Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 11 - Compound Interest

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Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 11 Compound Interest are provided here with simple step-by-step explanations. These solutions for Compound Interest are extremely popular among Class 8 students for Maths Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 144:

Answer:

$Principal for the first year = Rs . 2500 Interest for the first year = Rs . (\frac{2500 \times 10 \times 1}{100}) = Rs . 250 Amount at the end of the first year = Rs . (2500 + 250) = Rs . 2750 Principal for the second year = Rs . 2750 Interest for the second year = Rs . (\frac{2750 \times 10 \times 1}{100}) = Rs . 275 Amount at the end of the second year = Rs . (2750 + 275) = Rs . 3025 ∴ Compound interest = Rs . (3025 - 2500) = Rs . 525$

Page No 144:

Answer:

$Principal for the first year = Rs . 15625 Interest for the first year = Rs . (\frac{15625 \times 12 \times 1}{100}) = Rs . 1875 Amount at the end of the first year = Rs . (15625 + 1875) = Rs . 17500 Principal for the second year = Rs . 17500 Interest for the second year = Rs . (\frac{17500 \times 12 \times 1}{100}) = Rs . 2100 Amount at the end of the second year = Rs . (17500 + 2100) = Rs . 19600 Principal for the third year = Rs . 19600 Interest for the third year = Rs . (\frac{19600 \times 12 \times 1}{100}) = Rs . 2352 Amount at the end of the second year = Rs (19600 + 2352) = Rs . 21952 ∴ Compound interest = Rs . (21952 - 15625) = Rs . 6327$

Page No 144:

Answer:

$Principal amount = Rs . 5000 Simple interest = Rs . (\frac{5000 \times 2 \times 9}{100}) = Rs . 900 The compound interest can be calculated as follows : Principal for the first year = Rs . 5000 Interest for the first year = Rs . (\frac{5000 \times 9 \times 1}{100}) = Rs . 450 Amount at the end of the first year = Rs . (5000 + 450) = Rs . 5450 Principal for the second year = Rs . 5450 Interest for the second year = Rs . (\frac{5450 \times 9 \times 1}{100}) = Rs . 490.5 Amount at the end of the second year = Rs . (5450 + 490.5) = Rs . 5940.5 ∴ Compound interest = Rs . (5940.5 - 5000) = Rs . 940.5 Now, difference between the simple interest and the compound interest = (CI - SI) = Rs . (940.5 - 900) = Rs . 40.5$

Page No 144:

Answer:

$Principal for the first year = Rs . 25000 Interest for the first year = Rs . (\frac{25000 \times 8 \times 1}{100}) = Rs . 2000 Amount at the end of the first year = Rs . (25000 + 2000) = Rs . 27000 Principal for the second year = Rs . 27000 Interest for the second year = Rs . (\frac{27000 \times 8 \times 1}{100}) = Rs . 2160 Amount at the end of the second year = Rs . (27000 + 2160) = Rs . 29160 Therefore, Ratna has to pay Rs . 29160 after 2 years to discharge her debt .$

Page No 144:

Answer:

$Principal amount = Rs . 20000 Simple interest = Rs . (\frac{20000 \times 2 \times 12}{100}) = Rs . 4800 The compound interest can be calculated as follows : Principal for the first year = Rs . 20000 Interest for the first year = Rs . (\frac{20000 \times 12 \times 1}{100}) = Rs . 2400 Now, amount at the end of the first year = Rs . (20000 + 2400) = Rs . 22400 Principal for the second year = Rs . 22400 Interest for the second year = Rs . (\frac{22400 \times 12 \times 1}{100}) = Rs . 2688 Now, amount at the end of the second year = Rs . (22400 + 2688) = Rs . 25088 Hence, compound interest = Rs . (25088 - 20000) = Rs . 5088 Now, CI - SI = Rs . (5088 - 4800) = Rs . 288 ∴ The amount of money Harpreet will gain after two years is Rs 288 .$

Page No 144:

Answer:

$Principal for the first year = Rs . 64000 Interest for the first year = Rs . (\frac{64000 \times 15 \times 1}{100 \times 2}) = Rs . 4800 Now, amount at the end of the first year = Rs . (64000 + 4800) = Rs . 68800 Principal for the second year = Rs . 68800 Interest for the second year = Rs . (\frac{68800 \times 15 \times 1}{100 \times 2}) = Rs . 5160 Now, amount at the end of the second year = Rs . (68800 + 5160) = Rs . 73960 Principal for the third year = Rs . 73960 Interest for the third year = Rs . (\frac{73960 \times 15 \times 1}{100 \times 2}) = Rs . 5547 Now, amount at the end of the third year = Rs . (73960 + 5547) = Rs . 79507 ∴ Manoj will get an amount of Rs . 79507 after 3 years .$

Page No 144:

Answer:

$Principal amount = Rs . 6250 Rate of interest = 8 % per annum = 4 % for half year Time = 1 year = 2 half years Principal for the first half year = Rs . 6250 Interest for the first half year = Rs . (\frac{6250 \times 4 \times 1}{100}) = Rs . 250 Now, amount at the end of the first half year = Rs . (6250 + 250) = Rs . 6500 Principal for the second half year = Rs . 6500 Interest for the second half year = Rs . (\frac{6500 \times 4 \times 1}{100}) = Rs . 260 Now, amount at the end of the second half year = Rs (6500 + 260) = Rs . 6760 ∴ Compound interest = Rs (6760 - 6250) = Rs 510 Hence, Divakaran gets a compound interest of Rs 510 .$

Page No 145:

Answer:

$Principal amount = Rs . 16000 Rate of interest = 10 % per annum = 5 % for half year Time = 1 \frac{1}{2} years = 3 half years Principal for the first half year = Rs . 16000 Interest for the first half year = Rs . (\frac{16000 \times 5 \times 1}{100}) = Rs . 800 Now, amount at the end of the first half year = Rs . (16000 + 800) = Rs . 16800 Principal for the second half year = Rs . 16800 Interest for the second half year = Rs . (\frac{16800 \times 5 \times 1}{100}) = Rs . 840 Now, amount at the end of the second half year = Rs . (16800 + 840) = Rs . 17640 Principal for the third half year = Rs . 17640 Interest for the third half year = Rs . (\frac{17640 \times 5 \times 1}{100}) = Rs . 882 Now, amount at the end of the third half year = Rs . (17640 + 882) = Rs . 18522 ∴ The amount of money Michael has to pay the finance company after 1 \frac{1}{2} years is Rs 18522 .$

Page No 151:

Answer:

$Principal amount, P = Rs 6000 Rate of interest, R = 9 % per annum Time, n = 2 years . The formula for the amount including the compound interest is given below : A = Rs . P {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 6000 {(1 + \frac{9}{100})}^{2} \Rightarrow A = Rs . 6000 {(\frac{100 + 9}{100})}^{2} \Rightarrow A = Rs . 6000 {(\frac{109}{100})}^{2} \Rightarrow A = Rs . 6000 {(1.09 \times 1.09)}^{2} \Rightarrow A = Rs . 7128.6 i . e ., the amount including the compound interest is Rs 7128.6 . ∴ Compound interest = Rs (7128.6 - 6000) = Rs 1128.6$

