Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 3 Compound Interest

Question 1. Find the amount and the compound interest payable annually on:
(i) Rs 25000 for \( 1 \frac{1}{2} \) years at 10% per annum.
(ii) Rs 32000 for 2 years at \( 7 \frac{1}{2}\% \) per annum.
(iii) Rs 10000 for \( 2 \frac{1}{2} \) years at 6% per annum.
(iv) Rs 24000 for \( 1 \frac{1}{2} \) years at \( 7 \frac{1}{2}\% \) per annum.
Answer:
(i) For the first year:
Principal (\( P_1 \)) = Rs 25000, Rate (\( r \)) = 10%, Time (\( t \)) = 1 year.
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 25000 \left(1 + \frac{10}{100}\right) \)
\( = 25000 \left(\frac{11}{10}\right) \)
\( = \text{Rs } 27500 \)
Now, the principal for the next 6 months becomes Rs 27500.
Interest for the remaining 6 months = \( \frac{27500 \times 6 \times 10}{100 \times 12} = \text{Rs } 1375 \)
Hence, total Amount after \( 1 \frac{1}{2} \) years = \( 27500 + 1375 = \text{Rs } 28875 \)
Compound Interest (CI) = \( A - P = 28875 - 25000 = \text{Rs } 3875 \)

(ii) For the first year:
Principal (\( P_1 \)) = Rs 32000, Rate (\( r \)) = \( 7 \frac{1}{2}\% = \frac{15}{2}\% \).
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 32000 \left(1 + \frac{15}{2 \times 100}\right) \)
\( = 32000 \left(1 + \frac{3}{40}\right) \)
\( = 32000 \left(\frac{43}{40}\right) \)
\( = \text{Rs } 34400 \)
For the second year:
Principal (\( P_2 \)) = Rs 34400, Rate (\( r \)) = \( \frac{15}{2}\% \).
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 34400 \left(1 + \frac{15}{2 \times 100}\right) \)
\( = 34400 \left(\frac{43}{40}\right) \)
\( = \text{Rs } 36980 \)
Compound Interest (CI) = \( A - P = 36980 - 32000 = \text{Rs } 4980 \)

(iii) For the first year:
Principal (\( P_1 \)) = Rs 10000, Rate (\( r \)) = 6%.
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 10000 \left(1 + \frac{6}{100}\right) \)
\( = 10000 \times \frac{106}{100} \)
\( = \text{Rs } 10600 \)
For the second year:
Principal (\( P_2 \)) = Rs 10600, Rate (\( r \)) = 6%.
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 10600 \left(1 + \frac{6}{100}\right) \)
\( = 10600 \times \frac{106}{100} \)
\( = \text{Rs } 11236 \)
Now, the principal for the next 6 months is Rs 11236.
Interest for the remaining 6 months = \( \frac{11236 \times 6 \times 6}{100 \times 12} = \text{Rs } 337.08 \)
Hence, total Amount after \( 2 \frac{1}{2} \) years = \( 11236 + 337.08 = \text{Rs } 11573.08 \)
Compound Interest (CI) = \( A - P = 11573.08 - 10000 = \text{Rs } 1573.08 \)

(iv) For the first year:
Principal (\( P \)) = Rs 24000, Rate (\( r \)) = \( 7 \frac{1}{2}\% = \frac{15}{2}\% \), Time (\( t \)) = \( 1 \frac{1}{2} \) years.
Amount after 1 year = \( P \left(1 + \frac{r}{100}\right) \)
\( = 24000 \left(1 + \frac{15}{2 \times 100}\right) \)
\( = 24000 \left(\frac{43}{40}\right) \)
\( = \text{Rs } 25800 \)
Now, the principal for the next 6 months is Rs 25800.
Interest for the remaining 6 months = \( \frac{25800 \times 15 \times 6}{200 \times 12} = \text{Rs } 967.50 \)
Hence, total Amount after \( 1 \frac{1}{2} \) years = \( 25800 + 967.50 = \text{Rs } 26767.50 \)
Compound Interest (CI) = \( A - P = 26767.50 - 24000 = \text{Rs } 2767.50 \br />
In simple words: To calculate compound interest with a fractional year, determine the compounded amount for the complete years first. Use this value as the new principal to calculate simple interest for the remaining fractional months, then combine them.

Exam Tip: For periods like 1.5 years, calculate the compounding annually for 1 year first, and then find the simple interest on that amount for the next 6 months.

Question 2(a). Find the amount and the compound interest payable annually on Rs 16000 for 2 years at 15% and 12% for the successive years.
Answer:
For the first year:
Principal (\( P \)) = Rs 16000, Rate (\( R \)) = 15%, Time (\( T \)) = 1 year.
Interest = \( \frac{16000 \times 15 \times 1}{100} = \text{Rs } 2400 \)
Accumulated amount at the end of the first year = \( 16000 + 2400 = \text{Rs } 18400 \)

For the second year:
Principal (\( P \)) = Rs 18400, Rate (\( R \)) = 12%, Time (\( T \)) = 1 year.
Interest = \( \frac{18400 \times 12 \times 1}{100} = \text{Rs } 2208 \)
Accumulated amount at the end of the second year = \( 18400 + 2208 = \text{Rs } 20608 \)

Therefore, the final amount = Rs 20608
Compound Interest = \( A - P = 20608 - 16000 = \text{Rs } 4608 \)

In simple words: When interest rates change every year, compute the growth step-by-step. Work out the first year's interest, add it to the starting principal, and then use that new total to compute the second year's interest with the new rate.

Exam Tip: Ensure that you apply the correct interest rate for each specific year. Do not use the same rate for both years.

Question 2(b). Find the amount and the compound interest payable annually on Rs 17500 for 3 years at 8%, 10% and 12% for successive years.
Answer:
For the first year:
Principal (\( P \)) = Rs 17500, Rate (\( R \)) = 8%, Time (\( T \)) = 1 year.
Interest = \( \frac{17500 \times 8 \times 1}{100} = \text{Rs } 1400 \)
Accumulated amount at the end of the first year = \( 17500 + 1400 = \text{Rs } 18900 \)

For the second year:
Principal (\( P \)) = Rs 18900, Rate (\( R \)) = 10%, Time (\( T \)) = 1 year.
Interest = \( \frac{18900 \times 10 \times 1}{100} = \text{Rs } 1890 \)
Accumulated amount at the end of the second year = \( 18900 + 1890 = \text{Rs } 20790 \)

For the third year:
Principal (\( P \)) = Rs 20790, Rate (\( R \)) = 12%, Time (\( T \)) = 1 year.
Interest = \( \frac{20790 \times 12 \times 1}{100} = \text{Rs } 2494.80 \)
Accumulated amount at the end of the third year = \( 20790 + 2494.80 = \text{Rs } 23284.80 \)

Therefore, the final amount = Rs 23284.80
Total Compound Interest = \( A - P = 23284.80 - 17500 = \text{Rs } 5784.80 \)

In simple words: For a three-year period with changing rates, calculate the interest year-by-year. The final balance of each year becomes the starting principal for the next.

