Simple Interest Word Problems
Simple Interest Word Problems
Example 1: What is the simple interest and amount of $ 2000 at the rate of 8% per year for 2 1/2 years.
Solution: Principal amount = P = $ 2000
Rate of interest per year = R = 8%
Time for which it is borrowed = T = 2 1/2 year
Thus, simple interest for one year, SI = (P x R x T)/100
= (2000 x 8 x 5)/100 x 2
simple interest for one year = $400
Amount that has to pay at the end of the 2 1/2 year = Principal + Interest
= 2000 + 400
Amount that has to pay at the end of the 2 1/2 year = $2400
Answer: simple interest for 2 1/2 year is $400 and Amount has to pay at the end of the 2 1/2 year is $2400.
Example 2: Find the time, when Simple Interest = $600, Principal = $5000 and Rate = 6%. p.a.
Solution: Simple Interest = $600,
Principal amount = P = $ 5000
Rate of interest per year = R = 6%
Time T = ?
Formula for time T,
T = [{100 x (Simple Interest)}/P x R}]
put the value in the formula
T = [{100 x (600)}/5000 x 6}]
T = (100 x 600)/5000 x 6
T = 10/5
T = 2
T = 2 years.
Answer: Time is 2 years.
Example 3: Find the time, when Simple Interest = $1908, Principal = $9540 and Rate = 8%. p.a.
Solution: Simple Interest = $1908,
Principal amount = P = $ 9540
Rate of interest per year = R = 8%
Time T = ?
Formula for time T,
T = [{100 x (Simple Interest)}/P x R}]
put the value in the formula
T = [{100 x (1908)}/9540 x 8}]
T = (100 x 1908)/9540 x 8
T = 5/2
T = 2 1/2 years.
T = 2 years.
Answer: Time is 2 1/2 years.
Example 4: David takes a lone of $ 20000 at 10% per year as rate of interest from a bank. Find the interest and the amount he has to pay at the end of one year.
Solution: Principal amount = P = $ 20000
Rate of interest per year = R = 10%
Time for which it is borrowed = T = 1 year
Thus, simple interest for one year, SI = (P x R x T)/100
= 20000 x 10 x 1/100
simple interest for one year = $2000
Amount that David has to pay to the bank at the end of the year = Principal + Interest
= 20000 + 2000
Amount that David has to pay to the bank at the end of the year = $22000
Answer: Simple interest for one year is $2000 and Amount that David has to pay to the bank at the end of the year is $22000.
Example 5: John takes a lone of $ 40000 at 10% per year as rate of interest from a bank. Find the interest and the amount he has to pay at the end of one year.
Solution: Principal amount = P = $ 40000
Rate of interest per year = R = 10%
Time for which it is borrowed = T = 1 year
Thus, simple interest for one year, SI = (P x R x T)/100
= 40000 x 10 x 1/100
= $4000
Amount that John has to pay to the bank at the end of the year = Principal + Interest
= 40000 + 4000
= $44000
Answer: simple interest for one year is $4000 and Amount that john has to pay to the bank at the end of the year is $44000.
Example 6: Merry borrowed $50000 for 3 years at the rate of 5% per year. Find the interest accumulated at the end of 3 years.
Solution: Principal amount = P = $ 50000
Rate of interest per year = R = 5%
Time for which it is borrowed = T = 3 year
Thus, simple interest for one year, SI = (P x R x T)/100
= 50000 x 5 x 3/100
= $7500
Amount that Merry has to pay to the bank at the end of the year = Principal + Interest
= 50000 + 7500
= $57500
Answer: Simple interest for one year is $7500 and Amount that Merry has to pay to the bank at the end of the year is $57500.
Example 7: Riya pays $8000 as an amount on the sum of $3000 that he had borrowed for 5 years. Find the rate of interest.
Solution: Given that
A = $8000, P = $3000,
SI = A – P
SI = $8000 – $3000
SI = $5000
T = 5 years
R = ?
SI = (P x R X T)/100
R = SI x 100 / P x T
R = 5000 x 100 / 3000 x 5
R = 100/3
R = 33.3 %
Answer: Rate of interest for five years is 33.3%.