# GSEB Solutions Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

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Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Calculate the amount and compound interest on
(a) ā¹ 10,800 for 3 years at 12 $$\frac{1}{2}$$ % per annum compounded annually.
(b) ā¹ 18,000 for 2 $$\frac{1}{2}$$ years at 10% per annum compounded annually.
(c) ā¹ 62,500 for 1 $$\frac{1}{2}$$ years at 8% per annum compounded half yearly.
(d) ā¹ 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) ā¹ 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) Here, P = ā¹ 10800, T = 3 years,
R = 12 $$\frac{1}{2}$$% p.a.
We have: A = P[1 + $$\frac{R}{100}$$]n
= $$ā¹ 10800\left[1+\frac{12 \frac{1}{2}}{100}\right]^{3}$$
[āµ Interest compound annually, ā“n = 3]
= ā¹ 10800 $$\frac{225}{100}$$3
= ā¹ 10800 Ć $$\frac{225}{100}$$ Ć $$\frac{225}{100}$$ Ć $$\frac{225}{100}$$
= ā¹ $$\frac{675Ć9Ć9Ć9}{4Ć8}$$ = ā¹ $$\frac{492075}{32}$$
ā“ Amount = ā¹ 15377.34
Now, compound interest
= ā¹ 15377.34 – ā¹ 10800 = ā¹ 4577.34

(b) Here, P = ā¹ 18000, T = 2$$\frac{1}{2}$$years,
R = 10% p.a.
āµ Interest is compounded annually,
n = 2 + $$\frac{1}{2}$$

= ā¹ 9 Ć 11 Ć 11 Ć 21 = ā¹ 22869
ā“ Amount = ā¹ 22869
CI = ā¹ 22869 – ā¹ 18000 = ā¹ 4869

(c) Here P = ā¹ 62500, T = 1$$\frac{1}{2}$$ r = 8% p.a.
Compounded half yearly,
R = 8% p.a. = 4% per half year
T = 1$$\frac{1}{2}$$year ā n = 3 half years.
ā“ Amount = P(1 + $$\frac{R}{100}$$)n
= ā¹ 62500 (1 + $$\frac{4}{100}$$)3 = ā¹ 62500 ($$\frac{26}{25}$$)3
= ā¹ 62500 Ć $$\frac{26}{25}$$ Ć $$\frac{26}{25}$$ Ć $$\frac{26}{25}$$
= ā¹ 4 Ć 26 Ć 26 26 = ā¹ 70304
Amount = ā¹ 70304
CI = ā¹ 70304 – ā¹ 62500 = ā¹ 7804

(d) Here, P = ā¹ 8000, T = 1 year, R = 9% p.a.
Interest is compounded half yearly,
ā“ T = 1 year = 2 half years
R = 9% p.a. = $$\frac{9}{2}$$% half yearly
ā“ Amount = P(1 + $$\frac{R}{100}$$)n
= ā¹ 8000 Ć (1 + $$\frac{9}{200}$$)2
= ā¹ 8000 Ć $$\frac{209}{200}$$ Ć $$\frac{209}{200}$$ = ā¹ $$\frac{2 Ć 209 Ć 209}{10}$$
= ā¹ $$\frac{87362}{10}$$ = ā¹ 8736.20
CI = ā¹ 8736.20 – ā¹ 8000 = ā¹ 736.20

(e) Here, P = ā¹ 10000, T = 1 year
R = 8% p.a. compunded half yearly.
ā“ R = 8% p.a. = 4% per half yearly
T = 1 year ā n = 2 Ć 1 = 2
Now, amount = P(1 + $$\frac{R}{100}$$)n
= ā¹ 10,000 (1 + $$\frac{4}{100}$$)2 = ā¹ 10,000 ($$\frac{26}{25}$$)2
= ā¹ 10,000 Ć $$\frac{26}{25}$$ Ć $$\frac{26}{25}$$
CI = ā¹ 10816 – ā¹ 10000 = ā¹ 816.

Question 2.
Kamala borrowed ā¹ 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?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year for $$\frac{4}{12}$$ years.)
Solution:
Note: Here, we shall calculate the amount for 2 years using the CI formula. Then this amount will become the principal for next 4 months, i.e., $$\frac{4}{12}$$ years.
Here, P = ā¹ 26400, T = 2 years, R = 15% p.a
A = P[1 + $$\frac{R}{100}$$]n = ā¹ 26400[1 + $$\frac{15}{100}$$]2
= ā¹ 26400 Ć $$\frac{23}{20}$$2 = ā¹ 26400 Ć $$\frac{23}{20}$$ Ć $$\frac{23}{20}$$
= ā¹ 66 Ć 23 Ć 23 = ā¹ 34914
Again, P = 34914. T = 4 months = $$\frac{4}{12}$$ years,
R = 15% p.a.
ā“ Using, SI = $$\frac{PĆRĆT}{100}$$, we have
SI = ā¹ $$\frac{5819Ć3}{10}$$ = ā¹ $$\frac{17457}{10}$$ = ā¹ 1745.70
Amount = SI + P
= ā¹ (1745.70 + 34914) = ā¹ 36659.70
Thus, the required amount to be paid to the bank after 2 years 4 month = ā¹ 36659.70

