Compound Interest

 Problem : 1 [ Compound Interest ]       Solve this type of problem 1. Calculate Compound Interest and amount on Rs 4500 at 10 % per annum in 3 years. Show Answer A: Here P = "Rs. "4500, R = 10 %, N = 3 years.A = P (1 + R/100)^N= 4500 * ( 1 + 10 / 100 ) ^ 3= 4500 * ( 110 / 100 ) ^ 3= 5989.5:. CI = Rs. (5989.5 - 4500) = Rs. 1489.5

 Problem : 2 [ Compound Interest ]       Solve this type of problem 2. Calculate Compound Interest and amount on Rs 625 for 2 years at 9 %, compounded half-yearly. Show Answer A: Here P = "Rs." 625, N = 2 year = 4 half-yearsR = 9 % per year = 4.5 % half-yearlyA = P (1 + R/100)^N= 625 * (1 + 4.5 /100) ^ 4= 625 * ( 104.5 / 100) ^ 4= 745.3241:. CI = "Rs." (745.3241 - 625) = "Rs." 120.3241.

 Problem : 3 [ Compound Interest ]       Solve this type of problem 3. Calculate Compound Interest on Rs 8000 for 9 months at 20 % per annum, compounded quartely . Show Answer A: Here P = "Rs." 8000, N = 9 months = 3 quartersR = 20 % per year = 5 % per quarterA = P (1 + R/100)^N= 8000 * (1 + 5 / 100 ) ^ 3= 8000 * ( 105 / 100 ) ^ 3= 9261:. CI = "Rs." (9261 - 8000) = "Rs." 1261.

 Problem : 4 [ Compound Interest ]       Solve this type of problem 4. What sum of money becomes Rs. 9261 in 3 years at 5 % per annum, compounded annually ? Show Answer A: Here A = "Rs." 9261, R = 5%, N = 3 yearsA = P (1 + R/100)^N9261 = P ( 1 + 5 / 100 ) ^ 39261 = P ( 105 / 100 ) ^ 39261 * ( 100 / 105 ) ^ 3 = PP = 8000:. The sum is 8000.

 Problem : 5 [ Compound Interest ]       Solve this type of problem 5. What will Rs 25000 amounts to in 2 years at Compound Interest if the rates for successive years be 4 % and 5 % per year. Show Answer A: Here P = 25000, R_1 = 4 % for 1st year and R_2 = 5 % for 2nd year, N_1 = 1, N_2 = 1For Successvie ratesA = P * (1+R_1/100)^(N_1) * (1+R_2/100)^(N_2) ...A = 25000 * (1 + 4 /100) * (1 + 5 /100)= 25000 * ( 104 / 100 ) * ( 105 / 100 )= 27300.

 Problem : 6 [ Compound Interest ]       Solve this type of problem 6. A sum of money amounts to Rs 6690 after 3 years and to Rs 10035 after 6 years on compound interest. Calculate the sum of money. Show Answer A: For N = 3, A = "Rs." 6690A = P (1 + R/100)^NP (1 + R/100) ^ 3 = 6690 ->(1)For N = 6, A = "Rs." 10035A = P (1 + R/100)^NP (1 + R/100) ^ 6 = 10035 ->(1)On dividing (2) by (1), we get(1 + R/100) ^ 3 = 10035 / 6690 = 3 / 2putting this value in (1), we get=> P = 4460The sum of money is Rs. 4460 .

 Problem : 7 [ Compound Interest ]       Solve this type of problem 7. A man borrows Rs 2500 at 5 % simple interest for 2 years. He immediately lends this money at compound interest at the same rate for the same time. What is the additional amount he gets at the ends of the years correct to nearest rupee? Show Answer A: Here P = 2500, R = 5 %, N = 2SI = (P*R*N)/100 = (2500 * 5 * 2) / 100 = 6.25CI = P (1 + R/100)^N - P= 2500 * ( 1 + 5 / 100 ) ^ 2 - 2500= 2500 * ( 105 / 100 ) ^ 2 - 2500= 2500 * (( 105 / 100 ) ^ 2 - 1)= 256.25:. additional amounts he gets = CI - SI = 256.25 - 250 = 6.25 Rs.

 Problem : 8 [ Compound Interest ]       Solve this type of problem 8. The difference between the compound interest and the simple interest on a certain sum at 10 % per annum for 2 years is Rs. 52 . Find the sum. Show Answer A: Here R = 10 %, N = 2 years and CI - SI = 52 .Let the sum be Rs. X. Then,SI = (P*R*N)/100 = (X * 10 * 2) / 100 = 0.2 XCI = P (1 + R/100)^N - P= X ( 1 + 10 / 100 ) ^ 2 - X= X ( 110 / 100 ) ^ 2 - X= X (( 110 / 100 ) ^ 2 - 1 )= 0.21 X:. CI - SI = 0.21 X - 0.2 X = 0.01 XSo, 0.01 X = 52=> X = 5200 .Hence, Sum = Rs. 5200.

 Problem : 9 [ Compound Interest ]       Solve this type of problem 9. If the compound interest on a certain sum for 3 years at 10 % per annum be Rs. 331 , what would be the simple interest ? Show Answer A: Here R = 10 %, N = 3 years, CI = 331CI = P (1 + R/100)^N - P= P ( 1 + 10 / 100 ) ^ 3 - P= P ( 110 / 100 ) ^ 3 - P= P (( 110 / 100 ) ^ 3 - 1)= 0.331 PNow CI = 331=> 0.331 P = 331=> P = 1000Now, SI = (P*R*N)/100 = (1000 * 10 * 3) / 100 = 300.

 Problem : 10 [ Compound Interest ]       Solve this type of problem 10. A sum of money 2 times itself at compound interest in 15 years. In how many years, it will become 8 times of itself ? Show Answer A: Here Let P = X, N_1 = 15 then A = 2 XA = P (1 + R/100)^(N_1)=> 2 X = X (1 + R/100) ^ 15=> (1 + R/100) ^ 15 = 2 -> (1)After N_2 years it will become 8 times of the original sumA = P (1 + R/100)^(N_2)=> 8 X = X (1 + R/100)^(N_2)=> (1 + R/100)^(N_2) = 8 -> (2)Now, solving (1) and (2), we getN_2 = 45 yearsHence, the sum of money 8 times itself after 45 years.