OP Malhotra Class 9 Maths Solutions Chapter 2 Compound Interest Ex 2(B)

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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 (R₁) = 4% p.a.
and for the second year (R₂) = 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 (r₁) = 5%
for second year (r₂) = 10%
and for third year (r₃) = 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