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Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3
  • Difficulty Level : Medium
  • Last Updated : 04 Mar, 2021

Question 1. Calculate the amount and compound interest on

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

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

(iii) Rs 62,500 for 1\mathbf{\frac{1}{2}} years at 8% per annum compounded half-yearly.

(iv) Rs 8,000 for 1 year at 9% per annum compounded half-yearly.

(v) Rs 10,000 for 1 year at 8% per annum compounded half-yearly. 



Solution: 

(i) Given values are,

P = Rs 10,800

R = 12 \frac{1}{2} % per annum = \frac{25}{2} %

T = 3 Years

As it is compounded annually then, n = 3 times.

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n



A = 10,800 (1+ \frac{25}{2*100} )3 

A = 10,800 (1+ \frac{1}{8} )3

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

A = Rs 15,377.34

CI = A – P

CI = 15,377.34 – 10,800

CI = Rs 4,577.34

Hence, the amount = Rs 15,377.34 and 

Compound interest = Rs 4,577.34

(ii) Given values are,

P = Rs 18,000

R = 10 % per annum 

T = 2\frac{1}{2}  Years

As it is compounded annually then, n = 2\frac{1}{2}  times.

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 18,000 (1+ \frac{10}{100} )

What we will do here is Firstly we know 2\frac{1}{2}  Years is 2 years and 6 months which 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 has to be calculated : 

A = 18,000 (1+ \frac{1}{10} )2

A = 18,000 (\frac{11}{10} )2

A = Rs 21,780

CI = A – P

CI = 21,780 – 18,000

CI = Rs 3,780

Now, The amount for \frac{1}{2}  year has to be calculated: 

New P is equal to the amount after 2 Years. Hence, 

P = Rs 21,780

R = 10 % per annum 

T = \frac{1}{2}  year

SI = \frac{PRT}{100}

SI = \frac{21,780 × 10 × \frac{1}{2}}{100}

SI = \frac{21,780 × 10 × 1}{200}

SI = Rs 1,089

Hence, the Total amount = A + SI

                                           = 21,780 + 1,809

                                           = Rs 22,869  

Total compound interest = CI + SI

                                          = 3,780 + 1,809

                                          = Rs 4,869   

(iii) Given values are,

P = Rs 62,500

R = 8 % per annum hence 4% Half Yearly

T = 1\frac{1}{2}  Years

As it is compounded Half yearly then, n = 3 times. (1\frac{1}{2}  Years contains 3 half years)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 62,500 (1+ \frac{4}{100} )3

A = 62,500 (1+ \frac{1}{25} )3

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

A = Rs 70,304

CI = A – P

CI = 70,304 – 62,500

CI = Rs 7,804

Hence, the amount = Rs 70,304 and

Compound interest = Rs 7,804

(iv) Given values are,

P = Rs 8,000

R = 9 % per annum hence, \frac{9}{2}  % Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ \frac{9}{2*100} )2

A = 8,000 (1+ \frac{9}{200} )2

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

A = Rs 8,736.20

CI = A – P

CI = 8,736.20 – 8,000

CI = Rs 736.20

Hence, the amount = Rs 8,736.20 and

Compound interest = Rs 736.20

(v) Given values are,

P = Rs 10,000

R = 8 % per annum  hence, 4% Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

A = P (1+  \mathbf{\frac{R}{100}} )n

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

A =10,000 (1+ (\frac{1}{25} ))2

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

A = Rs 10,816

CI = A – P

CI = 10,816- 10,000

CI = Rs 816

Hence, the amount = Rs 10,816 and

Compound interest = Rs 816

Question 2. Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Solution:

Here, Given values are,

P = Rs 26,400

R = 15 % per annum 

T = 2 Years and 4 months, which is 2 \frac{1}{3}  years

As it is compounded annually then, n = 2\frac{1}{3}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 26,400 (1 + (\frac{15}{100} )2(1/3)

What we will do here is Firstly 2 years and 4 months which 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 has to be calculated:

A = 26,400 (1+ (\frac{15}{100} )2

A = 26,400 (1+ (\frac{3}{20} )2

 A = 26,400 (\frac{23}{20} )2

A = Rs 34,914

Now, The amount for (1/3) year (4 months) has to be calculated :

New P is equal to the amount after 2 Years. Hence,

P = Rs 34,914

R = 15 % per annum 

T = \frac{1}{3}  year

SI = \frac{P × R × T}{100}

SI = \frac{(34,914 × 15 × (1/3)}{100}

SI = \frac{(34,914 × 15 × 1)}{300}

SI = 1,745.70

Hence, the Total amount = A + SI

                                       = 34,914 + 1,745.70

                                       = Rs 36,659.70

Hence, the amount to be paid by Kamla = ₹ 36,659.70

Question 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?

