Simple interest
- A certain sum of money amounts to Rs. 1680 in 3 years and to 1800 in 5 years. Find the sum and the rate of interest.
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We know that , Amount = Principal + SI
So,P remains same in both cases. Only amount of interest is different in two cases because the time period are different.
P + Interest for 5 years = ₹ 1800 and P + Interest for 3 years = ₹ 1680
On subtraction we get,
Interest for 2 years = ₹ 120
Now, we solve for the case of 3 years.
Interest for 3 years₹ 120 × 3 = ₹ 180 2
And amount after 3 years = 1680
Principal (P) = Amount – SI = ₹ (1680 – 180) = ₹ 1500.Rate = SI × 100 Principal × Time
Correct Option: A
We know that , Amount = Principal + SI
So,P remains same in both cases. Only amount of interest is different in two cases because the time period are different.
P + Interest for 5 years = ₹ 1800 and P + Interest for 3 years = ₹ 1680
On subtraction we get,
Interest for 2 years = ₹ 120
Now, we solve for the case of 3 years.
Interest for 3 years₹ 120 × 3 = ₹ 180 2
And amount after 3 years = 1680
Principal (P) = Amount – SI = ₹ (1680 – 180) = ₹ 1500.Rate = SI × 100 Principal × Time R = 100 × 180 1500 × 3
⇒ R = 4%.
Note : Alternatively, we could have solved for 5 years too and got the same answer.
- In how many years will a sum of money double itself at 5% rate of interest?
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A sum doubles itself when amount of interest becomes equal to the principal.
So, I = P
Given, R = 5%
We can find the required answer with the help of given formula ,Time = 100I PR
On substitution we get,
Correct Option: B
A sum doubles itself when amount of interest becomes equal to the principal.
So, I = P
Given, R = 5%
We can find the required answer with the help of given formula ,Time = 100I PR
On substitution we get,T = 100 × P P × R
T = 20 years
- 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
- 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.
- 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
