# Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3

**Question 1. Calculate the amount and compound interest on**

**(i) Rs 10,800 for 3 years at 12****% per annum compounded annually.**

**(ii) Rs 18,000 for 2****years at 10% per annum compounded annually.**

**(iii) Rs 62,500 for 1****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 % per annum = %

T = 3 Years

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

We have,

A = P (1 +)^{n}A = 10,800 (1+ )

^{3}A = 10,800 (1+ )

^{3}A = 10,800 ()

^{3}

A = Rs 15,377.34

CI = A – PCI = 15,377.34 – 10,800

CI = Rs 4,577.34

Hence, the amount = Rs 15,377.34and

Compound interest = Rs 4,577.34

(ii) Given values are,P = Rs 18,000

R = 10 % per annum

T = 2 Years

As it is compounded annually then, n = 2 times.

We have,

A = P (1 +)^{n}A = 18,000 (1+ )

^{2Â½}What we will do here is Firstly we know 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+ )

^{2}A = 18,000 ()

^{2}

A = Rs 21,780

CI = A – PCI = 21,780 – 18,000

CI = Rs 3,780Now, The amount for 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 = year

SI =

SI =

SI =

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 Years

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

We have,

A = P (1 +)^{n}A = 62,500 (1+ )

^{3}A = 62,500 (1+ )

^{3}A = 62,500 ()

^{3}

A = Rs 70,304

CI = A – PCI = 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, % 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 +)^{n}A = 8,000 (1+ )

^{2}A = 8,000 (1+ )

^{2}A = 8,000 ()

^{2}

A = Rs 8,736.20

CI = A – PCI = 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+)^{n}A = 10,000 (1+ ())

^{2}A =10,000 (1+ ())

^{2}A = 10,000 ()

^{2}

A = Rs 10,816

CI = A – PCI = 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 years

As it is compounded annually then, n = 2 times

We have,

A = P (1 +)^{n}A = 26,400 (1 + ()

^{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+ ()

^{2}A = 26,400 (1+ ()

^{2}A = 26,400 ()

^{2}

A = Rs 34,914Now, 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 = year

SI =

SI =

SI =

SI = 1,745.70Hence, the Total amount =

A + SI= 34,914 + 1,745.70

= Rs 36,659.70

Hence, the amount to be paid by Kamla = â‚¹ 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 interestP = 12,500

R = 12% per annum

T = 3 Years

SI =

SI =

SI = Rs 4,500

Radha Case: at compound interestP = 12,500

R = 10% per annum

T = 3 Years

As it is compounded annually then, n = 3 times

We have,

A = P (1 +)^{n}A = 12,500 (1 + ())

^{3}A =12,500 (1 + )

^{3}A = 12,500 ()

^{3}A = Rs 16,637.5

CI = A – PCI = 16,637.5 – 12,500

CI = 4,137.5Clearly we can see that

Fabina paid more interest, and she paid4,500 – 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 interestP = 12,000

R = 6% per annum

T = 2 Years

SI =

SI =

SI = Rs 1,440

At compound interestP = 12,000

R = 6% per annum

T = 2 Years

As it is compounded annually then, n = 2 times

We have,

A = P (1 +)^{n}A = 12,000 (1+ ())

^{2}A =12,000 (1+ ())

^{2}A = 12,000 ()

^{2}

A = Rs 13,483.2CI = A – P

CI = 13,483.2 – 12,000

CI = 1,483.2Clearly we can see that,

1,483.2 – 1,440 = Rs 43.2

Hence,

the extra amount to be paid = â‚¹ 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 +)^{n}A =60,000 (1+ ())

^{1}A =60,000 (1+ ())

^{1}A = 60,000 ()

^{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 +)^{n}A = 60,000 (1+ ())

^{2}A = 60,000 (1+ ())

^{2}A = 60000 ()

^{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**** years if the interest is**

**(a) compounded annually.**

**(b) compounded half-yearly. **

**Solution:**

Let’s see each case

(a) Compounded AnnuallyP = 80,000

R = 10% per annum

T = 1 Year

As it is compounded annually then, n = 1 times

We have,

A = P (1 +)^{n}A = 80,000 (1 + ()

^{1Â½}What we will do here is Firstly we know 1 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+ ())

^{1}A = 80,000 (1+ ()

^{1}A = 80,000 ()

^{1}

A = Rs 88,000Now, The amount for 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 = Year

SI =

SI =

SI =

SI = 4,400Hence, the Total amount =

A + SI= 88,000 + 4,400

= Rs 92,400

(b) Compounded Half-yearlyP = 80,000

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

T = 1 Year

As it is compounded annually then, n = 3 times (as 1 Year is 3 half year)

We have,

A = P (1 +)^{n}A = 80,000 (1+ ()

^{3}A = 80,000 (1+ ()

^{3}A = 80,000 ()

^{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 +)^{n}A = 8,000 (1+ ())

^{2}A = 8,000 (1+ ())

^{2}A = 8,000 ()

^{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+)^{n}A = 8,000 (1+ ())

^{3}A = 8,000 (1+ ())

^{3}A = 8,000 ()

^{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 3

^{rd}year by having 2^{nd}Year Amount as P.P = 8,820

R = 5% per annum

T = 1 Year (2

^{nd}to 3^{rd}year)

SI =SI =

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**** 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 AnnuallyP = 10,000

R = 10% per annum

T = 1 Year

As it is compounded annually then, n = 1 times

We have,

A = P (1 +)^{n}A = 10,000 (1 + ()

^{1Â½}What we will do here is Firstly we know 1Â½ 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 + )

^{1}A = 10,000 (1+ )

^{1}A = 10,000 ()

^{1}A = Rs 11,000

CI = A – P

CI = 11,000-10,000

CI = 1,000Now, The amount for 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 = Year

SI =

SI =

SI =

SI = 550

Hence, the Total Interest (compounded annually)= CI + SI= 1,000 + 550

= Rs 1,550

Compounded Half-yearlyP = 10,000

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

T = 1 Year

As it is compounded annually then, n = 3 times (as 1 Year is 3 half year)

We have,

A = P (1 +)^{n}A = 10,000 (1 + ()

^{3}A = 10,000 (1+ )

^{3}A = 10,000 ()

^{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 11576.25Difference between the two interests = 1,576.25 â€“ 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 ****% per annum, interest being compounded **half-yearly**. **

**Solution:**

Let’s see this caseP = Rs 4,096

R = 12 % per annum ( % Half yearly)

T = 18 Months = 1 Year

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

We have,

A = P (1 +)^{n}A = 4,096 (1+ ()

^{3}A = 4,096 (1+ )

^{3}A = 4,096 (1+ ()

^{3}A = 4,096 ()

^{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 2001T = 2 Years (back)

n = 2

Population in 2003 = Population in 2001 (1 +)^{n}54,000 = P

_{1 }(1+())^{2}54,000 = P1 ()

^{2}54,000 = P1 ()

P1 = 54,000 ()

P1 = 48,979.59

P1 = 48,980 (approx.).

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

(b) Population in 2005T = 2 Years

n = 2

We have,

A = P (1 +)^{n}A = 54,000 (1+ )

^{2}A = 54,000 (1+ ()

^{2}A = 54,000 ()

^{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 +)^{n}A = 5,06,000 (1+ )

^{2}A = 5,06,000 (1+ )

^{2}A = 5,06,000 (1+ )

^{2}A = 5,06,000 ()

^{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 +)^{n}A = 42,000 (1- )

^{1}(negative sign because the price is reduced)A = 42,000 (1- ()

^{1}A = 42,000 ()

^{1}

A = Rs 38,640

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

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