Solution:

Let’s see each case 

Fabina Case: at simple interest

P = 12,500

R = 12% per annum  

T = 3 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,500 × 12 × 3)}{100}

SI = Rs 4,500

Radha Case: at compound interest

P = 12,500

R = 10% per annum  

T = 3 Years

As it is compounded annually then, n = 3 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 12,500 (1 + (\frac{10}{100} ))3

A =12,500 (1 + \frac{1}{10} )3

A = 12,500 (\frac{11}{10} )3

A = Rs 16,637.5

CI = A – P

CI = 16,637.5 – 12,500

CI = 4,137.5

Clearly we can see that Fabina paid more interest, and she paid

4,500 – 4,137.5 = Rs 362.5 more than Radha

Question 4. I borrowed Rs 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? 

Solution:

Lets see each case First

At simple interest

P = 12,000

R = 6% per annum  

T = 2 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,000 × 6 × 2)}{100}

SI = Rs 1,440

At compound interest

P = 12,000

R = 6% per annum  

T = 2 Years

As it is compounded annually then, n = 2 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 12,000 (1+ (\frac{6}{100} ))2

A =12,000 (1+ (\frac{3}{50} ))2

A = 12,000 (\frac{53}{50} )2

A = Rs 13,483.2 

CI = A – P

CI = 13,483.2 – 12,000

CI = 1,483.2

Clearly we can see that,

1,483.2 – 1,440 = Rs 43.2

Hence, the extra amount to be paid = ₹ 43.20

Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

(a) after 6 months?

(b) after 1 year?

Solution:

Let’s see each case 

(a) 

P = 60,000

R = 12% per annum (6% Half yearly)

T = 6 Months

As it is compounded Half Yearly then, n = 1 times (as 6 months is 1 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A =60,000 (1+ (\frac{6}{100} ))1

A =60,000 (1+ (\frac{3}{50} ))1

A = 60,000 (\frac{53}{25} )1

A = Rs 63,600

He would get Rs 63,600 after 6 Months.

(b)

P = 60,000

R = 12% per annum (6% Half yearly)

T = 1 Year

As it is compounded Half Yearly then, n = 2 times (as 1 Year is 2 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 60,000 (1+ (\frac{6}{100} ))2

A = 60,000 (1+ (\frac{3}{50} ))2

A = 60000 (\frac{53}{25} )2

A = Rs 67,416

He would get Rs 67,416 after 1 Year.

Question 6. Arif took a loan of Rs 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

(a) compounded annually.

(b) compounded half-yearly. 

Solution:

Let’s see each case 

(a) Compounded Annually

P = 80,000

R = 10% per annum

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 1 \frac{1}{2}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 80,000 (1 + (\frac{10}{100} )

What we will do here is Firstly we know 1\frac{1}{2}  Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 years has to be calculated :

A = 80,000 (1+ (\frac{10}{100} ))1

A = 80,000 (1+ (\frac{1}{10} )1

A = 80,000 (\frac{11}{10} )1

A = Rs 88,000

Now, The amount for \frac{1}{2}  Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 88,000

R = 10 % per annum

T =\frac{1}{2}  Year

SI = \frac{(P × R × T)}{ 100}

SI = \frac{(88,000 × 10 × \frac{1}{2})}{100}

SI = \frac{(88,000 × 10 × 1)}{200}

SI = 4,400

Hence, the Total amount = A + SI

                                        = 88,000 + 4,400

                                        = Rs 92,400

(b) Compounded Half-yearly

P = 80,000

R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2}  Year is 3 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 80,000 (1+ (\frac{5}{100} )3

A = 80,000 (1+ (\frac{1}{20} )3

A = 80,000 (\frac{21}{20} )3

A = Rs 92,610

Hence, the Total amount = Rs 92,610

Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(a) The amount credited against her name at the end of the second year.

(b) The interest for the 3rd year.

Solution:

Let’s see each case 

Here, 

P = 8,000

R = 5% Per annum

(a) The amount credited against Maria’s name at the end of the second year.

T = 2 Year

As it is compounded annually then, n = 2 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ (\frac{5}{100} ))2

A = 8,000 (1+ (\frac{1}{20} ))2

A = 8,000 (\frac{21}{20} )2

A = Rs 8,820

Hence, the amount credited against Maria’s name at the end of the second year = Rs 8,820

(b) The interest for the 3rd year.

