You can Download KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 1.

Calculate the amount and compound interest on

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

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

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

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

Solution:

(a) P = Rs. 10,800

Time = 3 years

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

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(10800\left(1+\frac{25}{200}\right)^{3}\)

= \(10800 \times \frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\)

= Rs. 15377.34

C.I. = A – P

= 15,377.34 – 10,800

= Rs. 4,577.34

(b) P = Rs. 18,000

R = 10% (p.a)

T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

= \(18000\left(1+\frac{10}{100}\right)^{\frac{5}{2}}\)

First calculate amount of 2 year then S.I. for \(\frac{1}{2}\) year

A = \(18000\left(1+\frac{10}{100}\right)^{2}\)

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

= 18 × 1210

= Rs. 21780

The amount of Rs. 21,780 is principal for next year

Hence, P = Rs. 21,780

Time = \(\frac{1}{2}\) year

Rate = 10% (p.a)

SI = \(\frac{21780 \times 10 \times 1}{2 \times 100}\) = 1089

C.I. for 2 years = A – P

= 21780 – 18000

= 3780

S.I. = Rs. 1089

Total Interest = 3780 + 1089 = Rs. 4869

Amount = 18000 + 4869 = Rs. 22,869

(c) P = Rs. 62,500

Time = \(\frac{3}{2}\) years

= \(\frac{3}{2}\) × 2

= 3 (half yearly)

R = 8% (p.a)

= \(\frac{8}{2}\)%

= 4% (half-yearly)

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

= \(62500\left(1+\frac{4}{100}\right)^{3}\)

= \(62500 \times \frac{104}{100} \times \frac{104}{100} \times \frac{104}{100}\)

= Rs. 70304

∴ Compound Interest = A – P

= 70304 – 62500

= Rs. 7804

(d) P = Rs. 8000

Time = 1 year = 1 × 2 = 2 (half-yearly)

Rate = 9% (p.a)

= \(\frac{9}{2}\)% (half-yearly)

A = \(8000\left(1+\frac{9}{2 \times 100}\right)^{2}\)

= \(8000 \times \frac{209}{100} \times \frac{209}{100}\)

= Rs. 8736.20

C.I. = A – P

= Rs. 8736.20 – 8000

= Rs. 736.20

(e) P = Rs. 10,000

Time = 1 year or 1 × 2 = 2 (half-yearly)

Rate = 8% p.a. or \(\frac{8}{2}\) = 4% (half-year)

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

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

= Rs. 10816

C.I. = A. – P.

= 10816 – 10,000

= Rs. 816

Question 2.

Kamla 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:

P = Rs. 26,400

R = 15% p.a.

Time = 2 years 4 months

Calculate amount for 2 years that take the amount as principal

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(26,400\left(1+\frac{15}{100}\right)^{2}\)

= \(26,400 \times \frac{115}{100} \times \frac{115}{100}\)

= Rs. 34914

Now, amount of Rs. 34914 is principal for the next 4 months or \(\frac{4}{12}=\frac{1}{3}\) year.

∴ SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{34,914 \times 15 \times 1}{100 \times 3}\)

= Rs. 1745.70

A = P + S.I.

= 34914 + 1740.70

= Rs. 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:

Fabina borrows = Rs. 12,500

R = 12% p.a.

T = 3 years

Since, Fabina borrowed this amount at S.I.

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

= \(\frac{12500 \times 12 \times 3}{100}\)

= Rs. 4500

Radha borrowed the amount at compound Interest

Rate = 10%

P = Rs. 12500

T = 3 years

= \(P\left(1-\frac{R}{100}\right)^{T}\)

= \(12500\left(1+\frac{10}{100}\right)^{3}\)

= \(12500 \times \frac{110}{100} \times \frac{110}{100} \times \frac{110}{100}\)

= Rs. 16637.50

CI = A – P

= 16637.50 – 12500

= Rs. 4137.50

Difference in the interest of both

= 4500 – 4137.50

= Rs. 362.50

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:

P = Rs. 12,000

R = 6% p.a.

T = 2 years

Since the amount is borrowed at S.I.

∴ SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{12000 \times 6 \times 2}{100}\)

= Rs. 1440

If it is borrowed at Compound Interest

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(12000\left(1+\frac{6}{100}\right)^{2}\)

= \(12000 \times \frac{106}{100} \times \frac{106}{100}\)

= Rs. 13,483.20

C.I = A – P

= 13,483.20 – 12000.00

= Rs. 1483.20

Difference in C.I. and S.I. = 1483.20 – 1440.00

= Rs. 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

(i) After 6 months?

(ii) After 1 year?

Solution:

(i) P = Rs. 60,000

R = 12% p.a. = \(\frac{12}{2}\) = 6% (half-yearly)

T = 6 months = 1 (half year)

A = \(\mathbf{P}\left(1+\frac{\mathbf{R}}{100}\right)^{\mathrm{T}}\)

= \(60000\left(1+\frac{6}{100}\right)^{1}\)

= \(60000 \times \frac{106}{100}\)

= 600 × 106

= Rs. 63600

(ii) When Time = 1 year = 1 × 2 = 2 (half years)

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(60,000\left(1+\frac{6}{100}\right)^{2}\)

= \(60000 \times \frac{106}{100}\)

= Rs. 67416

Question 6.

