Compound Interest-Chapter 9- Exercise 9(B)-Class 8

byAmlesh Mahato •June 13, 2020

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Exercise 9 (B)

1. Calculate the amount and the compound interest by using the formula for compound interest.

Q.N.	Principal	Rate Of interest	Time (In years)
1	Rs 625	4 % p.a.	2
2	Rs 1000	10 % p.a.	3
3	Rs 8000	10 % half-yearly	1.5

Q.1.Solution:

⇨ Given p=625 r=4 % n=2

⇨Then the annual Compound interest will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=625\left \{ \left ( 1+\frac{4}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=625\left \{ \left (\frac{100+4}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=625\left \{ \left (\frac{104}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=625\left \{\frac{10816}{10000}-1 \right \}$

$\large \Rightarrow C.I=625\left \{\frac{10816-10000}{10000} \right \}$

$\large \Rightarrow C.I=625\times \frac{816}{10000}$

$\large \Rightarrow C.I=\frac{625\times 816}{10000}$

$\large \Rightarrow C.I=\frac{510000}{10000}$

$\large \Rightarrow C.I=51 Rs$

⇨ Total Amount = P + C.I. = 625 + 51 =676 Rs

Q.2.Solution:

⇨ Given p=1000 r=10 % n=3

⇨ Then the annual Compound interest will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{ \left ( 1+\frac{10}{100} \right )^{3}-1 \right \}$ $\large \Rightarrow C.I=1000\left \{ \left (\frac{100+10}{100} \right )^{3}-1 \right \}$ $\large \Rightarrow C.I=1000\left \{ \left (\frac{110}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=1000\left \{\frac{1331000}{1000000}-1 \right \}$ $\large \Rightarrow C.I=1000\left \{\frac{1331000-1000000}{1000000}\right \}$

$\large \Rightarrow C.I=1000\times \frac{331000}{1000000}$

$\large \Rightarrow C.I=\frac{331000000}{1000000}$

$\large \Rightarrow C.I=331$

⇨ Total Amount = P + C.I. = 1000 + 331 =1331 Rs

Q.3.Solution:

⇨ Given p=8000 r=10 % n=1.5 or 3/2

⇨ Then the Half-yearly Compound interest will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{2\times 100} \right )^{2n}-1 \right \}$

$\large \large C.I=8000\left \{ \left ( 1+\frac{10}{2\times 100} \right )^{2\times \frac{3}{2}}-1 \right \}$

$\large \large C.I=8000\left \{ \left (\frac{200+10}{200} \right )^{3}-1 \right \}$

$\large \large C.I=8000\left \{ \left (\frac{210}{200} \right )^{3}-1 \right \}$

$\large \large C.I=8000\left \{\frac{9261000}{8000000}-1 \right \}$

$\large \large C.I=8000\left \{\frac{9261000-8000000}{8000000} \right \}$

$\large \large C.I=8000\times \frac{1261000}{8000000}$

$\large \large C.I=\frac{1261000}{1000}$

$\large \large C.I=1261$

⇨ Total Amount = P + C.I. = 8000 + 1261 =9261 Rs

Q.4. Reena and Ruchira borrowed Rs 60000 and Rs 50000 respectively for a period of 3 years. Reena paid simple interest at the rate of 10 % p.a., while Ruchira paid compound interest at the rate of 10 % p.a. Compounded annually.Who paid more interest and by how much ?

Solution:

⇨For Reena
⇨Given
⇨p = 60000
⇨r = 10 %
⇨n = 3 p.a.

⇨Therefore the Simple interest for Reena will be

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow S.I=\frac{60000\times 10\times 3}{100}$

$\large \Rightarrow S.I=\frac{600000\times 3}{100}$

$\large \Rightarrow S.I=6000\times 3$

$\large \Rightarrow S.I=18000$ Rs

⇨For Ruchira
⇨Given
⇨p = 50000
⇨r = 10 %
⇨n = 3 p.a.

⇨Therefore the Compound interest for Ruchira will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left ( 1+\frac{10}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left (\frac{100+10}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left (\frac{110}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \frac{1331000}{1000000}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \frac{1331000-1000000}{1000000} \right \}$

$\large \Rightarrow C.I=50000\times \frac{331000}{1000000}$

$\large \Rightarrow C.I=5\times 3310$ Rs

$\large \Rightarrow C.I=16550$

⇨From above Reena paid more interest than Ruchira by

= Interest paid by Reena - Interest paid by Ruchira

= 18000 - 16550
= 1450 Rs

Q.5. Sangeeta lent Rs 40960 to Amar to purchase a shop at 12.5 % per annum. If the interest is compounded semi-annually, find the interest paid by Amar after 1.5 or 3/2 years ?