Question 3.
Fabina borrows 12,500 at 12% per annum for 3 years al 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:
For Fabina
P = ā¹ 12500, T = 3 years, R = 12% p.a.
ā“ SI = $$\frac{PĆRĆT}{100}$$ = ā¹ $$\frac{12500Ć3Ć12}{100}$$
= ā¹ 125 Ć 3 Ć 12 = ā¹ 4500
For Radha P = ā¹ 12500, T = 3 years
R = 10% p.a. (Compounded annually)

ā“ CI = ā¹ 16637.5 – ā¹ 12500 = ā¹ 4137.50
Difference = ā¹ 4500 – ā¹ 4137.50 = ā¹ 362.50
Thus, Fabina pays ā¹ 362. 50 more.

Question 4.
I borrowed ā¹ 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:
For SI
Principal = ā¹ 12000; Time = 2 years
Rate = 6% p.a.
SI = $$\frac{PĆRĆT}{100}$$ = ā¹ $$\frac{12000Ć2Ć6}{100}$$
= ā¹ (120 Ć 2 Ć 6) = ā¹ 1440
For CI
Principal = ā¹ 12000; Time = 2 years
Rate = 6% p.a.

ā“ CI = ā¹ 13483. 20 – ā¹ 12000 = ā¹ 1483.20
Thus, excess amount = ā¹ 1483.20 – ā¹ 1440 = ā¹ 43.20

Question 5.
Vasudevan invested ā¹ 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) Amount after 6 months
P = ā¹ 60000, T = $$\frac{1}{2}$$ year, n = 1
[āµ Interest is compounded half yearly]
R = 12% p.a. = 6% per half year
ā“ Amount = P[1 + $$\frac{R}{100}$$]n
= ā¹ 6000 [1 + $$\frac{6}{100}$$]1 = ā¹ 60000 Ć $$\frac{53}{50}$$
= ā¹ 1200 Ć 53 = ā¹ 63600

(ii) Amount after 1 year
P = ā¹ 60000
T = 1 year; n = 2
R = 12% p.a. = 6% per half year
ā“ Amount = P[1 + $$\frac{R}{100}$$]n
= ā¹ 60000 Ć [1 + $$\frac{6}{100}$$]2
= ā¹ 60000 Ć $$\frac{53}{50}$$ Ć $$\frac{53}{50}$$ = ā¹ 24 Ć 53 Ć 53
= ā¹ 67416
Thus, amount after 6 months = ā¹ 63600 and amount after 1 year = ā¹ 67416

Question 6.
Arif took a loan of ā¹ 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) Compounded annually
P = 80000 R = 10% p.a.
T = $$\frac{1}{2}$$ years ā n = 1 + $$\frac{1}{2}$$ T = 1$$\frac{1}{2}$$ years
Amount for 1st year.
A = P[1 + $$\frac{R}{100}$$]n = ā¹ 80000 Ć [1 + $$\frac{10}{100}$$]1
= ā¹ 80000 Ć $$\frac{10}{100}$$ Ć $$\frac{1}{2}$$ = ā¹ 88000
SI on ā¹ 88000 for next $$\frac{1}{2}$$ year
= ā¹ 88000 Ć $$\frac{10}{100}$$ Ć $$\frac{1}{2}$$ = ā¹ 440 Ć 10 = ā¹ 4400
ā“ Amount = ā¹ 88000 + ā¹ 44000 = ā¹ 92400

(ii) Compounded half yearly
P = ā¹ 80000 P = ā¹ 80000
R = 10% p.a.
R = 10% p.a. = 5% per half year
T = 1$$\frac{1}{2}$$ years ā n = 1 + $$\frac{1}{2}$$ T = 1$$\frac{1}{2}$$ years
ā n = 3
Amount for 1st year.

ā“ Amount = ā¹ 88000 + ā¹ 44000 = ā¹ 92400
Thus, the difference between the two amounts = ā¹ 92610 – ā¹ 92400 = ā¹ 210

Question 7.
Maria invested ā¹ 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:
Principal = ā¹ 8000
Rate = 5% p.a. compounded annually
(i) Time = 2 years ā n = 2
ā“ Amount = P[1 + $$\frac{R}{100}$$]n
= ā¹ 8000[1 + $$\frac{5}{100}$$]2 = ā¹ 8000 Ć $$\left[\frac{21}{20}\right]^{2}$$
= ā¹ 8000 Ć $$\frac{21}{20}$$ Ć $$\frac{21}{20}$$ = ā¹ (20 Ć 21 Ć 21)
= ā¹ 8820
ā“ Amount credited against her name at the end of the two years = ā¹ 8820

(ii) Time = 3 years ā n = 3
ā“ Amount = ā¹ 8000[1 + $$\frac{5}{100}$$]3
= ā¹ 8000 $$\left[\frac{21}{20}\right]^{3}$$ = ā¹ 8000 Ć $$\frac{21}{20}$$ Ć $$\frac{21}{20}$$ Ć $$\frac{21}{20}$$
= ā¹ (21 Ć 21 Ć 21) = ā¹ 9261
āµ Interest paid during 3rd year
= [Amount at the end of 3rd year] – [Amount at the end of 2nd year]
= ā¹ 9261 – ā¹ 8820 = ā¹ 441

Question 8.
Find the amount and the compound interest on ā¹ 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:
Principal = ā¹ 10,000 Time = 1$$\frac{1}{2}$$%
Rate = 10% p.a.