Arif took a loan ofRs. 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.

Solution:

(i) P = Rs. 80,000

R = 10% p.a.

Time = 1\(\frac{1}{2}\) = 1 year and \(\frac{1}{2}\) year

A = \(\mathbf{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

= \(80,000\left(1+\frac{10}{100}\right)^{1}\)

= 80,000 × \(\frac{110}{100}\)

= 800 × 110

= Rs. 88000

The amount is principal for next year

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

= \(\frac{88000 \times 10 \times 1}{100 \times 2}\)

= Rs. 4400

Amount = 88,000 + 4400 = Rs. 92,400

(ii) If the interest is calculated half-yearly

Time = 1\(\frac{1}{2}\) (p.a.)

= \(\frac{1}{2}\) × 2

= 3 (half yearly)

Rate = 10% p.a. = \(\frac{10}{2}\) = 5% (half yearly)

A = \(80,000\left(1+\frac{5}{100}\right)^{2}\)

= \(80,000 \times \frac{105}{100} \times \frac{105}{100} \times \frac{105}{100}\)

= \(\frac{8 \times 105 \times 105 \times 105}{100}\)

= 2 × 21 × 21 × 105

= Rs. 92610

Difference in amount = 92610 – 92400 = Rs. 210

Question 7.

Maria invested Rs. 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 = Rs. 8,000

R = 5% (p.a.)

Time = 2 years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

= \(8000\left(1+\frac{5}{100}\right)^{2}\)

= \(8000 \times \frac{21}{20} \times \frac{21}{20}\)

= Rs. 8820

∴ Amount credited in Maria name after 2 years = Rs. 8820

(ii) If time = 3 years

= \(8000\left(1+\frac{5}{100}\right)^{3}\)

= \(80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)

= Rs. 9261

Interest for 3rd year = Amount of 3 years – Amount of 2 years

= 9261 – 8820

= 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:

P = Rs. 10,000

Time = \(\frac{3}{2}\) years

= \(\frac{3}{2}\) × 2

= 3 (half year)

R = 10% (p.a.) = \(\frac{10}{2}\) = 5 (half year)

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(10,000\left(1+\frac{5}{100}\right)^{3}\)

= \(10000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)

= Rs. 11,576.25

C.I. = A – P

= 11576.25 – 10,000

= Rs. 1576.25

If the interest is calculated compounded annually then

P = Rs. 10,000

Time = \(\frac{3}{2}\) years = 1\(\frac{1}{2}\) years = 1 year and \(\frac{1}{2}\) year

R = 10% p.a.

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

= \(10,000 \times \frac{110}{100}\)

= Rs. 11,000

The amount is principal for next time period

S.I = \(\frac{11000 \times 10 \times 1}{100 \times 2}\) = Rs. 550

Amount = 11000 + 550 = Rs. 11,550

C.I. = A – P

= Rs. 11,550 – 11,000

= Rs. 550

Yes the interest would be more when compounded half-yearly than it is calculated yearly.

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 compounded half-yearly.

Solution:

P = Rs. 4096

Time = 18 months 3 (half-yearly)

R = \(\frac{25}{2}\)% p.a.

= \(\frac{25}{2 \times 2}\)

= \(\frac{25}{4}\)% (half-yearly)

A = \(\mathbf{P}\left(1+\frac{\mathbf{R}}{100}\right)^{\mathbf{T}}\)

= \(4096\left(1+\frac{25}{4 \times 100}\right)^{3}\)

= \(4096 \times \frac{17}{16} \times \frac{17}{16} \times \frac{17}{16}\)

= Rs. 4913

Question 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?

Solution:

(i) Population in 2003 = 54000

Rate of increase in population = 5% (p.a.)

Let the population in 2001 = x

Since population increases compoundedly

∴ 54000 = \(x\left(1+\frac{5}{100}\right)^{2}\)

54000 = \(x \times \frac{21}{20} \times \frac{21}{20}\)

x = \(54000 \times \frac{21}{20} \times \frac{21}{20}\)

∴ x = 48979.59

Hence it was 48980 in 2001

(ii) Population in 2005 i.e. after 2 years = 54000 × \(\left(1+\frac{5}{100}\right)^{2}\)

= \(54000 \times \frac{21}{20} \times \frac{21}{20}\)

= 59535

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:

Starting count of bacteria 5,06,000

Rate of increase = 2.5% (half-hour)

Time = 2 hour

Bacteria count after 2 hours

= 5,06,000 \(\left(1+\frac{2.5}{100}\right)^{2}\)

= \(5,06,000 \times \frac{102.5}{100} \times \frac{102.5}{100}\)

= 531616.25 or 531616 (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:

Initial value = 42,000

Value depreciate = 8% (p.a)

Time = 1 year

Final value after 1 year = 42,000 \(\left(1-\frac{8}{100}\right)\)

= 42,000 × \(\frac{92}{100}\)

= 38,640