Solution:

⇨Given
⇨p=40960
⇨r=12.5 %
⇨n=1.5 or 3/2

⇨Then semi-annually Then the semi-annually Compound interest paid by Amar will be

$\large \Rightarrow C.I=40960\left \{ \left ( 1+\frac{12.5}{2\times 100} \right )^{^{2\times \frac{3}{2}}}-1 \right \}$ $\large \Rightarrow C.I=40960\left \{ \left ( 1+\frac{12.5}{200} \right )^{3}-1 \right \}$ $\large \Rightarrow C.I=40960\left \{ \left (\frac{200+12.5}{200} \right )^{3}-1 \right \}$ $\large \Rightarrow C.I=40960\left \{ \left (\frac{212.5}{200} \right )^{3}-1 \right \}$ $\large \Rightarrow C.I=40960\left \{\frac{9595703.125}{8000000}-1 \right \}$ $\large \Rightarrow C.I=40960\left \{\frac{9595703.125-8000000}{8000000} \right \}$ $\large \Rightarrow C.I=40960\times \frac{1595703.125}{8000000}$

$\large \Rightarrow C.I=\frac{65360000000}{8000000}$

$\large \Rightarrow C.I=\frac{65360}{8}$

$\large \Rightarrow C.I=8170$ Rs

Q.6. Sudhir lent Rs 2000 at compound interest at 10 % payable yearly, While Prashant lent Rs 2000 at compound interest at 10 % payable half yearly. Find the difference in the interest received by Sudhir and Prashant at the end of one year?

Solution:

⇨In case of Sudhir
⇨Given
⇨p=2000
⇨r=10 %
⇨n=1 p.a.
⇨Then the annual Compound interest paid by Sudhir will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \left ( 1+\frac{10}{100} \right )^{1}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \frac{100+10}{100}-1 \right \}$

$\large \Rightarrow C.I=2000\times \frac{110-100}{100}$

$\large \Rightarrow C.I=2000\times \frac{10}{100}$

$\large \Rightarrow C.I=\frac{200000}{100}$

$\large \Rightarrow C.I=200$ Rs

⇨In case of Prashant
⇨Given
⇨p=2000
⇨r=10 %
⇨n=1 Half-yearly.

⇨Then the half-yearly Compound interest paid by Prashant will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{2\times 100} \right )^{2n}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \left ( 1+\frac{10}{200} \right )^{2\times 1}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \left ( \frac{200+10}{200} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \left ( \frac{210}{200} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=2000\left \{ \frac{44100}{40000}-1 \right \}$

$\large \Rightarrow C.I=2000\times \frac{44100-40000}{40000}$

$\large \Rightarrow C.I=2000\times \frac{4100}{40000}$

$\large \Rightarrow C.I=\frac{8200000}{40000}$

$\large \Rightarrow C.I=\frac{820}{4}$

$\large \Rightarrow C.I=205$ Rs

⇨Therefore The difference in the interest received by Sudhir and Prashant at the end of one year Will be

= 205 - 200

= 5 Rs

Q.7. Find the difference between simple interest and compound interest on Rs 16000 for 1.5 rears or 3/2 years at 5 % per annum ,compound interest being reckoned half yearly.

Solution:

⇨Given
⇨p=16000
⇨r=5 %
⇨n=1.5 annul for S.I.
⇨The simple interest will be

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow S.I=\frac{16000\times 5\times 1\frac{1}{2}}{100}$

$\large \Rightarrow S.I=\frac{80000\times \frac{3}{2}}{100}$

$\large \Rightarrow S.I=800\times \frac{3}{2}$

$\large \Rightarrow S.I=400\times 3$

$\large \Rightarrow S.I=1200$ Rs

⇨Given
⇨p=16000
⇨r=5 %
⇨n=1.5 Half-yearly for C.I
⇨The compound interest will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{2\times 100} \right )^{2n}-1 \right \}$ $\large \Rightarrow C.I=16000\left \{ \left ( 1+\frac{5}{200} \right )^{2\times 1\frac{1}{2}}-1 \right \}$