Page No 151:

Answer:

$Principal amount, P = Rs . 10000 Rate of interest, R = 11 % per annum . Time, n = 2 years . The formula for the amount including the compound interest is given below : A = Rs . P {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 10000 {(1 + \frac{11}{100})}^{2} \Rightarrow A = Rs . 10000 {(\frac{100 + 11}{100})}^{2} \Rightarrow A = Rs . 10000 {(\frac{111}{100})}^{2} \Rightarrow A = Rs . 10000 {(1.11 \times 1.11)}^{2} \Rightarrow A = Rs . 12321 i . e ., the amount including the compound interest is Rs 12321 . ∴ Compound interest = Rs . (12321 - 10000) = Rs . 2321$

Page No 151:

Answer:

$Principal amount, P = Rs . 31250 Rate of interest, R = 8 % per annum . Time, n = 3 years . The formula for the amount including the compound interest is given below : A = Rs . P {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 31250 {(1 + \frac{8}{100})}^{3} \Rightarrow A = Rs . 31250 {(\frac{100 + 8}{100})}^{3} \Rightarrow A = Rs . 31250 {(\frac{108}{100})}^{3} \Rightarrow A = Rs . 31250 {(1.08 \times 1.08 \times 1.08)}^{3} \Rightarrow A = Rs . 39366 i . e ., the amount including the compound interest is Rs 39366 . ∴ Compound interest = Rs . (39366 - 31250) = Rs . 8116$

Page No 151:

Answer:

$Principal amount, P = Rs . 10240 Rate of interest, R = 12 \frac{1}{2} % p . a . Time, n = 3 years The formula for the amount including the compound interest is given below : A = Rs . P {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 10240 {(1 + \frac{25}{100 \times 2})}^{3} \Rightarrow A = Rs . 10240 {(1 + \frac{25}{200})}^{3} \Rightarrow A = Rs . 10240 {(1 + \frac{1}{8})}^{3} \Rightarrow A = Rs . 10240 {(\frac{8 + 1}{8})}^{3} \Rightarrow A = Rs . 10240 {(\frac{9}{8})}^{3} \Rightarrow A = Rs . 10240 {(1.125 \times 1.125 \times 1.125)}^{3} \Rightarrow A = Rs . 14580 i . e ., the amount including the compound interest is Rs 14580 . ∴ Compound interest = Rs (14580 - 10240) = Rs . 4340$

Page No 151:

Answer:

$Principal amount, P = Rs 62500 Rate of interest, R = 12 % p . a . Time, n = 2 years 6 months = \frac{5}{2} = 2 \frac{1}{2} years The formula for the amount including the compound interest is given below : A = Rs . P {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 62500 {(1 + \frac{12}{100})}^{2} \times (1 + \frac{\frac{1}{2} \times 12}{100}) \Rightarrow A = Rs . 62500 {(1 + \frac{12}{100})}^{2} \times (1 + \frac{6}{100}) \Rightarrow A = Rs . 62500 \times 1.12 \times 1.12 \times 1.06 \Rightarrow A = Rs . 83104 i . e ., the amount including the compound interest is Rs 83104 . ∴ Compound interest = Rs . (83104 - 62500) = Rs . 20604$

Page No 151:

Answer:

$Principal amount, P = Rs . 9000 Rate of interest, R = 10 % p . a . Time, n = 2 years 4 months = 2 \frac{1}{3} years = \frac{7}{3} years The formula for t h e amount including the compound interest is given below : A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . (9000 \times {(1 + \frac{10}{100})}^{2} \times (1 + \frac{\frac{1}{3} \times 10}{100})) = Rs . (9000 \times 1.10 \times 1.10 \times 1.033) = Rs . 11252.9 \approx 11253 i . e ., the amount including the compound interest is Rs 11253 . ∴ Compound interest = Rs . (11253 - 9000) = Rs . 2253$

Page No 151:

Answer:

$Principal amount, P = Rs . 8000 Rate of interest for the first year, p = 9 % p . a . Rate of interest for the second year, q = 10 % p . a . Time, n = 2 years . Formula for the amount including the compound interest for the first year : A = Rs . \{P \times (1 + \frac{p}{100}) \times (1 + \frac{q}{100})\} = Rs . \{8000 \times (1 + \frac{9}{100}) \times (1 + \frac{10}{100})\} = Rs . \{8000 \times (\frac{109}{100}) \times (\frac{110}{100})\} = Rs . \{8000 \times (1.09) \times (1.1)\} = Rs . 9592 i . e ., the amount including the compound interest for first year is Rs 9592 .$

Page No 151:

Answer:

$Principal amount, P = Rs . 125000 Rate of interest, R = 8 % p . a . Time, n = 3 year s The amount including the compound interest is calculated using the formula, A = Rs . P {(1 + \frac{R}{100})}^{n} = Rs . 125000 {(1 + \frac{8}{100})}^{3} = Rs . 125000 {(\frac{100 + 8}{100})}^{3} = Rs . 125000 {(\frac{108}{100})}^{3} = Rs . 125000 {(1.08)}^{3} = Rs . 125000 (1.08 \times 1.08 \times 1.08) = Rs . 157464 ∴ Anand has to pay Rs 157464 after 3 years to clear the debt .$

Page No 151:

Answer:

$Principal amount, P = Rs . 11000 Rate of interest, R = 10 % p . a . Time, n = 3 years The amount including the compound interest is calculated using the formula, A = Rs . P {(1 + \frac{R}{100})}^{n} = Rs . 11000 {(1 + \frac{10}{100})}^{3} = Rs . 11000 {(\frac{100 + 10}{100})}^{3} = Rs . 11000 {(\frac{110}{100})}^{3} = Rs . 11000 {(1.1)}^{3} = Rs . 11000 (1.1 \times 1.1 \times 1.1) = Rs . 14641 Therefore, Beeru has to pay Rs 14641 to clear the debt .$

Page No 151:

Answer:

$Principal amount, P = Rs . 18000 Rate of interest for the first year, p = 12 % p . a . Rate of interest for the second year, q = 12 \frac{1}{2} % p . a . Time, n = 2 years The formula for t h e amount including the compound interest for the first year is g i v e n b e l o w : A = \{P \times (1 + \frac{p}{100}) \times (1 + \frac{q}{100})\} = Rs . \{18000 \times (1 + \frac{12}{100}) \times (1 + \frac{25}{100 \times 2})\} = Rs . \{18000 \times (\frac{100 + 12}{100}) \times (1 + \frac{25}{200})\} = Rs . \{18000 \times (\frac{100 + 12}{100}) \times (1 + \frac{1}{8})\} = Rs . \{18000 \times (\frac{100 + 12}{100}) \times (\frac{8 + 1}{8})\} = Rs . \{18000 \times (\frac{112}{100}) \times (\frac{9}{8})\} = Rs . \{18000 \times (1.12) \times (1.125)\} = Rs . 22680 ∴ Shubhalaxmi has to pay Rs 22680 to the finance company after 2 years .$

