Compound Interest
- The least number of years in which a sum of money on 19% p.a. compound interest will be more than double is
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Let Principal be ₹ P ,Then
According to question ,
Amount ( A ) = ₹ 2P , Rate = 19% , Time = T years
Using the given formula ,A = P 
1 + R 
T 100 ⇒ 2P = P 1 + 19 T 100
Correct Option: B
Let Principal be ₹ P ,Then
According to question ,
Amount ( A ) = ₹ 2P , Rate = 19% , Time = T years
Using the given formula ,A = P 1 + R T 100 ⇒ 2P = P 1 + 19 T 100 ⇒ 2 = 119 T 100
⇒ 2 = (1.19)T
If T = 4 years, (1.19)4 > 2
Hence , Required time T = 4 years
- If the compound interest on a certain sum for 2 years at 3% per annum is ₹ 101.50, then the simple interest on the same sum at the same rate and for the same time will be
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Let the sum be P.
Given Here , Compound Interest ( CI ) = ₹ 101.50 , Rate ( R ) = 3% , Time = 2 years
Using the given formula ,C.I. = P 
1 + r n − 1 
100 101.50 = P 1 + 3 2 − 1 100 ⇒ 101.50 = P 103 2 − 1 100 ⇒ 101.50 = P 10609 - 10000 10000 ⇒ P = ₹ 101.50 × 10000 609 P = ₹ 1015000 609 ∴ S.I. = P × R × T 100 ⇒ S.I. = 1015000 × 2 × 3 = ₹ 100 609 × 100
We can find required answer with the help of given formula :
Here, C.I. = Rs 101.50 , R = 3% , S.I. = ?C.I. = S.I. 1 + R 200
Correct Option: C
Let the sum be P.
Given Here , Compound Interest ( CI ) = ₹ 101.50 , Rate ( R ) = 3% , Time = 2 years
Using the given formula ,C.I. = P 1 + r n − 1 100 101.50 = P 1 + 3 2 − 1 100 ⇒ 101.50 = P 103 2 − 1 100 ⇒ 101.50 = P 10609 - 10000 10000 ⇒ P = ₹ 101.50 × 10000 609 P = ₹ 1015000 609 ∴ S.I. = P × R × T 100 ⇒ S.I. = 1015000 × 2 × 3 = ₹ 100 609 × 100
We can find required answer with the help of given formula :
Here, C.I. = Rs 101.50 , R = 3% , S.I. = ?C.I. = S.I. 1 + R 200 101.50 = S.I. 1 + 3 200 S.I. = 101.50 × 200 = ₹ 100 203
- If the compound interest on a sum of money for 3 years at the rate of 5% per annum is ₹ 252.20, the simple interest on the same sum at the same rate and for the same time is
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Suppose principal be P .
Here , Principal ( P ) = ? , Compound Interest ( CI ) = ₹ 252.20 , Rate ( R ) = 5% , Time = 3 years⇒ P 
1 + 5 3 − 1 
= 252.20 100 ⇒ P 21 3 − 1 = 252.20 20 ⇒ P 21 × 21 × 21 − 20 × 20 × 20 = 252.20 20 × 20 × 20 ⇒ P × 1261 = 252.20 8000 ∴ P = 252.20 × 8000 = 1600 1261
Correct Option: B
Suppose principal be P .
Here , Principal ( P ) = ? , Compound Interest ( CI ) = ₹ 252.20 , Rate ( R ) = 5% , Time = 3 years⇒ P 1 + 5 3 − 1 = 252.20 100 ⇒ P 21 3 − 1 = 252.20 20 ⇒ P 21 × 21 × 21 − 20 × 20 × 20 = 252.20 20 × 20 × 20 ⇒ P × 1261 = 252.20 8000 ∴ P = 252.20 × 8000 = 1600 1261 ∴ SI = P × R × T 100 ⇒ SI = 1600 × 5 × 3 = ₹ 240 100
- On a certain sum of money the compound interest for 2 years is ₹ 282.15 and the simple interest for the same period of time is ₹ 270. The rate of interest per annum is
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Given that , C.I. = ₹ 282.15 , S.I. = ₹ 270 , Rate = r%
For two years ,C.I. = S.I. 1 + r 200 ⇒ 282.15= 270 1 + r 100 ⇒ 1 + r = 282.15 200 270 ⇒ r = 282.15 − 1 200 270 ⇒ r = 282.15 - 270 200 270
Correct Option: C
Given that , C.I. = ₹ 282.15 , S.I. = ₹ 270 , Rate = r%
For two years ,C.I. = S.I. 1 + r 200 ⇒ 282.15= 270 1 + r 100 ⇒ 1 + r = 282.15 200 270 ⇒ r = 282.15 − 1 200 270 ⇒ r = 282.15 - 270 200 270 ⇒ r = 12.15 200 270 ⇒ r = 12.15 × 200 = 9% 270
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is ₹ 510, the simple interest on the same sum at the same rate for the same period of time is :If the compound interest on a sum for 2 years at 12 1 % per annum 2
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Here , Principal ( P ) = ? , Compound Interest ( CI ) = ₹ 510 , Rate ( R ) = ( 25 / 2 )% , Time = 2 years
C.I. = P 1 + R T − 1 100 ⇒ 510 = P 1 + 25 2 − 1 200 ⇒ 510 = P 1 + 1 2 − 1 8 ⇒ 510 = P 81 − 1 64 ⇒ 510 = P 81 - 64 64 ⇒ P = 510 × 64 = ₹ 1920 17 ∴ S.I. = P × R × T 100 ∴ S.I. = 1920 × 2 × 25 = ₹ 480 100 × 2
Second Method to solve this question :Here, C.I. = ₹ 510 , R = 12 1 %, S.I. = ? 2
With the help of given formula ,C.I. = S.I. 1 + R 200 510 = S.I. 1 + 25 400
Correct Option: B
Here , Principal ( P ) = ? , Compound Interest ( CI ) = ₹ 510 , Rate ( R ) = ( 25 / 2 )% , Time = 2 years
C.I. = P 1 + R T − 1 100 ⇒ 510 = P 1 + 25 2 − 1 200 ⇒ 510 = P 1 + 1 2 − 1 8 ⇒ 510 = P 81 − 1 64 ⇒ 510 = P 81 - 64 64 ⇒ P = 510 × 64 = ₹ 1920 17 ∴ S.I. = P × R × T 100 ∴ S.I. = 1920 × 2 × 25 = ₹ 480 100 × 2
Second Method to solve this question :Here, C.I. = ₹ 510 , R = 12 1 %, S.I. = ? 2
With the help of given formula ,C.I. = S.I. 1 + R 200 510 = S.I. 1 + 25 400 510 = S.I. 425 400 S.I. = 510 × 400 = ₹ 480 425