Case I:
Interest is compounded half yearly
We have r = 10 p.a. = 5% per half yearly
T = 1$$\frac{1}{2}$$ years ā n = 3
ā“ A = P[1 + $$\frac{R}{100}$$]n = ā¹ 10000[1 + $$\frac{5}{100}$$]3
= ā¹ $$\frac{5Ć21Ć21Ć21}{4}$$ = ā¹ $$\frac{46305}{4}$$ = ā¹ 11576.25
ā“ Amount = ā¹ 11576.25
Now CI = Amount – Principal
= ā¹ 11576.25 – ā¹ 10,000 = ā¹ 1576.25

Case II:
Interest is compounded annually
We have R = 10% p.a. T = 1$$\frac{1}{2}$$year
Amount for 1 year = P[1 + $$\frac{R}{100}$$]n
= ā¹ 10000 Ć [1 + $$\frac{10}{100}$$]1 = ā¹ 10000 Ć $$\frac{11}{10}$$
= ā¹ 11000
ā“ Interest for 1st year = ā¹ 11000 – ā¹ 10,000
= ā¹ 1000
ā“ Interest for next $$\frac{1}{2}$$year on ā¹ 11000
= $$\frac{PĆRĆT}{100}$$ = ā¹ 11000 Ć $$\frac{10}{100}$$ Ć $$\frac{1}{2}$$ = ā¹ 550
ā“ Total interest = ā¹ 1000 + ā¹ 550 = ā¹ 1550
Since ā¹ 1576.25 > ā¹ 1550
ā“ Interest would be more in case if it is compounded half yearly.

Question 9.
Find the amount which Ram will get on ā¹ 4096, if he gave it for 18 months at 12$$\frac{1}{2}$$% per annum, interest being compounded half yearly?
Solution:
We have P = ā¹ 4096 T = 18 months
R = 12$$\frac{1}{2}$$% p.a.
āµ Interest is compounded half yearly.
T = 18 months ā n = $$\frac{18}{6}$$ = 3 six months
R = 12$$\frac{1}{2}$$% p.a. ā R = (12$$\frac{1}{2}$$ Ć· 2)% per half year.
= ($$\frac{25}{2}$$ Ć $$\frac{1}{2}$$)% = $$\frac{25}{4}$$% half yearly
Now, A = P[1 + $$\frac{R}{100}$$]n = ā¹ 4096 Ć [1 + $$\frac{25}{100}$$]3
= ā¹ 4096 Ć $$\frac{17}{16}$$3 = ā¹ 4096 Ć $$\frac{17}{16}$$ Ć $$\frac{17}{16}$$ Ć $$\frac{17}{16}$$
= ā¹ (17 Ć 17 Ć 17) = ā¹ 4913
Thus, the required amount = ā¹ 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 is 2005?
Solution:
Population in 2003 is P = 54000
(i) Let the population in 2001 (i.e., 2 years ago) be P
Since rate of increment in population = 5% p.a.
ā“ Present population = P(1 + $$\frac{5}{100}$$]2
or 54000 = P($$\frac{21}{20}$$]2 or 5400 = P($$\frac{441}{400}$$)
or P = $$\frac{54000 Ć 400}{441}$$ = 48979.59
= 48980 (approx)
Thus, the population in 2001 was about 48980

(ii) Initial population (in 2003). i.e., P = 54000
Rate of increment in population = 5% pa.
Time = 2 years ā n = 2
ā“ A = P[1 + $$\frac{R}{100}$$]2 = 54000 Ć $$\frac{21}{20}$$ Ć $$\frac{21}{20}$$
= 135 Ć 21 Ć 21 = 59535
Thus, the population in 2005 = 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 hours If the count was initially 5,06,000?
Solution:
Initial count of bactena (P) = 5,06,000
Increasing rate (R) = 2.5% per hour
Time (T) = 2 hours ā n = 2

= 53161625 or 531616 (approx.)
Thus, the count of bacteria after 2 hours will be 531616 (appiox.)

Question 12.
A scooter was bought at ā¹ 42,000 has value depredated at the raie of 8% per annum. Find its value after one year?
Solution:
Initial cost (value) of the scooter (P) = ā¹ 42000
Depreciation rate = 8% p.a.
Time = 1 year ā n = 1
Using A = P[1 + $$\frac{R}{100}$$]n, we have
A = ā¹ 42000 Ć [1 – $$\frac{8}{100}$$]1
= ā¹ 42000 Ć $$\frac{92}{100}$$ = ā¹ 420 Ć 92 = ā¹ 38640
Thus, the value of the scooter after 1 year will be ā¹ 38640.