Exam Tip: Be meticulous with calculations containing decimals, especially during the third year when working with fractional amounts.

Question 3. Calculate the amount and compound interest on Rs 20000 for 3 years at 10% per annum, interest being payable annually.
Answer:
First year principal (\( P_1 \)) = Rs 20000, Rate (\( r \)) = 10%.
Amount at the end of 1st year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 20000 \left(1 + \frac{10}{100}\right) \)
\( = 20000 \times \frac{110}{100} \)
\( = \text{Rs } 22000 \)
Second year principal (\( P_2 \)) = Rs 22000, Rate (\( r \)) = 10%.
Amount at the end of 2nd year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 22000 \left(1 + \frac{10}{100}\right) \)
\( = 22000 \times \frac{110}{100} \)
\( = \text{Rs } 24200 \)
Third year principal (\( P_3 \)) = Rs 24200, Rate (\( r \)) = 10%.
Amount at the end of 3rd year = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 24200 \left(1 + \frac{10}{100}\right) \)
\( = 24200 \times \frac{110}{100} \)
\( = \text{Rs } 26620 \)
Total Amount = Rs 26620
Compound Interest (CI) = \( A - P = 26620 - 20000 = \text{Rs } 6620 \)

In simple words: This problem tracks three consecutive years where the interest rate is 10%. Each year, the interest is added to the principal to form the new starting sum for the next year.

Exam Tip: You can choose to calculate using either the year-by-year simple interest method or the direct formula, but outlining it step-by-step helps show clear working.

Question 4. Compute the compound interest for the third year on Rs 5000 invested for 3 years at 10% per annum, the interest being payable annually.
Answer:
For the first year:
Principal (\( P \)) = Rs 5000, Rate (\( R \)) = 10%, Time (\( T \)) = 1 year.
Interest = \( \frac{5000 \times 10 \times 1}{100} = \text{Rs } 500 \)
Amount after the first year = \( 5000 + 500 = \text{Rs } 5500 \)

For the second year:
Principal (\( P \)) = Rs 5500, Rate (\( R \)) = 10%, Time (\( T \)) = 1 year.
Interest = \( \frac{5500 \times 10 \times 1}{100} = \text{Rs } 550 \)
Amount after the second year = \( 5500 + 550 = \text{Rs } 6050 \)

For the third year:
Principal (\( P \)) = Rs 6050, Rate (\( R \)) = 10%, Time (\( T \)) = 1 year.
Interest = \( \frac{6050 \times 10 \times 1}{100} = \text{Rs } 605 \)

Therefore, the compound interest specifically for the third year is Rs 605.

In simple words: To find the interest earned only during the third year, calculate the starting principal of that third year (which equals the amount at the end of the second year) and find its interest for one year.

Exam Tip: Do not calculate the total compound interest for all three years combined. The question asks exclusively for the interest accrued in the third year alone.

Question 5. Rakesh invests Rs 25600 at 5% per annum compound interest payable annually for 3 years. Find the amount standing to his credit at the end of the second year.
Answer:
For the first year:
Principal (\( P \)) = Rs 25600, Rate (\( R \)) = 5%, Time (\( T \)) = 1 year.
Interest = \( \frac{25600 \times 5 \times 1}{100} = \text{Rs } 1280 \)
Amount at the end of the first year = \( 25600 + 1280 = \text{Rs } 26880 \)

For the second year:
Principal (\( P \)) = Rs 26880, Rate (\( R \)) = 5%, Time (\( T \)) = 1 year.
Interest = \( \frac{26880 \times 5 \times 1}{100} = \text{Rs } 1344 \)
Amount at the end of the second year = \( 26880 + 1344 = \text{Rs } 28224 \)

Thus, the total amount credited at the end of the second year is Rs 28224.

In simple words: Rakesh's investment grows with interest added each year. After finding the interest for the first year, add it to the principal and repeat the process for the second year.

Exam Tip: Pay attention to what the question asks; since it asks for the amount at the end of the second year, do not waste time calculating the third year's values.

Question 6. Find the amount and compound interest on Rs 7500 for 1 \frac{1}{2} years at 8% per annum, payable semi-annually.
Answer:
Principal (\( P_1 \)) = Rs 7500, Rate for each half-year (\( r \)) = \( \frac{8\%}{2} = 4\% \).
Amount after the first half-year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 7500 \left(1 + \frac{4}{100}\right) \)
\( = 7500 \times \frac{104}{100} \)
\( = \text{Rs } 7800 \)

Principal for the second half-year (\( P_2 \)) = Rs 7800, Rate (\( r \)) = 4%.
Amount after 1 year (two half-years) = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 7800 \left(1 + \frac{4}{100}\right) \)
\( = 7800 \times \frac{104}{100} \)
\( = \text{Rs } 8112 \)

Principal for the third half-year (\( P_3 \)) = Rs 8112, Rate (\( r \)) = 4%.
Amount after \( 1 \frac{1}{2} \) years (three half-years) = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 8112 \left(1 + \frac{4}{100}\right) \)
\( = 8112 \times \frac{104}{100} \)
\( = \text{Rs } 8436.48 \)

Final Amount = Rs 8436.48
Total Compound Interest (CI) = \( A - P = 8436.48 - 7500 = \text{Rs } 936.48 \)

In simple words: When compounding occurs half-yearly, halve the annual rate to 4% and double the periods. For 1.5 years, we apply this rate over three half-yearly cycles.

Exam Tip: Remember to always divide the rate by 2 and multiply the number of years by 2 to determine the correct intervals for semi-annual compounding.

