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

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12\(\frac{1}{2}\)% compounded annually.
Answer:
Given, P = ₹ 10,800
R = 12\(\frac{1}{2}\) % p.a, = per annum and n = 3 years.
Amount after 3 years
= ₹ [10,800 × \(\left(1+\frac{25}{2 \times 100}\right)^{3}\)]
= ₹ (10,800 × \(\frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\))
= ₹ 15,377.34

∴ Compound interest
= ₹ (15,377.34 - 10,800)
= ₹ 4577.34

(b) ₹ 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
Answer:
Here, P = ₹ 18,000, R = 10% p.a. and T = years
Amount after 2\(\frac{1}{2}\) Years
= ₹ [18,000 × \(\left(1+\frac{10}{100}\right)^{2} \times\left(1+\frac{\frac{1}{2} \times 10}{100}\right)\)]
= ₹ (18,000 × \(\frac{110}{100} \times \frac{110}{100} \times \frac{105}{100}\))
= ₹ 22,869

∴ Compound interest = ₹ (22,869 - 18,000)
= ₹ 4869

(c) ₹ 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
Answer:
Here, Principal = ₹ 62,500
Rate = 8% p.a. = 4% per half-year,
Time = 1\(\frac{1}{2}\) years = 3 half years
Amount = ₹ [62,500 × \(\left(1+\frac{4}{100}\right)^{3}\)]
= ₹ (62,500 × \(\frac{104}{100} \times \frac{104}{100} \times \frac{104}{100}\))
= ₹ 70,304

∴ Compound interest = ₹ (70,304 - 62,500)
= ₹ 7804

(d) ₹ 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).
Answer:
Here, Principal = ₹ 8000
Time = 1 year = 2 half-year, Rate = 9% p.a.
= \(\left(\frac{9}{2}\right)\)% per half-year

Amount = ₹ [8000 × \(\left(1+\frac{9}{2 \times 100}\right)^{2}\)]
= ₹ (8000 × \(\frac{209}{200} \times \frac{209}{200}\))
= ₹ 8736.20 

∴ Compound interest = ₹ (8736.20 - 8000)
= ₹ 736.20

(e) ₹ 10,000 for 1 year at 8% per annum compounded Tialf yearly.
Answer:
We have, P = ₹ 10,000, R = 8% p.a.
= 4% per half-year
and n= 1 year = 2 half-years.
Amount after 1 year = ₹ 10,000 × \(\left(1+\frac{4}{100}\right)^{2}\)
= ₹ (10,000 × \(\frac{26}{25} \times \frac{26}{25}\))
= ₹ 10,816

Compound interest = ₹ (10,816 - 10,000)
= ₹ 816

Question 2.
Kamala borrowed ₹ 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?
Answer:
Here P = ₹ 26,400, R = 15% per
annum and n = 2 years and 4 months = 2\(\frac{1}{3}\) years

Amount after 2\(\frac{1}{3}\) years
= ₹ [26,400 × \(\left(1+\frac{15}{100}\right)^{2} \times\left(1+\frac{\frac{1}{3} \times 15}{100}\right)\)]
= ₹ (26,400 × \(\frac{115}{100} \times \frac{115}{100} \times \frac{105}{100}\))
= ₹ (26,400 × \(\frac{23}{20} \times \frac{23}{20} \times \frac{21}{20}\))
= ₹ 36,659.70
Hence, Kamala will pay ₹ 36,659.70 to the bank.

Question 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?
Answer:
In case of Fabina :
P = ₹ 12,500, R = 12% per annum and T = 3 years; Then
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\left(\frac{12,500 \times 12 \times 3}{100}\right)\)
= ₹ 4500

In case of Radha :
P = ₹ 12,500, R = 10% per annum and n = 3 years; Then
Amount = ₹ [12500 × \(\left(1+\frac{10}{100}\right)^{3}\)]
= ₹ (12500 × \(\frac{110}{100} \times \frac{110}{100} \times \frac{110}{100}\))
 
∴ Compound Interest
= ₹ (16,637.50 - 12,500)
= ₹ 4137.50
Hence, Fabina pays ₹ (4500 - 4137.50), i.e. 362.50 more as in interest.

Question 4.
1 borrowed  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?
Answer:
Here, P = ₹ 12,000, R = 6% per
annum and T = 2 years.
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\left(\frac{12,000 \times 6 \times 2}{100}\right)\)
= ₹ 1440

For Compound Interest-
Amount = ₹ [12,000 × \(\left(1+\frac{6}{100}\right)^{2}\)]
= ₹ (12,000 × \(\frac{106}{100} \times \frac{106}{100}\))
= ₹ 13,483.20

Compound Interest = ₹ (13,483.20- 12,000)
= ₹ 1483.20
So, I have to pay ₹ (1483.20 - 1440), i.e. ₹ 43.20 in excess.

