Compound Interest
- On a certain sum of money, the simple interest for 2 years is Rs. 350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?
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Given that , S.I. = Rs. 350 , Time = 2 years , Rate = 4%
As we know that ,Principal =
S.I. × 100 Time × Rate Principal = 350 × 100 = Rs. 4375 2 × 4 C.I. = P 
1 + R 
T − 1 
100 C.I. = 4375 1 + 4 2 − 1 100 C.I. = 4375 1 + 1 2 − 1 25
Correct Option: B
Given that , S.I. = Rs. 350 , Time = 2 years , Rate = 4%
As we know that ,Principal =
S.I. × 100 Time × Rate Principal = 350 × 100 = Rs. 4375 2 × 4 C.I. = P 1 + R T − 1 100 C.I. = 4375 1 + 4 2 − 1 100 C.I. = 4375 1 + 1 2 − 1 25 C.I. = 4375 26 2 − 1 25 C.I. = 4375 676 − 1 625 C.I. = 4375 × 51 = Rs. 357 625
Required difference = C.I. - S.I.
∴ Required difference = Rs. (357 – 350) = Rs. 7
- A sum of money at compound interest will amount to ₹ 650 at the end of the first year and ₹ 676 at the end of the second year. The amount of money is
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Given Here , A1 = ₹ 650 , T1 = 1 year and A2 = ₹ 676 , T2 = 2 years
Let Principal = ₹ P , Rate = R% per annum
We can find required answer with the help of given formula ,∴ A = P 
1 + R T 100 ⇒ 650 = P 1 + R 100 ⇒ 650 = 1 + R ...(i) P 100 Again, 676 = P 1 + R 2 100
Correct Option: D
Given Here , A1 = ₹ 650 , T1 = 1 year and A2 = ₹ 676 , T2 = 2 years
Let Principal = ₹ P , Rate = R% per annum
We can find required answer with the help of given formula ,∴ A = P 1 + R T 100 ⇒ 650 = P 1 + R 100 ⇒ 650 = 1 + R ...(i) P 100 Again, 676 = P 1 + R 2 100 ⇒ 676 = P 650 2 { Using (i) } P ⇒ 676 = P × 650 × 650 P2 ⇒ P = 650 × 650 = ₹ 625 676
- A sum becomes ₹ 4500 after two years and ₹ 6750 after four years at compound interest. The sum is
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Given in question , A1 = ₹ 4500 , T1 = 2 years and A2 = ₹ 6750 , T2 = 4 years
Suppose principal = PP 1 + r 2 = 4500 ..... (i) 100 P 1 + r 4 = 6750 ..... (ii) 100
On dividing equation (ii) by equation (i), we get1 + r 2 = 6750 100 4500
From equation (i), we getP × 6750 = 4500 4500 ⇒ P = 4500 × 4500 = ₹ 3,000 6750
Second Method to solve this question :
Here, b – a = 4 – 2 = 2 and B = ₹ 6750, A = ₹ 4500R% = B 1/2 − 1 × 100% A
Correct Option: C
Given in question , A1 = ₹ 4500 , T1 = 2 years and A2 = ₹ 6750 , T2 = 4 years
Suppose principal = PP 1 + r 2 = 4500 ..... (i) 100 P 1 + r 4 = 6750 ..... (ii) 100
On dividing equation (ii) by equation (i), we get1 + r 2 = 6750 100 4500
From equation (i), we getP × 6750 = 4500 4500 ⇒ P = 4500 × 4500 = ₹ 3,000 6750
Second Method to solve this question :
Here, b – a = 4 – 2 = 2 and B = ₹ 6750, A = ₹ 4500R% = B 1/2 − 1 × 100% A R% = 6750 1/2 − 1 × 100% 4500 R% = 3 1/2 − 1 × 100% 2 ⇒ 3 1/2 = 1 + R 2 100 ⇒ 3 = 1 + R 2 2 100
Using formula ,A = P 1 + R 2 100 4500 = P × 3 2
P = ₹ 3000
- Kamal took 6800 as a loan which along with interest is to be repaid in two equal
value of each instalment isannual instalments. If the rate of interest is 12 1 %, compounded annually, then the 2
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Let the annual instalment be y
Using the given formula ,A = P 1 + R T T y = P1 1 + 25 200 ⇒ y = P1 × 9 8 ⇒ P1 = 8 y 9 Similarly,P2 = 64 y 81
P1 + P2 = 6800⇒ 8 y + 64 y = 6800 9 81 ⇒ 72y + 64y = 6800 81 ⇒ 136y = 6800 81 ⇒ y = 6800 × 81 = ₹ 4050 136
We can find required answer with the help of given formula :Here, P = ₹ 6800, R = 25 % , n = 2 2 Each instalment = P 100 
+ 100 2 100 + r 100 + r Each instalment = 6800 100 + 100 2 100 + (25/2) 100 + (25/2)
Correct Option: C
Let the annual instalment be y
Using the given formula ,A = P 1 + R T T y = P1 1 + 25 200 ⇒ y = P1 × 9 8 ⇒ P1 = 8 y 9 Similarly,P2 = 64 y 81
P1 + P2 = 6800⇒ 8 y + 64 y = 6800 9 81 ⇒ 72y + 64y = 6800 81 ⇒ 136y = 6800 81 ⇒ y = 6800 × 81 = ₹ 4050 136
We can find required answer with the help of given formula :Here, P = ₹ 6800, R = 25 % , n = 2 2 Each instalment = P 100 + 100 2 100 + r 100 + r Each instalment = 6800 100 + 100 2 100 + (25/2) 100 + (25/2) Each instalment = 6800 200 + 200 2 225 225 Each instalment = 6800 200 1 + 200 225 225 Each instalment = 6800 × 225 × 225 = ₹ 4050 200 × 425
- A sum of money invested at compound interest amounts to ₹ 650 at the end of first year and ₹ 676 at the end of second year. The sum of money is :
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As per the given in question ,
Interest on ₹ 650 for 1 year = 676 – 650 = ₹ 26So, r = 26 × 100 650
⇒ r = 4% per annumP = A 1 + r t 100 P = 650 1 + 4 1 100 = 650 = 650 × 25 = ₹ 625 ( 26/25 ) 26
Using the given formula :
Here, b – a = 1 , B = Rs 676, A = ₹ 650R% = B − 1 × 100% A
Correct Option: C
As per the given in question ,
Interest on ₹ 650 for 1 year = 676 – 650 = ₹ 26So, r = 26 × 100 650
⇒ r = 4% per annumP = A 1 + r t 100 P = 650 1 + 4 1 100 = 650 = 650 × 25 = ₹ 625 ( 26/25 ) 26
Using the given formula :
Here, b – a = 1 , B = Rs 676, A = ₹ 650R% = B − 1 × 100% A R% = 676 − 1 × 100% 650 R% = 676 −650 × 100% 650 R% = 26 × 100% 650 R% = 100 = 4% 25 Amount = P 1 + R 1 100 650 = P 1 + 4 100 ⇒ P = 650 × 100 = ₹ 625 104
Note : A sum at a rate of interest compounded yearly becomes ₹ A, in n years and ₹ A2 in (n + 1) years,then P = A1 A1 n A2