Question 7. A man invests Rs 24000 for two years at compound interest. If his money amounts to Rs 27600 after one year, find the amount at the end of the second year.
Answer:
Let the interest rate be \( r\% \) per annum.
Amount after 1 year = \( P \left(1 + \frac{r}{100}\right) \)

\( \implies 27600 = 24000 \left(1 + \frac{r}{100}\right) \)

\( \implies 1 + \frac{r}{100} = \frac{27600}{24000} = \frac{23}{20} \)

\( \implies \frac{r}{100} = \frac{23}{20} - 1 = \frac{3}{20} \)

\( \implies r = \frac{100 \times 3}{20} = 15\% \)

Using this rate, we can determine the amount at the end of the second year:
Amount after 2 years = \( P \left(1 + \frac{r}{100}\right) \)
\( = 27600 \left(1 + \frac{15}{100}\right) \)
\( = 27600 \times \frac{115}{100} \)
\( = \text{Rs } 31740 \)

In simple words: Find the annual interest rate from the first year's growth first. Then, use that rate to compound the first year's final sum of Rs 27600 to find the second year's ending balance.

Exam Tip: Since compounding is annual, you can treat the first year's interest as simple interest on the initial principal to solve for the interest rate quickly.

Question 8. How much will Rs 14000 amount to in 2 years at compound interest, if the rates for the successive years be 5% and 8% respectively?
Answer:
Principal for the first year (\( P_1 \)) = Rs 14000, Rate (\( r \)) = 5%.
Amount after the first year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 14000 \left(1 + \frac{5}{100}\right) \)
\( = 14000 \times \frac{105}{100} \)
\( = \text{Rs } 14700 \)

Principal for the second year (\( P_2 \)) = Rs 14700, Rate (\( r \)) = 8%.
Amount after the second year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 14700 \left(1 + \frac{8}{100}\right) \)
\( = 14700 \times \frac{108}{100} \)
\( = \text{Rs } 15876 \)

Thus, the final amount is Rs 15876.

In simple words: Calculate the growth year-by-year since the rates are different. Apply 5% to the initial principal, and then apply 8% to that resulting amount.

Exam Tip: Ensure you apply the first year's rate to the first year and the second year's rate to the second year without swapping them.

Question 9. Find the amount and the compound interest on Rs 17500 for 3 years, if the rates for the successive years are 4%, 5% and 6% respectively, the interest is payable annually.
Answer:
Principal for the first year (\( P_1 \)) = Rs 17500, Rate (\( r \)) = 4%.
Amount after the first year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 17500 \left(1 + \frac{4}{100}\right) \)
\( = 17500 \times \frac{104}{100} \)
\( = \text{Rs } 18200 \)

Principal for the second year (\( P_2 \)) = Rs 18200, Rate (\( r \)) = 5%.
Amount after the second year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 18200 \left(1 + \frac{5}{100}\right) \)
\( = 18200 \times \frac{105}{100} \)
\( = \text{Rs } 19110 \)

Principal for the third year (\( P_3 \)) = Rs 19110, Rate (\( r \)) = 6%.
Amount after the third year = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 19110 \left(1 + \frac{6}{100}\right) \)
\( = 19110 \times \frac{106}{100} \)
\( = \text{Rs } 20256.60 \)

Final Amount = Rs 20256.60
Total Compound Interest (CI) = \( A - P = 20256.60 - 17500 = \text{Rs } 2756.60 \)

In simple words: This problem tracks three consecutive years where the interest rates increase from 4% to 5% and then to 6%. Each year's amount is compounded and forms the new principal for the next.

Exam Tip: Work out the multiplications carefully step-by-step to avoid decimal errors in the final year's calculation.

Question 10. A man borrows Rs 4000 at 14% per annum compound interest payable half-yearly. Find the amount he has to pay at the end of 1 \frac{1}{2} years.
Answer:
Since the interest is compounded half-yearly, we calculate the interest for each 6-month period (each half-year).
Here, Rate of interest (\( R \)) = 14% per annum, Time (\( T \)) = \( \frac{1}{2} \) year per interval.

For the first half-year:
Principal (\( P \)) = Rs 4000.
Interest = \( \frac{4000 \times 14 \times 1}{100 \times 2} = \text{Rs } 280 \)
Amount after the first half-year = \( 4000 + 280 = \text{Rs } 4280 \)

For the second half-year:
Principal (\( P \)) = Rs 4280.
Interest = \( \frac{4280 \times 14 \times 1}{100 \times 2} = \text{Rs } 299.60 \)
Amount after the second half-year = \( 4280 + 299.60 = \text{Rs } 4579.60 \)

For the third half-year:
Principal (\( P \)) = Rs 4579.60.
Interest = \( \frac{4579.60 \times 14 \times 1}{100 \times 2} = \text{Rs } 320.572 \)
Amount after the third half-year = \( 4579.60 + 320.572 = \text{Rs } 4900.172 \)

Therefore, the amount to be paid back at the end of \( 1 \frac{1}{2} \) years is Rs 4900.172.

In simple words: Since the compounding is half-yearly, we calculate interest for three 6-month periods, carrying over the accumulated amount as the starting principal for the next period.

Exam Tip: Be precise with your division by 2 in the denominator to account for the half-yearly intervals when calculating simple interest step-by-step.

Question 11. Calculate the amount and the compound interest to the nearest rupee on Rs 42000 for 2 years at 8% per annum, interest being payable half-yearly.
Answer:
Since the compounding is done semi-annually, the interest rate per half-year (\( r \)) = \( \frac{8\%}{2} = 4\% \).
The total time of 2 years contains 4 half-yearly periods.

For the first half-year:
Amount after \( \frac{1}{2} \) year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 42000 \left(1 + \frac{4}{100}\right) \)
\( = 42000 \times \frac{104}{100} \)
\( = \text{Rs } 43680 \)

For the second half-year:
Amount after 1 year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 43680 \left(1 + \frac{4}{100}\right) \)
\( = 43680 \times \frac{104}{100} \)
\( = \text{Rs } 45427.20 \)

For the third half-year:
Amount after \( 1 \frac{1}{2} \) years = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 45427.20 \left(1 + \frac{4}{100}\right) \)
\( = 45427.20 \times \frac{104}{100} \)
\( = \text{Rs } 47244.29 \)

For the fourth half-year:
Amount after 2 years = \( P_4 \left(1 + \frac{r}{100}\right) \)
\( = 47244.29 \left(1 + \frac{4}{100}\right) \)
\( = 47244.29 \times \frac{104}{100} \)
\( = \text{Rs } 49134.06 \)

Final Amount = Rs 49134.06
Compound Interest (CI) = \( A - P = 49134.06 - 42000 = \text{Rs } 7134.06 \)

In simple words: Since interest compounds every six months, we apply the 4% rate over four successive intervals to find the final total amount.