Question 5.
Yasudevan invested ₹ 60,000 at ah interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year? .
Answer:
Here Principal = ₹ 60,000, Rate = 12% per annum = 6% per half-year.
(i) Time = 6 month = 1 half-year
∴ Amount after 6 months = Amount after 1 half year
= ₹ [60,000 × \(\left(1+\frac{6}{100}\right)^{1}\)]
= ₹ (60,000 × \(\frac{106}{100}\))
= ₹ 63,600

(ii) Time = 1 year = 2 half-years
∴ Amount after 1 year
= ₹ [60,000 × \(\left(1+\frac{6}{100}\right)^{2}\)]
= ₹ (60,000 × \(\frac{106}{100} \times \frac{106}{100}\))
= ₹ 67,416

Question 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 1\(\frac{1}{2}\) years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Answer:
Here, P = ₹ 80,000, R = 10% per annum = 5% per half-year,
Time = 1\(\frac{1}{2}\) years = 3 half-years
(i) When compound interest is compounded annually :
Amount after 1\(\frac{1}{2}\) years
= ₹ [80,000 × \(\left(1+\frac{10}{100}\right)^{1} \times\left(1+\frac{\frac{1}{2} \times 10}{100}\right)\)]
= ₹ (80,000 × \(\frac{110}{100} \times \frac{105}{100}\))
= ₹ 92,400

(ii) When compound interest is compounded half yearly :
Amount after, 1\(\frac{1}{2}\) years
= ₹ [80,000 × \(\left(1+\frac{5}{100}\right)^{3}\)]
= ₹ (80,000 × \(\frac{105}{100} \times \frac{105}{100} \times \frac{105}{100}\)) = ₹ 92,610
∴ Difference in amounts = ₹ (92,610 - 92,400) = ₹ 210

Question 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.
Answer:
(i) Here, P = ₹ 8000, R = 5% per annum and n = 2 years
Amount after 2 years = P\(\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)
= ₹ [8000 × \(\left(1+\frac{5}{100}\right)^{2}\))
= ₹ (8000 × \(\frac{105}{100} \times \frac{105}{100}\)) = ₹ 8820

(ii) Principal for the 3rd year = ₹ 8820 Interest for the 3rd year
= ₹ \(\left(\frac{8820 \times 5 \times 1}{100}\right)\)
= ₹ 441

Question 8.
Find the amount and the compound interest on ₹ 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?
Answer:
Here, Principal = ₹ 10,000, Time = 1\(\frac{1}{2}\) years = 3 half years,
Rate = 10% per annum = 5% per half year.
Amount = ₹ [10,000 × \(\left(1+\frac{5}{100}\right)^{3}\)]
= ₹(10,000 × \(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\))
= ₹ 11,576.25 

and Compounded Interest
= ₹ (11,576.25 - 10000)
= ₹ 1576.25

When compound interest is compounded yearly
= 10,000 × \(\left(1+\frac{10}{100}\right) \times\left(1+\frac{\frac{1}{2} \times 10}{100}\right)\)
= 10,000 × \(\frac{110}{100} \times \frac{105}{100}\) = ₹ 11,550
Hence compound Interest
= ₹ 11,550 - 10,000 = ₹ 1550
Hence the interest compounded half yearly is more than the interest that he would get if it was compounded annually.

Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\frac{1}{2} % per annum, interest being compounded half yearly.
Answer:
Here, Principal = ₹ 4096
Time = 18 months = 3 half years
Rate = 12\(\frac{1}{2}\)% per annum = per half year
Amount = ₹ [4096 × \(\left(1+\frac{25}{4 \times 100}\right)^{3}\)]
= ₹ (4096 × \(\frac{17}{16} \times \frac{17}{16} \times \frac{17}{16}\)]
= ₹ 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?
Answer:
(i) Let P be the population in 2001, i.e., 2 years ago.
Then, Present population = P × \(\left(1+\frac{5}{100}\right)^{2}\)
or 54,000 = P × \(\frac{105}{100} \times \frac{105}{100}\)
⇒ P = \(\frac{54,000 \times 100 \times 100}{105 \times 105}\)
= 48,979.59
Hence, the population in the year 2001 is 48,980 (apppox.) 

(ii) Let P = 54,000 (Population in 2003)
∴ Population after 2 years, i.e., in 2005
= P\(\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)
= 54,000 × \(\left(1+\frac{5}{100}\right)^{2}\)
= 54,000 × \(\frac{105}{100} \times \frac{105}{100}\)
= 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.
Answer:
P = Original count of bacteria
= 5,06,000
Rate of increase = R = 2.5% per hour 
Time= 2 hours
∴ Bacteria count after 2 hours
= 5,06,000 × \(\left(1+\frac{2.5}{100}\right)^{2}\)
= 5,06,000 × \(\frac{102.5}{100} \times \frac{102.5}{100}\)
= 5,31,616.25
= 5,31,616 (Approx.)

Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Answer:
P = ₹ 42,000
Depreciation Rate (R) = 8% p.a.
T = 1 year, A = ?
Using A= P\(\left(1-\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)
⇒ A= ₹ 42,000 × \(\left(1-\frac{8}{100}\right)^{1}\)
= ₹ 42,000 × (1 - \(\frac{2}{25}\))
= ₹ (1680 × 23)
= ₹ 38,640