$\large \Rightarrow C.I=16000\left \{ \left ( \frac{200+5}{200} \right )^{2\times \frac{3}{2}}-1 \right \}$ $\large \Rightarrow C.I=16000\left \{ \left ( \frac{205}{200} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=16000\left \{ \frac{8615125}{8000000}-1 \right \}$ $\large \Rightarrow C.I=16000\left \{ \frac{8615125-8000000}{8000000} \right \}$

$\large \Rightarrow C.I=16000\times \frac{615125}{8000000}$

$\large \Rightarrow C.I=16\times \frac{615125}{8000}$

$\large \Rightarrow C.I=2\times \frac{615125}{1000}$

$\large \Rightarrow C.I=2\times 615.125$

$\large \Rightarrow C.I=1230.25$ Rs

⇨Therefore the difference between S.I and C.I will be

= C.I - S.I.
=1230.25 - 1200
=30.25 Rs

Q.8. Find the compound interest on 3125 for 3 years if the rates of interest for first, second and third year are respectively 4%, 5%, 6%.

Solution:

⇨Given
⇨p=3125
⇨r1=4 %
⇨r2=5%
⇨r3=6%
⇨n=3 years

⇨For different rates on successive years, Total amount 'A' will be

$\large A=p\left ( 1+\frac{r_{1}}{100} \right )\left ( 1+\frac{r_{2}}{100} \right )\left ( 1+\frac{r_{3}}{100} \right )$ $\large \Rightarrow A=3125\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{5}{100} \right )\left ( 1+\frac{6}{100} \right )$

$\large \Rightarrow A=3125\left (\frac{104}{100} \right )\left (\frac{105}{100} \right )\left (\frac{106}{100} \right )$

$\large \Rightarrow A=3125\times 1.04\times 1.05\times 1.06$

$\large \Rightarrow A=3617.25$ Rs

⇨Therefore The Compound interest will be

= A- P
=3617.25 - 3125
= 492.5 Rs

Q.9. A District contains 64000 inhabitants. If the population increases at the rate of 2.5 or 5/2 % per annum ,find the number of inhabitants at the end of 3 years.

Solution:

⇨Given
⇨p=64000
⇨r=2.5 %
⇨n=3

⇨The number of inhabitants at the end of 3 years

$\large A=p\left ( 1+\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=64000\left ( 1+\frac{2.5}{100} \right )^{3}$

$\large \Rightarrow A=64000\left (\frac{100+2.5}{100} \right )^{3}$

$\large \Rightarrow A=64000\left (\frac{102.5}{100} \right )^{3}$

$\large \Rightarrow A=64000\times \frac{1076890.625}{1000000}$

$\large \Rightarrow A=64\times \frac{1076890.625}{1000}$

$\large \Rightarrow A=64\times 1076.890625$

$\large \Rightarrow A=68921$ Rs

Q.10. The nagar palika of certain city started campaign to kill the stray dogs which numbered 1250 in the city as a result, the population of stray dogs started decreasing at the rate of 20 % per month . Calculate the number of stray dogs in the city three months after the campaign started.

Solution:

⇨Given
⇨p=1250

⇨r=20%

⇨n=3 months

⇨The number of stray dogs in the city three months after the campaign started Will be

$\large A=p\left ( 1-\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=1250\left ( 1-\frac{20}{100} \right )^{3}$

$\large \Rightarrow A=1250\left ( \frac{100-20}{100} \right )^{3}$

$\large \Rightarrow A=1250\left ( \frac{80}{100} \right )^{3}$

$\large \Rightarrow A=1250\times \frac{512000}{1000000}$

$\large \Rightarrow A=125\times \frac{512}{100}$

$\large \Rightarrow A=125\times 5.12$

$\large \Rightarrow A=640$ dogs

⇨Therefore no. of stray dogs=640

Q.11. The value of a machine depreciates each year by 10 % of it's value at the beginning of that year. It' value when new is Rs 750, find it's value when it is 2 years old.

