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### Compound Interest Part 1

 Question 1:  Suppose Ankit has 2300 Rs that he invests in an account that pays 7% interest compounded bi-annually. How much money does Ankit have at the end of 5 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 Rs is the amount being invested or P. The interest rate r is 7% which must be changed into a decimal and becomes r = 0.07. 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 = 2300 x ( 1 + 7/2)(2) x (5)  = 3244.38 Rs. So after 5 years, the account is worth 3244.38 Rs. Question 1:  Suppose Ankit has 2300 Rs that he invests in an account for 5 years at a certain rate compounded bi-annually. This becomes 3244.38 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 = (3244.38/2300)(1/(bi-annually x 57) - 1 = 7 %. Question 3:  Suppose John has a principal P that he invests in an account that pays 1% interest compounded monthly. John has 1751.75 \$ at the end of 3 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 1700 \$ 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 monthly, which tells us that n = 12. 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 = 1751.75/ ( 1 + 1/12)(12) x (3)  = 1700 \$ So after 5 years, the account is worth 1751.75 \$. Question 4:  Suppose Nina has 1500 Rs that she invests in an account that pays 15% interest compounded annually. In how many years will this become 2281.31 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1500 Rs is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded annually, which tells us that n = 1. The final amount is 2281.31 Rs.  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 (2281.31) - log (1500)) / (1* log(1 + 0.15/1)) => t = (LOG2281.31-LOG1500)/ (1* (1 + LOG0.15/1))  => t = 3 years So the time required is 3 years.  Question 4:  Suppose Nina has 1500 Rs that she invests in an account for 3 years at a certain rate compounded annually. This becomes 2281.31 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 = (2281.31/1500)(1/(annually x 315) - 1 = 15 %. Question 6:  Suppose Mary has 2900 \$ that she invests in an account that pays 7% interest compounded quarterly. How much money does Mary 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 2900 \$ is the amount being invested or P. The interest rate r is 7% which must be changed into a decimal and becomes r = 0.07. The interest is compounded quarterly, which tells us that n = 4. 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 = 2900 x ( 1 + 7/4)(4) x (3)  = 3571.17 \$. So after 5 years, the account is worth 3571.17 \$. Question 7:  Suppose Ankit has 100 Rs that he invests in an account that pays 15% interest compounded monthly. In how many years will this become 210.72 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 100 Rs is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded monthly, which tells us that n = 12. The final amount is 210.72 Rs.  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 (210.72) - log (100)) / (12* log(1 + 0.15/12)) => t = (LOG210.72-LOG100)/ (12* (1 + LOG0.15/12))  => t = 5 years So the time required is 5 years.  Question 8:  Suppose Martha has 900 £ that she invests in an account that pays 11% interest compounded annually. In how many years will this become 1108.89 £? 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 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 final amount is 1108.89 £.  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 (1108.89) - log (900)) / (1* log(1 + 0.11/1)) => t = (LOG1108.89-LOG900)/ (1* (1 + LOG0.11/1))  => t = 2 years So the time required is 2 years.  Question 9:  Suppose John has 200 \$ that he invests in an account that pays 14% interest compounded bi-annually. In how many years will this become 262.16 \$? 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 14% which must be changed into a decimal and becomes r = 0.14. The interest is compounded bi-annually, which tells us that n = 2. The final amount is 262.16 \$.  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 (262.16) - log (200)) / (2* log(1 + 0.14/2)) => t = (LOG262.16-LOG200)/ (2* (1 + LOG0.14/2))  => t = 2 years So the time required is 2 years.  Question 9:  Suppose John has 200 \$ that he invests in an account for 2 years at a certain rate compounded bi-annually. This becomes 262.16 \$ 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 = (262.16/200)(1/(bi-annually x 214) - 1 = 14 %. Question 10:  Suppose Nina has 1000 Rs that she invests in an account for 2 years at a certain rate compounded quarterly. This becomes 1104.49 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 = (1104.49/1000)(1/(quarterly x 25) - 1 = 5 %. Question 12:  Suppose Mary has 1300 \$ that she invests in an account that pays 3% interest compounded annually. In how many years will this become 1420.55 \$? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1300 \$ 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 annually, which tells us that n = 1. The final amount is 1420.55 \$.  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 (1420.55) - log (1300)) / (1* log(1 + 0.03/1)) => t = (LOG1420.55-LOG1300)/ (1* (1 + LOG0.03/1))  => t = 3 years So the time required is 3 years.  Question 13:  Suppose Ankit has 1600 Rs that he invests in an account that pays 4% interest compounded bi-annually. In how many years will this become 1731.89 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1600 Rs 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 bi-annually, which tells us that n = 2. The final amount is 1731.89 Rs.  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 (1731.89) - log (1600)) / (2* log(1 + 0.04/2)) => t = (LOG1731.89-LOG1600)/ (2* (1 + LOG0.04/2))  => t = 2 years So the time required is 2 years.  Question 14:  Suppose Martha has a principal P that she invests in an account that pays 15% interest compounded quarterly. Martha has 1252.89 £ 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 600 £ is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded quarterly, which tells us that n = 4. 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 = 1252.89/ ( 1 + 15/4)(4) x (5)  = 600 £ So after 5 years, the account is worth 1252.89 £. Question 15:  Suppose John has a principal P that he invests in an account that pays 5% interest compounded monthly. John has 946.05 \$ 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 900 \$ 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 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 = 946.05/ ( 1 + 5/12)(12) x (1)  = 900 \$ So after 5 years, the account is worth 946.05 \$. Question 15:  Suppose John has 900 \$ that he invests in an account for 1 years at a certain rate compounded monthly. This becomes 946.05 \$ 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 = (946.05/900)(1/(monthly x 15) - 1 = 5 %. Question 17:  Suppose Alex has 400 £ that he invests in an account that pays 14% interest compounded bi-annually. In how many years will this become 786.86 £? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 400 £ is the amount being invested or P. The interest rate r is 14% which must be changed into a decimal and becomes r = 0.14. The interest is compounded bi-annually, which tells us that n = 2. The final amount is 786.86 £.  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 (786.86) - log (400)) / (2* log(1 + 0.14/2)) => t = (LOG786.86-LOG400)/ (2* (1 + LOG0.14/2))  => t = 5 years So the time required is 5 years.  Question 18:  Suppose Mary has 200 \$ that she invests in an account that pays 15% interest compounded quarterly. How much money does Mary 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 200 \$ is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded quarterly, which tells us that n = 4. 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 = 200 x ( 1 + 15/4)(4) x (3)  = 311.09 \$. So after 5 years, the account is worth 311.09 \$. Question 18:  Suppose Mary has 200 \$ that she invests in an account for 3 years at a certain rate compounded quarterly. This becomes 311.09 \$ 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 = (311.09/200)(1/(quarterly x 315) - 1 = 15 %. Question 19:  Suppose Ankit has 2900 Rs that he invests in an account for 2 years at a certain rate compounded monthly. This becomes 3469.6 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 = (3469.6/2900)(1/(monthly x 29) - 1 = 9 %.