Page No 151:

Answer:

$Principal amount, P = Rs . 24000 Rate of interest, R = 10 % p . a . Time, n = 2 years 3 months = 2 \frac{1}{4} years The formula for the amount including the compound interest is g i v e n b e l o w : A = P \times {(1 + \frac{R}{100})}^{n} \times (1 + \frac{\frac{1}{4} R}{100}) = Rs . 24000 \times {(1 + \frac{10}{100})}^{2} \times (1 + \frac{\frac{1}{4} \times 10}{100}) = Rs . 24000 \times {(\frac{100 + 10}{100})}^{2} \times (\frac{100 + 2.5}{100}) = Rs . 24000 \times {(\frac{110}{100})}^{2} \times (\frac{100 + 2.5}{100}) = Rs . 24000 \times (1.1 \times 1.1 \times 1.025) = Rs . 24000 \times (1.250) = Rs . 29766 Therefore, Neha should pay Rs 29766 to the bank after 2 years 3 months .$

Page No 151:

Answer:

$Principal amount, P = Rs 16000 Rate of interest, R = \frac{15}{2} % p . a . Time, n = 2 years Now, simple interest = Rs (\frac{16000 \times 2 \times 15}{100 \times 2}) = Rs . 2400 Amount including the simple interest = Rs (16000 + 2400) = Rs 18400 The formula for t h e amount including the compound interest is g i v e n b e l o w : A = P {(1 + \frac{R}{100})}^{n} = Rs . 16000 {(1 + \frac{15}{100 \times 2})}^{2} = Rs . 16000 {(1 + \frac{15}{200})}^{2} = Rs . 16000 {(1 + \frac{3}{40})}^{2} = Rs . 16000 {(\frac{40 + 3}{40})}^{2} = Rs . 16000 {(\frac{43}{40})}^{2} = Rs . 16000 (1.075 \times 1.075) i . e ., the amount including the compound interest is Rs 18490 . Now, (CI - SI) = Rs . (18490 - 18400) = Rs . 90 Therefore, Abhay g a i n s Rs . 90 as profit at the end of 2 years .$

Page No 151:

Answer:

$Simple interest (SI) = Rs . 2400 Rate of interest, R = 8 % Time, n = 2 years The principal can be calculated using the formula : Sum = (\frac{100 \times SI}{R \times T}) \Rightarrow Sum = Rs . (\frac{100 \times 2400}{8 \times 2}) = Rs . 15000 i . e ., the principal is Rs . 15000 . The amount including the compound interest is calculated using the formula g i v e n b e l o w : A = P {(1 + \frac{R}{100})}^{n} = Rs . 15000 {(1 + \frac{8}{100})}^{2} = Rs . 15000 {(\frac{100 + 8}{100})}^{2} = Rs . 15000 {(\frac{108}{100})}^{2} = Rs . 15000 (1.08 \times 1.08) = Rs . 17496 i . e ., the amount including the compound interest is Rs . 17496 . ∴ Compound interest (CI) = Rs . (17496 - 15000) = Rs . 2496$

Page No 151:

Answer:

$Let Rs P b e t h e sum . Then SI = (\frac{P \times 2 \times 6}{100}) = Rs . \frac{12 P}{100} = Rs . \frac{3 P}{25} Also, CI = \{P \times {(1 + \frac{6}{100})}^{2} - P\} = Rs . \{P \times {(\frac{100 + 6}{100})}^{2} - P\} = Rs . \{P \times {(\frac{53}{50})}^{2} - P\} = Rs . \{(\frac{2809 P}{2500}) - P\} = Rs . \{\frac{2809 P - 2500 P}{2500}\} = Rs . \frac{309 P}{2500} N o w, (CI - SI) = Rs . (\frac{309 P}{2500} - \frac{3 P}{25}) = Rs . (\frac{309 P - 300 P}{2500}) = Rs . \frac{9 P}{2500} Now, Rs . 90 = \frac{9 P}{2500} \Rightarrow P = (\frac{90 \times 2500}{9}) = Rs . 25000 Hence, the r e q u i r e d sum is Rs . 25000 .$

Page No 151:

Answer:

$Let P b e the sum . Then SI = Rs (\frac{P \times 3 \times 10}{100}) = Rs \frac{30 P}{100} = Rs \frac{3 P}{10} A l s o, CI = Rs . \{P \times {(1 + \frac{10}{100})}^{3} - P\} = Rs . \{P \times {(\frac{100 + 10}{100})}^{3} - P\} = Rs . \{P \times {(\frac{11}{10})}^{3} - P\} = Rs . \{(\frac{1331 P}{1000}) - P\} = Rs . \{\frac{1331 P - 1000 P}{1000}\} = Rs . \frac{331 P}{1000} Now, (CI - SI) = Rs (\frac{331 P}{1000} - \frac{3 P}{10}) = Rs (\frac{331 P - 300 P}{1000}) = Rs \frac{31 P}{1000} Now, Rs . 93 = \frac{31 P}{1000} \Rightarrow P = (\frac{93 \times 1000}{31}) = Rs . 3000 Hence, the r e q u i r e d sum is Rs . 3000 .$

Page No 152:

Answer:

$Let P b e the sum . Rate of interest, R = 6 \frac{2}{3} % = \frac{20}{3} % Time, n = 2 years N o w, A = P \times {(1 + \frac{20}{100 \times 3})}^{2} = Rs . P \times {(1 + \frac{20}{300})}^{2} = Rs . P \times {(\frac{300 + 20}{300})}^{2} = Rs . P \times {(\frac{320}{300})}^{2} = Rs . P \times (\frac{16}{15} \times \frac{16}{15}) = Rs . \frac{256 P}{225} \Rightarrow Rs . 10240 = Rs . \frac{256 P}{225} \Rightarrow Rs . (\frac{10240 \times 225}{256}) = P ∴ P = Rs . 9000 Hence, the r e q u i r e d sum is Rs . 9000$

Page No 152:

Answer:

$Let P b e the sum . Rate of interest, R = 10 % Time, n = 3 years Now, A = P \times {(1 + \frac{10}{100})}^{3} = Rs . P \times {(\frac{100 + 10}{100})}^{3} = Rs . P \times {(\frac{110}{100})}^{3} = Rs . P \times (\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}) = Rs . \frac{1331 P}{1000} However, amount = Rs . 21296 Now, Rs . 21296 = Rs . \frac{1331 P}{1000} \Rightarrow Rs . (\frac{21296 \times 1000}{1331}) = P ∴ P = Rs . 16000 Hence, the r e q u i r e d sum is Rs . 16000 .$

Page No 152:

Answer:

$Let R % p . a . be the required rate . A = 4410 P = 4000 n = 2 years N o w, A = P {(1 + \frac{R}{100})}^{n} \Rightarrow 4410 = 4000 {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{4410}{4000} = {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{441}{400} = {(1 + \frac{R}{100})}^{2} \Rightarrow {(\frac{21}{20})}^{2} = {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{21}{20} - 1 = \frac{R}{100} \Rightarrow \frac{21 - 20}{20} = \frac{R}{100} \Rightarrow \frac{1}{20} = \frac{R}{100} \Rightarrow R = (\frac{1 \times 100}{20}) = 5 Hence, the required rate is 5 % p . a .$

Page No 152:

Answer:

$Let the required rate be R % p . a . A = 774.40 P = 640 n = 2 years N o w, A = P {(1 + \frac{R}{100})}^{n} \Rightarrow 774.40 = 640 {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{774.40}{640} = {(1 + \frac{R}{100})}^{2} \Rightarrow 1.21 = {(1 + \frac{R}{100})}^{2} \Rightarrow {(1.1)}^{2} = {(1 + \frac{R}{100})}^{2} \Rightarrow 1.1 - 1 = \frac{R}{100} \Rightarrow 0.1 = \frac{R}{100} \Rightarrow R = (0.1 \times 100) = 10 Hence, the required rate is 10 % p . a .$

Page No 152:

Answer:

$Let the required time be n years . Rate of interest, R = 10 % Principal amount, P = Rs . 1800 Amount with compound interest, A = Rs . 2178 Now, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 1800 \times {(1 + \frac{10}{100})}^{n} = Rs . 1800 \times {(\frac{100 + 10}{100})}^{n} = Rs . 1800 \times {(\frac{110}{100})}^{n} = Rs . 1800 \times {(\frac{11}{10})}^{n} However, amount = Rs . 2178 N o w, Rs . 2178 = Rs . 1800 \times {(\frac{11}{10})}^{n} \Rightarrow \frac{2178}{1800} = {(\frac{11}{10})}^{n} \Rightarrow \frac{121}{100} = {(\frac{11}{10})}^{n} \Rightarrow {(\frac{11}{10})}^{2} = {(\frac{11}{10})}^{n} \Rightarrow n = 2 ∴ Time, n = 2 years$

Page No 152:

Answer:

$Let the required time be n years . Rate of interest, R = 8 % Principal amount, P = Rs . 6250 Amount with compound interest, A = Rs . 7290 Then, A = P \times {(1 + \frac{R}{100})}^{n} \Rightarrow A = Rs . 6250 \times {(1 + \frac{8}{100})}^{n} = Rs . 6250 \times {(\frac{100 + 8}{100})}^{n} = Rs . 6250 \times {(\frac{108}{100})}^{n} = Rs . 6250 \times {(\frac{27}{25})}^{n} However, amount = Rs . 7290 N o w, Rs . 7290 = Rs . 6250 \times {(\frac{27}{25})}^{n} \Rightarrow \frac{7290}{6250} = {(\frac{27}{25})}^{n} \Rightarrow \frac{729}{625} = {(\frac{27}{25})}^{n} \Rightarrow {(\frac{27}{25})}^{2} = {(\frac{27}{25})}^{n} \Rightarrow n = 2 ∴ Time, n = 2 years$

Page No 152:

Answer:

$Population of the town, P = 125000 Rate of increase, R = 2 % Time, n = 3 years Then the population of the town after 3 years is g i v e n b y Population = P \times {(1 + \frac{R}{100})}^{3} = 125000 \times {(1 + \frac{2}{100})}^{3} = 125000 \times {(\frac{100 + 2}{100})}^{3} = 125000 \times {(\frac{102}{100})}^{3} = 125000 \times {(\frac{51}{50})}^{3} = 125000 \times (\frac{51}{50}) \times (\frac{51}{50}) \times (\frac{51}{50}) = (51 \times 51 \times 51) = 132651 Therefore, the population of the town after three years is 132651 .$

Page No 152:

Answer:

$Let the population of the town be 50000 . Rate of increase for the first year, p = 5 % Rate of increase for the second year, q = 4 % Rate of increase for the third year, r = 3 % Time = 3 years Now, present population = \{P \times (1 + \frac{p}{100}) \times (1 + \frac{q}{100}) \times (1 + \frac{r}{100})\} = \{50000 \times (1 + \frac{5}{100}) \times (1 + \frac{4}{100}) \times (1 + \frac{3}{100})\} = \{50000 \times (\frac{100 + 5}{100}) \times (\frac{100 + 4}{100}) \times (\frac{100 + 3}{100})\} = \{50000 \times (\frac{105}{100}) \times (\frac{104}{100}) \times (\frac{103}{100})\} = \{50000 \times (\frac{21}{20}) \times (\frac{26}{25}) \times (\frac{103}{100})\} = (21 \times 26 \times 103) = 56238 Therefore, the present population of the town is 56238 .$

Page No 152:

Answer:

$Population of the city in 2009, P = 120000 Rate of increase, R = 6 % Time, n = 3 years Then the population of the city in the year 2010 is g i v e n b y Population = P \times {(1 + \frac{R}{100})}^{n} = 120000 \times {(1 + \frac{6}{100})}^{1} = 120000 \times (\frac{100 + 6}{100}) = 120000 \times (\frac{106}{100}) = 120000 \times (\frac{53}{50}) = 2400 \times 53 = 127200 Therefore, the population of the city in 2010 is 127200 . A g a i n, p opulation of the city in 2010, P = 127200 Rate of decrease, R = 5 % Then the population of the city in the year 2011 is g i v e n b y Population = P \times {(1 - \frac{R}{100})}^{n} = 127200 \times {(1 - \frac{5}{100})}^{1} = 127200 \times (\frac{100 - 5}{100}) = 127200 \times (\frac{95}{100}) = 127200 \times (\frac{19}{20}) = 6360 \times 19 = 120840 Therefore, the population of the city in 2011 is 120840 .$

Page No 152:

Answer:

$Initial count of bacteria, P = 500000 Rate of increase, R = 2 % Time, n = 2 hours Then the count of bacteria at the end of 2 hours is g i v e n b y Count of bacteria = P \times {(1 + \frac{R}{100})}^{n} = 500000 \times {(1 + \frac{2}{100})}^{2} = 500000 \times {(\frac{100 + 2}{100})}^{2} = 500000 \times {(\frac{102}{100})}^{2} = 500000 \times {(\frac{51}{50})}^{2} = 500000 \times (\frac{51}{50}) \times (\frac{51}{50}) = (200 \times 51 \times 51) = 520200 Therefore, the count of bacteria at the end of 2 hours is 520200 .$

Page No 152:

Answer:

$Initial count of bacteria, P = 20000 Rate of increase, R = 10 % Time, n = 3 hours Then the count of bacteria at the end of the first hour is given by Count of bacteria = P \times {(1 + \frac{10}{100})}^{n} = 20000 \times {(1 + \frac{10}{100})}^{1} = 20000 \times (\frac{100 + 10}{100}) = 20000 \times (\frac{110}{100}) = 20000 \times (\frac{11}{10}) = 2000 \times 11 = 22000 Therefore, the count of bacteria at the end of the first hour is 22000 . The count of bacteria at the end of the second hour is given by Count of bacteria = P \times {(1 - \frac{10}{100})}^{n} = 22000 \times {(1 - \frac{10}{100})}^{1} = 22000 \times (\frac{100 - 10}{100}) = 22000 \times (\frac{90}{100}) = 22000 \times (\frac{9}{10}) = 2200 \times 9 = 19800 Therefore, the count of bacteria at the end of the second hour is 19800 . Then the count of bacteria at the end of t h e third hour is i s g i v e n b y Count of bacteria = P \times {(1 + \frac{10}{100})}^{n} = 19800 \times {(1 + \frac{10}{100})}^{1} = 19800 \times (\frac{100 + 10}{100}) = 19800 \times (\frac{110}{100}) = 19800 \times (\frac{11}{10}) = 1980 \times 11 = 21780 Therefore, the count of bacteria at the end of t h e first 3 hours is 21780 .$

Page No 152:

Answer:

$Initial value of the machine, P = Rs 625000 Rate of depreciation, R = 8 % Time, n = 2 years Then the value of the machine after two years is given by Value = P \times {(1 - \frac{R}{100})}^{n} = Rs 625000 \times {(1 - \frac{8}{100})}^{2} = Rs 625000 \times {(\frac{100 - 8}{100})}^{2} = Rs 625000 \times {(\frac{92}{100})}^{2} = Rs 625000 \times {(\frac{23}{25})}^{2} = Rs 625000 \times (\frac{23}{25}) \times (\frac{23}{25}) = Rs (1000 \times 23 \times 23) = Rs 529000 Therefore, the value of the machine after two years will be Rs . 529000 .$

Page No 152:

Answer:

$Initial value of the scooter, P = Rs 56000 Rate of depreciation, R = 10 % Time, n = 3 years Then the value of the scooter after three years is given by Value = P \times {(1 - \frac{R}{100})}^{n} = Rs . 56000 \times {(1 - \frac{10}{100})}^{3} = Rs . 56000 \times {(\frac{100 - 10}{100})}^{3} = Rs . 56000 \times {(\frac{90}{100})}^{3} = Rs . 56000 \times {(\frac{9}{10})}^{3} = Rs . 56000 \times (\frac{9}{10}) \times (\frac{9}{10}) \times (\frac{9}{10}) = Rs . (56 \times 9 \times 9 \times 9) = Rs . 40824 Therefore, the value of the scooter after three years will be Rs . 40824 .$

Page No 152:

Answer:

$Initial value of the car, P = Rs 348000 Rate of depreciation for the first year, p = 10 % Rate of depreciation for the second year, q = 20 % Time, n = 2 years . Then the value of the car after two years is given by Value = \{P \times (1 - \frac{p}{100}) \times (1 - \frac{q}{100})\} = Rs . \{348000 \times (1 - \frac{10}{100}) \times (1 - \frac{20}{100})\} = Rs . \{348000 \times (\frac{100 - 10}{100}) \times (\frac{100 - 20}{100})\} = Rs . \{348000 \times (\frac{90}{100}) \times (\frac{80}{100})\} = Rs . \{348000 \times (\frac{9}{10}) \times (\frac{8}{10})\} = Rs . (3480 \times 9 \times 8) = Rs . 250560 ∴ The value of the car after two years is Rs 250560 .$

Page No 152:

Answer:

$Let the initial value of the machine, P be Rs x . Rate of depreciation, R = 10 % Time, n = 3 years The present value of the machine is Rs 291600 . Then the initial value of the machine is given by Value = P \times {(1 - \frac{R}{100})}^{n} = Rs . x \times {(1 - \frac{10}{100})}^{3} = Rs . x \times {(\frac{100 - 10}{100})}^{3} = Rs . x \times {(\frac{90}{100})}^{3} = Rs . x \times {(\frac{9}{10})}^{3} ∴ Present value of the machine = Rs 291600 Now, Rs 291600 = Rs x \times (\frac{9}{10}) \times (\frac{9}{10}) \times (\frac{9}{10}) \Rightarrow x = Rs \frac{291600 \times 10 \times 10 \times 10}{9 \times 9 \times 9} \Rightarrow x = Rs \frac{291600000}{729} \Rightarrow x = Rs 400000 ∴ The initial value of the machine is Rs 400000 .$

Page No 154:

Answer:

$Principal, P = Rs . 8000 Time, n = 1 year = 2 half years Rate of interest per annum = 10 % Rate of interest for half year, R = \frac{10 %}{2} = 5 % The amount with the compound interest is given by Amount = Rs . P \times {(1 + \frac{R}{100})}^{2} = Rs . 8000 \times {(1 + \frac{5}{100})}^{2} = Rs . 8000 \times {(\frac{105}{100})}^{2} = Rs . 8000 \times {(\frac{21}{20})}^{2} = Rs . 8000 \times (\frac{21}{20}) \times (\frac{21}{20}) = Rs . (20 \times 21 \times 21) = Rs . 8820 ∴ Compound interest = amount - principal = Rs . (8820 - 8000) = Rs . 820$

Page No 154:

Answer:

$Principal, P = Rs . 31250 Annual rate of interest, R = 8 % Rate of interest for a half year = \frac{1}{2} \times 8 % = 4 % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 31250 \times {(1 + \frac{4}{100})}^{3} = 31250 \times {(1 + \frac{1}{25})}^{3} = 31250 \times {(\frac{25 + 1}{25})}^{3} = 31250 \times {(\frac{26}{25})}^{3} = 31250 \times (\frac{26}{25}) \times (\frac{26}{25}) \times (\frac{26}{25}) = Rs 2 \times 17576 = Rs 35152 Therefore, compound interest = amount - principal = Rs (35152 - 31250) = Rs 3902$

Page No 154:

Answer:

$Principal, P = Rs . 12800 Annual rate of interest, R = \frac{15}{2} % Rate of interest for a half year = \frac{1}{2} (\frac{15}{2}) % = \frac{15}{4} % Time, n = 1 year = 2 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 12800 \times {(1 + \frac{\frac{15}{4}}{100})}^{2} = 12800 \times {(1 + \frac{15}{100 \times 4})}^{2} = 12800 \times {(\frac{400 + 15}{400})}^{2} = 12800 \times {(\frac{415}{400})}^{2} = 12800 \times (\frac{83}{80}) \times (\frac{83}{80}) = (2 \times 83 \times 83) = Rs 13778 Therefore, compound interest = amount - principal = Rs (13778 - 12800) = Rs 978$

Page No 154:

Answer:

$Principal, P = Rs . 160000 Annual rate of interest, R = 10 % Rate of interest for a half year = \frac{10}{2} % = 5 % Time, n = 2 years = 4 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 160000 \times {(1 + \frac{5}{100})}^{4} = 160000 \times {(\frac{100 + 5}{100})}^{4} = 160000 \times {(\frac{105}{100})}^{4} = 160000 \times (\frac{21}{20}) \times (\frac{21}{20}) \times (\frac{21}{20}) \times (\frac{21}{20}) = (21 \times 21 \times 21 \times 21) = Rs 194481 Therefore, compound interest = amount - principal = Rs (194481 - 160000) = Rs 34481$

