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

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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

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

(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

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

(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

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

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

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

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

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

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

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

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

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

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.)

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

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