Solution:

⇨Given

⇨p=750

⇨r=10%

⇨n=2 months

⇨The value of a machine when it is 2 years old will be

$\large A=p\left ( 1-\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=750\left ( 1-\frac{10}{100} \right )^{2}$

$\large \Rightarrow A=750\left ( 1-\frac{1}{10} \right )^{2}$

$\large \Rightarrow A=750\left (\frac{10-1}{10} \right )^{2}$

$\large \Rightarrow A=750\left (\frac{9}{10} \right )^{2}$

$\large \Rightarrow A=750\times \frac{81}{100}$

$\large \Rightarrow A=75\times \frac{81}{10}$

$\large \Rightarrow A=75\times 8.1$

$\large \Rightarrow A=607.5$ Rs

Q.12. The national wealth of a country increases by 4% of it's value at the beginning of every year .Find the national wealth of the country in 1985, if it was estimated as Rs 3.125 x 10^12 in 1983.

Solution:

⇨Given

⇨p=3.125 X 10^12

⇨r=4%

⇨n=2 years

⇨The national wealth of the country in 1984

$\large A=p\left ( 1+\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=3.125\times 10^{12}\left ( 1+\frac{4}{100} \right )^{2}$

$\large \Rightarrow A=3.125\times 10^{12}\left (\frac{100+4}{100} \right )^{2}$

$\large \Rightarrow A=3.125\times 10^{12}\left (\frac{104}{100} \right )^{2}$

$\large \Rightarrow A=3.125\times 10^{12}\times \frac{10816}{10000}$

$\large \Rightarrow A=3.125\times 10^{12}\times 10816\times 10^{-4}$

$\large \Rightarrow A=3.125\times 10^{12}\times 1.0816$

$\large \Rightarrow A=3.38\times 10^{12}$

$\large \Rightarrow A=338\times 10^{10}$ Rs

⇨National wealth =3380 X 10^9 Rs

Q.13.The price of the scooter depreciates by 2% of it's at the beginning of each year. Find The sale value of the scooter after 3 years , if it's present sale value is Rs 12000.

Solution:

⇨Given

⇨p=12000

⇨r=2%

⇨n=3 years

⇨The sale value of the scooter after 3 years will be

$\large A=p\left ( 1-\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=12000\left ( 1-\frac{2}{100} \right )^{3}$

$\large \Rightarrow A=12000\left (\frac{100-2}{100} \right )^{3}$

$\large \Rightarrow A=12000\left (\frac{98}{100} \right )^{3}$

$\large \Rightarrow A=12000\times \frac{941192}{1000000}$

$\large \Rightarrow A=12\times \frac{941192}{1000}$

$\large \Rightarrow A=12\times941.192$

$\large \Rightarrow A=11294.304$ Rs

⇨value of scooter=11294.304 Rs

Q.14. The population of Andhra Pradesh was 5.4 X 10^7 in 1988. If the population was growing at a constant rate of 2.4 % per annum .What was the population in 2008 A.D.[Given that (1.024)^20=1.60694].

Solution:

⇨Given

⇨p=5.4 X 10^7

⇨r=2.4%

⇨n=2008 - 1988 =20 years

⇨The population in 2008 will be

$\large A=p\left ( 1+\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=5.4\times 10^{7}\left ( 1+\frac{2.4}{100} \right )^{20}$

$\large \Rightarrow A=5.4\times 10^{7}\left (\frac{100+2.4}{100} \right )^{20}$

$\large \Rightarrow A=5.4\times 10^{7}\left (\frac{102.4}{100} \right )^{20}$

$\large \Rightarrow A=5.4\times 10^{7}\left (1.024 \right )^{20}$

$\large \Rightarrow A=5.4\times 10^{7}\times 1.60693$

$\large \Rightarrow A=5.4\times 10^{7}\times 1.607$

$\large \Rightarrow A=8.6778\times 10^{7}$ (approx)

Multiple Choice Questions

Q.15. A steno- typist deposits Rs 10000 in a bank for a period of 3 years at 12 % per annum compound interest . The interest accrued to her after maturity will be

Solution:

⇨Given

⇨p=10000

⇨r=12%

⇨n=3 years

⇨The interest accrued to her after maturity will be

$\large C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( 1+\frac{12}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left (\frac{100+12}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left (\frac{112}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{\frac{1404928}{1000000}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{\frac{1404928-1000000}{1000000} \right \}$

$\large \Rightarrow C.I=10000\left \{\frac{404928}{1000000} \right \}$

$\large \Rightarrow C.I=\frac{404928}{100}$

$\large \Rightarrow C.I=4049.28$ Rs

Q.16. The present population of a city is 180000. If it increases at the rate of 8% per annum , It's population after 2 years will be

Solution:

⇨Given

⇨p=180000

⇨r=8%

⇨n=2 years

⇨Population after 2 years will be

$\large A=p\left ( 1+\frac{r}{100} \right )^{n}$

$\large \Rightarrow A=180000\left ( 1+\frac{8}{100} \right )^{2}$

$\large \Rightarrow A=180000\left (\frac{100+8}{100} \right )^{2}$

$\large \Rightarrow A=180000\left (\frac{108}{100} \right )^{2}$

$\large \Rightarrow A=180000\times \frac{11664}{10000}$

$\large \Rightarrow A=18\times 11664$

$\large \Rightarrow A=209952$

High Order Thinking

Q.17. Umesh set up a small factory by investing ₹40,000. During the first three successive years his profits were 5%,10% and 15% respectively. If each year the profit was on previous years' capital, calculate his total profit.

Solution:

⇨Given

⇨p=40000

⇨r1=5%

⇨r2=10%

⇨r3=15%

⇨Simple interest for 1st year will be

     $\large S.I=\frac{p\times r_{1}\times n}{100}$

$\large \Rightarrow S.I=\frac{40000\times 5\times 1}{100}$

$\large \Rightarrow S.I=400\times 5\times 1$

$\large \Rightarrow S.I=2000$ Rs

⇨Total money invested in 2nd year=40000 + 2000 = 42000 Rs

⇨A/Q

⇨Profit in 2nd year =10%

⇨Then Simple interest for 2nd year will be

       $\large S.I=\frac{p\times r_{2}\times n}{100}$

$\large \Rightarrow S.I=\frac{42000\times 10\times 1}{100}$

$\large \Rightarrow S.I=420\times 10\times 1$

$\large \Rightarrow S.I=4200$ Rs

⇨Total money invested in 3rd year

= 42000 + 4200

= 46200 Rs

⇨Profit in 2nd year =15%

⇨Then Simple interest for 3rd year will be

       $\large S.I=\frac{p\times r_{3}\times n}{100}$

$\large \Rightarrow S.I=\frac{46200\times 15\times 1}{100}$

$\large \Rightarrow S.I=462\times 15\times 1$

$\large \Rightarrow S.I=6930$ Rs

⇨Total Money After 3 years

= 46200 + 6930

= 53130 Rs

⇨Therefore total profit will be

= 53130 - 40000

= 13130 Rs

Q.18.The price value of the share of a company increase the the rate of 15% in the first year suffered a loss of 10% in the next year and again increased by 20% the third year if the percent value of a share is Rs 2000 what will be it's value after 3 years.

Solution:

⇨Given
⇨p=2000

⇨r1=15% increase

⇨r2=10% decrease

⇨r3=20% increase

⇨Simple interest after 1 year will be

$\large S.I=\frac{p\times r_{1}\times n}{100}$

$\large S.I=\frac{p\times r_{2}\times n}{100}$

$\large S.I=\frac{p\times r_{3}\times n}{100}$

$\large \Rightarrow S.I=\frac{2000\times 15\times 1}{100}$

$\large \Rightarrow S.I=20\times 15$

$\large \Rightarrow S.I=300$ Rs

⇨Share value after 1 year will be

= 2000 + 300

= 2300 Rs

⇨Simple interest after 2nd year will be

$\large \Rightarrow S.I=\frac{2300\times 10\times 1}{100}$

$\large \Rightarrow S.I=23\times 10$

$\large \Rightarrow S.I=230$ Rs

⇨Share value after 2nd year will be

= 2300 - 230

= 2070 Rs

⇨Simple interest after 3rd year will be

$\large \Rightarrow S.I=\frac{2070\times 20\times 1}{100}$

$\large \Rightarrow S.I=207\times 2$

$\large \Rightarrow S.I=414$ Rs

⇨Share value after 2nd year will be

= 2070 + 414

= 2484 Rs

Tags: Math-Class-8

Compound Interest-Chapter 9- Exercise 9(B)-Class 8

Q.N.

Principal

Rate Of interest

Time (In years)

1

Rs 625

4 % p.a.

2

2

Rs 1000

10 % p.a.

3

3

Rs 8000

10 % half-yearly

1.5

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Compound Interest-Chapter 9- Exercise 9(B)-Class 8

Q.N.

Principal

Rate Of interest

Time (In years)

1

Rs 625

4 % p.a.

2

2

Rs 1000

10 % p.a.

3

3

Rs 8000

10 % half-yearly

1.5

dogs ⇨Therefore no. of stray dogs=640

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