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) ā¹ 10,800 for 3 years at 12 \(\frac{1}{2}\) % per annum compounded annually.

(b) ā¹ 18,000 for 2 \(\frac{1}{2}\) years at 10% per annum compounded annually.

(c) ā¹ 62,500 for 1 \(\frac{1}{2}\) years at 8% per annum compounded half yearly.

(d) ā¹ 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) ā¹ 10,000 for 1 year at 8% per annum compounded half yearly.

Solution:

(a) Here, P = ā¹ 10800, T = 3 years,

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

We have: A = P[1 + \(\frac{R}{100}\)]^{n}

= \(ā¹ 10800\left[1+\frac{12 \frac{1}{2}}{100}\right]^{3}\)

[āµ Interest compound annually, ā“n = 3]

= ā¹ 10800 \(\frac{225}{100}\)^{3}

= ā¹ 10800 Ć \(\frac{225}{100}\) Ć \(\frac{225}{100}\) Ć \(\frac{225}{100}\)

= ā¹ \(\frac{675Ć9Ć9Ć9}{4Ć8}\) = ā¹ \(\frac{492075}{32}\)

ā“ Amount = ā¹ 15377.34

Now, compound interest

= ā¹ 15377.34 – ā¹ 10800 = ā¹ 4577.34

(b) Here, P = ā¹ 18000, T = 2\(\frac{1}{2}\)years,

R = 10% p.a.

āµ Interest is compounded annually,

n = 2 + \(\frac{1}{2}\)

= ā¹ 9 Ć 11 Ć 11 Ć 21 = ā¹ 22869

ā“ Amount = ā¹ 22869

CI = ā¹ 22869 – ā¹ 18000 = ā¹ 4869

(c) Here P = ā¹ 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}

= ā¹ 62500 (1 + \(\frac{4}{100}\))^{3} = ā¹ 62500 (\(\frac{26}{25}\))^{3}

= ā¹ 62500 Ć \(\frac{26}{25}\) Ć \(\frac{26}{25}\) Ć \(\frac{26}{25}\)

= ā¹ 4 Ć 26 Ć 26 26 = ā¹ 70304

Amount = ā¹ 70304

CI = ā¹ 70304 – ā¹ 62500 = ā¹ 7804

(d) Here, P = ā¹ 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}

= ā¹ 8000 Ć (1 + \(\frac{9}{200}\))^{2}

= ā¹ 8000 Ć \(\frac{209}{200}\) Ć \(\frac{209}{200}\) = ā¹ \(\frac{2 Ć 209 Ć 209}{10}\)

= ā¹ \(\frac{87362}{10}\) = ā¹ 8736.20

CI = ā¹ 8736.20 – ā¹ 8000 = ā¹ 736.20

(e) Here, P = ā¹ 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}

= ā¹ 10,000 (1 + \(\frac{4}{100}\))^{2} = ā¹ 10,000 (\(\frac{26}{25}\))^{2}

= ā¹ 10,000 Ć \(\frac{26}{25}\) Ć \(\frac{26}{25}\)

CI = ā¹ 10816 – ā¹ 10000 = ā¹ 816.

Question 2.

Kamala borrowed ā¹ 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 = ā¹ 26400, T = 2 years, R = 15% p.a

A = P[1 + \(\frac{R}{100}\)]^{n} = ā¹ 26400[1 + \(\frac{15}{100}\)]^{2}

= ā¹ 26400 Ć \(\frac{23}{20}\)^{2} = ā¹ 26400 Ć \(\frac{23}{20}\) Ć \(\frac{23}{20}\)

= ā¹ 66 Ć 23 Ć 23 = ā¹ 34914

Again, P = 34914. T = 4 months = \(\frac{4}{12}\) years,

R = 15% p.a.

ā“ Using, SI = \(\frac{PĆRĆT}{100}\), we have

SI = ā¹ \(\frac{5819Ć3}{10}\) = ā¹ \(\frac{17457}{10}\) = ā¹ 1745.70

Amount = SI + P

= ā¹ (1745.70 + 34914) = ā¹ 36659.70

Thus, the required amount to be paid to the bank after 2 years 4 month = ā¹ 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 = ā¹ 12500, T = 3 years, R = 12% p.a.

ā“ SI = \(\frac{PĆRĆT}{100}\) = ā¹ \(\frac{12500Ć3Ć12}{100}\)

= ā¹ 125 Ć 3 Ć 12 = ā¹ 4500

For Radha P = ā¹ 12500, T = 3 years

R = 10% p.a. (Compounded annually)

ā“ CI = ā¹ 16637.5 – ā¹ 12500 = ā¹ 4137.50

Difference = ā¹ 4500 – ā¹ 4137.50 = ā¹ 362.50

Thus, Fabina pays ā¹ 362. 50 more.

Question 4.

I borrowed ā¹ 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 = ā¹ 12000; Time = 2 years

Rate = 6% p.a.

SI = \(\frac{PĆRĆT}{100}\) = ā¹ \(\frac{12000Ć2Ć6}{100}\)

= ā¹ (120 Ć 2 Ć 6) = ā¹ 1440

For CI

Principal = ā¹ 12000; Time = 2 years

Rate = 6% p.a.

ā“ CI = ā¹ 13483. 20 – ā¹ 12000 = ā¹ 1483.20

Thus, excess amount = ā¹ 1483.20 – ā¹ 1440 = ā¹ 43.20

Question 5.

