myhelper: RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11B

Exercise 11B

Page-151

Q1 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 1:

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

Q2 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 2:

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

Q3 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 3:

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

Q4 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 4:

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

Q5 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 5:

By using the formula, find the amount and compound interest on:

62500 for 2 years 6 months at 12% per annum compounded annually.

Answer 5:

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

Q6 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 6:

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

Q7 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 7:

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

Q8 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 8:

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

Q9 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 9:

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

Q10 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 10:

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

Q11 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 11:

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

Q12 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 12:

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

Q13 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 13:

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

Q14 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 14:

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

Q15 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 15:

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

Page-152

Q16 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 16:

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

Q17 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 17:

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

Answer 17:

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

Q18 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 18:

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

Answer 18:

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

Q19 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 19:

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

Answer 19:

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

Q20 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 20:

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

Answer 20:

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

Q21 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 21:

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

Answer 21:

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

Q22 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 22:

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

Q23 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 23:

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

Q24 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 24:

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

Q25 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 25:

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

Q26 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 26:

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

Q27 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 27:

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

Q28 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 28:

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

Q29 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 29:

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

Q30 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

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 30:

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

RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11B

Exercise 11B

Page-151

Q1 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 1:

Answer 1:

Q2 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 2:

Answer 2:

Q3 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 3:

Answer 3:

Q4 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 4:

Answer 4:

Q5 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 5:

62500 for 2 years 6 months at 12% per annum compounded annually.Answer 5:

Q6 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 6:

Answer 6:

Q7 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 7:

Answer 7:

Q8 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 8:

Answer 8:

Q9 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 9:

Answer 9:

Q10 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 10:

Answer 10:

Q11 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 11:

Answer 11:

Q12 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 12:

Answer 12:

Q13 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 13:

Answer 13:

Q14 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 14:

Answer 14:

Q15 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 15:

Answer 15:

Page-152

Q16 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 16:

Answer 16:

Q17 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 17:

Answer 17:

Q18 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 18:

Answer 18:

Q19 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 19:

Answer 19:

Q20 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 20:

Answer 20:

Q21 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 21:

Answer 21:

Q22 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 22:

Answer 22:

Q23 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 23:

Answer 23:

Q24 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

Question 24:

Answer 24:

Q25 | Ex-11B | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

62500 for 2 years 6 months at 12% per annum compounded annually.

Answer 5: