Rs Aggarwal 2017 for Class 8 Math Chapter 11 - Compound Interest

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Rs Aggarwal 2017 Solutions for Class 8 Math 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 Math Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 144:

Question 1:

Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum, compounded annually.

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:

Question 2:

Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum, compounded annually.

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:

Question 3:

Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum.

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:

Question 4:

Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?

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:

Question 5:

Harpreet borrowed Rs 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years.

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:

Question 6:

Manoj deposited a sum of Rs 64000 in a post office for 3 years, compounded annually at $7 \frac{1}{2} %$ per annum. What amount will he get on maturity?

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:

Question 7:

Divakaran deposited a sum of Rs 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets.

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:

Question 8:

Michael borrowed Rs 16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after $1 \frac{1}{2}$ years?

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:

Question 1:

By using the formula, find the amount and compound interest on:
Rs 6000 for 2 years at 9% per annum compounded annually.

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:

Question 2:

By using the formula, find the amount and compound interest on:
Rs 10000 for 2 years at 11% per annum compounded annually.

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:

Question 3:

By using the formula, find the amount and compound interest on:
Rs 31250 for 3 years at 8% per annum compounded annually.

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:

Question 4:

By using the formula, find the amount and compound interest on:
Rs 10240 for 3 years at $12 \frac{1}{2} %$ per annum compounded annually.

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:

Question 5:

By using the formula, find the amount and compound interest on:
Rs 10240 for 3 years at $12 \frac{1}{2} %$ per annum compounded annually.

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:

Question 6:

By using the formula, find the amount and compound interest on:
Rs 9000 for 2 years 4 months at 10% per annum compounded annually.

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:

Question 7:

Find the amount of Rs 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.

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:

Question 8:

Anand obtained a loan of Rs 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount wil he have to pay after 3 years to clear the debt?

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:

Question 9:

Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually?

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:

Question 10:

Shubhalaxmi took a loan of Rs 18000 from surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and $12 \frac{1}{2} %$ per annum during the second year, how much will she have to pay after 2 years?

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:

Question 11:

Neha borrowed Rs 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

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:

Question 12:

Abhay borrowed Rs 16000 at $7 \frac{1}{2} %$ per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years?

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:

Question 13:

The simple interest on a sum of money for 2 years at 8% per annum is Rs 2400. What will be the compound interest on that sum at the same rate and for the same period?

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:

Question 14:

The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs 90. Find the sum.

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:

Question 15:

The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 93. Find the sum.

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:

Question 16:

A sum of money amounts to Rs 10240 in 2 years at $6 \frac{2}{3} %$ per annum, compounded annually. Find the sum.

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:

Question 17:

What sum of money will amount to Rs 21296 in 3 years at 10% per annum, compounded annually?

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:

Question 18:

At what rate per cent per annum will Rs 4000 amount to Rs 4410 in 2 years when compounded annually?

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:

Question 19:

At what rate per cent per annum will Rs 640 amount to Rs 774.40 in 2 years when compounded annually?

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:

Question 20:

In how many years will Rs 1800 amount to Rs 2178 at 10% per annum when compounded annually?

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:

Question 21:

In how many years will Rs 6250 amount to Rs 7290 at 8% per annum, compounded annually?

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:

Question 22:

The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years?

Answer:

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

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Question 23:

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population?

Answer:

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

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Question 24:

The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011?

Answer:

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

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Question 25:

The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.

Answer:

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

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Question 26:

The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.

Answer:

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

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Question 27:

A machine is purchased for Rs 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?

Answer:

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

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Question 28:

A scooter is bought at Rs 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years?

Answer:

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

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Question 29:

A car is purchased for Rs 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?

Answer:

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

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Question 30:

The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased?

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:

Question 1:

Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly.

Answer:

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

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Question 2:

Find the amount and the compound interest on Rs 31250 for $1 \frac{1}{2}$ years at 8% per annum, compounded half-yearly.