Exam Tip: Carefully compute each decimal multiplication step as any minor rounding mistake early on will accumulate through the four cycles.

Question 12. A man lends Rs 15000 at 10.5% per annum compound interest, interest reckoned yearly, and another man lends the same sum at 10% per annum, interest being reckoned half-yearly. Who is the gainer at the end of one year and by how much?
Answer:
Case I: Compounded annually at 10.5% for 1 year.
Principal (\( P \)) = Rs 15000, Rate (\( r \)) = 10.5%.
Amount after 1 year = \( P \left(1 + \frac{r}{100}\right) \)
\( = 15000 \left(1 + \frac{10.5}{100}\right) \)
\( = 15000 \times \frac{110.5}{100} \)
\( = \text{Rs } 16575 \)

Case II: Compounded half-yearly at 10% for 1 year.
Principal (\( P_1 \)) = Rs 15000, Rate for each half-year (\( r \)) = \( \frac{10\%}{2} = 5\% \).
Amount after the first half-year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 15000 \left(1 + \frac{5}{100}\right) \)
\( = 15000 \times \frac{105}{100} \)
\( = \text{Rs } 15750 \)

Principal for the second half-year (\( P_2 \)) = Rs 15750, Rate (\( r \)) = 5%.
Amount after 1 year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 15750 \left(1 + \frac{5}{100}\right) \)
\( = 15750 \times \frac{105}{100} \)
\( = \text{Rs } 16537.50 \)

Comparing Case I and Case II:
Difference in amounts = \( 16575 - 16537.50 = \text{Rs } 37.50 \)
Hence, the first lender gains Rs 37.50 more than the second lender.

In simple words: The first option compounding yearly at 10.5% yields a slightly higher amount than the second option compounding half-yearly at 10%. The difference in earnings is Rs 37.50.

Exam Tip: Clearly write out "Case I" and "Case II" in your sheet so that the examiner can follow both calculations easily before you compare the final results.

Question 13. Find the difference between the compound interest and the simple interest on Rs 20000 at 12% per annum for 3 years, the compound interest being payable annually.
Answer:
Case I: Compound Interest (CI) calculation
Principal for the first year (\( P_1 \)) = Rs 20000, Rate (\( r \)) = 12%.
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 20000 \left(1 + \frac{12}{100}\right) \)
\( = 20000 \times \frac{112}{100} \)
\( = \text{Rs } 22400 \)

Principal for the second year (\( P_2 \)) = Rs 22400, Rate (\( r \)) = 12%.
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 22400 \left(1 + \frac{12}{100}\right) \)
\( = 22400 \times \frac{112}{100} \)
\( = \text{Rs } 25088 \)

Principal for the third year (\( P_3 \)) = Rs 25088, Rate (\( r \)) = 12%.
Amount after 3 years = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 25088 \left(1 + \frac{12}{100}\right) \)
\( = 25088 \times \frac{112}{100} \)
\( = \text{Rs } 28098.56 \)

Total Compound Interest (CI) = \( A - P = 28098.56 - 20000 = \text{Rs } 8098.56 \)

Case II: Simple Interest (SI) calculation
Simple Interest = \( \frac{20000 \times 12 \times 3}{100} = \text{Rs } 7200 \)

Difference:
Difference between CI and SI = \( 8098.56 - 7200 = \text{Rs } 898.56 \)

In simple words: Compound interest earns more because interest from previous years gets added to the principal. The difference between compounding and simple growth over three years here is Rs 898.56.

Exam Tip: Be sure to compute both the complete Compound Interest and the Simple Interest before attempting to find the difference between them.

Question 14. The simple interest on a certain sum of money at 4% per annum for 2 years is Rs 1500. What will be the compound interest on the same sum for the same time at the same rate, interest being payable annually?
Answer:
Let the principal sum be \( P \).
Given Simple Interest (SI) = Rs 1500, Rate (\( r \)) = 4% per annum, Time (\( t \)) = 2 years.
Using the Simple Interest formula:
\( \text{Simple Interest} = \frac{P \times r \times t}{100} \)

\( \implies 1500 = \frac{P \times 4 \times 2}{100} \)

\( \implies P = \frac{150000}{8} = \text{Rs } 18750 \)

Now, using this principal (\( P_1 = \text{Rs } 18750 \)) to calculate Compound Interest for 2 years at 4%:
For the first year:
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 18750 \left(1 + \frac{4}{100}\right) \)
\( = 18750 \times \frac{104}{100} \)
\( = \text{Rs } 19500 \)

For the second year:
Principal for the second year (\( P_2 \)) = Rs 19500.
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 19500 \left(1 + \frac{4}{100}\right) \)
\( = 19500 \times \frac{104}{100} \)
\( = \text{Rs } 20280 \)

Compound Interest (CI) = \( A - P = 20280 - 18750 = \text{Rs } 1530 \)

In simple words: First find the starting principal from the given simple interest. Once we have the principal, compute its compound interest over two years at the same rate.

Exam Tip: Make sure not to mix up simple and compound interest formulas. Solve for the principal using the simple interest formula first before switching to compounding.

Question 15. Find the difference between the simple and the compound interest on Rs 5000 invested for 3 years at 6% per annum, interest payable yearly.
Answer:
Case I: Compound Interest (CI) calculation
Principal for the first year (\( P_1 \)) = Rs 5000, Rate (\( r \)) = 6%.
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 5000 \left(1 + \frac{6}{100}\right) \)
\( = 5000 \times \frac{106}{100} \)
\( = \text{Rs } 5300 \)

Principal for the second year (\( P_2 \)) = Rs 5300, Rate (\( r \)) = 6%.
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 5300 \left(1 + \frac{6}{100}\right) \)
\( = 5300 \times \frac{106}{100} \)
\( = \text{Rs } 5618 \)

Principal for the third year (\( P_3 \)) = Rs 5618, Rate (\( r \)) = 6%.
Amount after 3 years = \( P_3 \left(1 + \frac{r}{100}\right) \)
\( = 5618 \left(1 + \frac{6}{100}\right) \)
\( = 5618 \times \frac{106}{100} \)
\( = \text{Rs } 5955.08 \)

Total Compound Interest (CI) = \( A - P = 5955.08 - 5000 = \text{Rs } 955.08 \)

Case II: Simple Interest (SI) calculation
Simple Interest = \( \frac{5000 \times 6 \times 3}{100} = \text{Rs } 900 \)

Difference:
Difference between CI and SI = \( 955.08 - 900 = \text{Rs } 55.08 \)

In simple words: Simple interest gives Rs 900 over three years, while compounding interest gives Rs 955.08. Subtracting the two values gives a difference of Rs 55.08.

