Premium Online Home Tutors

3 Tutor System

Starting just at 265/hour

(a)₹ 10,800 for 3 years at \(12\frac { 1 }{ 2 }\) % per annum compounded annually.

(b)₹ 18,000 for \(2\frac { 1 }{ 2 }\) years at 10% per annum compounded annually.

(c)₹ 62,500 for \(1\frac { 1 }{ 2 }\) years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(e)₹ 10,000 for 1 year at 8% per annum compounded half yearly.

Answer :

(a) Given, P=₹ 10,800,n=3 years.

We have, R=\(12\frac{1}2\)% = \(\frac{25}2\)%p.a.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,800(1+\frac{25}{100})^3\)

\(=10,800(\frac{9}{8})^3\)

\(=10,800\times(\frac{9}{8})\times(\frac{9}{8})\times(\frac{9}{8})\)

\(=\frac{4,92,075}{32}=₹ \; 15,377.34\)

Thus, we have Compound interest=A - P

=₹ 15,377.34-₹ 10,800 = ₹ 4,577.35

Hence we have, amount = ₹ 15,377.34 and CI = ₹ 4,577.34

(b)We have: P = ₹ 18,000, n = \(2\frac { 1 }{ 2 }\) years and R = 10% p.a.

The amount for \(2\frac { 1 }{ 2 }\) years, can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years:

Amount = \(18000(1+\frac{10}{100})^2\)

= \(18000\times(\frac{110}{100})^2\)

= \(18000\times(\frac{11}{10})\times(\frac{11}{10})=₹ \; 21,780\)

∴ Interest after 2 years = Amount - P

=21,780-18,000=₹ 3,780

So, now taking principal amount as ₹ 21,780, the SI for the next \(\frac{1}2\)years will be as:

SI=\(\frac{P\times R\times n}{100}\)

=\(\frac{21,780\times10\times1}{100\times2}\)

=\(₹\; 1,089\)

Therefore,Total CI = ₹ 3780 +₹ 1,089= ₹ 4,869

Amount = P + I = ₹ 21,780 + ₹ 1,089 =₹ 22,869

Hence, the amount = ₹ 22,869 and CI = ₹ 4,869

(c)Given, P=₹ 62,500,n=\(1\frac{1}2=\frac{3}2\) years per annum compounded half yearly i.e., \(\frac{3}2\times 2\) years=3 half years.

We have, R=\(8\)%=\(\frac{8}2\)% = \(4\)%half yearly

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=62,500(1+\frac{4}{100})^3\)

\(=62,500(\frac{26}{25})^3\)

\(=62,500\times(\frac{26}{25})\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=4 \times 26\times 26\times 26=₹ \; 70,304\)

Thus, we have Compound interest=A - P

=₹ 70,304-₹ 62,500 = ₹ 7,804

Hence we have, amount = ₹ 70,304 and CI = ₹ 7804

(d)Given, P=₹ 8,000,n=1 years and R = 9% per annum compounded half yearly

Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{9}2\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=8,000(1+\frac{9}{2\times100})^2\)

\(=8,000(\frac{209}{200})^2\)

\(=8,000\times(\frac{209}{200})\times(\frac{209}{200})\)

\(=₹ \;(\frac{8,7362}{10})\)

\(=₹ \; 8,736.20\)

Thus, we have Compound interest=A - P

=₹ 8,736.20-₹ 8,000 = ₹ 736.20

Hence we have, amount = ₹ 8736.20 and CI = ₹ 736.20

(e)Given, P=₹ 10,000,n=1 years and R = 8% per annum compounded half yearly

Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{8}2=4\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,000(1+\frac{4}{100})^2\)

\(=10,000(\frac{26}{25})^2\)

\(=10,000\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=16\times 26\times 26\)

\(=₹ \; 10,816\)

Thus, we have Compound interest=A - P

=₹ 10,816 -₹ 10,000 = ₹ 816

Hence we have, amount = ₹ 10,816 and CI = ₹ 816

- 2.Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac { 4 }{ 12 }\) years).
- 3. Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
- I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
- 5.Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get (i) after 6 months? (ii) after 1 year?
- 6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after \(1\frac { 1 }{ 2 }\) years if the interest is (i) compounded annually. (ii) compounded half yearly.
- 7.Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (i) The amount credited against her name at the end of the second year. (ii) The interest for the third year.
- 8. Find the amount and the compound interest on ₹ 10,000 for \(1\frac { 1 }{ 2 }\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
- 9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at \(12\frac { 1 }{ 2 }\) per annum, interest being compounded half yearly.
- 10.The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. (i) Find the population in 2001. (ii) What would be its population in 2005?
- 11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
- 12.A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

- NCERT solutions for class 8 maths chapter 1 Factorisation and Mensuration
- NCERT solutions for class 8 maths chapter 1 Rational Numbers
- NCERT solutions for class 8 maths chapter 1 Factorization
- NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable
- NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals
- NCERT solutions for class 8 maths chapter 4 Practical Geometry
- NCERT solutions for class 8 maths chapter 5 Data Handling
- NCERT solutions for class 8 maths chapter 6 Squares and Square Roots
- NCERT solutions for class 8 maths chapter 7 Cube and Cube Roots
- NCERT solutions for class 8 maths chapter 8 Comparing Quantities
- NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities
- NCERT solutions for class 8 maths chapter 10 Visualising Solid Shapes
- NCERT solutions for class 8 maths chapter 11 Mensuration
- NCERT solutions for class 8 maths chapter 12 Exponents and Powers
- NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions
- NCERT solutions for class 8 maths chapter 14 Factorization
- NCERT solutions for class 8 maths chapter 15 Introduction to Graphs
- NCERT solutions for class 8 maths chapter 16 Playing with Numbers

- NCERT solutions for class 8 science chapter 1 Crop Production & Management
- NCERT solutions for class 8 science chapter 2 Microorganisms: Friend and Foe
- NCERT solutions for class 8 science chapter 3 Synthetic Fibers and Plastics
- NCERT solutions for class 8 science chapter 4 Materials: Metals and Nonmetals
- NCERT solutions for class 8 science chapter 5 Coal and Petroleum
- NCERT solutions for class 8 science chapter 6 Combustion and Flame
- NCERT solutions for class 8 science chapter 7 Conservation of Plants and Animals
- NCERT solutions for class 8 science chapter 8 Cell-Structure and Function
- NCERT solutions for class 8 science chapter 9 Reproduction in Animals
- NCERT solutions for class 8 science chapter 10 Reaching the Age of Adolsence
- NCERT solutions for class 8 science chapter 11 Force and Pressure
- NCERT solutions for class 8 science chapter 12 Friction
- NCERT solutions for class 8 science chapter 13 sound
- NCERT solutions for class 8 science chapter 14 Chemical Effects of Electric Current
- NCERT solutions for class 8 science chapter 15 Some Natural Phenomena
- NCERT solutions for class 8 science chapter 16 Light
- NCERT solutions for class 8 science chapter 17 Stars and the Solar System
- NCERT solutions for class 8 science chapter 18 Pollution of Air and Water