Page No 154:

Answer:

$Principal, P = Rs . 40960 Annual rate of interest, R = \frac{25}{2} % Rate of interest for half year = \frac{25}{4} % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 40960 \times {(1 + \frac{25}{100 \times 4})}^{3} = 40960 \times {(\frac{400 + 25}{400})}^{3} = 40960 \times {(\frac{425}{400})}^{3} = 40960 \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) = (10 \times 17 \times 17 \times 17) = Rs 49130 Therefore, compound interest = amount - principal = Rs (49130 - 40960) = Rs 8170 Therefore, Swati has to pay Rs . 49130, which includes an interest of Rs . 8170, to the bank after 1 \frac{1}{2} years .$

Page No 154:

Answer:

$Let the principal amount be P = Rs . 125000 . Annual rate of interest, R = 12 % Rate of interest for a half year = 6 % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 125000 \times {(1 + \frac{6}{100})}^{3} = Rs . 125000 \times {(\frac{100 + 6}{100})}^{3} = Rs . 125000 \times {(\frac{106}{100})}^{3} = Rs . 125000 \times (\frac{53}{50}) \times (\frac{53}{50}) \times (\frac{53}{50}) = Rs . (53 \times 53 \times 53) = Rs . 148877 Now, C I = A - P = Rs . (148877 - 125000) = Rs . 23877 Therefore, Aslam has to pay an interest of Rs . 23877 to the bank after 1 \frac{1}{2} years .$

Page No 154:

Answer:

$Let the principal amount be P = Rs . 20000 . Annual rate of interest, R = 6 % Rate of interest for half year = 3 % Time, n = 1 year = 2 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 20000 \times {(1 + \frac{3}{100})}^{2} = Rs . 20000 \times {(\frac{100 + 3}{100})}^{2} = Rs . 20000 \times {(\frac{103}{100})}^{2} = Rs . 20000 \times (\frac{103}{100}) \times (\frac{103}{100}) = Rs . (2 \times 103 \times 103) = Rs . 21218 Therefore, Sheela gets Rs . 21218 after 1 year .$

Page No 154:

Answer:

$Let the principal amount be P = Rs . 65536 . Annual rate of interest, R = \frac{25}{2} % Rate of interest for a half year = \frac{25}{4} % Time, n = 2 years = 4 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 65536 \times {(1 + \frac{25}{100 \times 4})}^{4} = Rs . 65536 \times {(\frac{400 + 25}{400})}^{4} = Rs . 65536 \times {(\frac{425}{400})}^{4} = Rs . 65536 \times {(\frac{17}{16})}^{4} = Rs . 65536 \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16}) = Rs . (17 \times 17 \times 17 \times 17) = Rs . 83521 Now, CI = A - P = Rs . (83521 - 65536) = Rs . 17985 Therefore, interest earned when compounded half yearly = Rs . 17985 Amount when the interest is compounded yearly is given by A = P \times {(1 + \frac{R}{100})}^{n} = R s . 65536 \times {(1 + \frac{25}{100 \times 2})}^{2} = Rs . 65536 \times {(\frac{200 + 25}{200})}^{2} = Rs . 65536 \times {(\frac{225}{200})}^{2} = Rs . 65536 \times {(\frac{9}{8})}^{2} = Rs . 65536 \times (\frac{9}{8}) \times (\frac{9}{8}) = Rs . 82944 Therefore, CI = A - P = Rs . (82944 - 65536) = Rs . 17408 ∴ Difference between the interests compounded half yearly and yearly = Rs . (17985 - 17408) = Rs . 577$

Page No 154:

Answer:

$Let the principal amount be P = Rs 32000 . Annual rate of interest, R = 5 % Rate of interest for a quarter year = \frac{5}{4} % Time, n = 6 months = 2 quarter years Then the amount with the compound interest is given by A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . 32000 \times {(1 + \frac{5}{100 \times 4})}^{2} = Rs . 32000 \times {(\frac{400 + 5}{400})}^{2} = Rs . 32000 \times {(\frac{405}{400})}^{2} = Rs . 32000 \times {(\frac{81}{80})}^{2} = Rs . 32000 \times (\frac{81}{80}) \times (\frac{81}{80}) = Rs . (5 \times 81 \times 81) = Rs . 32805 Therefore, Sudershan will receive an amount of Rs . 32805 after 6 months .$

Page No 154:

Answer:

$Let the principal amount be P = Rs 390625 . Annual rate of interest, R = 16 % Rate of interest for a quarter year = \frac{16}{4} % = 4 % Time, n = 1 year = 4 quarter years Then the amount with the compound interest is given by A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . 390625 \times {(1 + \frac{4}{100})}^{4} = Rs . 390625 \times {(\frac{100 + 4}{100})}^{4} = Rs . 390625 \times {(\frac{104}{100})}^{4} = Rs . 390625 \times {(\frac{26}{25})}^{4} = Rs . 390625 \times (\frac{26}{25}) \times (\frac{26}{25}) \times (\frac{26}{25}) \times (\frac{26}{25}) = Rs . (26 \times 26 \times 26 \times 26) = Rs . 456976 Therefore, Arun has to pay Rs 456976 after 1 year .$

Page No 154:

Answer:

(c) Rs. 832

$A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 5000 \times {(1 + \frac{8}{100})}^{2} = Rs . 5000 \times {(\frac{108}{100})}^{2} = Rs . 5000 \times {(\frac{27}{25})}^{2} = Rs . 5000 \times (\frac{27}{25}) \times (\frac{27}{25}) = Rs . (8 \times 27 \times 27) = Rs . 5832 ∴ Interest = amount - principal = Rs (5832 - 5000) = Rs 832$

Page No 154:

Answer:

(b) Rs. 3310

$A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 10000 \times {(1 + \frac{10}{100})}^{3} = Rs . 10000 \times {(\frac{110}{100})}^{3} = Rs . 10000 \times {(\frac{11}{10})}^{3} = Rs . 10000 \times (\frac{11}{10}) \times (\frac{11}{10}) \times (\frac{11}{10}) = Rs . (10 \times 11 \times 11 \times 11) = Rs . 13310 ∴ Compound interest = amount - principal = Rs (13310 - 10000) = Rs 3310$

Page No 154:

Answer:

(a) Rs 1872

$Here, A = P \times {(1 + \frac{R}{100})}^{1} \times (1 + \frac{\frac{1}{2} R}{100}) = Rs 10000 \times (1 + \frac{12}{100}) \times (1 + \frac{\frac{1}{2} \times 12}{100}) = Rs 10000 \times (\frac{100 + 12}{100}) \times (\frac{100 + 6}{100}) = Rs 10000 \times (\frac{112}{100}) \times (\frac{106}{100}) = Rs 10000 \times (\frac{28}{25}) \times (\frac{53}{50}) = Rs (8 \times 28 \times 53) = Rs 11872 ∴ Compound interest = amount - principal = Rs (11872 - 10000) = Rs 1872$