Exam Tip: Since both methods are used for the same principal, rate, and time, you can verify your calculation knowing that CI must always be greater than SI.

Question 16. Simple interest on a sum of money for 2 years at 4% per annum is Rs 450. Find the compound interest on the same sum for 1 year at the same rate, interest being payable half-yearly.
Answer:
First, find the principal using Simple Interest.
Given SI = Rs 450, Rate (\( R \)) = 4%, Time (\( T \)) = 2 years.
Using Simple Interest formula:
\( \text{Simple Interest} = \frac{P \times R \times T}{100} \)

\( \implies 450 = \frac{P \times 4 \times 2}{100} \)

\( \implies P = \frac{45000}{8} = \text{Rs } 5625 \)

Now, using this principal (\( P_1 = \text{Rs } 5625 \)), we calculate compound interest for 1 year, compounded half-yearly at 4% per annum.
Rate of interest for each half-year (\( r \)) = \( \frac{4\%}{2} = 2\% \).
Time of 1 year contains 2 half-yearly periods.

For the first half-year:
Amount after \( \frac{1}{2} \) year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 5625 \left(1 + \frac{2}{100}\right) \)
\( = 5625 \times \frac{102}{100} \)
\( = \text{Rs } 5737.50 \)

For the second half-year:
Principal for the second half-year (\( P_2 \)) = Rs 5737.50.
Amount after 1 year = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 5737.50 \left(1 + \frac{2}{100}\right) \)
\( = 5737.50 \times \frac{102}{100} \)
\( = \text{Rs } 5852.25 \)

Total Compound Interest (CI) = \( A - P = 5852.25 - 5625 = \text{Rs } 227.25 \)

In simple words: Find the principal from the simple interest first. Then compute its compound interest for one year using two half-yearly compounding steps at half the annual rate.

Exam Tip: Be careful with the compounding frequency: for 1 year compounded half-yearly, calculate two compounding steps using half the given annual interest rate.

Question 17. A man borrows Rs 62500 at 8% per annum simple interest for 2 years. He immediately lends this money out at compound interest payable annually at the same rate and for the same time. What is his gain at the end of 2 years?
Answer:
Case I: Simple Interest (SI) paid
Principal (\( P \)) = Rs 62500, Rate (\( r \)) = 8%, Time (\( t \)) = 2 years.
Simple Interest = \( \frac{62500 \times 8 \times 2}{100} = \text{Rs } 10000 \)
Amount to be repaid under simple interest = \( 62500 + 10000 = \text{Rs } 72500 \)

Case II: Compound Interest (CI) received
Principal for the first year (\( P_1 \)) = Rs 62500, Rate (\( r \)) = 8%.
Amount after 1 year = \( P_1 \left(1 + \frac{r}{100}\right) \)
\( = 62500 \left(1 + \frac{8}{100}\right) \)
\( = 62500 \times \frac{108}{100} \)
\( = \text{Rs } 67500 \)

Principal for the second year (\( P_2 \)) = Rs 67500, Rate (\( r \)) = 8%.
Amount after 2 years = \( P_2 \left(1 + \frac{r}{100}\right) \)
\( = 67500 \left(1 + \frac{8}{100}\right) \)
\( = 67500 \times \frac{108}{100} \)
\( = \text{Rs } 72900 \)

Calculation of net gain:
Gain = Amount received (CI) - Amount repaid (SI)
\( = 72900 - 72500 = \text{Rs } 400 \)

In simple words: The man borrows at simple interest and lends at compound interest. Since compounding accumulates more value than simple interest over two years, he earns a net gain of Rs 400.

Exam Tip: You can directly compare the final amounts from both cases to find the profit, as the initial borrowed and lent principal sums are the same.

Question 18. What sum will amount to Rs 10120 in 2 years at compound interest payable annually, if the rates are 10% and 15% for the successive years?
Answer:
Let us assume an initial Principal of Rs 100.

For the first year:
Principal (\( P \)) = Rs 100, Rate (\( R \)) = 10%, Time (\( T \)) = 1 year.
Interest = \( \frac{100 \times 10 \times 1}{100} = \text{Rs } 10 \)
Amount after the first year = \( 100 + 10 = \text{Rs } 110 \)

For the second year:
Principal (\( P \)) = Rs 110, Rate (\( R \)) = 15%, Time (\( T \)) = 1 year.
Interest = \( \frac{110 \times 15 \times 1}{100} = \text{Rs } 16.50 \)
Amount after the second year = \( 110 + 16.50 = \text{Rs } 126.50 \)

By applying the unitary method:
When the final accumulated amount is Rs 126.50, the starting Principal is Rs 100.
Therefore, when the final accumulated amount is Rs 10120, the actual Principal (\( P \)) is:
\( P = \frac{10120 \times 100}{126.50} = \text{Rs } 8000 \)

Hence, the required starting sum is Rs 8000.

In simple words: Assuming a starting sum of Rs 100, it grows to Rs 126.50 in two years. Comparing this ratio with the target amount of Rs 10120 gives a starting principal of Rs 8000.

Exam Tip: Assuming Rs 100 as a temporary starting principal is an efficient way to solve problems where the starting sum is unknown.

