Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 2 Compound Interest Without Using Formula

Exercise 2(A)

Question 1. Determine the compound interest in each of the following cases without utilizing any formula:
(i) On Rs. 3500 for 2 years at 10% per annum.
(ii) On Rs. 6000 for 3 years at 5% per annum.
Answer:
(i) First year principal = Rs. 3500
Rate of interest = 10%
Interest for the first year = \( \frac{3500 \times 10 \times 1}{100} = \text{Rs. } 350 \)
Sum at the end of the first year = Rs. 3500 + Rs. 350 = Rs. 3850
Second year starting principal = Rs. 3850
Interest for the second year = \( \frac{3850 \times 10 \times 1}{100} = \text{Rs. } 385 \)
Sum at the end of the second year = Rs. 3850 + Rs. 385 = Rs. 4235
Total compound interest = Rs. 350 + Rs. 385 = Rs. 735

(ii) First year starting principal = Rs. 6000
Rate of interest = 5%
Interest for the first year = \( \frac{6000 \times 5 \times 1}{100} = \text{Rs. } 300 \)
Sum at the end of the first year = Rs. 6000 + Rs. 300 = Rs. 6300
Second year starting principal = Rs. 6300
Interest for the second year = \( \frac{6300 \times 5 \times 1}{100} = \text{Rs. } 315 \)
Sum at the end of the second year = Rs. 6300 + Rs. 315 = Rs. 6615
Third year starting principal = Rs. 6615
Interest for the third year = \( \frac{6615 \times 5 \times 1}{100} = \frac{33075}{100} = \text{Rs. } 330.75 \)
Sum at the end of the third year = Rs. 6615 + Rs. 330.75 = Rs. 6945.75
Total compound interest = Rs. 300 + Rs. 315 + Rs. 330.75 = Rs. 945.75
In simple words: To find the interest without a formula, calculate the simple interest for one year at a time. Add that interest to the principal before starting the next year.

Exam Tip: Be careful to use the updated amount from the previous year as the principal for the next year. Do not use the original starting principal for subsequent years.

Question 2. Find the compound interest accumulated in each of the following scenarios without using a direct formula:
(i) On Rs. 8000 for \( 2\frac{1}{2} \) years at 15% per annum.
(ii) On Rs. 20000 for \( 2\frac{1}{4} \) years at 10% per annum.
Answer:
(i) First year principal = Rs. 8000
Rate of interest = 15%
Time period = 1 year
Interest for the first year = \( \frac{8000 \times 15 \times 1}{100} = \text{Rs. } 1200 \)
Total sum after first year = Rs. 8000 + Rs. 1200 = Rs. 9200
Second year principal = Rs. 9200, Rate = 15%, Time = 1 year
Interest for the second year = \( \frac{9200 \times 15 \times 1}{100} = \text{Rs. } 1380 \)
Total sum after second year = Rs. 9200 + Rs. 1380 = Rs. 10580
For the remaining \( \frac{1}{2} \) year:
Starting principal = Rs. 10580, Rate = 15%, Time = \( \frac{1}{2} \) year
Interest for this period = \( \frac{10580 \times 15 \times 1}{100 \times 2} = \frac{79350}{100} = \text{Rs. } 793.50 \)
Total amount accumulated after \( 2\frac{1}{2} \) years = Rs. 10580 + Rs. 793.50 = Rs. 11373.50
Total compound interest = Rs. 11373.50 - Rs. 8000 = Rs. 3373.50

(ii) First year starting principal = Rs. 20000, Rate = 10%, Time = 1 year
Interest for the first year = \( \frac{20000 \times 10 \times 1}{100} = \text{Rs. } 2000 \)
Total amount after first year = Rs. 20000 + Rs. 2000 = Rs. 22000
Second year principal = Rs. 22000, Rate = 10%, Time = 1 year
Interest for the second year = \( \frac{22000 \times 10 \times 1}{100} = \text{Rs. } 2200 \)
Total amount after second year = Rs. 22000 + Rs. 2200 = Rs. 24200
For the remaining \( \frac{1}{4} \) year:
Starting principal = Rs. 24200, Rate = 10%, Time = \( \frac{1}{4} \) year
Interest for this fractional period = \( \frac{24200 \times 10 \times 1}{100 \times 4} = \frac{60500}{100} = \text{Rs. } 605 \)
Total amount accumulated after \( 2\frac{1}{4} \) years = Rs. 24200 + Rs. 605 = Rs. 24805
Total compound interest = Rs. 24805 - Rs. 20000 = Rs. 4805
In simple words: When the time is a fraction of a year, calculate the interest for the whole years first. Then use the final amount as the principal for the fractional year.

Exam Tip: For fractional years, remember to put the denominator of the fraction in the denominator of the interest calculation alongside 100.

Question 3. Determine the final amount and compound interest for the following:
(i) On Rs. 4600 for 2 years, where the interest rates are 10% for the first year and 12% for the second year.
(ii) On Rs. 16000 for 3 years, where the interest rates are 10% for the first year, 14% for the second year, and 15% for the third year.
Answer:
(i) First year starting principal = Rs. 4600, Rate = 10%, Time = 1 year
Interest for the first year = \( \frac{4600 \times 10 \times 1}{100} = \text{Rs. } 460 \)
Total amount after the first year = Rs. 4600 + Rs. 460 = Rs. 5060
Second year starting principal = Rs. 5060, Rate = 12%, Time = 1 year
Interest for the second year = \( \frac{5060 \times 12 \times 1}{100} = \frac{60720}{100} = \text{Rs. } 607.20 \)
Total amount after 2 years = Rs. 5060 + Rs. 607.20 = Rs. 5667.20
Total compound interest = Rs. 5667.20 - Rs. 4600 = Rs. 1067.20

(ii) First year starting principal = Rs. 16000, Rate = 10%, Time = 1 year
Interest for the first year = \( \frac{16000 \times 10 \times 1}{100} = \text{Rs. } 1600 \)
Total amount after the first year = Rs. 16000 + Rs. 1600 = Rs. 17600
Second year starting principal = Rs. 17600, Rate = 14%, Time = 1 year
Interest for the second year = \( \frac{17600 \times 14 \times 1}{100} = \frac{246400}{100} = \text{Rs. } 2464 \)
Total amount after the second year = Rs. 17600 + Rs. 2464 = Rs. 20064
Third year starting principal = Rs. 20064, Rate = 15%, Time = 1 year
Interest for the third year = \( \frac{20064 \times 15 \times 1}{100} = \text{Rs. } 3009.60 \)
Total amount after 3 years = Rs. 20064 + Rs. 3009.60 = Rs. 23073.60
Total compound interest = Rs. 23073.60 - Rs. 16000 = Rs. 7073.60
In simple words: When rates change every year, just use the specific rate given for each year to find that year's interest.

Exam Tip: Be careful not to use the first-year interest rate for the second or third years. Double-check which rate corresponds to which year.

Question 4. Find the final amount and compound interest on Rs. 2400 for \( 2\frac{1}{2} \) years at 5% interest per annum compounded annually.
Answer:
First year starting principal = Rs. 2400, Rate = 5%, Time = 1 year
Interest for the first year = \( \frac{2400 \times 5 \times 1}{100} = \text{Rs. } 120 \)
Total amount after the first year = Rs. 2400 + Rs. 120 = Rs. 2520
Second year starting principal = Rs. 2520, Rate = 5%, Time = 1 year
Interest for the second year = \( \frac{2520 \times 5 \times 1}{100} = \text{Rs. } 126 \)
Total amount after the second year = Rs. 2520 + Rs. 126 = Rs. 2646
For the final \( \frac{1}{2} \) year:
Starting principal = Rs. 2646, Rate = 5%, Time = \( \frac{1}{2} \) year
Interest for the last half-year = \( \frac{2646 \times 5 \times 1}{100 \times 2} = \text{Rs. } 66.15 \)
Total amount after \( 2\frac{1}{2} \) years = Rs. 2646 + Rs. 66.15 = Rs. 2712.15
Total compound interest = Rs. 2712.15 - Rs. 2400 = Rs. 312.15
In simple words: Work out the interest for the first and second years in full. Then, find the interest for the remaining six months using the final amount as the starting point.

Exam Tip: Keep your decimal additions clean when summing the final year's fractional interest to avoid simple calculation errors.

Question 5. Calculate the compound interest for the second year on Rs. 8000 at 10% interest per annum compounded annually.
Answer:
First year principal = Rs. 8000, Rate = 10%, Time = 1 year
Interest for the first year = \( \frac{8000 \times 10 \times 1}{100} = \text{Rs. } 800 \)
Amount at the end of the first year = Rs. 8000 + Rs. 800 = Rs. 8800
Second year principal = Rs. 8800, Rate = 10%, Time = 1 year
Interest for the second year = \( \frac{8800 \times 10 \times 1}{100} = \text{Rs. } 880 \)
Thus, the compound interest specifically for the second year is Rs. 880.
In simple words: The interest of the second year is computed on the total sum (principal + first-year interest) at the start of the second year.

Exam Tip: If the question asks only for the "interest for the second year", do not add the first-year interest to it. Provide only the second-year interest as the final answer.

Question 6. A borrowed Rs. 2500 from B at 12% per annum compound interest. After 2 years, A paid back Rs. 2936 and a watch to clear the debt. Find the value of the watch.
Answer:
First year principal = Rs. 2500, Rate = 12%, Time = 1 year
Interest for the first year = \( \frac{2500 \times 12 \times 1}{100} = \text{Rs. } 300 \)
Amount at the end of the first year = Rs. 2500 + Rs. 300 = Rs. 2800
Second year principal = Rs. 2800, Rate = 12%, Time = 1 year
Interest for the second year = \( \frac{2800 \times 12 \times 1}{100} = \text{Rs. } 336 \)
Total amount due after 2 years = Rs. 2800 + Rs. 336 = Rs. 3136
Cash amount paid by A = Rs. 2936
Value of the watch = Total amount due - Cash amount paid
\( \implies \) Value of the watch = Rs. 3136 - Rs. 2936 = Rs. 200
In simple words: Find the total amount A owes after two years. Subtract the cash paid from this total to find the value of the watch.

Exam Tip: Every \( \implies \) must be preceded by a line break. Ensure the units (Rs.) are written alongside each term in the final steps.

Question 7. Find the final amount of Rs. 50,000 after 3 years, if the rate of interest for the consecutive years is 6%, 8% and 10% per annum respectively.
Answer:
Interest for the first year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{50000 \times 6 \times 1}{100} = \text{Rs. } 3000 \)
Total amount after the first year = Rs. 50000 + Rs. 3000 = Rs. 53000
Interest for the second year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{53000 \times 8 \times 1}{100} = \text{Rs. } 4240 \)
Total amount after the second year = Rs. 53000 + Rs. 4240 = Rs. 57240
Interest for the third year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{57240 \times 10 \times 1}{100} = \text{Rs. } 5724 \)
Total amount after the third year = Rs. 57240 + Rs. 5724 = Rs. 62964
Hence, the final amount will be Rs. 62,964.
In simple words: Work out the interest for each year sequentially using that year's specific rate, and update the principal after every year.

Exam Tip: Clearly mention the formula \( \text{Simple Interest} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \) at least once when solving without a direct compound interest formula.

Question 8. Meenal invests Rs. 75,000 for 3 years at compound interest. The rate of interest for the first two years is 15% per annum and for the third year, it is 16% per annum. Find the total sum she will receive at the end of the third year.
Answer:
Interest for the first year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{75000 \times 15 \times 1}{100} = \text{Rs. } 11250 \)
Amount at the end of the first year = Rs. 75000 + Rs. 11250 = Rs. 86250
Interest for the second year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{86250 \times 15 \times 1}{100} = \text{Rs. } 12937.50 \)
Amount at the end of the second year = Rs. 86250 + Rs. 12937.50 = Rs. 99187.50
Interest for the third year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{99187.50 \times 16 \times 1}{100} = \text{Rs. } 15870 \)
Amount at the end of the third year = Rs. 99187.50 + Rs. 15870 = Rs. 115057.50
So, the total sum Meenal will receive at the end of the third year is Rs. 1,15,057.50.
In simple words: Calculate the interest year-by-year. Remember to change the rate from 15% to 16% in the third year.

Exam Tip: Be meticulous with calculations involving decimals (like Rs. 12,937.50) to ensure the subsequent steps remain entirely accurate.

Question 9. Find the difference between the compound interest compounded half-yearly and the simple interest on Rs. 18,000 for 1 year at 10% per annum.
Answer:
1. To calculate the Simple Interest (S.I.):
Principal (P) = Rs. 18000, Rate (R) = 10%, Time (T) = 1 year
S.I. = \( \frac{18000 \times 10 \times 1}{100} = \text{Rs. } 1800 \)

2. To calculate the Compound Interest (C.I.) compounded half-yearly:
For the first half-year:
Principal (P) = Rs. 18000, Rate (R) = 10%, Time (T) = \( \frac{1}{2} \) year
Interest = \( \frac{18000 \times 10 \times 1}{100 \times 2} = \text{Rs. } 900 \)
Amount after the first half-year = Rs. 18000 + Rs. 900 = Rs. 18900

For the second half-year:
Principal (P) = Rs. 18900, Rate (R) = 10%, Time (T) = \( \frac{1}{2} \) year
Interest = \( \frac{18900 \times 10 \times 1}{100 \times 2} = \text{Rs. } 945 \)
Amount after 1 year = Rs. 18900 + Rs. 945 = Rs. 19845
Total Compound Interest (C.I.) = Rs. 19845 - Rs. 18000 = Rs. 1845

Difference between C.I. and S.I. = Rs. 1845 - Rs. 1800 = Rs. 45
Thus, the gain is Rs. 45.
In simple words: Simple interest is calculated on the original amount once. Half-yearly compound interest is calculated twice in a year, with the interest from the first six months added to the principal.

Exam Tip: Compounding half-yearly means calculating interest every 6 months (Time = 1/2). Ensure you multiply the denominator of the interest formula by 2.

Question 10. Meenal lends Rs. 4,000 for 3 years at compound interest. The rate of interest is 8% for the first year and 10% per annum for the next two years. Find the compound interest at the end of the third year.
Answer:
Interest for the first year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{4000 \times 8 \times 1}{100} = \text{Rs. } 320 \)
Amount at the end of the first year = Rs. 4000 + Rs. 320 = Rs. 4320
Interest for the second year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{4320 \times 10 \times 1}{100} = \text{Rs. } 432 \)
Amount at the end of the second year = Rs. 4320 + Rs. 432 = Rs. 4752
Interest for the third year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{4752 \times 10 \times 1}{100} = \text{Rs. } 475.20 \)
Amount at the end of the third year = Rs. 4752 + Rs. 475.20 = Rs. 5227.20
Total compound interest accumulated = Rs. 5227.20 - Rs. 4000 = Rs. 1227.20
In simple words: Work out the interest for each of the three years one-by-one. Subtract the original Rs. 4000 from the final total amount to find the total interest.

Exam Tip: Be careful to subtract the initial principal from the final amount to get the compound interest, as the amount itself is not the compound interest.

Exercise 2(B)

Question 1. Calculate the difference between the compound interest and the simple interest on Rs. 4000 for 2 years at 8% per annum.
Answer:
First year starting principal = Rs. 4000, Rate = 8%, Time = 1 year
Interest for the first year = \( \frac{4000 \times 8 \times 1}{100} = \text{Rs. } 320 \)
Amount at the end of the first year = Rs. 4000 + Rs. 320 = Rs. 4320
Second year starting principal = Rs. 4320, Rate = 8%, Time = 1 year
Interest for the second year = \( \frac{4320 \times 8 \times 1}{100} = \text{Rs. } 345.60 \)
Amount at the end of the second year = Rs. 4320 + Rs. 345.60 = Rs. 4665.60
Total Compound Interest (C.I.) = Rs. 4665.60 - Rs. 4000 = Rs. 665.60

Simple Interest (S.I.) for 2 years = \( \frac{4000 \times 8 \times 2}{100} = \text{Rs. } 640 \)

Difference between C.I. and S.I. = Rs. 665.60 - Rs. 640 = Rs. 25.60
In simple words: Find the compound interest by stepping through each year, then calculate the standard simple interest for the whole period. Subtract the two to find the difference.

Exam Tip: Double-check your simple interest calculation using the direct formula \( \text{S.I.} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \) to save time.

Question 2. Find the difference between the compound interest of the third year and the first year on Rs. 12,500 for 3 years, when the rates of interest for the three successive years are 12%, 15% and 18% respectively.
Answer:
First year principal = Rs. 12500, Rate = 12%, Time = 1 year
Interest for the first year = \( \frac{12500 \times 12 \times 1}{100} = \text{Rs. } 1500 \)
Amount at the end of the first year = Rs. 12500 + Rs. 1500 = Rs. 14000
Second year principal = Rs. 14000, Rate = 15%, Time = 1 year
Interest for the second year = \( \frac{14000 \times 15 \times 1}{100} = \text{Rs. } 2100 \)
Amount at the end of the second year = Rs. 14000 + Rs. 2100 = Rs. 16100
Third year principal = Rs. 16100, Rate = 18%, Time = 1 year
Interest for the third year = \( \frac{16100 \times 18 \times 1}{100} = \text{Rs. } 2898 \)
Amount at the end of the third year = Rs. 16100 + Rs. 2898 = Rs. 18998
Difference between the interest of the third year and the first year = Rs. 2898 - Rs. 1500 = Rs. 1398
In simple words: Calculate the interest specifically for year 1 and year 3. Then find the difference between these two values.

Exam Tip: Keep the interest amounts for year 1 and year 3 separate from the accumulated sums so you can subtract them directly at the end.

Question 3. The compound interest on a certain sum of money for the second year exceeds the interest for the first year by Rs. 96. If the rate of interest is 8% per annum, find the sum of money.
Answer:
Let us assume the starting principal is Rs. 100.
For the first year:
Principal (P) = Rs. 100, Rate (R) = 8%, Time (T) = 1 year
Interest for the first year = \( \frac{100 \times 8 \times 1}{100} = \text{Rs. } 8 \)
Amount at the end of the first year = Rs. 100 + Rs. 8 = Rs. 108

For the second year:
Principal (P) = Rs. 108, Rate (R) = 8%, Time (T) = 1 year
Interest for the second year = \( \frac{108 \times 8 \times 1}{100} = \text{Rs. } 8.64 \)

Difference in interest between the second year and the first year = Rs. 8.64 - Rs. 8 = Rs. 0.64
If the difference in interest is Rs. 0.64, then the principal is Rs. 100.
When the difference in interest is Rs. 96, the principal = \( \text{Rs. } \frac{96 \times 100}{0.64} = \text{Rs. } 15,000 \)
In simple words: Start by assuming a simple principal of Rs. 100 to find the difference in interest. Use a ratio to scale up this difference to the real value of Rs. 96 to find the actual principal.

Exam Tip: The "assume Rs. 100" method is highly reliable for questions where the principal is unknown. Show the ratio step clearly to gain full marks.

Question 4. A person borrows Rs. 5000 at an interest rate of 12% per half-year compounded semi-annually. He repays Rs. 1800 at the end of every 6 months. Calculate the third payment he has to make at the end of 18 months to completely clear the loan.
Answer:
Principal borrowed at start = Rs. 5000
Rate of interest = 12% per half-year
Interest for the first 6 months = \( \frac{12}{100} \times \text{Rs. } 5000 = \text{Rs. } 600 \)
Amount due at the end of the first 6 months = Rs. 5000 + Rs. 600 = Rs. 5600
Amount remaining after the first payment = Rs. 5600 - Rs. 1800 = Rs. 3800

Interest on Rs. 3800 for the second 6 months = \( \frac{12}{100} \times \text{Rs. } 3800 = \text{Rs. } 456 \)
Amount due at the end of 12 months = Rs. 3800 + Rs. 456 = Rs. 4256
Amount remaining after the second payment = Rs. 4256 - Rs. 1800 = Rs. 2456

Interest on Rs. 2456 for the third 6 months = \( \frac{12}{100} \times \text{Rs. } 2456 = \text{Rs. } 294.72 \)
Amount due at the end of 18 months = Rs. 2456 + Rs. 294.72 = Rs. 2750.72
Therefore, the third payment required to clear the entire debt is Rs. 2750.72.
In simple words: Add the interest earned at the end of each half-year to the balance, then subtract the Rs. 1800 payment. Calculate the next interest on this new balance.

Exam Tip: Be sure to subtract the repayment amount (Rs. 1800) at the end of each period before calculating interest for the next period.

Question 5. A man borrows Rs. 6,000 at 5% per annum compound interest. If he repays Rs. 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.
Answer:
Initial principal borrowed = Rs. 6000
Rate of interest per annum = 5%
Interest for the first year = \( \frac{5}{100} \times \text{Rs. } 6000 = \text{Rs. } 300 \)
Amount accumulated at the end of the first year = Rs. 6000 + Rs. 300 = Rs. 6300
Balance remaining after the first repayment = Rs. 6300 - Rs. 1200 = Rs. 5100

Interest on Rs. 5100 for the second year = \( \frac{5}{100} \times \text{Rs. } 5100 = \text{Rs. } 255 \)
Amount accumulated at the end of the second year = Rs. 5100 + Rs. 255 = Rs. 5355
Balance remaining after the second repayment = Rs. 5355 - Rs. 1200 = Rs. 4155
Hence, the outstanding loan amount at the beginning of the third year is Rs. 4,155.
In simple words: Calculate the interest at the end of the year, add it to the balance, subtract the annual payment, and repeat this for the second year to find the remaining debt.

Exam Tip: The loan outstanding at the beginning of the third year is equal to the amount remaining at the end of the second year after the repayment has been deducted.

Question 6. The difference between the simple interest and the compound interest compounded half-yearly on a certain sum of money for 1 year at 10% per annum is Rs. 180. Find the sum.
Answer:
Let us assume the principal (P) is Rs. 100.
Rate (R) = 10%, Time (T) = 1 year
Simple Interest (S.I.) = \( \frac{100 \times 10 \times 1}{100} = \text{Rs. } 10 \)

Now, calculating the Compound Interest (C.I.) compounded half-yearly:
Rate = 5% per half-year, Time = 1 half-year (for each step)
For the first half-year:
Interest = \( \frac{100 \times 5 \times 1}{100} = \text{Rs. } 5 \)
Amount = Rs. 100 + Rs. 5 = Rs. 105

For the second half-year:
Principal = Rs. 105
Interest = \( \frac{105 \times 5 \times 1}{100} = \text{Rs. } 5.25 \)
Total Compound Interest (C.I.) = Rs. 5 + Rs. 5.25 = Rs. 10.25
Difference between C.I. and S.I. = Rs. 10.25 - Rs. 10 = Rs. 0.25

If the difference is Rs. 0.25, the principal is Rs. 100.
If the difference is Rs. 180, the principal = \( \frac{100}{0.25} \times 180 = \text{Rs. } 72000 \)
Hence, the required sum of money is Rs. 72,000.
In simple words: Find the difference between compound and simple interest on Rs. 100. Scale that difference up to the actual difference of Rs. 180 to find the total sum.

Exam Tip: Remember that compounding half-yearly means halving the annual interest rate (from 10% to 5%) and calculating interest over two six-month intervals.

Question 7. The value of a machine depreciates at the rate of 15% per annum. If the depreciation during the second year is Rs. 5,355, find the original cost of the machine.
Answer:
Let us assume the original cost of the machine is Rs. 100.
Depreciation during the first year = 15% of Rs. 100 = Rs. 15
Value of the machine at the beginning of the second year = Rs. 100 - Rs. 15 = Rs. 85
Depreciation during the second year = 15% of Rs. 85 = Rs. 12.75

When the second-year depreciation is Rs. 12.75, the original cost is Rs. 100.
When the second-year depreciation is Rs. 5355, the original cost = \( \frac{100}{12.75} \times 5355 = \text{Rs. } 42,000 \)
Hence, the original cost of the machine is Rs. 42,000.
In simple words: Find the second-year depreciation starting with a cost of Rs. 100. Use ratios to find the actual original price based on the real depreciation amount.

Exam Tip: Depreciation works like reverse compound interest. Ensure you subtract the first-year depreciation from the cost before finding the second year's depreciation.

Question 8. A sum of Rs. 5600 is lent at 14% per annum compound interest. Find:
(i) The interest for the first year.
(ii) The amount at the end of the first year.
(iii) The interest for the second year (nearly).
Answer:
(i) Principal (P) = Rs. 5600, Rate (R) = 14%, Time (T) = 1 year
Interest for the first year = \( \frac{5600 \times 14 \times 1}{100} = \text{Rs. } 784 \)

(ii) Amount at the end of the first year = Rs. 5600 + Rs. 784 = Rs. 6384

(iii) Principal for the second year = Rs. 6384, Rate = 14%, Time = 1 year
Interest for the second year = \( \frac{6384 \times 14 \times 1}{100} = \text{Rs. } 893.76 \approx \text{Rs. } 894 \text{ (nearly)} \)
In simple words: Solve each sub-part step-by-step. Use the sum from part (ii) as the principal to find the interest in part (iii).

Exam Tip: Pay attention to the instruction "(nearly)". Round your final decimal answer to the nearest whole rupee if requested.

Question 9(i). Find the difference between the interest of the third year and the second year on a sum of Rs. 48,000 for 3 years at 10% per annum compound interest.
Answer:
Starting principal (P) = Rs. 48,000
Interest for the first year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{48000 \times 10 \times 1}{100} = \text{Rs. } 4800 \)
Amount at the end of the first year = Rs. 48,000 + Rs. 4,800 = Rs. 52,800
Interest for the second year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{52800 \times 10 \times 1}{100} = \text{Rs. } 5,280 \)
Amount at the end of the second year = Rs. 52,800 + Rs. 5,280 = Rs. 58,080
Interest for the third year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{58080 \times 10 \times 1}{100} = \text{Rs. } 5,808 \)
Difference between the interest of the third year and the second year = Rs. 5,808 - Rs. 5,280 = Rs. 528
In simple words: Calculate the interest values for both the second and third years. Subtract the second year's interest from the third year's interest to find the difference.

Exam Tip: Be sure not to subtract the final amounts; the question asks for the difference between the interest of the two years.

Question 9(ii). A sum of Rs. 50,000 is invested at compound interest for three years, with interest rates being 10%, 12%, and 14% for the first, second, and third years respectively. Calculate the total interest earned during the first and third years.
Answer:
Interest for the first year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{50000 \times 10 \times 1}{100} = \text{Rs. } 5,000 \)
Amount at the end of the first year = Rs. 50,000 + Rs. 5,000 = Rs. 55,000
Interest for the second year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{55000 \times 12 \times 1}{100} = \text{Rs. } 6,600 \)
Amount at the end of the second year = Rs. 55,000 + Rs. 6,600 = Rs. 61,600
Interest for the third year = \( \frac{\text{P} \times \text{R} \times \text{T}}{100} = \frac{61600 \times 14 \times 1}{100} = \text{Rs. } 8,624 \)
Total interest earned during the first and third years = Rs. 5,000 + Rs. 8,624 = Rs. 13,624
In simple words: Work out the interest for each of the three years. Then add together the interest amounts from year 1 and year 3.

Exam Tip: Skip adding the second year's interest to your final sum, as the question specifically requests the sum of the first and third years' interests.

Question 10. A man saves Rs. 3,000 at the end of each year in a bank that pays interest at 10% per annum compounded annually. Find his total savings at the end of the third year.
Answer:
Savings deposited at the end of the first year = Rs. 3000
For the second year:
Starting principal = Rs. 3000, Rate = 10%, Time = 1 year
Interest earned = \( \frac{3000 \times 10 \times 1}{100} = \text{Rs. } 300 \)
Value at the end of the second year = Rs. 3000 + Rs. 300 = Rs. 3300
Deposit added at the end of the second year = Rs. 3000
Total principal for the third year = Rs. 3300 + Rs. 3000 = Rs. 6300

For the third year:
Starting principal = Rs. 6300, Rate = 10%, Time = 1 year
Interest earned = \( \frac{6300 \times 10 \times 1}{100} = \text{Rs. } 630 \)
Amount at the end of the third year (before the last deposit) = Rs. 6300 + Rs. 630 = Rs. 6930
Deposit added at the end of the third year = Rs. 3000
Total accumulated savings at the end of the third year = Rs. 6930 + Rs. 3000 = Rs. 9930
In simple words: The savings grow by 10% interest each year. Add a new deposit of Rs. 3000 at the end of each year to find the total sum.

Exam Tip: Be careful with the timing of deposits. Since deposits are made at the "end of each year", the first deposit earns interest for two years, the second deposit earns interest for one year, and the third deposit earns no interest.

Question 11. A man borrows Rs. 10,000 at 5% per annum compound interest. At the end of the first year, he pays back 35% of the total amount outstanding. At the end of the second year, he pays back 42% of the amount then outstanding. Find the amount he must pay at the end of the third year to clear his entire debt.
Answer:
The initial principal borrowed is Rs. 10,000.
Interest accrued for the first year is calculated as:
Interest = \( \frac{P \times R \times T}{100} \)
= \( \frac{10,000 \times 5 \times 1}{100} \)
= Rs. 500
Hence, the cumulative debt at the end of the first year is:
Rs. 10,000 + Rs. 500 = Rs. 10,500
At this stage, the borrower pays back 35% of Rs. 10,500:
Amount repaid = \( \frac{35}{100} \times 10,500 \) = Rs. 3,675
The remaining debt to be carried forward is:
Rs. 10,500 - Rs. 3,675 = Rs. 6,825
For the second year, this remaining balance of Rs. 6,825 serves as the principal.
Interest for the second year is computed as:
Interest = \( \frac{6,825 \times 5 \times 1}{100} \)
= Rs. 341.25
Therefore, the total outstanding amount at the end of the second year is:
Rs. 6,825 + Rs. 341.25 = Rs. 7,166.25
At this point, the borrower pays back 42% of Rs. 7,166.25:
Amount repaid = \( \frac{42}{100} \times 7,166.25 \) = Rs. 3,009.825
The outstanding balance is:
Rs. 7,166.25 - Rs. 3,009.825 = Rs. 4,156.425
For the third year, interest is calculated on Rs. 4,156.425:
Interest = \( \frac{4,156.425 \times 5 \times 1}{100} \)
= Rs. 207.82125
The total amount due at the end of the third year is:
Rs. 4,156.425 + Rs. 207.82125 = Rs. 4,364.24625
Thus, to completely clear his debt at the end of the third year, he must pay Rs. 4,364.24625.
In simple words: To find the final repayment, calculate the interest and add it to the loan each year. Then, subtract the percentage portion he paid back to find the new starting principal for the next year.

Exam Tip: Always remember that the percentage paid back is calculated on the total outstanding amount at the end of that year, not on the original sum borrowed.

Question 12. A man saves Rs. 8,000 at the beginning of each year in a bank that pays 10% compound interest per annum. Find his total savings at the beginning of the third year.
Answer:
During the 1st year:
Principal (P) = Rs. 8,000, Rate (R) = 10%, Time (T) = 1 year.
Interest accrued = \( \frac{8000 \times 10 \times 1}{100} \) = Rs. 800.
Amount at year-end = Rs. 8,000 + Rs. 800 = Rs. 8,800.
During the 2nd year:
A new saving of Rs. 8,000 is added at the start of the year.
New Principal (P) = Rs. 8,800 + Rs. 8,000 = Rs. 16,800.
Interest accrued = \( \frac{16800 \times 10 \times 1}{100} \) = Rs. 1,680.
Amount at year-end = Rs. 16,800 + Rs. 1,680 = Rs. 18,480.
Total savings at the beginning of the 3rd year:
Another deposit of Rs. 8,000 is made at the start of the third year.
Total savings = Rs. 18,480 + Rs. 8,000 = Rs. 26,480.
In simple words: The man adds Rs. 8,000 to his savings account at the start of every year. We calculate interest on the updated balance at the end of each year to find how much money accumulates.

Exam Tip: Be careful with the phrase "at the beginning of the third year". You must add the third year's deposit of Rs. 8,000 to the accumulated balance of the first two years to find the final savings.

Exercise 2(C)

Question 1. The compound interest on a sum of money for two consecutive years are Rs. 5,700 and Rs. 7,410 respectively. Find the rate of interest.
Answer:
The difference in interest between two consecutive years helps us find the rate of interest.
Difference in interest = Rs. 7,410 - Rs. 5,700 = Rs. 1,710.
This difference of Rs. 1,710 is the interest earned on Rs. 5,700 for one year.
Using the formula for rate of interest:
Rate of interest = \( \frac{\text{Difference in interest} \times 100}{\text{C.I. of preceding year} \times \text{Time}} \% \)
= \( \frac{1710 \times 100}{5700 \times 1} \% \)
= 30% per annum.
In simple words: The interest increase from one year to the next is just the interest earned on the previous year's interest. We divide this difference by the first year's interest and multiply by 100 to get the rate.

Exam Tip: When interests for two consecutive years are given, the interest on the first year's interest is the difference between the two interests. This is a very quick way to find the rate of interest.

Question 2. The compound interest on a certain sum of money for two successive half-years are Rs. 650 and Rs. 760.50 respectively. Find the rate of interest per annum.
Answer:
The difference in interest between two consecutive half-year periods is:
Rs. 760.50 - Rs. 650 = Rs. 110.50.
This extra amount of Rs. 110.50 is the interest earned on Rs. 650 for a duration of one half-year.
Let R be the rate of interest per annum.
Interest = \( \frac{P \times R \times T}{100} \)
Substituting the values:
110.50 = \( \frac{650 \times R \times \frac{1}{2}}{100} \)

\implies R = \( \frac{110.50 \times 100 \times 2}{650} \)

\implies R = 34% per annum.
In simple words: The difference in interest between two consecutive half-years is the interest on the first half-year's interest. We use the simple interest formula with a time period of half a year to find the annual rate.

Exam Tip: Remember to set the time period to half a year (\( T = \frac{1}{2} \)) when calculating interest compounded half-yearly, so that you find the annual rate of interest correctly.

Question 3. A sum of money compounded annually amounts to Rs. 5,292 in two years and to Rs. 5,556.60 in three years. Find:
(i) the rate of interest,
(ii) the original sum.
Answer:
(i) Let's determine the rate of interest:
Amount after 2 years = Rs. 5,292
Amount after 3 years = Rs. 5,556.60
The interest earned in the third year is the difference between these two amounts:
Interest for 3rd year = Rs. 5,556.60 - Rs. 5,292 = Rs. 264.60
This Rs. 264.60 is the interest earned on the principal of Rs. 5,292 for 1 year.
Rate of interest = \( \frac{\text{Interest} \times 100}{\text{Principal} \times \text{Time}} \% \)
= \( \frac{264.60 \times 100}{5292 \times 1} \% \)
= 5% per annum.

(ii) Let's find the original sum:
Assume the initial sum of money is Rs. 100.
Interest for the 1st year = 5% of Rs. 100 = Rs. 5.
Amount after 1 year = Rs. 100 + Rs. 5 = Rs. 105.
Interest for the 2nd year = 5% of Rs. 105 = Rs. 5.25.
Amount after 2 years = Rs. 105 + Rs. 5.25 = Rs. 110.25.
If the amount after 2 years is Rs. 110.25, the original sum is Rs. 100.
If the amount after 2 years is Rs. 5,292, the original sum is:
Sum = \( \frac{100 \times 5292}{110.25} \) = Rs. 4,800.
In simple words: First, find the rate by calculating the interest earned on the 2nd-year amount during the 3rd year. Then, work backwards from Rs. 100 to find the original sum that grows to the given 2-year amount.

Exam Tip: When using the unitary method to find the principal, ensure your calculations for the 2-year amount of Rs. 100 are completely accurate to avoid errors in the final step.

Question 4. The compound interest on a certain sum of money for the second and third years are Rs. 1,089 and Rs. 1,197.90 respectively. Find:
(i) the rate of interest per annum,
(ii) the sum of money.
Answer:
(i) Interest earned in the 2nd year = Rs. 1,089
Interest earned in the 3rd year = Rs. 1,197.90
Difference in interest between consecutive years = Rs. 1,197.90 - Rs. 1,089 = Rs. 108.90
This difference of Rs. 108.90 is the interest on Rs. 1,089 for 1 year.
Rate of interest = \( \frac{100 \times \text{Interest}}{\text{Principal} \times \text{Time}} \% \)
= \( \frac{100 \times 108.90}{1089 \times 1} \% \)
= 10% per annum.

(ii) Let the original principal be Rs. 100.
Interest for the 1st year = 10% of Rs. 100 = Rs. 10.
Amount at the end of the 1st year = Rs. 100 + Rs. 10 = Rs. 110.
Interest for the 2nd year = 10% of Rs. 110 = Rs. 11.
If the 2nd year interest is Rs. 11, the sum is Rs. 100.
If the 2nd year interest is Rs. 1,089, then the sum is:
Sum = \( \frac{100 \times 1089}{11} \) = Rs. 9,900.
In simple words: The interest increase between consecutive years helps find the rate. Then, using a base of Rs. 100, we find what the interest would be in the second year and scale it up to find our original sum.

Exam Tip: Do not mistake the second year's interest for the first year's interest. The second year's interest is calculated on the amount at the end of the first year.

Question 5. A sum of Rs. 8,000 amounts to Rs. 9,440 in one year, interest being compounded annually. Find the rate of interest and the compound interest for the third year.
Answer:
For the first year:
Principal (P) = Rs. 8,000, Amount (A) = Rs. 9,440, Time (T) = 1 year.
Interest earned = Rs. 9,440 - Rs. 8,000 = Rs. 1,440.
Rate of interest = \( \frac{\text{Interest} \times 100}{\text{Principal} \times \text{Time}} \% \)
= \( \frac{1440 \times 100}{8000 \times 1} \% \)
= 18% per annum.

For the second year:
New Principal (P) = Rs. 9,440, Rate (R) = 18%, Time (T) = 1 year.
Interest = Rs. \( \frac{9440 \times 18 \times 1}{100} \) = Rs. 1,699.20.
Amount at year-end = Rs. 9,440 + Rs. 1,699.20 = Rs. 11,139.20.

For the third year:
New Principal (P) = Rs. 11,139.20, Rate (R) = 18%, Time (T) = 1 year.
Interest = Rs. \( \frac{11139.20 \times 18 \times 1}{100} \) = Rs. 2,005.06.
In simple words: Find the rate from the first year's interest. Then use that rate to compute the compound interest year by year until you reach the third year.

Exam Tip: Be sure to calculate the principal for each year step-by-step by adding the previous year's interest, rather than using the original principal.

Question 6. A sum of Rs. 15,000 amounts to Rs. 15,600 in 6 months, interest being compounded half-yearly. Find:
(i) the rate of interest per annum,
(ii) the amount after 1.5 years.
Answer:
(i) For the 1st half-year:
Principal (P) = Rs. 15,000, Amount (A) = Rs. 15,600, Time (T) = \( \frac{1}{2} \) year.
Interest earned = Rs. 15,600 - Rs. 15,000 = Rs. 600.
Rate of interest = \( \frac{\text{Interest} \times 100}{\text{Principal} \times \text{Time}} \% \)
= \( \frac{600 \times 100}{15000 \times \frac{1}{2}} \% \)
= 8% per annum.

(ii) For the 2nd half-year:
New Principal (P) = Rs. 15,600, Rate (R) = 8%, Time (T) = \( \frac{1}{2} \) year.
Interest = \( \frac{15600 \times 8 \times \frac{1}{2}}{100} \) = Rs. 624.
Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224.

For the 3rd half-year (which completes 1.5 years):
New Principal (P) = Rs. 16,224, Rate (R) = 8%, Time (T) = \( \frac{1}{2} \) year.
Interest = \( \frac{16224 \times 8 \times \frac{1}{2}}{100} \) = Rs. 648.96.
Amount after 1.5 years = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.
In simple words: Since interest is compounded half-yearly, we work in steps of 6 months. Calculate the interest for three consecutive 6-month periods using the updated principal each time to find the final amount.

Exam Tip: Remember that 1.5 years has three 6-month periods. Be careful to use \( T = \frac{1}{2} \) in each step of your compound interest calculation.

Question 7. Find the amount of Rs. 12,800 after 3 years at 10% compound interest per annum.
Answer:
Let's compute the amount year-by-year:
For the 1st year:
Principal (P) = Rs. 12,800, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{12800 \times 10 \times 1}{100} \) = Rs. 1,280.
Amount at the end of the 1st year = Rs. 12,800 + Rs. 1,280 = Rs. 14,080.

For the 2nd year:
New Principal (P) = Rs. 14,080, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{14080 \times 10 \times 1}{100} \) = Rs. 1,408.
Amount at the end of the 2nd year = Rs. 14,080 + Rs. 1,408 = Rs. 15,488.

For the 3rd year:
New Principal (P) = Rs. 15,488, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{15488 \times 10 \times 1}{100} \) = Rs. 1,548.80.
Amount at the end of the 3rd year = Rs. 15,488 + Rs. 1,548.80 = Rs. 17,036.80.
In simple words: We calculate the 10% interest for each of the three years, adding it back to the principal at the end of every year before starting the next.

Exam Tip: Ensure that you do not round off intermediate decimals, as small differences can affect the accuracy of the final compound interest amount.

Question 8. The compound interest on a certain sum of money for the second and third years are Rs. 864 and Rs. 933.12 respectively. Find:
(i) the rate of interest,
(ii) the sum of money,
(iii) the interest for the fourth year.
Answer:
(i) Interest for the 2nd year = Rs. 864
Interest for the 3rd year = Rs. 933.12
Difference in interest = Rs. 933.12 - Rs. 864 = Rs. 69.12
This extra interest of Rs. 69.12 is earned on Rs. 864 in 1 year.
Rate of interest = \( \frac{\text{Difference in interest} \times 100}{\text{Preceding year interest} \times \text{Time}} \% \)
= \( \frac{69.12 \times 100}{864 \times 1} \% \)
= 8% per annum.

(ii) Let's assume the sum is Rs. 100.
Interest for the 1st year = 8% of Rs. 100 = Rs. 8.
Amount at the end of 1st year = Rs. 100 + Rs. 8 = Rs. 108.
Interest for the 2nd year = 8% of Rs. 108 = Rs. 8.64.
If the 2nd year interest is Rs. 8.64, the original sum is Rs. 100.
If the 2nd year interest is Rs. 864, the sum is:
Sum = \( \frac{100 \times 864}{8.64} \) = Rs. 10,000.

(iii) To find the interest for the 4th year:
Interest for the 1st year = \( \frac{10000 \times 8 \times 1}{100} \) = Rs. 800.
The principal for the 4th year is the sum of the initial principal and the interests of the first three years:
Principal for the 4th year = Rs. 10,000 + Rs. 800 + Rs. 864 + Rs. 933.12 = Rs. 12,597.12.
Interest for the 4th year = 8% of Rs. 12,597.12 = Rs. 1,007.77.
In simple words: Find the rate of interest from the second and third year values. Then use Rs. 100 to find the original principal and calculate the interest earned in the fourth year.

Exam Tip: The fourth year's principal is obtained by adding the initial sum and the interest from the first, second, and third years. Double-check this summation before applying the rate.

Question 9. A sum of money compounded annually amounts to Rs. 20,160 in three years and to Rs. 24,192 in four years. Find:
(i) the rate of interest per annum,
(ii) the amount in two years,
(iii) the amount in five years.
Answer:
(i) Amount in 3 years = Rs. 20,160
Amount in 4 years = Rs. 24,192
The interest earned in the fourth year is the difference:
Interest for 4th year = Rs. 24,192 - Rs. 20,160 = Rs. 4,032
This interest of Rs. 4,032 is earned on the 3rd-year amount of Rs. 20,160 in 1 year.
Rate of interest = \( \frac{4032 \times 100}{20160 \times 1} \% \)
= 20% per annum.

(ii) Let's find the amount at the end of 2 years:
Assume the amount at the end of 2 years is Rs. 100.
Interest for the 3rd year = 20% of Rs. 100 = Rs. 20.
Therefore, the amount at the end of 3 years would be Rs. 100 + Rs. 20 = Rs. 120.
If the amount in 3 years is Rs. 120, the amount in 2 years is Rs. 100.
If the amount in 3 years is Rs. 20,160, the amount in 2 years is:
Amount = \( \frac{100 \times 20160}{120} \) = Rs. 16,800.

(iii) To find the amount at the end of 5 years:
Amount in 5 years = Amount in 4 years + 20% of Amount in 4 years
= Rs. 24,192 + 20% of Rs. 24,192
= Rs. 24,192 + Rs. 4,838.40 = Rs. 29,030.40.
In simple words: First find the rate using the interest earned between years 3 and 4. Work backwards to find the 2-year amount, and work forwards to find the 5-year amount.

Exam Tip: Be comfortable with both forward and backward compound interest calculations. Setting up a base of Rs. 100 helps simplify backward calculations immensely.

Question 10. A man borrows Rs. 8,000 at 7% per annum compound interest. At the end of the first year, he pays back Rs. 3,560. Find:
(i) the interest paid for the second year,
(ii) the total interest paid in two years,
(iii) the total amount of money paid in two years to clear the debt.
Answer:
(i) For the 1st year:
Principal (P) = Rs. 8,000, Rate (R) = 7%, Time (T) = 1 year.
Interest = \( \frac{8000 \times 7 \times 1}{100} \) = Rs. 560.
Amount outstanding at the end of 1st year = Rs. 8,000 + Rs. 560 = Rs. 8,560.
Amount repaid = Rs. 3,560.
Balance principal for the 2nd year = Rs. 8,560 - Rs. 3,560 = Rs. 5,000.
Interest for the 2nd year = \( \frac{5000 \times 7 \times 1}{100} \) = Rs. 350.

(ii) Total interest paid in two years:
Total interest = Interest of 1st year + Interest of 2nd year
= Rs. 560 + Rs. 350 = Rs. 910.

(iii) Total money paid to clear the debt:
Total money paid = Original loan amount + Total interest accrued
= Rs. 8,000 + Rs. 910 = Rs. 8,910.
In simple words: Calculate the first year's interest, add it, subtract the payment, and find the interest for the second year. The total debt is cleared by paying back the total borrowed amount plus all interests.

Exam Tip: Ensure that you subtract the repaid amount from the total amount outstanding (Principal + Interest) at the end of the year, not just from the starting principal.

Question 11. The value of a machine depreciates by a certain percentage every year. If the depreciation in its value during the first and second years are Rs. 4,000 and Rs. 3,600 respectively, calculate:
(i) the rate of depreciation per annum,
(ii) the original cost of the machine,
(iii) the value of the machine at the end of the third year.
Answer:
(i) Depreciation in the 1st year = Rs. 4,000
Depreciation in the 2nd year = Rs. 3,600
Difference in depreciation = Rs. 4,000 - Rs. 3,600 = Rs. 400
This decrease in depreciation represents the rate applied to the first year's depreciation.
Rate of depreciation = \( \frac{\text{Difference in depreciation} \times 100}{\text{1st year depreciation}} \% \)
= \( \frac{400}{4000} \times 100 \% \)
= 10% per annum.

(ii) Let the original cost of the machine be Rs. 100.
Depreciation during the 1st year = 10% of Rs. 100 = Rs. 10.
When the first-year depreciation is Rs. 10, the original cost is Rs. 100.
When the first-year depreciation is Rs. 4,000, the original cost is:
Original cost = \( \frac{100}{10} \times 4000 \) = Rs. 40,000.

(iii) To find the final value at the end of the 3rd year:
Total depreciation over three years = 1st year depreciation + 2nd year depreciation + 3rd year depreciation
Value after 2 years = Rs. 40,000 - (Rs. 4,000 + Rs. 3,600) = Rs. 32,400.
Depreciation for the 3rd year = 10% of Rs. 32,400 = Rs. 3,240.
Total depreciation = Rs. 4,000 + Rs. 3,600 + Rs. 3,240 = Rs. 10,840.
Value of the machine at the end of the third year = Rs. 40,000 - Rs. 10,840 = Rs. 29,160.
In simple words: Find the depreciation rate from the difference between the first and second year depreciation. Work backward to find the starting cost, then apply the rate for three years.

Exam Tip: Depreciation is calculated on the reducing balance, so each year's depreciation is less than the previous year's. Double-check your 3rd-year depreciation using the value after 2 years.

Question 12. Find the total depreciation in the value of a machine costing Rs. 32,000 in two years, if the rate of depreciation is 5% per annum.
Answer:
Original cost of machine = Rs. 32,000.
Depreciation rate = 5% per annum.
Depreciation for the 1st year = 5% of Rs. 32,000 = Rs. 1,600.
Value of the machine after 1 year = Rs. 32,000 - Rs. 1,600 = Rs. 30,400.
Depreciation for the 2nd year = 5% of Rs. 30,400 = Rs. 1,520.
Value of the machine after 2 years = Rs. 30,400 - Rs. 1,520 = Rs. 28,880.
Total depreciation in two years = Original cost - Final value
= Rs. 32,000 - Rs. 28,880 = Rs. 3,120.
In simple words: Calculate the 5% depreciation for the first year, subtract it to find the new value, and then calculate another 5% depreciation on this new value to get the total loss in value.

Exam Tip: Never calculate total depreciation as simply 10% (5% + 5%) of the original value, because the second year's depreciation is calculated on the reduced value.

Question 13. The difference between the compound interest for the third year and the first year on a certain sum of money at 10% per annum is Rs. 252. Find the sum.
Answer:
Let's assume the principal sum is Rs. 100.
Rate of interest = 10% per annum.
Interest for the 1st year = 10% of Rs. 100 = Rs. 10.
Amount at the end of the 1st year = Rs. 100 + Rs. 10 = Rs. 110.
Interest for the 2nd year = 10% of Rs. 110 = Rs. 11.
Amount at the end of the 2nd year = Rs. 110 + Rs. 11 = Rs. 121.
Interest for the 3rd year = 10% of Rs. 121 = Rs. 12.10.
Difference between the interest of the 3rd year and the 1st year is:
Rs. 12.10 - Rs. 10 = Rs. 2.10.
When the difference is Rs. 2.10, the principal is Rs. 100.
When the difference is Rs. 252, the principal is:
Principal = \( \frac{100 \times 252}{2.10} \) = Rs. 12,000.
In simple words: Assume a principal of Rs. 100 and find the first and third year interests. Scale up the principal by comparing the assumed difference to the actual given difference.

Exam Tip: Be sure to find the difference between the third-year interest and the first-year interest, rather than the total interest or the final amounts.

Question 14. A man borrows Rs. 10,000 at 10% compound interest per annum. At the end of each year, he pays back 30% of the sum borrowed. Find his outstanding principal at the beginning of the third year.
Answer:
For the 1st year:
Principal (P) = Rs. 10,000, Rate (R) = 10%, Time (T) = 1 year.
Interest accrued = \( \frac{10000 \times 10 \times 1}{100} \) = Rs. 1,000.
Amount at the end of the 1st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000.
At the end of the first year, he pays back 30% of the original sum borrowed:
Amount paid = 30% of Rs. 10,000 = Rs. 3,000.
Outstanding principal at the start of the 2nd year = Rs. 11,000 - Rs. 3,000 = Rs. 8,000.

For the 2nd year:
New Principal (P) = Rs. 8,000, Rate (R) = 10%, Time (T) = 1 year.
Interest accrued = \( \frac{8000 \times 10 \times 1}{100} \) = Rs. 800.
Amount at the end of the 2nd year = Rs. 8,000 + Rs. 800 = Rs. 8,800.
At the end of the second year, he pays back another 30% of the borrowed sum:
Amount paid = 30% of Rs. 10,000 = Rs. 3,000.
Outstanding principal at the start of the 3rd year = Rs. 8,800 - Rs. 3,000 = Rs. 5,800.
In simple words: Find the year-end balance by adding interest to the principal, then subtract Rs. 3,000 (which is 30% of the original Rs. 10,000 loan) each year.

Exam Tip: Read carefully whether the repayment percentage is calculated on the original sum borrowed or on the outstanding balance at year-end. Here, it is based on the borrowed sum.

Question 15. A man borrows Rs. 10,000 at 10% compound interest per annum. At the end of each year, he pays back 20% of the amount outstanding at that time. Find the amount he has to pay at the beginning of the third year.
Answer:
For the 1st year:
Principal (P) = Rs. 10,000, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{10000 \times 10 \times 1}{100} \) = Rs. 1,000.
Outstanding amount at the end of the 1st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000.
At the end of the first year, he pays back 20% of the outstanding amount:
Amount repaid = 20% of Rs. 11,000 = Rs. 2,200.
Principal for the 2nd year = Rs. 11,000 - Rs. 2,200 = Rs. 8,800.

For the 2nd year:
Principal (P) = Rs. 8,800, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{8800 \times 10 \times 1}{100} \) = Rs. 880.
Outstanding amount at the end of the 2nd year = Rs. 8,800 + Rs. 880 = Rs. 9,680.
At the end of the second year, he pays back 20% of the outstanding amount:
Amount repaid = 20% of Rs. 9,680 = Rs. 1,936.
Amount he has to pay at the beginning of the 3rd year = Rs. 9,680 - Rs. 1,936 = Rs. 7,744.
In simple words: Calculate the interest, add it, and subtract 20% of the total at the end of each year. The remainder is the balance due at the start of the next year.

Exam Tip: Be sure to compute the 20% repayment on the total amount outstanding (Principal + Interest) rather than on the original loan value.

Exercise 2(D)

Question 1. A sum of money amounts to Rs. 6,593.40 in two years, the rates of interest for the first and second years being 10% and 11% respectively. Find the sum.
Answer:
Let's assume the principal sum is Rs. 100.
For the 1st year:
Principal (P) = Rs. 100, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{100 \times 10 \times 1}{100} \) = Rs. 10.
Amount at the end of the 1st year = Rs. 100 + Rs. 10 = Rs. 110.

For the 2nd year:
Principal (P) = Rs. 110, Rate (R) = 11%, Time (T) = 1 year.
Interest = \( \frac{110 \times 11 \times 1}{100} \) = Rs. 12.10.
Amount at the end of the 2nd year = Rs. 110 + Rs. 12.10 = Rs. 122.10.
If the amount after 2 years is Rs. 122.10, the principal is Rs. 100.
If the amount after 2 years is Rs. 6,593.40, the principal is:
Principal = \( \frac{100 \times 6593.40}{122.10} \) = Rs. 5,400.
In simple words: Find what Rs. 100 would grow to over two years with different rates of interest each year. Then, compare this to the actual amount to find the original principal.

Exam Tip: Apply the rates of interest in the correct chronological order as different rates apply to different years.

Question 2. The value of a machine depreciates by 10% per annum for the first two years, and by 15% per annum during the third year. Express the total depreciation of the machine in three years as a single percentage of its original value.
Answer:
Let the initial cost of the machine be Rs. 100.
1st year depreciation = 10% of Rs. 100 = Rs. 10.
Value at the end of 1st year = Rs. 100 - Rs. 10 = Rs. 90.
2nd year depreciation = 10% of Rs. 90 = Rs. 9.
Value at the end of 2nd year = Rs. 90 - Rs. 9 = Rs. 81.
3rd year depreciation = 15% of Rs. 81 = Rs. 12.15.
Value at the end of 3rd year = Rs. 81 - Rs. 12.15 = Rs. 68.85.
Total depreciation = Original value - Final value
= Rs. 100 - Rs. 68.85 = Rs. 31.15.
Thus, the total depreciation expressed as a single percentage is 31.15%.
In simple words: We assume the machine's initial value is Rs. 100. We decrease it by 10% twice, and then by 15% once. The total drop from 100 is our total percentage depreciation.

Exam Tip: Do not just add the percentages (10% + 10% + 15% = 35%). Depreciation compounds on the declining balance, resulting in a lower actual percentage change.

Question 3. A man borrows Rs. 12,000 at 10% per annum compound interest, interest being compounded half-yearly. At the end of every six months, he pays back Rs. 4,000. Find the amount outstanding at the end of 1.5 years.
Answer:
Since the interest is compounded half-yearly, calculations are done for 6-month steps.
For the 1st half-year:
Principal (P) = Rs. 12,000, Rate (R) = 10%, Time (T) = \( \frac{1}{2} \) year.
Interest = \( \frac{12000 \times 10 \times 1}{100 \times 2} \) = Rs. 600.
Amount outstanding = Rs. 12,000 + Rs. 600 = Rs. 12,600.
Repayment at end of 1st half-year = Rs. 4,000.
Balance principal for 2nd half-year = Rs. 12,600 - Rs. 4,000 = Rs. 8,600.

For the 2nd half-year:
Principal (P) = Rs. 8,600, Rate (R) = 10%, Time (T) = \( \frac{1}{2} \) year.
Interest = \( \frac{8600 \times 10 \times 1}{100 \times 2} \) = Rs. 430.
Amount outstanding = Rs. 8,600 + Rs. 430 = Rs. 9,030.
Repayment at end of 2nd half-year = Rs. 4,000.
Balance principal for 3rd half-year = Rs. 9,030 - Rs. 4,000 = Rs. 5,030.

For the 3rd half-year:
Principal (P) = Rs. 5,030, Rate (R) = 10%, Time (T) = \( \frac{1}{2} \) year.
Interest = \( \frac{5030 \times 10 \times 1}{100 \times 2} \) = Rs. 251.50.
Amount outstanding at the end of 1.5 years = Rs. 5,030 + Rs. 251.50 = Rs. 5,281.50.
In simple words: Since interest is calculated every six months, we compute the interest, add it, and subtract the Rs. 4,000 payment. We repeat this three times to find the remaining balance.

Exam Tip: Be sure to compute the interest using a time period of \( T = \frac{1}{2} \) (6 months) for each of the three compounding intervals.

Question 4. The sum of the compound interest for the first year and the third year on a certain sum of money at 10% per annum compounded annually is Rs. 2,652. Find the sum.
Answer:
Let's assume the principal sum is Rs. 100.
For the 1st year:
Principal (P) = Rs. 100, Rate (R) = 10%, Time (T) = 1 year.
Interest for 1st year = \( \frac{100 \times 10 \times 1}{100} \) = Rs. 10.
Amount at the end of 1st year = Rs. 100 + Rs. 10 = Rs. 110.

For the 2nd year:
Principal (P) = Rs. 110, Rate (R) = 10%, Time (T) = 1 year.
Interest for 2nd year = \( \frac{110 \times 10 \times 1}{100} \) = Rs. 11.
Amount at the end of 2nd year = Rs. 110 + Rs. 11 = Rs. 121.

For the 3rd year:
Principal (P) = Rs. 121, Rate (R) = 10%, Time (T) = 1 year.
Interest for 3rd year = \( \frac{121 \times 10 \times 1}{100} \) = Rs. 12.10.

Sum of interest for the 1st year and 3rd year:
Total assumed interest = Interest of 1st year + Interest of 3rd year
= Rs. 10 + Rs. 12.10 = Rs. 22.10.
If the sum of these interests is Rs. 22.10, the principal is Rs. 100.
If the sum of these interests is Rs. 2,652, the principal is:
Principal = \( \frac{100 \times 2652}{22.10} \) = Rs. 12,000.
In simple words: Using an assumed principal of Rs. 100, calculate the first and third year interests. Find their sum, then scale up the principal proportionally using the actual sum given.

Exam Tip: Ensure you sum the interest of the first and third years, ignoring the second year interest, as specified in the question.

Question 5. The value of a machine depreciates by 12% per annum. If the depreciation in its value during the second year is Rs. 2,640, find its original value.
Answer:
Let the original cost of the machine be Rs. 100.
Depreciation for the 1st year = 12% of Rs. 100 = Rs. 12.
Value of the machine at the end of 1st year = Rs. 100 - Rs. 12 = Rs. 88.
Depreciation for the 2nd year = 12% of Rs. 88 = Rs. 10.56.
If the depreciation in the 2nd year is Rs. 10.56, the original cost is Rs. 100.
If the depreciation in the 2nd year is Rs. 2,640, the original cost is:
Original value = \( \frac{100 \times 2640}{10.56} \) = Rs. 25,000.
In simple words: Find what the second-year depreciation would be if the machine initially cost Rs. 100. Scale up this initial value by comparing your result to the actual depreciation value.

Exam Tip: Remember that the second-year depreciation is calculated on the reduced value after the first year, not on the original starting cost.

Question 6. The difference between the simple interest and compound interest on a certain sum of money at 8% per annum for 2 years is Rs. 64. Find the sum.
Answer:
Let the principal sum be Rs. x.
Simple Interest (SI) for 2 years at 8% per annum is:
SI = \( \frac{x \times 8 \times 2}{100} \) = 0.16x.

For compound interest (compounded annually):
Interest for the 1st year = \( \frac{x \times 8 \times 1}{100} \) = 0.08x.
Principal for the 2nd year = x + 0.08x = 1.08x.
Interest for the 2nd year = \( \frac{1.08x \times 8 \times 1}{100} \) = 0.0864x.
Total compound interest (CI) for 2 years = Interest of 1st year + Interest of 2nd year
= 0.08x + 0.0864x = 0.1664x.

The difference between compound interest and simple interest is Rs. 64:
CI - SI = 64

\implies 0.1664x - 0.16x = 64

\implies 0.0064x = 64

\implies x = \frac{64}{0.0064}

\implies x = 10,000.
Hence, the principal sum is Rs. 10,000.
In simple words: Express both simple and compound interests in terms of a variable x. Find their algebraic difference, set it equal to Rs. 64, and solve for x.

Exam Tip: For a 2-year period, the difference between CI and SI is always equal to the simple interest on the first year's interest. This can serve as a quick way to verify your equations.

Question 7. Find the compound interest for the second year on Rs. 13,500 at 16% per annum compounded annually.
Answer:
For the 1st year:
Principal (P) = Rs. 13,500, Rate (R) = 16%, Time (T) = 1 year.
Interest accrued = \( \frac{13500 \times 16 \times 1}{100} \) = Rs. 2,160.
Amount outstanding at the end of the 1st year = Rs. 13,500 + Rs. 2,160 = Rs. 15,660.

For the 2nd year:
New Principal (P) = Rs. 15,660, Rate (R) = 16%, Time (T) = 1 year.
Interest accrued for the 2nd year = \( \frac{15660 \times 16 \times 1}{100} \) = Rs. 2,505.60.
In simple words: Calculate the first-year interest and add it to the principal. Then, calculate 16% of this new total to find only the interest earned in the second year.

Exam Tip: Pay close attention to the question. It asks for the interest of the *second year* only, not the total compound interest or the final amount.

Question 8. Find the compound interest for the third year on Rs. 48,000 at 10% per annum compounded annually.
Answer:
For the 1st year:
Principal (P) = Rs. 48,000, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{48000 \times 10 \times 1}{100} \) = Rs. 4,800.
Amount at the end of 1st year = Rs. 48,000 + Rs. 4,800 = Rs. 52,800.

For the 2nd year:
New Principal (P) = Rs. 52,800, Rate (R) = 10%, Time (T) = 1 year.
Interest = \( \frac{52800 \times 10 \times 1}{100} \) = Rs. 5,280.
Amount at the end of 2nd year = Rs. 52,800 + Rs. 5,280 = Rs. 58,080.

For the 3rd year:
New Principal (P) = Rs. 58,080, Rate (R) = 10%, Time (T) = 1 year.
Interest for the 3rd year = \( \frac{58080 \times 10 \times 1}{100} \) = Rs. 5,808.
In simple words: Step through each year's compounding by adding interest to find the next principal. The interest for the third year is computed directly on the second year's ending amount.

Exam Tip: Be sure to write out each step clearly to show the examiner how you arrived at the third year's principal of Rs. 58,080.

Question 9. Ashok borrowed Rs. 12,000 at a certain rate of compound interest per annum. At the end of the first year, he paid back Rs. 4,000. If the interest for the second year is Rs. 920, find:
(i) the rate of interest,
(ii) the total debt outstanding at the end of the second year.
Answer:
(i) Let the annual compound interest rate be x%.
For the 1st year:
Principal (P) = Rs. 12,000.
Interest earned = \( \frac{12000 \times x \times 1}{100} \) = 120x.
Amount outstanding at the end of 1st year = Rs. 12,000 + 120x.
Ashok repays Rs. 4,000 at the end of the 1st year.
Remaining principal for the 2nd year = Rs. 12,000 + 120x - Rs. 4,000 = Rs. 8,000 + 120x.
Interest accrued during the 2nd year is:
Interest = \( \frac{(8000 + 120x) \times x \times 1}{100} \) = \( 80x + 1.2x^2 \).
Given that the interest for the second year is Rs. 920:
1.2x² + 80x = 920
Multiply the entire equation by 5 to clear decimals:
6x² + 400x - 4600 = 0
Divide the equation by 2:
3x² + 200x - 2300 = 0
Factoring the quadratic equation:
3x² + 230x - 30x - 2300 = 0

\implies x(3x + 230) - 10(3x + 230) = 0

\implies (3x + 230)(x - 10) = 0
This yields x = 10 or x = \( -\frac{230}{3} \).
Since the rate of interest cannot be negative, we have x = 10.
Hence, the rate of interest is 10% per annum.

(ii) To find the total debt outstanding at the end of the second year:
Total debt at the end of the second year is equal to the principal of the second year plus the second year's interest.
Principal for the second year = Rs. 8,000 + 120(10) = Rs. 8,000 + Rs. 1,200 = Rs. 9,200.
Interest for the second year = Rs. 920.
Total debt outstanding = Rs. 9,200 + Rs. 920 = Rs. 10,120.
In simple words: Set up an equation for the second year's interest in terms of the rate x. Solve the quadratic equation to find the interest rate, then calculate the outstanding debt.

Exam Tip: Be comfortable with algebraic manipulations and factoring quadratics in compound interest problems, as they frequently appear in board examinations.

Question 10. The compound interest on a sum of money for three consecutive years are Rs. 1,500, Rs. 1,725, and Rs. 2,070 respectively. Find the rate of interest for the second and third years.
Answer:
The interest earned in the second year is Rs. 1,725, while the interest in the first year is Rs. 1,500.
The increase in interest from the first year to the second year is:
Rs. 1,725 - Rs. 1,500 = Rs. 225.
This Rs. 225 is the interest earned on the first year's interest (Rs. 1,500) for one year.
Rate of interest for the second year = \( \frac{225}{1500} \times 100 \% \) = 15%.

Similarly, the interest earned in the third year is Rs. 2,070, while the interest in the second year is Rs. 1,725.
The increase in interest from the second year to the third year is:
Rs. 2,070 - Rs. 1,725 = Rs. 345.
This Rs. 345 is the interest earned on the second year's interest (Rs. 1,725) for one year.
Rate of interest for the third year = \( \frac{345}{1725} \times 100 \% \) = 20%.
In simple words: Since each year's interest grows by the interest earned on the preceding interest, we find the percentage increase of interest between successive years to obtain the interest rates.

Exam Tip: Remember that interest on interest is the fundamental principle of compounding. You can find the rate of interest for any period by dividing the increase in interest by the interest of the previous period.