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

Exercise 11D

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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 1:

(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 2:

(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 3:

(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 4:

(c) Rs 961

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

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Question 5:

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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 5:

(b) Rs. 5051

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

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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 6:

(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 7:

(a) Rs.1209

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

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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 8:

(b) 26460

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

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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 9:

(c) Rs. 43740

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

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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 10:

(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 11:

(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 12:

(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 13:

(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 14:

(d) Rs 4096

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

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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 15:

(c) 6%

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

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

Exercise 11D

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

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

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

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

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