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## S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Ex 2(B)

Calculate the amount and the compound interest by using the formula for compound interest :

Question 1.

Principal = ₹ 625, Rate of interest = 4 % p.a., Time (in years) = 2.

Solution:

Principal (P) = ₹ 625

Rate (R) = 4% p.a.

Period (n) = 2 years

∴ Amount (A) = P(1 + \(\frac { R }{ 100 }\))^{n}

= ₹ 625 (1 + \(\frac { 4 }{ 100 }\))^{2}

= ₹ 625 (1 + \(\frac { 1 }{ 25 }\))^{2} = 625 (\(\frac { 26 }{ 25 }\))^{2}

= ₹ 625 x \(\frac { 26 }{ 25 }\) x \(\frac { 26 }{ 25 }\) = ₹ 676

and C.I. = A – P = ₹ 676 – 625 = ₹ 51

Question 2.

Principal = ₹ 8000, Rate of interest = 15 % p.a., Time (in years) = 3.

Solution:

Principal (P) = ₹ 8000

Rate (R) = 15% p.a.

Period (n) = 3 years

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

= ₹ 8000 \(\left(1+\frac{\mathrm{15}}{100}\right)^3\)

= ₹ 8000 \(\left(\frac{23}{20}\right)^3\)

= ₹ 8000 x \(\frac {23 }{ 20 }\) x \(\frac {23 }{ 20 }\) x \(\frac {23 }{ 20 }\) = ₹ 12167

∴ C.I. = A – P = ₹ 12167 – 8000 = ₹ 4167

Question 3.

Principal = ₹ 1000, Rate of interest = 10 % p.a., Time (in years) = 3.

Solution:

Principal (P) = ₹ 1000

Rate (R) = 10% p.a.

Period (n) = 3 years

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

= ₹ 1000 \(\left(1+\frac{10}{100}\right)^3\)

= ₹ 1000 \(\left(\frac{11}{10}\right)^3\)

= ₹ 1000 x \(\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\) = ₹ 1331

∴ C.I. = A – P = ₹ 1331 – 1000 = ₹ 331

Question 4.

Principal = ₹ 8000, Rate of interest = 10 % half yearly, Time (in years) = 1\(\frac { 1 }{ 2 }\).

Solution:

Principal (P) = ₹ 8000

Rate (R) = 10% p.a. (half-yearly) or 5% half year

Period (n) = 1 \(\frac { 1 }{ 2 }\) years = 3 half years

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

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

= ₹ 8000 x \(\left(\frac{21}{20}\right)^3\)

= ₹ 8000 x \(\frac { 21 }{ 20 }\) x latex]\frac { 21 }{ 20 }[/latex] x latex]\frac { 21 }{ 20 }[/latex] = ₹ 9261

∴ C.I. = A – P = ₹ 9261 – 8000 = ₹ 1261

Question 5.

Principal = ₹ 700, Rate of interest = 20 % half yearly, Time (in years) = 1\(\frac { 1 }{ 2 }\).

Solution:

Principal (P) = ₹ 700

Rate (R) = 20% p.a. (half yearly) or 10% half yearly

Period (n)= 1 \(\frac { 1 }{ 2 }\) years or 3 half years

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

= ₹ 700 \(\left(1+\frac{10}{100}\right)^3\)

= ₹ 700 \(\left(\frac{11}{10}\right)^3\)

= ₹ 700 x \(\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\)

= ₹ \(\frac{9317}{10}\) = ₹ 931.70

and C.l. = A – P = ₹ 931.70 – 700 = ₹ 231.70

Question 6.

Sangeeta lent ₹ 40960 to Amar to purchase a shop at 12.5% per annum. If the interest is compounded semi-annually, find the interest paid by Amar after 1 \(\frac { 1 }{ 2 }\) years.

Solution:

Amount of loan (P) = ₹ 40960

Rate (R) = 12.5% = 12 \(\frac { 1 }{ 2 }\) = \(\frac { 25 }{ 2 }\) % p.a.

= \(\frac { 25 }{ 4 }\) % semi annually

Period (n) = 1 \(\frac { 1 }{ 2 }\) years or 3 half years

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

= ₹ 40960 \(\left(1+\frac{25}{4 \times 100}\right)^3\)

= ₹ 40960 \(\left(\frac{17}{16}\right)^3\)

= ₹ 40960 x \(\frac { 17 }{ 16 }\) x \(\frac { 17 }{ 16 }\) x \(\frac { 17 }{ 16 }\)

= ₹ 49130

∴ C. Interest = A – P

= ₹ 49130 – 40960 = ₹ 8170

Question 7.

Sudhir lent ₹ 2000 at compound interest at 10% payable yearly, while Prashant lent ₹ 2000 at compound interest at 10% payable half-yearly. Find the difference in the interest received by Sudhir and Prashant at the end of one year.

Solution:

Incase of Sudhir,

Amount lent (P) = ₹ 2000

Rate (R) = 10% p.a.

