Simple Interest Examples

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Home > Word Problems > Simple Interest Word Problems example

Simple Interest examples ( Enter your problem )

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1. Examples

Problem : 1 / 13 [ Simple Interest ] Enter your problem 1. Find the simple interest on Rs 730 for 184 days at 25/4 % per annum.


Solution:	Here `P = `Rs `730, R = 25/4` % and Time `= 184` days = `184/365` years `SI = (PRN)/100 = ( 730 * 25/4 * 184 /365) / 100 = 23` Simple Interest is Rs `23` .

Problem : 2 / 13 [ Simple Interest ] Enter your problem 2. Find the simple interest on Rs 4660 for 5 years at 7/2 % per annum.


Solution:	Here `P = ` Rs `4660, N = 5` years and `R = 7/2` % `SI = (PRN)/100 = ( 4660 * 7/2 * 5 ) / 100 = 815.5` Simple Interest is Rs `815.5` .

Problem : 3 / 13 [ Simple Interest ] Enter your problem 3. The interest on a certain amount of money at 8 % per year for a period of 4 years is Rs 512 . Find the sum of money.


Solution:	Here `SI = `Rs `512, N = 4` years, `R = 8` % `SI = (PRN)/100` `P = (SI100) / (RN)` `P = (512 * 100 ) / ( 8 * 4 ) = 1600` The sum of money is Rs `1600` .

Problem : 4 / 13 [ Simple Interest ] Enter your problem 4. Simple Interest on Rs 972 at 14 % per annum for a certain time is Rs 476.28 . Find the time ?


Solution:	Here `P = 972, SI = `Rs `476.28, R = 14`% `SI = (PRN)/100 ` `N = (SI100) / (PR)` `N = ( 476.28 * 100 ) / ( 972 * 14 ) = 3.5` The time is `3.5` years

Problem : 5 / 13 [ Simple Interest ] Enter your problem 5. Simple Interest on Rs 972 at a certain rate % per annum for 3.5 years is Rs 476.28 . Find the rate ?


Solution:	Here `P = 972, SI = ` Rs `476.28, N = 3.5` `SI = (PRN)/100` `R = (SI100) / (PN) = ( 476.28 * 100 ) / ( 972 * 3.5 ) = 14` The rate is `14` %.

Problem : 6 / 13 [ Simple Interest ] Enter your problem 6. A sum of money lent at simple interest amounts to Rs 1008 in 2 years and to 1112 in 3 years.Find the sum and the rate of interest.


Solution:	Principal + Interest for `3` years = `1112` Principal + Interest for `2` years = `1008` `=>` Interest for `1` year = `1112 - 1008 = 104` `=>` Interest for `2` years = `2 * 104 = 208` Now, Principal + Interest for `2` years = `1008` `=>` Principal + `208 = 1008` `=>` Principal = `1008 - 208 = 800` `SI = (PRN)/100` `R = (SI100) / (PN) = ( 208 * 100 ) / ( 800 * 2 ) = 13` The Sum of money is Rs `800` and the rate of interest is `13` %

Problem : 7 / 13 [ Simple Interest ] Enter your problem 7. What sum of money lent out at simple interest at 9 % p.a. for 3/2 years will produce the same interest as Rs 2250 lent at 6 % p.a. for 5 years.


Solution:	Here `P = ` Rs `2250, N = 5` years, `R = 6`% `SI = (PRN)/100` `SI = (2250 * 6 * 5 ) / 100 = 675` Now, `SI = ` Rs `675, N = 3/2` years, `R = 9`% `SI = (PRN)/100` `P = (SI * 100) / (R * N) = ( 675 * 100) / ( 9 * 3/2 ) = 5000` The sum of money is Rs `5000`

Problem : 8 / 13 [ Simple Interest ] Enter your problem 8. A man puts out Rs 500 for 4 years on simple interest and Rs 600 for 3 years. The total interest he receives is Rs 190 . What is the rate percent per annum?


Solution:	Let the rate(`R`) % per annum = `X` `SI` on Rs `500 = (PRN)/100 = (500 * X * 4) / 100 = 20 X` `SI` on Rs `600 = (PRN)/100 = (600 * X * 3) / 100 = 18 X` Total `SI` = Rs `190` `20 X + 18 X = 190` `38 X = 190` `X = 190 / 38 = 5` Rate is `5` %

Problem : 9 / 13 [ Simple Interest ] Enter your problem 9. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3 % higher rate, it whould have fetched Rs 300 more. Find the sum.


Solution:	Let sum = Rs `X` and original rate = `R` % Then, `(X * ( R + 3 ) * 2)/100 - (X * R * 2)/100 = 300` `2 R X + 6 X - 2 R X = 300 * 100` `6 X = 300 * 100` `X = 5000` The sum is Rs `5000` .

Problem : 10 / 13 [ Simple Interest ] Enter your problem 10. Divide Rs 16875 into two parts such that the interest at 7/2 % p.a. for 2 years on one part is equal to the other at 4 % p.a. for 5 years.


Solution:	Let the Principal of one part = `X` Rs The Principal of second part = `( 16875 - X )` Rs For one part, `P = X` Rs, `R = 7/2 %, N = 2` years `SI = (PRN)/100` `SI = ( X * 7/2 * 2 ) / 100 = (7 * X)/100 ->(1)` Now for second part, `P = ( 16875 - X ) Rs, R = 4 %, N = 5` years `SI = (PRN)/100` `SI = (( 16875 - X ) * 4 * 5 ) / 100 = (20 * ( 16875 - X ))/100 ->(2)` Interest is the same as `(1) = (2)` `=> (7 * X)/100 = (20 * ( 16875 - X ))/100` `=> 7 X = 20 * ( 16875 - X )` `=> 7 X = 20 * 16875 - 20 X` `=> 27 X = 20 * 16875` `=> X = (20 * 16875) / 27` `=> X = 12500` `:.`One part of Principal = Rs `12500` and the second part = Rs `( 16875 - 12500 )` = Rs `4375` The two parts are Rs `12500` and Rs `4375` .

Problem : 11 / 13 [ Simple Interest ] Enter your problem 11. A shopkeeper borrowed Rs 20000 from two money lendeRs For one loan he paid 12 % and for the other 14 % per annum. After 1 year, he paid Rs 2560 as interest. How much did he borrow at each rate ?


Solution:	Let money at `12 % = X` and that at `14 % = ( 20000 - X )` After one year total interest = `2560` `(X * 12 * 1) / 100 + ((20000 - X) * 14 * 1) / 100 = 2560` `12 X + 14 * 20000 - 14 X = 2560 * 100` `-2 X = -24000` `X = 12000` `:.`Money at `12` % = Rs `12000` and Money at `14` % = Rs `8000` .

Problem : 12 / 13 [ Simple Interest ] Enter your problem 12. At what rate percent per annum will sum of money double in 8 years?


Solution:	Let, Sum = `X`, Time = `8` years and Amount = `2X`. `:. SI = X. ` `SI = (PRN)/100` `R = (SI * 100)/(P * N) = (X * 100)/( X * 8 ) = 25/2` % `:.` sum of money double at `25/2` % in `8` years.

Problem : 13 / 13 [ Simple Interest ] Enter your problem 13. Rajeev deposited money in the post office which is doubled in 20 years at a simple rate of interest. In how many years will the original sum triple itself ?


Solution:	Let, Sum = `X`, Time = `20` years and Amount = `2X`. `:. SI = X. ` `SI = (PRN)/100` `R = (SI * 100)/(P * N) = (X * 100)/( X * 20 ) = 5` % `:.` sum of money double at `5` % in `20` years.

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