Question 19. Sunil borrowed Rs. 50,000 at 10% per annum simple interest. He immediately lent it to another person at the same rate of interest for the same period, compounded annually. Find his gain at the end of 1 1/2 years.
Answer:
First, let us find the simple interest Sunil has to pay:
\( P = \text{Rs. } 50000 \), \( R = 10\% \), and \( T = 1\frac{1}{2}\text{ years} = \frac{3}{2}\text{ years} \)
\( \text{S.I.} = \text{Rs. } \frac{50000 \times 10 \times 3}{100 \times 2} = \text{Rs. } 7500 \)

Next, we calculate the compound interest Sunil will earn:
For the initial year:
\( P = \text{Rs. } 50000 \), \( R = 10\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{50000 \times 10 \times 1}{100} = \text{Rs. } 5000 \)
Thus, the sum at the end of year one is:
\( \text{Amount} = \text{Rs. } 50000 + \text{Rs. } 5000 = \text{Rs. } 55000 \)

For the subsequent six months:
\( P = \text{Rs. } 55000 \), \( R = 10\% \), and \( T = \frac{1}{2}\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{55000 \times 10 \times 1}{100 \times 2} = \text{Rs. } 2750 \)
The total accumulated value is:
\( \text{Amount} = \text{Rs. } 55000 + \text{Rs. } 2750 = \text{Rs. } 57750 \)

Therefore, the overall compound interest accrued is:
\( \text{Total C.I. earned} = \text{Rs. } 57750 - \text{Rs. } 50000 = \text{Rs. } 7750 \)

The net profit Sunil makes over 1.5 years is:

\( \implies \text{Sunil's gain} = \text{C.I. earned} - \text{S.I. paid} \)
\( = \text{Rs. } 7750 - \text{Rs. } 7500 \)
\( = \text{Rs. } 250 \)
In simple words: Sunil borrows money at simple interest but lends it at compound interest. Because compound interest grows faster, he earns Rs. 7,750 while only paying Rs. 7,500, giving him a clean profit of Rs. 250.

Exam Tip: For fractional time periods like 1 1/2 years under annual compounding, calculate the interest for the full year first, and then use the resulting amount as the principal for the remaining half-year.

Question 20. A machine depreciates at the rate of 5% in the first year, 5% in the second year, and 10% in the third year. Find the net percentage depreciation of the machine at the end of three years.
Answer:
Suppose we start with an initial value of Rs. 100 for the machine.
During the first year, the value drop is:
\( \text{Depreciation} = 5\% \text{ of Rs. } 100 = \frac{5}{100} \times 100 = \text{Rs. } 5 \)
The machine's value at the start of year two becomes:
\( \text{Value} = \text{Rs. } 100 - \text{Rs. } 5 = \text{Rs. } 95 \)

During the second year, the value drop is:
\( \text{Depreciation} = 5\% \text{ of Rs. } 95 = \frac{5}{100} \times 95 = \text{Rs. } 4.75 \)
The machine's value at the start of year three becomes:
\( \text{Value} = \text{Rs. } 95 - \text{Rs. } 4.75 = \text{Rs. } 90.25 \)

During the third year, the value drop is:
\( \text{Depreciation} = 10\% \text{ of Rs. } 90.25 = \frac{10}{100} \times 90.25 = \text{Rs. } 9.025 \)
The final machine value after three years is:
\( \text{Value} = \text{Rs. } 90.25 - \text{Rs. } 9.025 = \text{Rs. } 81.225 \)

The overall drop in value is calculated as:
\( \text{Net depreciation} = \text{Rs. } 100 - \text{Rs. } 81.225 = \text{Rs. } 18.775 \text{ or } 18.775\% \)
In simple words: By assuming the machine starts at Rs. 100, we apply each year's depreciation rate step-by-step. By the end, the value falls to Rs. 81.225, meaning it has lost 18.775% of its starting worth.

Exam Tip: Assuming an initial value of Rs. 100 makes percentage depreciation problems much easier to calculate compared to using a variable like x.

Question 21. A man borrows Rs. 6,500 at 10% per annum compound interest, interest being compounded half-yearly. He repays Rs. 2,000 at the end of each half-year. Find the amount outstanding after the third payment.
Answer:
For the first six-month period:
\( P = \text{Rs. } 6500 \), \( R = 10\% \), and \( T = \frac{1}{2}\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{6500 \times 10 \times 1}{100 \times 2} = \text{Rs. } 325 \)
\( \text{Amount} = \text{Rs. } 6500 + \text{Rs. } 325 = \text{Rs. } 6825 \)
Amount repaid at the end of the first six months = Rs. 2000
Remaining balance for the second six-month period = Rs. 6825 - Rs. 2000 = Rs. 4825

For the second six-month period:
\( P = \text{Rs. } 4825 \), \( R = 10\% \), and \( T = \frac{1}{2}\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{4825 \times 10 \times 1}{100 \times 2} = \text{Rs. } 241.25 \)
\( \text{Amount} = \text{Rs. } 4825 + \text{Rs. } 241.25 = \text{Rs. } 5066.25 \)
Amount repaid at the end of the second six months = Rs. 2000
Remaining balance for the third six-month period = Rs. 5066.25 - Rs. 2000 = Rs. 3066.25

For the third six-month period:
\( P = \text{Rs. } 3066.25 \), \( R = 10\% \), and \( T = \frac{1}{2}\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{3066.25 \times 10 \times 1}{100 \times 2} = \text{Rs. } 153.3125 \)
\( \text{Amount} = \text{Rs. } 3066.25 + \text{Rs. } 153.3125 = \text{Rs. } 3219.5625 \)
Amount repaid at the end of the third six months = Rs. 2000
Outstanding balance after making the third installment:
\( = \text{Rs. } 3219.5625 - \text{Rs. } 2000 \)
\( = \text{Rs. } 1219.5625 \)
\( \approx \text{Rs. } 1220 \text{ (to the nearest rupee)} \)
In simple words: Every six months, interest is added to the balance, and then Rs. 2,000 is paid off. Repeating this process three times leaves a final unpaid balance of Rs. 1,220.

Exam Tip: Be meticulous with calculations containing decimals. Do not round off intermediate steps; only round off the final answer to the nearest rupee.