Answer:

$Principal, P = Rs . 31250 Annual rate of interest, R = \frac{3}{2} % Rate of interest for a half year = \frac{1}{2} (\frac{3}{2}) % = \frac{3}{4} % Time, n = 1 \frac{1}{2} years = 3 half years Then the amount with the compound interest is given by A = P \times {(1 + \frac{R}{100})}^{n} = 31250 \times {(1 + \frac{\frac{3}{4}}{100})}^{3} = 31250 \times {(1 + \frac{3}{100 \times 4})}^{3} = 31250 \times {(\frac{400 + 3}{400})}^{3} = 31250 \times {(\frac{403}{400})}^{3} = 31250 \times (\frac{403}{400}) \times (\frac{403}{400}) \times (\frac{403}{400}) = Rs 31250 \times (1.0075 \times 1.0075 \times 1.0075) = Rs 31958.41 Therefore, compound interest = amount - principal = Rs (31958.41 - 31250) = Rs 708.41$

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Question 3:

Find the amount and the compound interest on Rs 12800 for 1 year at $7 \frac{1}{2} %$ per annum, compounded semi-annually.

Answer:

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

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Question 4:

Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.

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:

Question 5:

Swati borrowed Rs 40960 from a bank to buy a piece of land. If the bank charges $12 \frac{1}{2} %$ per annum, compounded half-yearly, what amount will she have to pay after $1 \frac{1}{2}$ years? Also, find the interest paid by her.

Answer:

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

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Question 6:

Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.

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:

Question 7:

Sheela deposited Rs 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?

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:

Question 8:

Neeraj lent Rs 65536 for 2 years at $12 \frac{1}{2} %$ per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?

Answer:

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

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Question 9:

Sudershan deposited Rs 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?

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:

Question 10:

Arun took a loan of Rs 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?

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:

Question 1:

Tick (✓) the correct answer:
The compound interest on Rs 5000 at 8% per annum for 2 years, compounded annually, is
(a) Rs 800
(b) Rs 825
(c) Rs 832
(d) Rs 850

Answer:

(c) Rs. 832

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

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Question 2:

Tick (✓) the correct answer:
The compound interest on Rs 10000 at 10% per annum for 3 years, compounded annually, is
(a) Rs 1331
(b) Rs 3310
(c) Rs 3130
(d) Rs 13310

Answer:

(b) Rs. 3310

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

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Question 3:

Tick (✓) the correct answer:
The compound interest on Rs 10000 at 12% per annum for $1 \frac{1}{2}$ years, compounded annually, is
(a) Rs 1872
(b) Rs 1720
(c) Rs 1910.16
(d) Rs 1782

Answer:

(a) Rs 1872

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

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Question 4:

Tick (✓) the correct answer:
The compound interest on Rs 4000 at 10% per annum for 2 years 3 months, compounded annually, is
(a) Rs 916
(b) Rs 900
(c) Rs 961
(d) Rs 896

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:

Question 5:

Tick (✓) the correct answer:
A sum of Rs 25000 was given as loan on compound interest for 3 years compounded annually at 5% per annum during the first year, 6% per annum during the second year and 8% per annum during the third year. The compound interest is
(a) Rs 5035
(b) Rs 5051
(c) Rs 5072
(d) Rs 5150

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:

Question 6:

Tick (✓) the correct answer:
The compound interest on Rs 6250 at 8% per annum for 1 year, compounded half yearly, is
(a) Rs 500
(b) Rs 510
(c) Rs 550
(d) Rs 512.50

Answer:

(b) Rs. 510

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

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Question 7:

Tick (✓) the correct answer:
The compound interest on Rs 40000 at 6% per annum for 6 months, compounded quarterly, is
(a) Rs 1209
(b) Rs 1902
(c) Rs 1200
(d) Rs 1306

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:

Question 8:

Tick (✓) the correct answer:
The present population of a town is 24000. If it increases at the rate of 5% per annum, what will be its population after 2 years?
(a) 26400
(b) 26460
(c) 24460
(d) 26640

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:

Question 9:

Tick (✓) the correct answer:
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago for Rs 60000. What is the present value of the machine?
(a) Rs 53640
(b) Rs 51680
(c) Rs 43740
(d) Rs 43470

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:

Question 10:

Tick (✓) the correct answer:
The value of a machine depreciates at the rate of 20% per annum. It was purchased 2 years ago. If its present value is Rs 40000, for how much was it purchased?
(a) Rs 56000
(b) Rs 62500
(c) Rs 65200
(d) Rs 56500

Answer:

(b) Rs. 62500

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

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Question 11:

Tick (✓) the correct answer:
The annual rate of growth in population of a town is 10%. If its present population is 33275, what was it 3 years ago?
(a) 25000
(b) 27500
(c) 30000
(d) 26000

Answer:

(a) 25000

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

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Question 12:

Tick (✓) the correct answer:
If the simple interest on a sum of money at 5% per annum for 3 years is Rs 1200 then the compound interest on the same sum for the same period at the same rate will be
(a) Rs 1225
(b) Rs 1236
(c) Rs 1248
(d) Rs 1261

Answer:

(d) Rs 1261

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

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Question 13:

Tick (✓) the correct answer:
If the compound interest on a sum for 2 years at $12 \frac{1}{2} %$ per annum is Rs 510, the simple interest on the same sum at the same rate for the same period of time is
(a) Rs 400
(b) Rs 450
(c) Rs 460
(d) Rs 480

Answer:

(d) Rs 480

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

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Question 14:

Tick (✓) the correct answer:
The sum that amounts to Rs 4913 in 3 years at $6 \frac{1}{4} %$ per annum compounded annually, is
(a) Rs 3096
(b) Rs 4076
(c) Rs 4085
(d) Rs 4096

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:

Question 15:

Tick (✓) the correct answer:
At what rate per cent per annum will a sum of Rs 7500 amount to Rs 8427 in 2 years, compounded annually?
(a) 4%
(b) 5%
(c) 6%
(d) 8%

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:

Question 1:

Find the amount and the compound interest on Rs 3000 for 2 years at 10% per annum.

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:

Question 2:

Find the amount of Rs 10000 after 2 years compounded annually; the rate of interest being 10% per annum during the first year and 12% per annum during the second year. Also, find the compound interest.

Answer:

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

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Question 3:

Find the amount and the compound interest on Rs 6000 for 1 year at 10% per annum compounded half-yearly.

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:

Question 4:

A sum amounts to Rs 23762 in 2 years at 9% per annum, compounded annually. Find the sum.

Answer:

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

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Question 5:

A scooter is bought for Rs 32000. Its value depreciates at 10% per annum. What will be its value after 2 years?

Answer:

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

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Question 6:

Mark (✓) against the correct answer:
The compound interest on Rs 5000 at 10% per annum for 2 years is
(a) Rs 550
(b) Rs 1050
(c) Rs 950
(d) Rs 825

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:

Question 7:

Mark (✓) against the correct answer:
The annual rate of growth in population of a town is 5%. If its present population is 4000, what will be its population after 2 years?
(a) 4441
(b) 4400
(c) 4410
(d) 4800

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:

Question 8:

Mark (✓) against the correct answer:
At what rate per cent per annum will Rs 5000 amount to Rs 5832 in 2 years, compounded annually?
(a) 11%
(b) 10%
(c) 9%
(d) 8%

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:

Question 9:

Mark (✓) against the correct answer:
If the simple interest on a sum of money at 10% per annum for 3 years is Rs 1500, then the compound interest on the same sum at the same rate for the same period is
(a) Rs 1655
(b) Rs 1155
(c) Rs 1555
(d) Rs 1855

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:

Question 10:

Mark (✓) against the correct answer:
If the compound interest on a certain sum for 2 years at 10% per annum is Rs 1050, the sum is
(a) Rs 3000
(b) Rs 4000
(c) Rs 5000
(d) Rs 6000

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:

Question 11:

Fill in the blanks:
(i) $A = P {(1 + \frac{. . . . . . . .}{100})}^{n} .$
(ii) (Amount) - (Principal) = .........
(iii) If the value of a machine is Rs P and it depreciates at R% per annum, then its value after 2 years is .........
(iv) If the population P of a town increases at R% per annum, then its population after 5 years is .........

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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