Exercise 5.3
1. Sudhakar borrows ₹15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments?
Solution:
Simple Interest = Principal × Rate × Time
= ₹
= ₹
Total Amount to be repaid = Principal + Simple Interest
= ₹
= ₹
Number of monthly repayments =
=
Monthly repayment =
= ₹
= ₹
Therefore, Sudhakar's monthly repayment is ₹268.75.
2. A TV was bought at a price of ₹21000. After 1 year the value of the TV was depreciated by 5%. Find the value of the TV after 1 year.
Solution:
Original price of the TV = ₹
Depreciation percentage = 5 %
Depreciation amount = 5 % of ₹21000
=
= ₹
Value of the TV after 1 year = Original price - Depreciation amount
= ₹
= ₹
Therefore, the value of the TV after 1 year is ₹19950.
3. Find the amount and the compound interest on ₹8000 at 5% per annum, for 2 years compounded annually.
Solution:
Principal (P) = ₹8000
Rate (R) = 5 % per annum
Time (T) = 2 years
Amount (A) = P ×
= ₹
= ₹ 8000 ×
= ₹ 8000 ×
= ₹
Compound Interest (CI) = Amount - Principal
= ₹
= ₹
Therefore, the amount is ₹8820 and the compound interest is ₹820.
4. Find the amount and the compound interest on ₹6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.
Solution:
Principal (P) = ₹6500
Year 1:
Rate (R1) = 5%
Amount after 1 year (
= ₹ 6500 × (1 +
= ₹ 6500 ×
= ₹
Year 2:
Principal for year 2 =
Rate (
Amount after 2 years (
= ₹ 6825 × (1 +
= ₹ 6825 ×
= ₹
Compound Interest (CI) =
= ₹
= ₹
Therefore, the amount after 2 years is ₹7234.50, and the compound interest is ₹734.50.
5. Prathibha borrows ₹47000 from a finance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5-year period. Find: (a) How much amount Prathibha should repay the finance company at the end of five years. (b) Her equal monthly repayments.
Solution:
(a) Total amount to repay:
Principal (P) = ₹ 47000
Rate (R) = 17 %
Time (T) = 5 years
Simple Interest (SI) =
= ₹
= ₹
Total Amount (A) = P + SI
= ₹
= ₹
(b) Equal monthly repayments:
Number of months =
=
Monthly repayment =
= ₹
= ₹
Therefore:
(a) Prathibha should repay ₹86950 at the end of five years.
(b) Her equal monthly repayment is approximately ₹1449.17.
6. The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015?
Solution:
Initial population (2011) = 68,09,000
Annual growth rate = 4.7 %
Number of years = 2015 - 2011 =
Final Population = Initial Population ×
= 6809000 ×
= 6809000 ×
= 6809000 ×
= 6809000 ×
=
Therefore, the estimated population of Hyderabad at the end of 2015 will be approximately 81,82,199.
7. Find the Compound interest pa> id when a sum of ₹10000 is invested for 1 year and 3 months at 8 1/2 % per annum compounded annually.
Solution:
Time = 1 year and 3 months = 1 +
Amount after 1 year = Principal × (1 +
= ₹
= ₹ 10000 ×
= ₹
Interest for the remaining 3 months (0.25 years) on the amount at the end of year 1:
Interest = ₹ 10850 ×
= ₹ 10850 ×
= ₹
Final Amount = Amount after 1 year + Interest for 3 months
= ₹
= ₹
Compound Interest = Final Amount - Principal
= ₹
= ₹
Therefore, the compound interest pa> id is approximately ₹1080.31.
8. 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 1/2 years, if the interest is compounded annually and compounded half-yearly.
Solution:
1. Compounded Annually:
Principal (P) = ₹ 80,000
Rate (R) = 10 % per annum
Time (T) = 1.5 years
Amount after 1 year = ₹ 80,000 × (1 +
Interest for the next 0.5 years = ₹ 88,000 ×
Total Amount after 1.5 years = ₹ 88000 + ₹ 4400 = ₹
2. Compounded Half-Yearly:
Principal (P) = ₹ 80,000
Rate (R) =
Time (T) =
Amount after 1.5 years = ₹ 80,000 ×
= ₹ 80,000 ×
= ₹ 80,000 ×
= ₹
3. Difference:
Difference in amounts = ₹
Therefore, the difference in amounts Arif would be paying is ₹ 210.