Period = 1 year

∴ Interest = \(\frac { PRT }{ 100 }\)

= ₹ \(\frac{2000 \times 10 \times 1}{100}\) = ₹ 200

Incase of Prashant,

Principal (P) = ₹ 2000

Rate (R) = 10% p.a. or 5% half-yearly

Period (n) = 1 year or 2 half years

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

= ₹ 2000 \(\left(1+\frac{5}{100}\right)^2\)

= ₹ 2000 x \(\left(\frac{21}{20}\right)^2\)

= ₹ 2000 x \(\frac { 21 }{ 20 }\) x \(\frac { 21 }{ 20 }\) = ₹ 2205

∴ C. Interest = A – P = ₹ 2205 – 2000 = ₹ 205

Difference in their interest paid = ₹ 205 – 200 = ₹ 5

Question 8.

Flow much will ₹ 25000 amount to in 2 years at compound interest, if the rates for the successive years be 4 and 5 per cent per year ?

Solution:

Principal (P) = ₹ 25000

Period (n) = 2 years

Rate for the first year (R_{1}) = 4% p.a.

and for the second year (R_{2}) = 5%

∴ Amount = P \(\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\)

= ₹ 25000 \(\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\)

= ₹ 25000 x \(\frac { 26 }{ 25 }\) x \(\frac { 21 }{ 20 }\) = ₹ 27300

Question 9.

Umesh set up a small factory by investing ₹ 40,000. During the first three successive years his profits were 5%, 10% and 15% respectively. If each year the profit was on previous year’s capital, calculate his total profit.

Solution:

Investment of Umesh (P) = ₹ 40,000

Period (n) = 3 years

Rate of profit for the first year (r_{1}) = 5%

for second year (r_{2}) = 10%

and for third year (r_{3}) = 15%

∴ Total amount after 3 years

= P \(\left(1+\frac{r_1}{100}\right)+\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)\)

= 40000 \(\left(1+\frac{5}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)\)

₹ 40,000 x \(\frac { 21 }{ 20 }\) x \(\frac { 11 }{ 10 }\) x \(\frac { 23 }{ 20 }\) = ₹ 53130

Profit = A – P = ₹ 53130 – 40000

= ₹ 13130

Question 10.

Himachal Pradesh State Electricity Board issued in July 1988, 20 year bonds worth ₹ 6.25 crore. The issue price of each bond is ₹ 100 and it carries an annual interest of 11.5%, compounded half-yearly. Jasbir invested ₹ 5000 in these bonds. Find the amount that he gets on maturity of the bonds in 2008.

[Given that (1.0575)^{40} = 9.35869]

Solution:

(1.0575)^{40} = 9.35869 is given amount of total bonds = ₹ 6.25 crore

and price of each bond = ₹ 100

Rate of interest (r) = 11.5% p.a. (half-yearly)

or 5.75% half-yearly

Jasbir’s investment = ₹ 5000

Period (n) = 2008 – 1988 = 20 years = 40 half years

∴Amount of the bonds after 40 half years

= P \(\left(1+\frac{r}{100}\right)^n=₹ 5000\left(1+\frac{5.75}{100}\right)^{40}\)

= ₹ 5000 (1 + 1.0575)^{40}

= ₹ 5000 x 9.35869 = ₹ 46793.45

∴ Maturity value of the bonds = ₹ 46793.45

Question 11.

A district contains 64000 inhabitants. If the population increases at the rate of 2 \(\frac { 1 }{ 2 }\) % per annum, find the number of inhabitants at the end of 3 years.

Solution:

Population of a district (P) = 64000

Rate of increase (R) = 2 \(\frac { 1 }{ 2 }\) % p.a.

Period (n) = 3 years

∴ Population after 3 years = P \(\left(1+\frac{\mathrm{R}}{100}\right)^n\)

= 64000 \(\left(1+\frac{5}{2 \times 100}\right)^3\)

= 64000 x \(\left(\frac{41}{40}\right)^3\)

= 64000 x \(\frac{41}{40} \times \frac{41}{40} \times \frac{41}{40}\)

Question 12.

The Nagar Palika of a certain city started campaign to kill stray dogs which numbered 1250 in the city. As a result, the population of stray dogs started decreasing at the rate of 20% per month. Calculate the number of stray dogs in the city three months after the campaign started.

Solution:

In a city no. of stray dogs in the beginning (P) = 1250

Rate of decrease (R) = 20% per month

Period (n) = 3 months

∴ Number of stray dogs after 3 months

= P \(\left(1-\frac{\mathrm{R}}{100}\right)^n=1250\left(1-\frac{20}{100}\right)^3\)

= 1250(\(\frac { 1 }{ 2 }\))³ = 1250 x \(\frac { 4 }{ 5 }\) x \(\frac { 4 }{ 5 }\) x \(\frac { 4 }{ 5 }\) = 640

Question 13.

8000 blood donors were registered with a charitable hospital. Some student organi-zations started mobilizing people for this noble cause. As a result, the number of donors registered, increased at the rate of 20% per half year. Find the total number of new registrants during 1 \(\frac { 1 }{ 2 }\) years.

Solution:

No. of blood donors in the beginning (P) = 8000

Rate of Increase = 20% per half year

Period (n) = 1 \(\frac { 1 }{ 2 }\) years or 3 half year

∴ Increased donors after 3 half years

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

= 800 \(\left(1+\frac{20}{100}\right)^3\)

= 8000 x \(\left(\frac{6}{5}\right)^3\) = 8000 x \(\frac { 6 }{ 5 }\) x \(\frac { 6 }{ 5 }\) x \(\frac { 6 }{ 5 }\)

= 13824

∴ Net increase in donors = A – P = 13824 – 8000 = 5824