Compound Interest
- A sum borrowed under compound interest doubles itself in 10 years. When will it become fourfold of itself at the same rate of interest ?
-
View Hint View Answer Discuss in Forum
Let the sum be P which becomes 2P in 10 years. Hence, 4P in 20 years .
We can find required answer with the help of given formula :
Unitary Method can also be used.
Here, m = 2, t = 10Correct Option: B
Let the sum be P which becomes 2P in 10 years. Hence, 4P in 20 years .
We can find required answer with the help of given formula :
Unitary Method can also be used.
Here, m = 2, t = 10
Time taken to become 4 times = 22 times
∴ Time taken to become 4 times = t × n = 10 × 2 = 20 years
- A sum of money doubles itself in 4 years at compound interest. It will amount to 8 times itself at the same rate of interest in :
-
View Hint View Answer Discuss in Forum
As per the given in question ,
A sum of ₹ P becomes ₹ 2P in 4 years.
Similarly, ₹ 2P will become 2 × 2P = ₹ 4P in next 4 years and ₹ 4P will become 2 × 4P = ₹ 8P in yet another 4 years. So, the total time = 4 + 4 + 4 = 12 years
Second Method to solve this question :
Here, m = 2, t = 4
Time taken to become 23 = n × t yearsCorrect Option: B
As per the given in question ,
A sum of ₹ P becomes ₹ 2P in 4 years.
Similarly, ₹ 2P will become 2 × 2P = ₹ 4P in next 4 years and ₹ 4P will become 2 × 4P = ₹ 8P in yet another 4 years. So, the total time = 4 + 4 + 4 = 12 years
Second Method to solve this question :
Here, m = 2, t = 4
Time taken to become 23 = n × t years
Required Time = 3 × 4 = 12 years
Note : - If a sum of money becomes n times in t years, it will become t1 = ny times at the same rate of interest in t1 years given by,
[t1 = yt]
- If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is :
-
View Hint View Answer Discuss in Forum
Suppose P = ₹ 100 and amount A = ₹ 225
Time = 2 years , Rate = r%A = P 
1 + r 
t 100 ⇒ 225 = 100 1 + r 2 100 ⇒ 225 = 
1 + r 
2 100 100 ⇒ 1 + r = 15 100 10 ⇒ 100 + r = 15 100 10
⇒ 100 + r = 150 ⇒ r = 150 - 100
⇒ r = 50%
We can find required answer with the help of given formula :
Here, n = 2.25 , t = 2 years
R% = (n1/t − 1) × 100%
Correct Option: D
Suppose P = ₹ 100 and amount A = ₹ 225
Time = 2 years , Rate = r%A = P 1 + r t 100 ⇒ 225 = 100 1 + r 2 100 ⇒ 225 = 1 + r 2 100 100 ⇒ 1 + r = 15 100 10 ⇒ 100 + r = 15 100 10
⇒ 100 + r = 150 ⇒ r = 150 - 100
⇒ r = 50%
We can find required answer with the help of given formula :
Here, n = 2.25 , t = 2 years
R% = (n1/t − 1) × 100%
R% = [(2.25)1/2 − 1] × 100%
R% = [1.5 − 1] × 100%
R% = 0.5 × 100% = 50%
- If the difference between the compound interest and the simple interest on a certain sum at the rate of 5% per annum for 2 years is Rs. 20, then the sum is :
-
View Hint View Answer Discuss in Forum
Here , Principal = ? , C.I. - S.I. = Rs. 20 , Time = 2 years , Rate = 5%
For 2 years, C.I. – S.I. = PR2 10000 ⇒ 20 = P × 5 × 5 10000
Correct Option: D
Here , Principal = ? , C.I. - S.I. = Rs. 20 , Time = 2 years , Rate = 5%
For 2 years, C.I. – S.I. = PR2 10000 ⇒ 20 = P × 5 × 5 10000 ⇒ P = 20 400
⇒ P = Rs. (20 × 400) = Rs. 8000
- A sum of money invested at compound interest amounts in 3 years to ₹ 2,400 and in 4 years to ₹ 2,520. The interest rate per annum is :
-
View Hint View Answer Discuss in Forum
On the basis of given details in question ,
S.I. on ₹ 2400 for 1 year = ₹ (2,520 – 2,400) = ₹ 120∴ Rate = S.I. × 100 % Principal × Time Rate = 120 × 100 = 5% 2400 × 1
Second Method to solve this question :
Here, b – a = 4 – 3 = 1 and B = Rs 2520, A = ₹ 2400R% = B − 1 × 100% A
Correct Option: A
S.I. on ₹ 2400 for 1 year = ₹ (2,520 – 2,400) = ₹ 120
∴ Rate = S.I. × 100 % Principal × Time Rate = 120 × 100 = 5% 2400 × 1
Second Method to solve this question :
Here, b – a = 4 – 3 = 1 and B = Rs 2520, A = ₹ 2400R% = B − 1 × 100% A R% = 2520 − 1 × 100% 2400 R% = 2520 − 2400 × 100% 2400 R% = 120 × 100% = 5% 2400