Question 22. A man borrows Rs. 20,000 at 10% per annum compound interest. He repays Rs. 5,000 at the end of the first year and Rs. 10,000 at the end of the second year. How much should he pay at the end of the third year to clear his debt?
Answer:
For the first year:
\( P = \text{Rs. } 20000 \), \( R = 10\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{20000 \times 10 \times 1}{100} = \text{Rs. } 2000 \)
\( \text{Amount} = \text{Rs. } 20000 + \text{Rs. } 2000 = \text{Rs. } 22000 \)
Payment made at the first year's end = Rs. 5000
Outstanding balance for the second year = Rs. 22000 - Rs. 5000 = Rs. 17000

For the second year:
\( P = \text{Rs. } 17000 \), \( R = 10\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{17000 \times 10 \times 1}{100} = \text{Rs. } 1700 \)
\( \text{Amount} = \text{Rs. } 17000 + \text{Rs. } 1700 = \text{Rs. } 18700 \)
Payment made at the second year's end = Rs. 10000
Outstanding balance for the third year = Rs. 18700 - Rs. 10000 = Rs. 8700

For the third year:
\( P = \text{Rs. } 8700 \), \( R = 10\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{8700 \times 10 \times 1}{100} = \text{Rs. } 870 \)
\( \text{Amount} = \text{Rs. } 8700 + \text{Rs. } 870 = \text{Rs. } 9570 \)
To completely settle the loan at the third year's end, the person must pay Rs. 9570.
In simple words: The borrower pays back part of the loan at the end of each year. Each year, interest is added to the remaining balance before subtracting his payment. At the end of the third year, he pays the full final amount of Rs. 9,570 to completely clear the debt.

Exam Tip: Remember that interest is calculated on the remaining balance at the beginning of each year, not on the original starting principal of Rs. 20,000.

Question 23. The value of a ring appreciates by 10% per annum. If the total appreciation in its value at the end of two years is Rs. 6,300, find the original value of the ring.
Answer:
Let us assume the initial value of the ring (\( P_1 \)) is Rs. 100.
Appreciation for the first year is:
\( \text{Appreciation} = \text{Rs. } \frac{100 \times 10 \times 1}{100} = \text{Rs. } 10 \)
Value of the ring at the first year's end (\( A_1 \)) = Rs. 100 + Rs. 10 = Rs. 110

Starting value of the ring for the second year (\( P_2 \)) = Rs. 110
Appreciation for the second year is:
\( \text{Appreciation} = \text{Rs. } \frac{110 \times 10 \times 1}{100} = \text{Rs. } 11 \)

Combined appreciation over both years:
\( = \text{Rs. } (10 + 11) \)
\( = \text{Rs. } 21 \)

If the combined appreciation is Rs. 21, the original value (\( P_1 \)) is Rs. 100.
When the combined appreciation is Rs. 6,300, the actual original value is:
\( = \text{Rs. } \frac{100 \times 6300}{21} \)
\( = \text{Rs. } 30000 \)

Thus, the initial value of the ring is Rs. 30,000.
In simple words: By assuming the ring started at Rs. 100, we find a total gain of Rs. 21 over two years. Since the actual gain is Rs. 6,300 (which is 300 times larger), the starting value must also be 300 times larger than Rs. 100, which is Rs. 30,000.

Exam Tip: Using the unitary method with an assumed value of Rs. 100 is an elegant way to solve appreciation or depreciation problems without dealing with algebraic variables.

Question 24. A sum of Rs. 15,500 is lent at compound interest for 3 years. The rates of interest for the first, second, and third years are 10%, 15%, and 20% respectively. Find the difference between the compound interest of the second year and the third year.
Answer:
For the first year:
\( P = \text{Rs. } 15500 \), \( R = 10\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{15500 \times 10 \times 1}{100} = \text{Rs. } 1550 \)
\( \text{Amount} = \text{Rs. } 15500 + \text{Rs. } 1550 = \text{Rs. } 17050 \)

For the second year:
\( P = \text{Rs. } 17050 \), \( R = 15\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{17050 \times 15 \times 1}{100} = \text{Rs. } 2557.50 \)
\( \text{Amount} = \text{Rs. } 17050 + \text{Rs. } 2557.50 = \text{Rs. } 19607.50 \)

For the third year:
\( P = \text{Rs. } 19607.50 \), \( R = 20\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{19607.50 \times 20 \times 1}{100} = \text{Rs. } 3921.50 \)
\( \text{Amount} = \text{Rs. } 19607.50 + \text{Rs. } 3921.50 = \text{Rs. } 23529 \)

The difference between the compound interest of the second year and the third year is:
\( = \text{Rs. } (3921.50 - 2557.50) \)
\( = \text{Rs. } 1364 \)
In simple words: Since the interest rates increase each year, the amount of interest earned grows too. The interest earned in the third year is Rs. 1,364 more than what was earned in the second year.

Exam Tip: Be careful to calculate the interest of each specific year individually, rather than finding the total interest over the three years, as the question specifically asks for the difference between the second and third year's interest.

Question 25. Samidha borrowed Rs. 7,500 from Shreya at 30% per annum compound interest. At the end of 2 years, she cleared her debt by paying Rs. 10,000 and a juicer. Find the cost of the juicer.
Answer:
For the first year:
\( P = \text{Rs. } 7500 \), \( R = 30\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{7500 \times 30 \times 1}{100} = \text{Rs. } 2250 \)
\( \text{Amount} = \text{Rs. } 7500 + \text{Rs. } 2250 = \text{Rs. } 9750 \)

For the second year:
\( P = \text{Rs. } 9750 \), \( R = 30\% \), and \( T = 1\text{ year} \)
\( \text{Interest} = \text{Rs. } \frac{9750 \times 30 \times 1}{100} = \text{Rs. } 2925 \)
\( \text{Amount} = \text{Rs. } 9750 + \text{Rs. } 2925 = \text{Rs. } 12675 \)

Thus, the total amount that Samidha has to pay Shreya is Rs. 12675.
Since Samidha gave Rs. 10000 and a juicer to settle this debt:

\( \implies \text{Rs. } 10000 + \text{Cost of juicer} = \text{Rs. } 12675 \)

\( \implies \text{Cost of juicer} = \text{Rs. } 12675 - \text{Rs. } 10000 = \text{Rs. } 2675 \)

Therefore, the value of the juicer is Rs. 2675.
In simple words: Samidha owes a total of Rs. 12,675 after two years. Since she pays Rs. 10,000 in cash, the juicer must cover the remaining balance of Rs. 2,675.

Exam Tip: First find the final amount due using compound interest. Then set up a simple equation where Cash + Item Value = Total Amount to find the item's cost.

