NCERT Solutions for Class 8 Maths Chapter 8 - Comparing Quantities Exercise 8.3

The NCERT exercise question and answers given in PDF form will help students frame a perfect solution for the Math Exam. The solutions are aimed at helping students master the concepts by practising the problems. NCERT Solutions for Class 8 Maths Chapter 8, Comparing Quantities Exercise 8.3, are prepared by BYJU’S subject matter experts to help the students enhance their grip on the different concepts related to Compound Interest. Download NCERT Solutions for Maths Chapter 8 to strengthen the basic foundation.



NCERT Solutions for Class 8 Maths Chapter 8 – Comparing Quantities Exercise 8.3

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Access Answers of Maths NCERT Class 8 Chapter 8 – Comparing Quantities Exercise 8.3 Page Number 133

Exercise 8.3 Page No: 133

1. Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12½ % per annum compounded annually.

Solution:

Principal (P) = ₹ 10,800

Rate (R) = 12½ % = 25/2 % (annual)

Number of years (n) = 3

Amount (A) = P(1 + R/100)n

= 10800(1 + 25/200)3

= 10800(225/200)3

= 15377.34375

= ₹ 15377.34 (approximately)

C.I. = A – P = ₹ (15377.34 – 10800) = ₹ 4,577.34

(b) ₹ 18000 for 2½ years at 10% per annum compounded annually.

Solution:

Principal (P) = ₹ 18,000

Rate (R) = 10% annual

Number of years (n) = 2½

The amount for 2 years and 6 months can be calculated by calculating the amount for 2 years using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.

First, the amount for 2 years has to be calculated

Amount, A = P(1 + R/100)n

= 18000(1 + 1/10)2

= 18000(11/10)2

= ₹ 21780

By taking ₹ 21780 as principal, the S.I. for the next ½ year will be calculated

S.I. = (21780 x ½ x 10)/100

= ₹ 1089

Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780

And, interest for the next ½ year = ₹ 1089

Total C.I. = ₹ 3780 + ₹ 1089

= ₹ 4,869

Therefore,

Amount, A = P + C.I.

= ₹ 18000 + ₹ 4869

= ₹ 22,869

(c) ₹ 62500 for 1½ years at 8% per annum compounded half yearly.

Solution:

Principal (P) = ₹ 62,500

Rate = 8% per annum or 4% per half-year

Number of years = 1½

There will be 3 half-years in 1½ years

Amount, A = P(1 + R/100)n

= 62500(1 + 4/100)3

= 62500(104/100)3

= 62500(26/25)3

= ₹ 70304

C.I. = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804

(d) ₹ 8000 for 1 year at 9% per annum compound half yearly.

(You can use the year-by-year calculation using S.I. formula to verify)

Solution:

Principal (P) = ₹ 8000

Rate of interest = 9% per annum or 9/2% per half-year

Number of years = 1 year

There will be 2 half-years in 1 year

Amount, A = P(1 + R/100)n

= 8000(1 + 9/200)2

= 8000(209/200)2

= 8736.20

C.I. = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20

(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.

Solution:

Principal (P) = ₹ 10,000

Rate = 8% per annum or 4% per half-year

Number of years = 1 year

There are 2 half-years in 1 year

Amount, A = P(1 + R/100)n

= 10000(1 + 4/100)2

= 10000(1 + 1/25)2

= 10000(26/25)2

= ₹ 10816

C.I. = A – P = ₹ 10816 – ₹ 10000 = ₹ 816

2. Kamala borrowed ₹ 26400 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?

(Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)

Solution:

Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, then calculating the simple interest for 4 months on the amount obtained at the end of 2 years.

First, the amount for 2 years has to be calculated

Amount, A = P(1 + R/100)n

= 26400(1 + 15/100)2

= 26400(1 + 3/20)2

= 26400(23/20)2

= ₹ 34914

By taking ₹ 34,914 as principal, the S.I. for the next 1/3 years will be calculated

S.I. = (34914 × 1/3 x 15)/100 = ₹ 1745.70

Interest for the first two years = ₹ (34914 – 26400) = ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

3. Fabina borrows ₹ 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:

Interest paid by Fabina = (P x R x T)/100

= (12500 x 12 x 3)/100

= 4500

Amount paid by Radha at the end of 3 years = A = P(1 + R/100)n

A = 12500(1 + 10/100)3

= 12500(110/100)3

= ₹ 16637.50

C.I. = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Thus, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Hence, Fabina will have to pay ₹ 362.50 more.

4. I borrowed ₹ 12000 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:

P = ₹ 12000

R = 6% per annum

T = 2 years

S.I. = (P x R x T)/100

= (12000 x 6 x 2)/100

= ₹ 1440

To find the compound interest, the amount (A) has to be calculated

Amount, A = P(1 + R/100)n

= 12000(1 + 6/100)2

= 12000(106/100)2

= 12000(53/50)2

= ₹ 13483.20

∴ C.I. = A − P

= ₹ 13483.20 − ₹ 12000

= ₹ 1,483.20

C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440

= ₹ 43.20

Therefore, the extra amount to be paid is ₹ 43.20.

5. Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Solution:

(i) P = ₹ 60,000

Rate = 12% per annum = 6% per half-year

n = 6 months = 1 half-year

Amount, A = P(1 + R/100)n

= 60000(1 + 6/100)1

= 60000(106/100)

= 60000(53/50)

= ₹ 63600

(ii) There are 2 half-years in 1 year

So, n = 2

Amount, A = P(1 + R/100)n

= 60000(1 + 6/100)2

= 60000(106/100)2

= 60000(53/50)2

= ₹ 67416

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 1½ years if the interest is

(i) Compounded annually

(ii) Compounded half yearly

Solution:

(i) P = ₹ 80,000

R = 10% per annum

n = 1½ years

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

First, the amount for 1 year has to be calculated

Amount, A = P(1 + R/100)n

= 80000(1 + 10/100)1

= 80000 x 11/100

= ₹ 88000

By taking ₹ 88,000 as principal, the S.I. for the next ½ year will be calculated as

S.I. = (P x R x T)/100

= (88000 x 10 x ½)/100

= ₹ 4400

Interest for the first year = ₹ 88000 – ₹ 80000 = ₹ 8000

And interest for the next ½ year = ₹ 4,400

Total C.I. = ₹ 8,000 + ₹ 4,400 = ₹ 12,400

A = P + C.I.= ₹ (80000 + 12400)

= ₹ 92,400

(ii) The interest is compounded half yearly

Rate = 10% per annum = 5% per half-year

There will be three half-years in 1½ years

Amount, A = P(1 + R/100)n

= 80000(1 + 5/100)3

= 80000(105/100)3

= ₹ 92610

Thus, the difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

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 3rd year

Solution:

(i) P = ₹ 8,000

R = 5% per annum

n = 2 years

Amount, A = P(1 + R/100)n

= 8000(1 + 5/100)2

= 8000(105/100)2

= ₹ 8820

(ii) The interest for the next year, i.e. the third year, has to be calculated. By taking ₹ 8,820 as principal, the S.I. for the next year will be calculated.

S.I. = (P x R x T)/100

= (8820 x 5 x 1)/100

= ₹ 441

8. Find the amount and the compound interest on ₹ 10,000 for 1½ 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:

P = ₹ 10,000

Rate = 10% per annum = 5% per half-year

n = 1½ years

There will 3 half-years in 1½ years

Amount, A = P(1 + R/100)n

= 10000(1 + 5/100)3

= 10000(105/100)3

= ₹ 11576.25

C.I. = A − P

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

Amount, A = P(1 + R/100)n

= 10000(1 + 10/100)1

= 10000(110/100)

= ₹ 11000

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

S.I. = (P x R x T)/100

= (11000 x 10 x ½)/100

= ₹ 550

So, the interest for the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Hence, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So the difference between two interests = 1576.25 – 1550 = 26.25

Therefore, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.

9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.

Solution:

P = ₹ 4,096

R = 12½ per annum = 25/2 per annum = 25/4 per half-year

n = 18 months

There will be 3 half-years in 18 months

Therefore, amount A = P(1 + R/100)n

= 4096(1 + 25/(4 x 100))3

= 4096 x (1 + 1/16)3

= 4096 x (17/16)3

= ₹ 4913

Therefore, the required amount is ₹ 4,913.

10. The population of a place increased to 54000 in 2003 at a rate of 5% per annum

(i) find the population in 2001

(ii) what would be its population in 2005?

Solution:

(i) It’s given that population in the year 2003 = 54,000

54,000 = (Population in 2001) (1 + 5/100)2

54,000 = (Population in 2001) (105/100)2

Population in 2001 = 54000 x (100/105)2

= 48979.59

Therefore, the population in the year 2001 was approximately 48,980

(ii) Population in 2005 = 54000(1 + 5/100)2

= 54000(105/100)2

= 54000(21/20)2

= 59535

Therefore, the population in the year 2005 would be 59,535.

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:

The initial count of bacteria is given as 5,06,000

Bacteria at the end of 2 hours = 506000(1 + 2.5/100)2

= 506000(1 + 1/40)2

= 506000(41/40)2

= 531616.25

Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).

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

Solution:

Principal = Cost price of the scooter = ₹ 42,000

Depreciation = 8% of ₹ 42,000 per year

= (P x R x T)/100

= (42000 x 8 x 1)/100

= ₹ 3360

Thus, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

Access other exercise solutions of Class 8 Maths Chapter 8 – Comparing Quantities

Exercise 8.1 Solutions 6 Questions (1 Long Answer Question, 5 Short Answer Questions)

Exercise 8.2 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions)



NCERT Solutions for Class 8 Maths Chapter 8 – Comparing Quantities Exercise 8.3

Exercise 8.3 of NCERT Solutions for Class 8 Maths Chapter 8 – Comparing Quantities is based on

  1. Compound Interest
  2. Deducing a Formula for Compound Interest
  3. Rate Compounded Annually or Half Yearly (Semi-annually)
  4. Applications of Compound Interest Formula

Also, explore – 

NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8

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