The Learning Point‎ > ‎Mathematics‎ > ‎

Compound Interest - Basic Examples and Problems





Question 1: 

Suppose Ankit has a principal P that he invests in an account that pays 13% interest compounded bi-annually. Ankit has 1877.14 Rs at the end of 5 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 1000 Rs is the amount being invested or P. The interest rate r is 13% which must be changed into a decimal and becomes r = 0.13. The interest is compounded bi-annually, which tells us that n = 2. The money will stay in the account for 5 years so t = 5. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 1877.14/ ( 1 + 13/2)(2) x (5)  = 1000 Rs

So after 5 years, the account is worth 1877.14 Rs.


Question 2: 

Suppose Martha has 500 £ that she invests in an account that pays 1% interest compounded quarterly. In how many years will this become 520.38 £?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 500 £ is the amount being invested or P. The interest rate r is 1% which must be changed into a decimal and becomes r = 0.01. The interest is compounded quarterly, which tells us that n = 4. The final amount is 520.38 £. 
We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'.

Taking logarithms: log A = log P + nt log (1 + r/n) => t =  (log (520.38) - log (500)) / (4* log(1 + 0.01/4))

=> t = (LOG520.38-LOG500)/ (4* (1 + LOG0.01/4)) 

=> t = 4 years

So the time required is 4 years. 


Question 3: 

Suppose John has 1400 $ that he invests in an account that pays 5% interest compounded monthly. In how many years will this become 1709.25 $?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 1400 $ is the amount being invested or P. The interest rate r is 5% which must be changed into a decimal and becomes r = 0.05. The interest is compounded monthly, which tells us that n = 12. The final amount is 1709.25 $. 
We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'.

Taking logarithms: log A = log P + nt log (1 + r/n) => t =  (log (1709.25) - log (1400)) / (12* log(1 + 0.05/12))

=> t = (LOG1709.25-LOG1400)/ (12* (1 + LOG0.05/12)) 

=> t = 4 years

So the time required is 4 years. 


Question 4: 

Suppose Nina has a principal P that she invests in an account that pays 11% interest compounded annually. Nina has 2997.0 Rs at the end of 1 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2700 Rs is the amount being invested or P. The interest rate r is 11% which must be changed into a decimal and becomes r = 0.11. The interest is compounded annually, which tells us that n = 1. The money will stay in the account for 1 years so t = 1. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 2997.0/ ( 1 + 11/1)(1) x (1)  = 2700 Rs

So after 5 years, the account is worth 2997.0 Rs.


Question 5: 

Suppose Alex has 2300 £ that he invests in an account that pays 11% interest compounded bi-annually. How much money does Alex have at the end of 3 years?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2300 £ is the amount being invested or P. The interest rate r is 11% which must be changed into a decimal and becomes r = 0.11. The interest is compounded bi-annually, which tells us that n = 2. The money will stay in the account for 3 years so t = 3. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 2300 x ( 1 + 11/2)(2) x (3)  = 3171.34 £.

So after 5 years, the account is worth 3171.34 £.

Question 6: 

Suppose Mary has 2500 $ that she invests in an account that pays 10% interest compounded quarterly. In how many years will this become 3711.26 $?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2500 $ is the amount being invested or P. The interest rate r is 10% which must be changed into a decimal and becomes r = 0.1. The interest is compounded quarterly, which tells us that n = 4. The final amount is 3711.26 $. 
We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'.

Taking logarithms: log A = log P + nt log (1 + r/n) => t =  (log (3711.26) - log (2500)) / (4* log(1 + 0.1/4))

=> t = (LOG3711.26-LOG2500)/ (4* (1 + LOG0.1/4)) 

=> t = 4 years

So the time required is 4 years. 


Question 7: 

Suppose Ankit has a principal P that he invests in an account that pays 2% interest compounded monthly. Ankit has 2289.71 Rs at the end of 2 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2200 Rs is the amount being invested or P. The interest rate r is 2% which must be changed into a decimal and becomes r = 0.02. The interest is compounded monthly, which tells us that n = 12. The money will stay in the account for 2 years so t = 2. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 2289.71/ ( 1 + 2/12)(12) x (2)  = 2200 Rs

So after 5 years, the account is worth 2289.71 Rs.


Question 8: 

Suppose Martha has a principal P that she invests in an account that pays 2% interest compounded annually. Martha has 2318.57 £ at the end of 5 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2100 £ is the amount being invested or P. The interest rate r is 2% which must be changed into a decimal and becomes r = 0.02. The interest is compounded annually, which tells us that n = 1. The money will stay in the account for 5 years so t = 5. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 2318.57/ ( 1 + 2/1)(1) x (5)  = 2100 £

So after 5 years, the account is worth 2318.57 £.



Question 8: 


Suppose Martha has 2100 £ that she invests in an account for 5 years at a certain rate compounded annually. This becomes 2318.57 £ at the end of the period. What was the Rate%?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



Let's plug in the values to compute the rate 'r'

r = (A/P)(1/nt)-1 = (2318.57/2100)(1/(annually x 52) - 1 = 2 %.


