ML Aggarwal Class 9 Solutions Chapter 2 delves into the intricacies of compound interest, unraveling its principles and applications. Through this comprehensive guide, students can grasp the essence of compound interest and its relevance in the world of mathematics and beyond.
ML Aggarwal Class 9 Solutions Chapter 2: Compound Interest PDF download
Ml aggarwal class 9 exercise 2.1 solutions bothiya pdf
Question: Find the amount and the compound interest on ₹8000 at 5% per annum for 2 years.
Solution:
- Principal (P) = ₹8000Rate of interest (r) = 5% p.a.Period (n) = 2 years
- Interest for the first year = (8000 * 5 * 1) / 100 = ₹400
- Principal for the second year = 8000 + 400 = ₹8400
- Interest for the second year = (8400 * 5 * 1) / 100 = ₹420
- Amount after the second year = 8400 + 420 = ₹8820
Total compound interest = 8820 – 8000 = ₹820
Question: A man invests ₹46875 at 4% per annum compound interest for 3 years. Calculate: (i) the amount standing to his credit at the end of the second year. (ii) the interest for the third year.
Solution:
- Principal (P) = ₹46875Rate of interest (r) = 4% p.a.Period (n) = 3 years
- Interest for the first year = (46875 * 4 * 1) / 100 = ₹1875
- Principal for the second year = 46875 + 1875 = ₹48750
- Interest for the second year = (48750 * 4 * 1) / 100 = ₹1950
- Amount at the end of the second year = 48750 + 1950 = ₹50700
Interest for the third year = (50700 * 4 * 1) / 100 = ₹2028
Question: Calculate the compound interest for the second year on ₹8000 for three years at 10% p.a. Also find the sum due at the end of the third year.
Solution:
- Principal (P) = ₹8000Rate of interest (r) = 10% p.a.Period (n) = 3 years
- Interest for the first year = (8000 * 10 * 1) / 100 = ₹800
- Principal for the second year = 8000 + 800 = ₹8800
- Interest for the second year = (8800 * 10 * 1) / 100 = ₹880
- Principal for the third year = 8800 + 880 = ₹9680
- Interest for the third year = (9680 * 10 * 1) / 100 = ₹968
Amount at the end of the third year = 9680 + 968 = ₹10648
Question: Ramesh invests ₹12800 for three years at the rate of 10% per annum compound interest. Find: (i) the sum due to Ramesh at the end of the first year. (ii) the interest he earns for the second year.
Solution:
- Principal (P) = ₹12800Rate of interest (r) = 10% p.a.Period (n) = 3 years
- Interest for the first year = (12800 * 10 * 1) / 100 = ₹1280
- Sum due at the end of the first year = 12800 + 1280 = ₹14080
- Interest for the second year = (14080 * 10 * 1) / 100 = ₹1408
- Sum due at the end of the second year = 14080 + 1408 = ₹15488
- Interest for the third year = (15488 * 10 * 1) / 100 = ₹1548.80
Total amount due = 15488 + 1548.80 = ₹17036.80
Question: The simple interest on a sum of money for 2 years at 12% per annum is ₹1380. Find:(i) the sum of money. (ii) the compound interest on this sum for one year payable half-yearly at the same rate.
Solution:
- Simple Interest (SI) = ₹1380Rate of interest (R) = 12% p.a.Period (T) = 2 years
- Sum of money (P) = (SI * 100) / (R * T)Sum of money (P) = (1380 * 100) / (12 * 2) = ₹5750
- Rate of interest (R) = 12% p.a. = 6% half-yearlyPeriod (n) = 1 year = 2 half-years
- Amount (A) = P * (1 + R/100)^nAmount (A) = 5750 * (1 + 6/100)^2
Compound Interest = A – P
Compound Interest = (A – P) = 6460.70 – 5750 = ₹710.70
Question: A person invests ₹10000 for two years at a certain rate of interest, compounded annually. At the end of one year, this sum amounts to ₹11200. Calculate: (i) the rate of interest per annum. (ii) the amount at the end of the second year.
Solution:
- Principal (P) = ₹10000Amount after 1 year = ₹11200Period (n) = 2 years
- Amount (A) = P * (1 + r/100)^nr = (100 * (A/P)^(1/n)) – 100r = (100 * (11200/10000)^(1/1)) – 100 = 12%
Amount after 2 years = 10000 * (1 + 12/100)^2 = ₹12544
Question: Mr. Dubey borrows ₹100000 from State Bank of India at 11% per annum compound interest. He repays ₹41000 at the end of the first year and ₹47700 at the end of the second year. Find the amount outstanding at the beginning of the third year.
Solution:
- Principal (P) = ₹100000Rate of interest (r) = 11% p.a.Period (n) = 3 years
- Interest for the first year = (100000 * 11 * 1) / 100 = ₹11000
- Amount at the end of the first year = 100000 + 11000 – 41000 = ₹70000
- Interest for the second year = (70000 * 11 * 1) / 100 = ₹7700
- Amount at the end of the second year = 70000 + 7700 – 47700 = ₹30500
Amount outstanding = 30500 * (1 + 11/100) = ₹33855
Question: Jaya borrowed ₹50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹33000 at the end of the first year. Find the amount she must pay at the end of the second year to clear her debt.
Solution:
- Principal (P) = ₹50000Rate of interest for the first year = 12% p.a.Rate of interest for the second year = 15% p.a.Period (n) = 2 years
- Interest for the first year = (50000 * 12 * 1) / 100 = ₹6000
- Amount at the end of the first year = 50000 + 6000 – 33000 = ₹206000
- Interest for the second year = (206000 * 15 * 1) / 100 = ₹30900
Amount at the end of the second year = 206000 + 30900 = ₹236900
Introduction to Compound Interest
Compound interest is a phenomenon where interest is not only earned on the initial amount, known as the principal, but also on the accumulated interest over time. This compounding effect can lead to exponential growth in savings or debt, depending on whether you’re saving money or borrowing it.
Understanding the Basics
Before delving into complex calculations, it’s essential to understand the basic components of compound interest: principal amount, interest rate, time period, and the final amount. These elements interplay to determine the overall growth or cost of an investment or loan.
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