Page No 155:

Answer:

(c) Rs 961

$Here, A = P \times {(1 + \frac{R}{100})}^{2} \times (1 + \frac{\frac{1}{4} R}{100}) = Rs . 4000 \times {(1 + \frac{10}{100})}^{2} \times (1 + \frac{\frac{1}{4} \times 10}{100}) = Rs . 4000 \times {(\frac{100 + 10}{100})}^{2} \times (\frac{40 + 1}{40}) = Rs . 4000 \times {(\frac{110}{100})}^{2} \times (\frac{41}{40}) = Rs . 4000 \times (\frac{11}{10}) \times (\frac{11}{10}) \times (\frac{41}{40}) = Rs . (11 \times 11 \times 41) = Rs . 4961 ∴ Compound interest = amount - principal = Rs (4961 - 4000) = Rs 961$

Page No 155:

Answer:

(b) Rs. 5051

$Here, A = Rs . P \times (1 + \frac{p}{100}) \times (1 + \frac{q}{100}) \times (1 + \frac{r}{100}) = Rs . 25000 \times (1 + \frac{5}{100}) \times (1 + \frac{6}{100}) \times (1 + \frac{8}{100}) = Rs . 25000 \times (\frac{105}{100}) \times (\frac{106}{100}) \times (\frac{108}{100}) = Rs . 25000 \times (\frac{21}{20}) \times (\frac{53}{50}) \times (\frac{27}{25}) = Rs . (21 \times 53 \times 27) = Rs . 30051 ∴ Compound interest = amount - principal = Rs . (30051 - 25000) = Rs . 5051$

Page No 155:

Answer:

(b) Rs. 510

$Rate of interest compounded half yearly = \frac{8}{2} % = 4 % Time = 1 year = 2 half years Now, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 6250 \times {(1 + \frac{4}{100})}^{2} = Rs . 6250 \times {(\frac{104}{100})}^{2} = Rs . 6250 \times (\frac{26}{25}) \times (\frac{26}{25}) = Rs . (10 \times 26 \times 26) = Rs . 6760 ∴ Compound interest = amount - principal = Rs . (6760 - 6250) = Rs . 510$

Page No 155:

Answer:

(a) Rs.1209

$Time = 6 months = 2 quater years Rate compounded quarter yearly = \frac{6}{4} % = \frac{3}{2} % Now, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 40000 \times {(1 + \frac{3}{100 \times 2})}^{2} = Rs . 40000 \times {(\frac{203}{200})}^{2} = Rs . 40000 \times (\frac{203}{200}) \times (\frac{203}{200}) = Rs . (203 \times 203) = Rs . 41209 ∴ Compound interest = amount - principal = Rs . 41209 - Rs . 40000 = Rs . 1209$

Page No 155:

Answer:

(b) 26460

$Here, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 24000 \times {(1 + \frac{5}{100})}^{2} = Rs . 24000 \times {(\frac{105}{100})}^{2} = Rs . 24000 \times (\frac{21}{20}) \times (\frac{21}{20}) = Rs . (60 \times 21 \times 21) = Rs . 26460$

Page No 155:

Answer:

(c) Rs. 43740

$Here, A = Rs . P \times {(1 - \frac{R}{100})}^{n} = Rs . 60000 \times {(1 - \frac{10}{100})}^{3} = Rs . 60000 \times {(\frac{90}{100})}^{3} = Rs . 60000 \times (\frac{9}{10}) \times (\frac{9}{10}) \times (\frac{9}{10}) = Rs . (60 \times 9 \times 9 \times 9) = Rs . 43740$

Page No 155:

Answer:

(b) Rs. 62500

$Here, A = P \times {(1 - \frac{R}{100})}^{n} = P \times {(1 - \frac{20}{100})}^{2} = P \times {(\frac{80}{100})}^{2} = P \times (\frac{4}{5}) \times (\frac{4}{5}) \Rightarrow 40000 = \frac{16 P}{25} ∴ P = \frac{40000 \times 25}{16} = Rs 62500$

Page No 155:

Answer:

(a) 25000

$Let P be the popultion 3 years ago . Now, present population = 33275 \Rightarrow 33275 = P \times {(1 + \frac{10}{100})}^{3} \Rightarrow 33275 = P \times {(\frac{110}{100})}^{3} \Rightarrow 33275 = P \times (\frac{11}{10}) \times (\frac{11}{10}) \times (\frac{11}{10}) \Rightarrow 33275 = \frac{1331 P}{1000} ∴ P = \frac{33275 \times 1000}{1331} = 25000$

Page No 155:

Answer:

(d) Rs 1261

$Here, SI = \frac{P \times 5 \times 3}{100} \Rightarrow 1200 = \frac{P \times 5 \times 3}{100} \Rightarrow P = \frac{1200 \times 100}{5 \times 3} = Rs 8000 Amount at the end of 3 years = Rs 8000 \times {(1 + \frac{5}{100})}^{3} = Rs 8000 \times {(\frac{105}{100})}^{3} = Rs 8000 \times (\frac{21}{20}) \times (\frac{21}{20}) \times (\frac{21}{20}) = Rs (21 \times 21 \times 21) = Rs 9261 ∴ CI = A - P = Rs (9261 - 8000) = Rs 1261$

Page No 155:

Answer:

(d) Rs 480

$We have : 510 = \{P \times {(1 + \frac{25}{100 \times 2})}^{2}\} - P \Rightarrow 510 = \Rightarrow \{P \times {(\frac{8 + 1}{8})}^{2}\} - P \Rightarrow 510 = \{P \times (\frac{9}{8}) \times (\frac{9}{8})\} - P \Rightarrow 510 = (\frac{81 P}{64} - P) \Rightarrow 510 = (\frac{81 P - 64 P}{64}) \Rightarrow 510 = \frac{17 P}{64} ∴ P = \frac{510 \times 64}{17} = Rs 1920 Now, SI = \frac{P \times R \times T}{100} = Rs \frac{1920 \times 2 \times 25}{100 \times 2} = Rs 480$

Page No 155:

Answer:

(d) Rs 4096

$We have Rs 4913 = \{P \times {(1 + \frac{25}{100 \times 4})}^{3}\} \Rightarrow Rs 4913 = \{P \times {(\frac{16 + 1}{16})}^{3}\} \Rightarrow Rs 4913 = \{P \times (\frac{17}{16}) \times (\frac{17}{16}) \times (\frac{17}{16})\} \Rightarrow Rs 4913 = \frac{4913 P}{4096} \Rightarrow P = Rs \frac{4913 \times 4096}{4913} = Rs 4096$

Page No 155:

Answer:

(c) 6%

$Here, A = P \times (1 + \frac{R}{100}) = Rs . 7500 \times {(1 + \frac{R}{100})}^{2} = Rs . 7500 \times {(1 + \frac{R}{100})}^{2} However, amount = Rs . 8427 Now, Rs . 8427 = = Rs . 7500 \times {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{Rs . 8427}{Rs . 7500} = {(1 + \frac{R}{100})}^{2} \Rightarrow {(\frac{53}{50})}^{2} = {(1 + \frac{R}{100})}^{2} \Rightarrow (1 + \frac{R}{100}) = (\frac{53}{50}) \Rightarrow \frac{R}{100} = \frac{53}{50} - 1 \Rightarrow \frac{R}{100} = \frac{53 - 50}{50} = \frac{3}{50} ∴ R = \frac{300}{50} = 6 %$

Page No 157:

Answer:

$Here, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 3000 \times {(1 + \frac{10}{100})}^{2} = Rs . 3000 \times {(\frac{110}{100})}^{2} = Rs . 3000 \times (\frac{11}{10}) \times (\frac{11}{10}) = Rs . (30 \times 11 \times 11) = Rs . 3630 ∴ C I = A - P = Rs . (3630 - 3000) = Rs . 630 Hence, the amount is Rs . 3630 and the CI is Rs . 630 .$

Page No 157:

Answer:

$Here, A = P \times (1 + \frac{p}{100}) \times (1 + \frac{r}{100}) = Rs . 10000 \times (1 + \frac{10}{100}) \times (1 + \frac{12}{100}) = Rs . 10000 \times (\frac{110}{100}) \times (\frac{112}{100}) = Rs . 10000 \times (\frac{11}{10}) \times (\frac{28}{25}) = Rs . (40 \times 11 \times 28) = Rs . 12320 ∴ CI = A - P = Rs (12320 - 10000) = Rs . 2320 Hence, the amount is Rs 12320 and the CI is Rs 2320 .$

Page No 157:

Answer:

$Let the principal amount be P = Rs 6000 . Rate of interest = 10 % p . a . = 5 % for half yearly Time (n) = 1 year = 2 half years Now, A = P \times {(1 + \frac{R}{100})}^{n} = Rs 6000 \times {(1 + \frac{5}{100})}^{2} = Rs 6000 \times {(\frac{105}{100})}^{2} = Rs 6000 \times (\frac{21}{20}) \times (\frac{21}{20}) = Rs (15 \times 21 \times 21) = Rs 6615 ∴ CI = A - P = Rs (6615 - 6000) = Rs 615 Hence, the amount is Rs 6615 and the CI is Rs 615 .$

Page No 157:

Answer:

$Amount (A) = Rs 23762 Rate of interest (R) = 9 % Time (n) = 2 years Now, A = P \times {(1 + \frac{R}{100})}^{n} \Rightarrow Rs 23762 = P \times {(1 + \frac{9}{100})}^{2} \Rightarrow Rs 23762 = P \times (\frac{109}{100}) \times (\frac{109}{100}) \Rightarrow P = Rs \frac{23762 \times 100 \times 100}{109 \times 109} \Rightarrow P = Rs 20000 ∴ The principal amount is Rs 20000 .$

Page No 157:

Answer:

$Let the principal amount be P = Rs 32000 . Rate of interest (R) = 10 % Time (n) = 2 years Now, A = Rs . P \times {(1 - \frac{R}{100})}^{n} = Rs . 32000 \times {(1 - \frac{10}{100})}^{2} = Rs . 32000 \times {(\frac{90}{100})}^{2} = Rs . 32000 \times (\frac{9}{10}) \times (\frac{9}{10}) = Rs . (320 \times 9 \times 9) = Rs . 25920 ∴ The value of the scooter after 2 years is Rs 25920 .$

Page No 157:

Answer:

(b) Rs. 1050

$P = Rs . 5000 R = 10 % n = 2 years Now, A = Rs . P \times {(1 + \frac{R}{100})}^{n} = Rs . 5000 \times {(1 + \frac{10}{100})}^{2} = Rs . 5000 \times {(\frac{110}{100})}^{2} = Rs . 5000 \times (\frac{11}{10}) \times (\frac{11}{10}) = Rs . (50 \times 11 \times 11) = Rs . 6050 ∴ CI = A - P = Rs . (6050 - 5000) = Rs . 1050$

Page No 157:

Answer:

(c) 4410

Present population =4000
Rate of growth = 5%

To find the population of the town after 2 years, we have:

$A = P \times {(1 + \frac{R}{100})}^{n} = 4000 \times {(1 + \frac{5}{100})}^{2} = 4000 \times {(\frac{105}{100})}^{2} = 4000 \times (\frac{21}{20}) \times (\frac{21}{20}) = (10 \times 21 \times 21) = 4410$

Page No 157:

Answer:

(d) 8%

$Here, A = Rs . P \times {(1 + \frac{R}{100})}^{n} \Rightarrow Rs . 5832 = Rs . 5000 \times {(1 + \frac{R}{100})}^{2} \Rightarrow \frac{Rs . 5832}{Rs . 5000} = {(1 + \frac{R}{100})}^{2} \Rightarrow {(\frac{27}{25})}^{2} = {(1 + \frac{R}{100})}^{2} \Rightarrow 1 + \frac{R}{100} = \frac{27}{25} \Rightarrow \frac{R}{100} = \frac{27}{25} - 1 = \frac{27 - 25}{25} = \frac{2}{25} ∴ R = \frac{100 \times 2}{25} = 8 %$

Page No 157:

Answer:

(a) Rs. 1655

$Here, SI = \frac{P \times R \times T}{100} \Rightarrow Rs . 1500 = \frac{P \times 10 \times 3}{100} \Rightarrow P = \frac{1500 \times 100}{10 \times 3} = Rs . 5000 Now, A = P \times {(1 + \frac{R}{100})}^{n} = Rs . 5000 \times {(1 + \frac{10}{100})}^{3} = Rs . 5000 \times {(\frac{110}{100})}^{3} = Rs . 5000 \times (\frac{11}{10}) \times (\frac{11}{10}) \times (\frac{11}{10}) = Rs . (5 \times 11 \times 11 \times 11) = Rs . 6655 ∴ CI = A - P = Rs (6655 - 5000) = Rs 1655$

Page No 157:

Answer:

(c) Rs. 5000

$Here, A = P \times {(1 + \frac{R}{100})}^{n} = P \times {(1 + \frac{10}{100})}^{2} = P \times {(\frac{110}{100})}^{2} = P \times (\frac{11}{10}) \times (\frac{11}{10}) N o w, C I = A - P \Rightarrow Rs . 1050 = \frac{121 p}{100} - P = \frac{121 P - 100 P}{100} = \frac{21 P}{100} ∴ P = Rs \frac{1050 \times 100}{21} = Rs 5000$

Page No 157:

Answer:

$(i) A = P {(1 + \frac{R}{100})}^{n} (ii) Compound interest (iii) A = P {(1 - \frac{R}{100})}^{2}, where A is the value of the machine after 2 years (iv) A = P {(1 + \frac{R}{100})}^{5}, where A is the population of the town after 5 years$

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