9. I borrowed ₹12000 from Prasad at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compounded annually, what extra amount would I have to pay?
Solution:
1. Simple Interest Calculation:
Principal (P) = ₹ 12,000
Rate (R) = 6% per annum
Time (T) = 2 years
Simple Interest (SI) =
=
Total amount to be paid with simple interest = P + SI
= ₹
= ₹
2. Compound Interest Calculation:
Principal (P) = ₹ 12,000
Rate (R) = 6% per annum
Time (T) = 2 years
Amount (A) with compound interest = P ×
= ₹ 12,000 ×
= ₹ 12,000 ×
= ₹ 12,000 ×
= ₹
3. Difference:
Extra amount to be paid = Amount with compound interest - Amount with simple interest
= ₹
= ₹
Therefore, you would have to pay ₹43.20 extra if the interest were compounded annually.
10. 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
Solution:
Initial count = 5,06,000
Growth rate = 2.5 % per hour
Time = 2 hours
Final Count = Initial Count ×
= 506000 ×
= 506000 ×
= 506000 × (
= 506000 ×
=
Therefore, the bacteria count at the end of 2 hours will be approximately 5,31,616.
11. Kamala borrowed ₹26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
Solution:
Amount after 2 years:
Principal (P) = ₹26,400
Rate (R) = 15%
Time (T) = 2 years
Amount after 2 years = P ×
= ₹ 26,400 ×
= ₹ 26,400 × (
= ₹ 26,400 ×
= ₹
Simple interest for the remaining 4 months:
Principal for 4 months = ₹
Time = 4 months =
Simple Interest =
=
= ₹
Final amount:
Final Amount = Amount after 2 years + Simple Interest for 4 months
= ₹
= ₹
Therefore, Kamala will pay ₹36,554.70 at the end of 2 years and 4 months to clear the loan.
12. Bharathi borrows an amount of ₹12500 at 12% per annum for 3 years at simple interest, and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
Bharathi (Simple Interest):
Principal (P) = ₹ 12,500
Rate (R) = 12 % per annum
Time (T) = 3 years
Simple Interest (SI) =
=
= ₹
Madhuri (Compound Interest):
Principal (P) = ₹ 12,500
Rate (R) = 10 % per annum
Time (T) = 3 years
Amount (A) = P ×
= ₹ 12,500 ×
= ₹ 12,500 × (
= ₹ 12,500 ×
= ₹
Compound Interest (CI) = A - P
= ₹
= ₹
Comparison:
Bharathi's interest = ₹ 4,500
Madhuri's interest = ₹ 4,137.50
Thus,
Difference = Bharathi's interest - Madhuri's interest
= ₹ 4,500 - ₹ 4,137.50
= ₹
Therefore, Bharathi pays ₹362.50 more interest than Madhuri.
13. Machinery worth ₹10000 depreciated by 5%. Find its value after 1 year.
Solution:
Original value of machinery = ₹ 10000
Depreciation rate = 5%
Depreciation amount = 5% of ₹10000
=
= ₹
Value of machinery after 1 year = Original value - Depreciation amount
= ₹
= ₹
Therefore, the value of the machinery after 1 year is ₹ 9500.
14. Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.
Solution:
Present population = 12 lakh
Rate of increase = 4%
Time = 2 years
Population after 2 years = Present population ×
= 1200000 ×
= 1200000 × (
= 1200000 ×
=
Therefore, the population of the city after 2 years will be 12,97,920.
15. Calculate compound interest on ₹1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly.
Solution:
Principal (P) = ₹ 1000
Annual interest rate = 10 %
Quarterly interest rate =
Time = 1 year =
Amount (A) = P ×
= ₹ 1000 ×
= ₹ 1000 × (
= ₹ 1000 ×
= ₹
Compound Interest (CI) = Amount - Principal
= ₹
= ₹
Therefore, the compound interest is approximately ₹ 103.81.