T = 3 Year

As it is compounded annually then, n = 3 times

We have,

A = P (1+ \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ (\frac{5}{100} ))3

A = 8,000 (1+ (\frac{1}{20} ))3

A = 8,000 (\frac{21}{20} )3

A = Rs 9,261

The interest for the 3rd year = Amount after 3 years – Amount after 2 Years

                                                 = 9,261 – 8,820

                                                 = Rs 441

Another Solution for (b)

As we can calculate interest of 3rd year by having 2nd Year Amount as P.

P = 8,820

R = 5% per annum

T = 1 Year (2nd to 3rd year)

SI = \frac{(P × R × T)}{100}

SI = \frac{(8,820 × 5 × 1)}{100}

SI = Rs 441

The interest for the 3rd year = Rs 441

Question 8. Find the amount and the compound interest on Rs 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?

Solution:

Let’s see each cases

Compounded Annually

P = 10,000

R = 10% per annum

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 1 \frac{1}{2}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 10,000 (1 + (\frac{10}{100} )

What we will do here is Firstly we know 1½ Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 year has to be calculated:

A = 10,000 (1 + \frac{10}{100} )1

A = 10,000 (1+ \frac{1}{10} )1

A = 10,000 (\frac{11}{10} )1

A = Rs 11,000

CI = A – P

CI = 11,000-10,000

CI = 1,000

Now, The amount for \frac{1}{2}  Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 11,000

R = 10 % per annum

T =\frac{1}{2}  Year

SI = \frac{(P × R × T)}{100}

SI = \frac{(11,000 × 10 × \frac{1}{2})}{100}

SI = \frac{11,000 × 10}{200}

SI = 550

Hence, the Total Interest (compounded annually)= CI + SI

                                       = 1,000 + 550

                                       = Rs 1,550

Compounded Half-yearly

P = 10,000

R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2}  Year is 3 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 10,000 (1 + (\frac{5}{100} )3

A = 10,000 (1+ \frac{1}{20} )3

A = 10,000 (\frac{21}{20} )3

A = Rs 11,576.25

CI = A – P

CI = 11,576.25 – 10,000

CI = 1,576.25

Hence, the Total Interest (compounded Half Yearly) = Rs 11576.25

Difference between the two interests = 1,576.25 – 1,550 = Rs 26.25

Hence, the interest will be Rs 26.25 more when compounded half-yearly than the interest when compounded annually.

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 \frac{1}{2} % per annum, interest being compounded half-yearly

Solution:

Let’s see this case

P = Rs 4,096

R = 12 \frac{1}{2}  % per annum (\frac{25}{4}  % Half yearly)

T = 18 Months = 1\frac{1}{2}  Year

As it is compounded Half yearly then, n = 3 Times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 4,096 (1+ (\frac{\frac{25}{4}}{100} )3

A = 4,096 (1+ \frac{25}{400} )3

A = 4,096 (1+ (\frac{1}{16} )3

A = 4,096 (\frac{17}{16} )3

A = Rs 4,913

Ram will get the amount = Rs 4,913

Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(a) find the population in 2001.

(b) what would be its population in 2005?

Solution:

Here,

P = 54,000 (in 2003)

R = 5% per annum

(a) Population in 2001

T = 2 Years (back)

n = 2

Population in 2003 = Population in 2001 (1 + \mathbf{\frac{R}{100}} )n

54,000 = P1 (1+(\frac{5}{100} ))2

54,000 = P1 (\frac{21}{20} )2

54,000 = P1 (\frac{441}{400} )

P1 = 54,000 (\frac{441}{400} )

P1 = 48,979.59

P1 = 48,980 (approx.).

Population in 2001 was 48,980 (approx.).

(b) Population in 2005

T = 2 Years

n = 2

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 54,000 (1+ \frac{5}{100} )2

A = 54,000 (1+ (\frac{1}{20} )2

A = 54,000 (\frac{21}{20} )2

A = 59,535

Population in 2005 will be 59,535

Question 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.

Solution:

Here,

P = 5,06,000

R = 2.5% per hour

T = 2 hours

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 5,06,000 (1+ \frac{2.5}{100} )2

A = 5,06,000 (1+ \frac{25}{1000} )2

A = 5,06,000 (1+ \frac{1}{40} )2

A = 5,06,000 (\frac{41}{40} )2

A = 5,31,616.25

A = 5,31,616 (approx.)

Bacteria at the end of 2 hours = 5,31,616 (approx.)

Question 12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Here,

P = 42,000

R = 8% per annum (depreciated)

T = 1 Year

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 42,000 (1- \frac{8}{100} )1 (negative sign because the price is reduced)

A = 42,000 (1- (\frac{2}{25} )1

A = 42,000 (\frac{23}{25} )1

A = Rs 38,640

The value of scooter after one year will be = Rs 38,640

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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