myhelper: RS Aggarwal solution class 8 chapter 11 Compound Interest Test Paper 11

Test Paper 11

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Q1 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Question 1:

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

Answer 1:

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

Q2 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 2:

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

Q3 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 3:

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

Q4 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Question 4:

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

Answer 4:

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

Q5 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 5:

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

Q6 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 6:

(b) Rs. 1050

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

Q7 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 7:

(c) 4410

Present population =4000
Rate of growth = 5%

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

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

Q8 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 8:

(d) 8%

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

Q9 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 9:

(a) Rs. 1655

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

Q10 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 10:

(c) Rs. 5000

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

Q11 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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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 11:

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

RS Aggarwal solution class 8 chapter 11 Compound Interest Test Paper 11

Test Paper 11

Page-157

Q1 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Question 1:

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Q2 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q3 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q4 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q5 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q6 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q7 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q8 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q9 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q10 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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Q11 | Test Paper 11 | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest

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