Compound Interest
- The sum of money which becomes ₹ 2420 at 10 % rate of compound interest after two years is
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Given Here , Amount ( A ) = ₹ 2420 , P = ? , Rate ( R ) = 10% , Time = 2 years
Using the given formula ,A = P 
1 + R 
T 100 ⇒ 2420 = P 1 + 10 2 100 ⇒ 2420 = P 
1 + 1 2 = P 11 2 10 10
Correct Option: A
Given Here , Amount ( A ) = ₹ 2420 , P = ? , Rate ( R ) = 10% , Time = 2 years
Using the given formula ,A = P 1 + R T 100 ⇒ 2420 = P 1 + 10 2 100 ⇒ 2420 = P 1 + 1 2 = P 11 2 10 10 ⇒ P = 2420 × 10 × 10 = Rs. 2000 11 × 11
- When principal = ₹ S, rate of interest = 2r % p.a, then a person will get after 3 years at compound interest
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Here , Amount ( A ) = ? , Principal (P) = ₹ S , Rate ( R ) = 2r% per annum , Time = 3 years
As we know that ,∴ Amount = P 1 + R T 100
Correct Option: C
Here , Amount ( A ) = ? , Principal (P) = ₹ S , Rate ( R ) = 2r% per annum , Time = 3 years
As we know that ,∴ Amount = P 1 + R T 100 Amount = S 1 + 2r 3 = S 1 + r 3 100 50
- Rekha invested a sum of ₹ 12000 at 5% per annum compound interest. She received an amount of ₹ 13230 after n years. Find n.
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Given that , Amount ( A ) = ₹ 13230 , P = ₹ 12000 , Rate ( R ) = 5% , Time = n years
Using the given formula ,A = P 1 + R T 100 ⇒ 13230 = 12000 1 + 5 n 100 ⇒ 13230 = 1 + 1 n 12000 20 ⇒ 441 = 21 n 400 20
Correct Option: D
Given that , Amount ( A ) = ₹ 13230 , P = ₹ 12000 , Rate ( R ) = 5% , Time = n years
Using the given formula ,A = P 1 + R T 100 ⇒ 13230 = 12000 1 + 5 n 100 ⇒ 13230 = 1 + 1 n 12000 20 ⇒ 441 = 21 n 400 20 ⇒ 21 n = 21 2 20 20
⇒ n = 2 years
- A sum of Rs. 2420 is accumulated in 2 years at 10% compund interest on a certain amount. Then the original amount is :
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Let the principal be Rs. P.
Given in question , Amount ( A ) = Rs. 2420 , Rate ( R ) = 10% , Time = 2 years
Using the given formula ,∴ A = P 1 + R T 100 ⇒ 2420 = P 1 + 10 2 100 ⇒ 2420 = P × 1 + 10 2 100 ⇒ 2420 = P 11 2 10
Correct Option: B
Let the principal be Rs. P.
Given in question , Amount ( A ) = Rs. 2420 , Rate ( R ) = 10% , Time = 2 years
Using the given formula ,∴ A = P 1 + R T 100 ⇒ 2420 = P 1 + 10 2 100 ⇒ 2420 = P × 1 + 10 2 100 ⇒ 2420 = P 11 2 10 ⇒ P = 2420 × 10 × 10 = Rs. 2000 11 × 11
- Two years ago, the value of my motorbike was ₹ 62500. If the value depreciates by 4% every year, now its value is
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Here , P = ₹ 62500 , Rate ( R ) = 4% , Time = 2 years
Using the given formula ,Present worth of bike = P 1 − R T 100 Present worth of bike = 62500 1 − 4 2 100 Present worth of bike = 62500 1 − 1 2 25
Correct Option: B
Here , P = ₹ 62500 , Rate ( R ) = 4% , Time = 2 years
Using the given formula ,Present worth of bike = P 1 − R T 100 Present worth of bike = 62500 1 − 4 2 100 Present worth of bike = 62500 1 − 1 2 25 Present worth of bike = 62500 25 − 1 2 25 ⇒ Present worth of bike = 62500 × 24 × 24 = ₹ 57600 25 × 25
