Simple interest
- Find the annual instalment that will discharge a debt of ₹ 12900 due in 4 years at 5% per annum simple interest.
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Let each equal annual instalment be p.
First instalment is paid after 1 year and hence will remain with the lender for the remaining (4 – 1) = 3 years. Similarly, second instalment will remain with the lender for 2 years, third instalment for 1 year and the final fourth instalment remain p as such.
A = A1 + A2 + A3 + A4
We can find the required answer with the help of given formula ,A = P 
100 + RT 
100 ⇒ A = p 
100 + 5 × 3 + 100 + 5 × 2 + 100 + 5 × 1 + 100 + 5 × 0 
100 100 100 100 ⇒ 12900 = p 115 + 110 + 105 + 100 100
Correct Option: D
Let each equal annual instalment be p.
First instalment is paid after 1 year and hence will remain with the lender for the remaining (4 – 1) = 3 years. Similarly, second instalment will remain with the lender for 2 years, third instalment for 1 year and the final fourth instalment remain p as such.
A = A1 + A2 + A3 + A4
We can find the required answer with the help of given formula ,A = P 100 + RT 100 ⇒ A = p 100 + 5 × 3 + 100 + 5 × 2 + 100 + 5 × 1 + 100 + 5 × 0 100 100 100 100 ⇒ 12900 = p 115 + 110 + 105 + 100 100 ⇒ 12900 = 430 p 100 ⇒ p = 12900× 100 430
⇒ p = ₹ 3000
- A man borrowed ₹ 16000 from two persons. He paid 6% interest to one and 10% per annum to the other. In one year he paid total interest ₹1120. How much did he borrow at each rate?
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Let the sum borrowed at 6% be ₹ p = P1
Then the sum borrowed at 10% = ₹ (16000 – p ) = P2
Time is one year in both cases
R1 = 6%
R2 = 10%
SI = SI1 + SI2SI = P1R1T + P2R2T = 2200 100 100 SI = T (P1R1 + P2R2) 100 Or, P1R1 + P2R2 = 100 SI T
On substitution we get,
Correct Option: B
Let the sum borrowed at 6% be ₹ p = P1
Then the sum borrowed at 10% = ₹ (16000 – p ) = P2
Time is one year in both cases
R1 = 6%
R2 = 10%
SI = SI1 + SI2SI = P1R1T + P2R2T = 2200 100 100 SI = T (P1R1 + P2R2) 100 Or, P1R1 + P2R2 = 100 SI T
On substitution we get,(p × 6) + (16000 – p)10 = 100 × 1120 1
⇒ 160000 – 4p = 112000
⇒ 4p = 48000
⇒ p = 12000 and 16000 – p = ₹ 4000.
- A man lends a certain sum of money and gets an interest equal to 1/16 th of the principal. The time for which money was lent is equal to the rate of interest. Find the rate of interest per annum.
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As we know that ,
SI = PRT 100 Given : SI = P and T = R 16
So, on substitution we getP = P × R × R 16 100 R2 = 100 16
Correct Option: D
As we know that ,
SI = PRT 100 Given : SI = P and T = R 16
So, on substitution we getP = P × R × R 16 100 R2 = 100 16 ⇒ R = 10 % = 5 % = 2 1 % 4 2 2
- At what rate per annum will a sum of ₹ 5000 amount to ₹ 6000 in 4 years?
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Here, P = ₹ 5000 , A = ₹ 6000 , T = 4 years
So, SI = Amount – Principal = ₹ (6000 – 5000) = ₹ 1000
As we know that ,Rate = SI × 100 Principal × Time
Correct Option: C
Here, P = ₹ 5000 , A = ₹ 6000 , T = 4 years
So, SI = Amount – Principal = ₹ (6000 – 5000) = ₹ 1000
As we know that ,Rate = SI × 100 Principal × Time ⇒ R = 100 × 1000 ⇒ R = 5%. 5000 × 4
- A person made a fixed deposit of Rs. 30,000 in a bank for 5 years at 10% simple interest per annum. He had to withdraw the whole amount after 3 years to meet the expenses of his daughter’s marriage and he received Rs. 7800 less than what he would have got after 5 years. What is the rate of simple interest per annum paid by the bank for this premature encashment ?
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As we know that ,
SI = Principal × Time × Rate 100
Let the required rate of interest be R% per annum.
According to question ,∴ 30000 × 5 × 10 - 30000 × 3 × R = 7800 100 100
⇒ 15000 - 900R = 7800
⇒ 900R = 15000 - 7800 = 7200
Correct Option: C
As we know that ,
SI = Principal × Time × Rate 100
Let the required rate of interest be R% per annum.
According to question ,∴ 30000 × 5 × 10 - 30000 × 3 × R = 7800 100 100
⇒ 15000 - 900R = 7800
⇒ 900R = 15000 - 7800 = 7200⇒ R = 7200 = 8% per annum 900
