 = 3 years
If the principal be ₹ P and rate = r% , thenA = P 1 + R T 100 ⇒ 1440 = P 1 + r 2 ..... (i) 100 and 1728 = P 1 + r 3 ..... (ii) 100
On dividing equation (ii) by (i),1728 = 1 + r 1440 100 ∴ r = 1728 − 1 100 1440 = 1728 − 1440 = 288 1440 1440 ⇒ r = 288 × 100 1440
∴ r = 20% per annum
We can find required answer with the help of given formula :
Here, b – a = 3 – 2 = 1 and B = Rs 1728 , A = Rs. 1440R% = B − 1 × 100% A
Correct Option: D
Given in question , A1 = ₹ 1440 , T1 = 2 years
and A2 = ₹ 1728 , T2 = 3 years
If the principal be ₹ P and rate = r% , thenA = P 1 + R T 100 ⇒ 1440 = P 1 + r 2 ..... (i) 100 and 1728 = P 1 + r 3 ..... (ii) 100
On dividing equation (ii) by (i),1728 = 1 + r 1440 100 ∴ r = 1728 − 1 100 1440 = 1728 − 1440 = 288 1440 1440 ⇒ r = 288 × 100 1440
∴ r = 20% per annum
We can find required answer with the help of given formula :
Here, b – a = 3 – 2 = 1 and B = Rs 1728 , A = Rs. 1440R% = B − 1 × 100% A R% = 1728 − 1 × 100% 1440 R% = 1728 − 1440 × 100% 1440 R% = 288 × 100% = 20% 1440
- An amount of money at compound interest grows up to ₹ 3,840 in 4 years and up to ₹ 3,936 in 5 years. Find the rate of interest.
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Here , A1 = ₹ 3,840 , T1 = 4 years and A2 = ₹ 3,936 , T2 = 5 years
Using the given formula , we haveA = P 1 + R T 100 ∴ 3840 = P 1 + R 4 ..... (i) 100
3936 = P1 + R 5 ..... (ii) 100
Dividing equation (ii) by equation (i),3936 = 1 + R 3840 100 ⇒ R = 3936 − 1 100 3840 = 3936 − 3840 = 96 3840 3840 ⇒ R = 96 × 100% = 2.5% 3840
Second Method tom solve this question :
Here, b – a = 5 – 4 = 1 and B = Rs. 3,936, A = Rs. 3,840R% = B − 1 × 100% A
Correct Option: A
Here , A1 = ₹ 3,840 , T1 = 4 years and A2 = ₹ 3,936 , T2 = 5 years
Using the given formula , we haveA = P 1 + R T 100 ∴ 3840 = P 1 + R 4 ..... (i) 100
3936 = P1 + R 5 ..... (ii) 100
Dividing equation (ii) by equation (i),3936 = 1 + R 3840 100 ⇒ R = 3936 − 1 100 3840 = 3936 − 3840 = 96 3840 3840 ⇒ R = 96 × 100% = 2.5% 3840
Second Method tom solve this question :
Here, b – a = 5 – 4 = 1 and B = Rs. 3,936, A = Rs. 3,840R% = B − 1 × 100% A R% = 3936 − 1 × 100% 3840 R% = 3936 − 3840 × 100% 3840 R% = 96 × 100% 3840 R% = 10 % = 2.5% 4
- A certain sum of money amounts to ₹ 2,420 in 2 years and ₹ 2,662 in 3 years at some rate of compound interest, compounded annually. The rate of interest per annum is
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Let the rate of interest = R% per annum.
We know that ,A = P 1 + R T 100 ⇒ 2420 = P 1 + R 2 ..... (i) 100 ⇒ 2662 = P 1 + R 3 ..... (ii) 100
Dividing equation (ii) by (i), we get1 + R = 2662 100 2420 ⇒ R = 2662 − 1 100 2420 ⇒ R = 2662 − 2420 = 242 = 1 100 2420 2420 10 ⇒ R = 1 × 100% = 10% 10
We can find required answer with the help of given formula :
Here, a = 2 years , b = 3 years
b – a = 3 – 2 = 1 and B = Rs. 2,662 , A = Rs,2,420R% = B − 1 × 100% A
Correct Option: D
Let the rate of interest = R% per annum.
We know that ,A = P 1 + R T 100 ⇒ 2420 = P 1 + R 2 ..... (i) 100 ⇒ 2662 = P 1 + R 3 ..... (ii) 100
Dividing equation (ii) by (i), we get1 + R = 2662 100 2420 ⇒ R = 2662 − 1 100 2420 ⇒ R = 2662 − 2420 = 242 = 1 100 2420 2420 10 ⇒ R = 1 × 100% = 10% 10
We can find required answer with the help of given formula :
Here, a = 2 years , b = 3 years
b – a = 3 – 2 = 1 and B = Rs. 2,662 , A = Rs,2,420R% = B − 1 × 100% A R% = 2662 − 1 × 100% 2420 R% = 2662 − 2420 × 100% 2420 R% = 242 × 100% = 10% 2420
