Compound Interest
- The compound interest on a certain sum of money for 2 years at 5% is ₹ 328, then the sum is
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Let the principal be Rs. P
Given in question , Compound Interest ( CI ) = ₹ 328 , P = ? , Rate ( R ) = 5% , Time = 2 years∴ C.I. = P 
1 + R 
T − 1 
100 ⇒ 328 = P 1 + 5 2 − 1 100 ⇒ 328 = P 21 2 − 1 20 ⇒ 328 = P 441 − 1 400 ⇒ 328 = P 441 − 400 400
Correct Option: C
Let the principal be Rs. P
Given in question , Compound Interest ( CI ) = ₹ 328 , P = ? , Rate ( R ) = 5% , Time = 2 years∴ C.I. = P 1 + R T − 1 100 ⇒ 328 = P 1 + 5 2 − 1 100 ⇒ 328 = P 21 2 − 1 20 ⇒ 328 = P 441 − 1 400 ⇒ 328 = P 441 − 400 400 ⇒ 328 = 41P 400 ⇒ P = 328 × 400 = ₹ 3200 41
- A sum of ₹ 3,200 invested at 10% p.a. compounded quarterly amounts to 3,362. Compute the time period.
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Here , Amount ( A ) = ₹ 3,362 , P = ₹ 3,200 , Rate ( R ) = 10% , Time = quarterly = 4t
A = P 1 + R T 100
Interest is compounded quarterly ,3,362 = 3200 1 + 10 4t 400 ⇒ 3362 = 1 + 1 4t 3200 40 ⇒ 1681 = 41 4t 1600 40
Correct Option: A
Here , Amount ( A ) = ₹ 3,362 , P = ₹ 3,200 , Rate ( R ) = 10% , Time = quarterly = 4t
A = P 1 + R T 100
Interest is compounded quarterly ,3,362 = 3200 1 + 10 4t 400 ⇒ 3362 = 1 + 1 4t 3200 40 ⇒ 1681 = 41 4t 1600 40 ⇒ 41 2 = 41 4t 40 40 ⇒ 4t = 2 ⇒ t = 1 year 2
- The compound interest on ₹ 5,000 for 3 years at 10% p. a. will amount to
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Here , Compound Interest ( CI ) = ? , P = ₹ 5,000 , Rate ( R ) = 10% , Time = 3 years
C.I. = P 1 + R T − 1 100 C.I. = 5000 1 + 10 3 − 1 100 C.I. = 5000 11 3 − 1 10 C.I. = 5000 1331 − 1 1000
Correct Option: B
Here , Compound Interest ( CI ) = ? , P = ₹ 5,000 , Rate ( R ) = 10% , Time = 3 years
C.I. = P 1 + R T − 1 100 C.I. = 5000 1 + 10 3 − 1 100 C.I. = 5000 11 3 − 1 10 C.I. = 5000 1331 − 1 1000 C.I. = 5000 1331 - 1000 1000 C.I. = 5000 × 331 = ₹ 1655 1000
- ₹ 800 at 5% per annum compounded annually will amount to ₹ 882 in
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Given that , Amount ( A ) = ₹ 882 , P = ₹ 800 , Rate ( R ) = 5% , Time = T years
Using the given formula ,A = P 1 + R T 100 ⇒ 882 = 800 1 + 5 T 100 ⇒ 882 = 21 T 800 20
Correct Option: B
Given that , Amount ( A ) = ₹ 882 , P = ₹ 800 , Rate ( R ) = 5% , Time = T years
Using the given formula ,A = P 1 + R T 100 ⇒ 882 = 800 1 + 5 T 100 ⇒ 882 = 21 T 800 20 ⇒ 441 = 21 2 = 21 T 400 20 20
Equating powers on both sides ,
∴ T = 2 years
- A man borrows ₹ 21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years ?
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If each instalment be y, then
∴Present worth of first instalment = y = 10y [ 1 + ( 11 / 100 ) ] 11 Present worth of second instalment = y = 100y [ 1 + ( 11 / 100 ) ]² 121 ∴ 10 y + 100 y = 21000 11 121 ⇒ 110y + 100y = 21000 121
⇒ 210y = 21000 × 121⇒ y = 21000 × 121 = ₹ 12100 210
Second Method to solve this question :
Here, n = 2 , P = ₹ 21000 , r = 10%Each annual instalment = P 100 + 100 2 100 + r 100 + r Each annual instalment = 21000 100 + 100 2 110 110
Correct Option: B
If each instalment be y, then
∴Present worth of first instalment = y = 10y [ 1 + ( 11 / 100 ) ] 11 Present worth of second instalment = y = 100y [ 1 + ( 11 / 100 ) ]² 121 ∴ 10 y + 100 y = 21000 11 121 ⇒ 110y + 100y = 21000 121
⇒ 210y = 21000 × 121⇒ y = 21000 × 121 = ₹ 12100 210
Second Method to solve this question :
Here, n = 2 , P = ₹ 21000 , r = 10%Each annual instalment = P 100 + 100 2 100 + r 100 + r Each annual instalment = 21000 100 + 100 2 110 110 Each annual instalment = 21000 100 + 10000 110 12100 Each annual instalment = 21000 10 + 100 11 121 Each annual instalment = 21000 × 121 110 + 100 Each annual instalment = 21000 × 121 = 12100 210
