Compound Interest Example-1

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Compound Interest examples ( Enter your problem )

( Enter your problem )

5. Simple Interest
(Previous method)

7. Installment
(Next method)

1. Example-1

Problem : 1 / 10 [ Compound Interest ] Enter your problem 1. Calculate Compound Interest and amount on Rs 4500 at 10 % per annum in 3 years.


Solution:	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 / 10 [ Compound Interest ] Enter your problem 2. Calculate Compound Interest and amount on Rs 625 for 2 years at 9 %, compounded half-yearly.


Solution:	Here $P = "Rs" 625, N = 2$ year = $4$ half-years $R = 9$ % per year = $4.5$ % half-yearly $A = P (1 + R/100)^N$ $= 625 * (1 + 4.5 /100) ^ 4$ $= 625 * ( 104.5 / 100) ^ 4$ $= 745.3241$ $:. CI = "Rs" (745.3241 - 625) = "Rs" 120.3241$ .

Problem : 3 / 10 [ Compound Interest ] Enter your problem 3. Calculate Compound Interest on Rs 8000 for 9 months at 20 % per annum, compounded quarterly .


Solution:	Here $P = "Rs" 8000, N = 9$ months = $3$ quarters $R = 20$ % per year = $5$ % per quarter $A = P (1 + R/100)^N$ $= 8000 * (1 + 5 / 100 ) ^ 3$ $= 8000 * ( 105 / 100 ) ^ 3$ $= 9261$ $:. CI = "Rs" (9261 - 8000) = "Rs" 1261$ .

Problem : 4 / 10 [ Compound Interest ] Enter your problem 4. What sum of money becomes Rs 9261 in 3 years at 5 % per annum, compounded annually ?


Solution:	Here $A = "Rs" 9261, R = 5%, N = 3$ years $A = P (1 + R/100)^N$ $9261 = P ( 1 + 5 / 100 ) ^ 3$ $9261 = P ( 105 / 100 ) ^ 3$ $9261 * ( 100 / 105 ) ^ 3 = P$ $P = 8000$ $:.$ The sum is $8000$ .

Problem : 5 / 10 [ Compound Interest ] Enter your problem 5. What will Rs 25000 amounts to in two years at Compound Interest if the rates for successive years be 4 % and 5 % per year.


Solution:	Here $P = 25000, R_1 = 4$ % for 1st year and $R_2 = 5$ % for 2nd year, $N_1 = 1, N_2 = 1$ For Successvie rates $A = P * (1+R_1/100)^(N_1) * (1+R_2/100)^(N_2) ...$ $A = 25000 * (1 + 4 /100) * (1 + 5 /100)$ $= 25000 * ( 104 / 100 ) * ( 105 / 100 )$ $= 27300$ .

Problem : 6 / 10 [ Compound Interest ] Enter your 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.


Solution:	For $N = 3, A = "Rs" 6690$ $A = P (1 + R/100)^N$ $P (1 + R/100) ^ 3 = 6690 ->(1)$ For $N = 6, A = "Rs" 10035$ $A = P (1 + R/100)^N$ $P (1 + R/100) ^ 6 = 10035 ->(1)$ On dividing $(2)$ by $(1)$ , we get $(1 + R/100) ^ 3 = 10035 / 6690 = 3 / 2$ putting this value in $(1)$ , we get $=> P = 4460$ The sum of money is Rs $4460$ .

Problem : 7 / 10 [ Compound Interest ] Enter your 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?


Solution:	Here $P = 2500, R = 5 %, N = 2$ $SI = (PRN)/100 = (2500 * 5 * 2) / 100 = 6.25$ $CI = P (1 + R/100)^N - P$ $= 2500 * ( 1 + 5 / 100 ) ^ 2 - 2500$ $= 2500 * ( 105 / 100 ) ^ 2 - 2500$ $= 2500 * (( 105 / 100 ) ^ 2 - 1)$ $= 256.25$ $:.$ additional amounts he gets $= CI - SI = 256.25 - 250 = 6.25$ Rs

Problem : 8 / 10 [ Compound Interest ] Enter your 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.


Solution:	Here $R = 10 %, N = 2$ years and $CI - SI = 52$ . Let the sum be Rs $X$ . Then, $SI = (PRN)/100 = (X * 10 * 2) / 100 = 0.2 X$ $CI = P (1 + R/100)^N - P$ $= X ( 1 + 10 / 100 ) ^ 2 - X$ $= X ( 110 / 100 ) ^ 2 - X$ $= X (( 110 / 100 ) ^ 2 - 1 )$ $= 0.21 X$ $:. CI - SI = 0.21 X - 0.2 X = 0.01 X$ So, $0.01 X = 52$ => $X = 5200$ . Hence, Sum = Rs $5200$ .

Problem : 9 / 10 [ Compound Interest ] Enter your 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 ?


Solution:	Here $R = 10 %, N = 3$ years, $CI = 331$ $CI = P (1 + R/100)^N - P$ $= P ( 1 + 10 / 100 ) ^ 3 - P$ $= P ( 110 / 100 ) ^ 3 - P$ $= P (( 110 / 100 ) ^ 3 - 1)$ $= 0.331 P$ Now $CI = 331$ $=> 0.331 P = 331$ $=> P = 1000$ Now, $SI = (PRN)/100 = (1000 * 10 * 3) / 100 = 300$ .

Problem : 10 / 10 [ Compound Interest ] Enter your 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 ?


Solution:	Here Let $P = X, N_1 = 15$ then $A = 2 X$ $A = P (1 + R/100)^(N_1)$ $=> 2 X = X (1 + R/100) ^ 15$ $=> (1 + R/100) ^ 15 = 2 -> (1)$ After $N_2$ years it will become $8$ times of the original sum $A = P (1 + R/100)^(N_2)$ $=> 8 X = X (1 + R/100)^(N_2)$ $=> (1 + R/100)^(N_2) = 8 -> (2)$ Now, solving $(1)$ and $(2)$ , we get $N_2 = 45$ years Hence, the sum of money $8$ times itself after $45$ years.

5. Simple Interest
(Previous method)

7. Installment
(Next method)

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