Question 10: 

Suppose Nina has 1200 Rs that she invests in an account that pays 2% interest compounded quarterly. How much money does Nina have at the end of 4 years?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 1200 Rs is the amount being invested or P. The interest rate r is 2% which must be changed into a decimal and becomes r = 0.02. The interest is compounded quarterly, which tells us that n = 4. The money will stay in the account for 4 years so t = 4. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 1200 x ( 1 + 2/4)(4) x (4)  = 1299.69 Rs.

So after 5 years, the account is worth 1299.69 Rs.

Question 11: 

Suppose Alex has a principal P that he invests in an account that pays 13% interest compounded monthly. Alex has 381.77 £ at the end of 5 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 200 £ is the amount being invested or P. The interest rate r is 13% which must be changed into a decimal and becomes r = 0.13. The interest is compounded monthly, which tells us that n = 12. The money will stay in the account for 5 years so t = 5. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 381.77/ ( 1 + 13/12)(12) x (5)  = 200 £

So after 5 years, the account is worth 381.77 £.


Question 12: 

Suppose Mary has a principal P that she invests in an account that pays 4% interest compounded annually. Mary has 584.93 $ at the end of 4 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 500 $ is the amount being invested or P. The interest rate r is 4% which must be changed into a decimal and becomes r = 0.04. The interest is compounded annually, which tells us that n = 1. The money will stay in the account for 4 years so t = 4. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 584.93/ ( 1 + 4/1)(1) x (4)  = 500 $

So after 5 years, the account is worth 584.93 $.



Question 12: 


Suppose Mary has 500 $ that she invests in an account for 4 years at a certain rate compounded annually. This becomes 584.93 $ at the end of the period. What was the Rate%?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



Let's plug in the values to compute the rate 'r'

r = (A/P)(1/nt)-1 = (584.93/500)(1/(annually x 44) - 1 = 4 %.


Question 14: 

Suppose Martha has a principal P that she invests in an account that pays 4% interest compounded quarterly. Martha has 728.42 £ at the end of 1 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 700 £ is the amount being invested or P. The interest rate r is 4% which must be changed into a decimal and becomes r = 0.04. The interest is compounded quarterly, which tells us that n = 4. The money will stay in the account for 1 years so t = 1. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 728.42/ ( 1 + 4/4)(4) x (1)  = 700 £

So after 5 years, the account is worth 728.42 £.


Question 15: 

Suppose John has a principal P that he invests in an account that pays 3% interest compounded monthly. John has 1690.99 $ at the end of 4 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 1500 $ is the amount being invested or P. The interest rate r is 3% which must be changed into a decimal and becomes r = 0.03. The interest is compounded monthly, which tells us that n = 12. The money will stay in the account for 4 years so t = 4. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 1690.99/ ( 1 + 3/12)(12) x (4)  = 1500 $

So after 5 years, the account is worth 1690.99 $.


Question 16: 

Suppose Nina has 1700 Rs that she invests in an account that pays 9% interest compounded annually. How much money does Nina have at the end of 2 years?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 1700 Rs is the amount being invested or P. The interest rate r is 9% which must be changed into a decimal and becomes r = 0.09. The interest is compounded annually, which tells us that n = 1. The money will stay in the account for 2 years so t = 2. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 1700 x ( 1 + 9/1)(1) x (2)  = 2019.77 Rs.

So after 5 years, the account is worth 2019.77 Rs.


Question 16: 


Suppose Nina has 1700 Rs that she invests in an account for 2 years at a certain rate compounded annually. This becomes 2019.77 Rs at the end of the period. What was the Rate%?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



Let's plug in the values to compute the rate 'r'

r = (A/P)(1/nt)-1 = (2019.77/1700)(1/(annually x 29) - 1 = 9 %.


Question 18: 

Suppose Mary has 900 $ that she invests in an account that pays 8% interest compounded quarterly. In how many years will this become 1235.51 $?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 900 $ is the amount being invested or P. The interest rate r is 8% which must be changed into a decimal and becomes r = 0.08. The interest is compounded quarterly, which tells us that n = 4. The final amount is 1235.51 $. 
We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'.

Taking logarithms: log A = log P + nt log (1 + r/n) => t =  (log (1235.51) - log (900)) / (4* log(1 + 0.08/4))

=> t = (LOG1235.51-LOG900)/ (4* (1 + LOG0.08/4)) 

=> t = 4 years

So the time required is 4 years. 



Question 18: 


Suppose Mary has 900 $ that she invests in an account for 4 years at a certain rate compounded quarterly. This becomes 1235.51 $ at the end of the period. What was the Rate%?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



Let's plug in the values to compute the rate 'r'

r = (A/P)(1/nt)-1 = (1235.51/900)(1/(quarterly x 48) - 1 = 8 %.


Question 20: 

Suppose Martha has a principal P that she invests in an account that pays 5% interest compounded annually. Martha has 3087.0 £ at the end of 2 years. What was the starting principal P?

Let’s look at the compound interest formula and see how many values for the variables we are given in the problem.



The 2800 £ is the amount being invested or P. The interest rate r is 5% which must be changed into a decimal and becomes r = 0.05. The interest is compounded annually, which tells us that n = 1. The money will stay in the account for 2 years so t = 2. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. 

The required amount = 3087.0/ ( 1 + 5/1)(1) x (2)  = 2800 £

So after 5 years, the account is worth 3087.0 £.