Vasudevan invested ā¹ 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 = ā¹ 60000, T = \(\frac{1}{2}\) year, n = 1

[āµ Interest is compounded half yearly]

R = 12% p.a. = 6% per half year

ā“ Amount = P[1 + \(\frac{R}{100}\)]^{n}

= ā¹ 6000 [1 + \(\frac{6}{100}\)]^{1} = ā¹ 60000 Ć \(\frac{53}{50}\)

= ā¹ 1200 Ć 53 = ā¹ 63600

(ii) Amount after 1 year

P = ā¹ 60000

T = 1 year; n = 2

R = 12% p.a. = 6% per half year

ā“ Amount = P[1 + \(\frac{R}{100}\)]^{n}

= ā¹ 60000 Ć [1 + \(\frac{6}{100}\)]^{2}

= ā¹ 60000 Ć \(\frac{53}{50}\) Ć \(\frac{53}{50}\) = ā¹ 24 Ć 53 Ć 53

= ā¹ 67416

Thus, amount after 6 months = ā¹ 63600 and amount after 1 year = ā¹ 67416

Question 6.

Arif took a loan of ā¹ 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} = ā¹ 80000 Ć [1 + \(\frac{10}{100}\)]^{1}

= ā¹ 80000 Ć \(\frac{10}{100}\) Ć \(\frac{1}{2}\) = ā¹ 88000

SI on ā¹ 88000 for next \(\frac{1}{2}\) year

= ā¹ 88000 Ć \(\frac{10}{100}\) Ć \(\frac{1}{2}\) = ā¹ 440 Ć 10 = ā¹ 4400

ā“ Amount = ā¹ 88000 + ā¹ 44000 = ā¹ 92400

(ii) Compounded half yearly

P = ā¹ 80000 P = ā¹ 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 = ā¹ 88000 + ā¹ 44000 = ā¹ 92400

Thus, the difference between the two amounts = ā¹ 92610 – ā¹ 92400 = ā¹ 210

Question 7.

Maria invested ā¹ 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 = ā¹ 8000

Rate = 5% p.a. compounded annually

(i) Time = 2 years ā n = 2

ā“ Amount = P[1 + \(\frac{R}{100}\)]^{n}

= ā¹ 8000[1 + \(\frac{5}{100}\)]^{2} = ā¹ 8000 Ć \(\left[\frac{21}{20}\right]^{2}\)

= ā¹ 8000 Ć \(\frac{21}{20}\) Ć \(\frac{21}{20}\) = ā¹ (20 Ć 21 Ć 21)

= ā¹ 8820

ā“ Amount credited against her name at the end of the two years = ā¹ 8820

(ii) Time = 3 years ā n = 3

ā“ Amount = ā¹ 8000[1 + \(\frac{5}{100}\)]^{3}

= ā¹ 8000 \(\left[\frac{21}{20}\right]^{3}\) = ā¹ 8000 Ć \(\frac{21}{20}\) Ć \(\frac{21}{20}\) Ć \(\frac{21}{20}\)

= ā¹ (21 Ć 21 Ć 21) = ā¹ 9261

āµ Interest paid during 3rd year

= [Amount at the end of 3rd year] – [Amount at the end of 2nd year]

= ā¹ 9261 – ā¹ 8820 = ā¹ 441

Question 8.

Find the amount and the compound interest on ā¹ 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 = ā¹ 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} = ā¹ 10000[1 + \(\frac{5}{100}\)]^{3}

= ā¹ \(\frac{5Ć21Ć21Ć21}{4}\) = ā¹ \(\frac{46305}{4}\) = ā¹ 11576.25

ā“ Amount = ā¹ 11576.25

Now CI = Amount – Principal

= ā¹ 11576.25 – ā¹ 10,000 = ā¹ 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}

= ā¹ 10000 Ć [1 + \(\frac{10}{100}\)]^{1} = ā¹ 10000 Ć \(\frac{11}{10}\)

= ā¹ 11000

ā“ Interest for 1st year = ā¹ 11000 – ā¹ 10,000

= ā¹ 1000

ā“ Interest for next \(\frac{1}{2}\)year on ā¹ 11000

= \(\frac{PĆRĆT}{100}\) = ā¹ 11000 Ć \(\frac{10}{100}\) Ć \(\frac{1}{2}\) = ā¹ 550

ā“ Total interest = ā¹ 1000 + ā¹ 550 = ā¹ 1550

Since ā¹ 1576.25 > ā¹ 1550

ā“ Interest would be more in case if it is compounded half yearly.

Question 9.

Find the amount which Ram will get on ā¹ 4096, if he gave it for 18 months at 12\(\frac{1}{2}\)% per annum, interest being compounded half yearly?

Solution:

We have P = ā¹ 4096 T = 18 months

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

āµ 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} = ā¹ 4096 Ć [1 + \(\frac{25}{100}\)]^{3}

= ā¹ 4096 Ć \(\frac{17}{16}\)^{3} = ā¹ 4096 Ć \(\frac{17}{16}\) Ć \(\frac{17}{16}\) Ć \(\frac{17}{16}\)

= ā¹ (17 Ć 17 Ć 17) = ā¹ 4913

Thus, the required amount = ā¹ 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 ā¹ 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) = ā¹ 42000

Depreciation rate = 8% p.a.

Time = 1 year ā n = 1

Using A = P[1 + \(\frac{R}{100}\)]^{n}, we have

A = ā¹ 42000 Ć [1 – \(\frac{8}{100}\)]^{1}

= ā¹ 42000 Ć \(\frac{92}{100}\) = ā¹ 420 Ć 92 = ā¹ 38640

Thus, the value of the scooter after 1 year will be ā¹ 38640.