Exercise 3.2

Question 1. Find the amount and compound interest on:
(i) Rs 8,000 for 3 years at 10% per annum compounded annually.
(ii) Rs 15,000 for 2 years at 8% per annum compounded semi-annually.
(iii) Rs 12,000 for 1 1/2 years at 5% per annum compounded annually.
(iv) Rs 25,000 for 2 years at 6% per annum compounded semi-annually.
(v) Rs 16,000 for 3 years at 10%, 8%, and 6% for successive years.
Answer:
(i) Here \( P = \text{Rs. } 8000 \), \( t = 3\text{ years} \), and \( r = 10\% \)
Amount is calculated as:
\( \text{Amount} = P \left(1 + \frac{r}{100}\right)^t = 8000 \left(1 + \frac{10}{100}\right)^3 \)
\( = 8000 \left(\frac{11}{10}\right)^3 \)
\( = 8000 \times \frac{1331}{1000} = \text{Rs. } 10648 \)
Hence, the accumulated Amount = Rs. 10648
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 10648 - \text{Rs. } 8000 = \text{Rs. } 2648 \)

(ii) Here \( P = \text{Rs. } 15000 \), \( t = 2\text{ years} \), and \( r = 8\% \)
Since compounding is semi-annual, we use:
\( \text{Amount} = P \left(1 + \frac{r}{200}\right)^{2t} = 15000 \left(1 + \frac{8}{200}\right)^4 \)
\( = 15000 \left(\frac{26}{25}\right)^4 \)
\( = 15000 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} \approx \text{Rs. } 17547.88 \)
Hence, the accumulated Amount = Rs. 17547.88
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 17547.88 - \text{Rs. } 15000 = \text{Rs. } 2547.88 \)

(iii) Here \( P = \text{Rs. } 12000 \), \( t = 1\frac{1}{2}\text{ years} \), and \( r = 5\% \)
Since compounding is annual:
The value after the first year is:
\( \text{Amount after 1 year} = P \left(1 + \frac{r}{100}\right)^t = 12000 \left(1 + \frac{5}{100}\right) \)
\( = 12000 \left(\frac{105}{100}\right) = \text{Rs. } 12600 \)
The interest for the remaining half-year is calculated as:
\( \text{Interest for next half year} = \frac{12600 \times 5 \times 1}{100 \times 2} = \text{Rs. } 315 \)
Hence, the total Amount = Rs. 12600 + Rs. 315 = Rs. 12915
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 12915 - \text{Rs. } 12000 = \text{Rs. } 915 \)

(iv) Here \( P = \text{Rs. } 25000 \), \( t = 2\text{ years} \), and \( r = 6\% \)
Since compounding is semi-annual, we have:
\( \text{Amount} = P \left(1 + \frac{r}{200}\right)^{2t} = 25000 \left(1 + \frac{6}{200}\right)^4 \)
\( = 25000 \left(\frac{103}{100}\right)^4 \approx \text{Rs. } 28137.72 \)
Hence, the accumulated Amount = Rs. 28137.72
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 28137.72 - \text{Rs. } 25000 = \text{Rs. } 3137.72 \)

(v) Here \( P = \text{Rs. } 16000 \), \( t = 3\text{ years} \), with successive interest rates of \( r_1 = 10\% \), \( r_2 = 8\% \), and \( r_3 = 6\% \)
We use the formula for successive rates:
\( \text{Amount} = P \left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right)\left(1 + \frac{r_3}{100}\right) \)
\( = 16000 \left(1 + \frac{10}{100}\right)\left(1 + \frac{8}{100}\right)\left(1 + \frac{6}{100}\right) \)
\( = 16000 \left(\frac{11}{10}\right)\left(\frac{108}{100}\right)\left(\frac{106}{100}\right) \approx \text{Rs. } 20148.48 \)
Hence, the accumulated Amount = Rs. 20148.48
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 20148.48 - \text{Rs. } 16000 = \text{Rs. } 4148.48 \)
In simple words: To find the compound interest and total amount, we use different formulas depending on whether compounding is annual, semi-annual, or if the interest rate changes every year.

Exam Tip: Remember to modify the interest rate and time period based on compounding: for half-yearly compounding, divide the annual rate by 2 (or use r/200 in the formula) and multiply the years by 2 for the exponent.

Question 2. Find the compound interest and amount on Rs 15,000 for 2 1/2 years at 10% per annum compounded annually.
Answer:
For this problem, \( P = \text{Rs. } 15000 \), \( t = 2\frac{1}{2}\text{ years} \), and \( r = 10\% \).
The value after the first 2 years is calculated as:
\( \text{Amount after 2 years} = P \left(1 + \frac{r}{100}\right)^2 = 15000 \left(1 + \frac{10}{100}\right)^2 \)
\( = 15000 \left(\frac{11}{10}\right)^2 = \text{Rs. } 18150 \)

The interest earned over the next six months is:
\( \text{Interest for next half year} = \frac{18150 \times 10 \times 1}{100 \times 2} = \text{Rs. } 907.50 \)
Thus, the total accumulated amount is:
\( \text{Amount} = \text{Rs. } 18150 + \text{Rs. } 907.50 = \text{Rs. } 19057.50 \)

Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 19057.50 - \text{Rs. } 15000 = \text{Rs. } 4057.50 \)
In simple words: Since we compound annually for 2.5 years, we find the compound interest for the first 2 full years, then calculate the simple interest for the remaining half-year on that new amount.

Exam Tip: For fractional years under annual compounding, do not use a fractional exponent. Split the time into whole years (using the compound interest formula) and the remaining fraction of a year (using the simple interest formula on the accumulated amount).

Question 3. Find the amount on Rs 36,000 for 2 years at 15% per annum compounded annually.
Answer:
For this case, \( P = \text{Rs. } 36000 \), \( t = 2\text{ years} \), and \( r = 15\% \).
The final sum is calculated as follows:
\( \text{Amount} = P \left(1 + \frac{r}{100}\right)^t = 36000 \left(1 + \frac{15}{100}\right)^2 \)
\( = 36000 \left(\frac{115}{100}\right)^2 = \text{Rs. } 47610 \)

Hence, the final Amount = Rs. 47610.
In simple words: We apply the compound interest formula directly to find the total money accumulated at the end of 2 years.

Exam Tip: Simplify the fraction inside the parentheses before squaring to make the multiplication much faster and reduce chances of error.

Question 4. Find the amount and compound interest on Rs 50,000 for 1 1/2 years at 8% per annum compounded half-yearly.
Answer:
For this problem, \( P = \text{Rs. } 50000 \), \( t = 1\frac{1}{2}\text{ years} \), and \( r = 8\% \).
Because interest is compounded half-yearly, we use the modified formula:
\( \text{Amount} = P \left(1 + \frac{r}{200}\right)^{2t} = 50000 \left(1 + \frac{8}{200}\right)^3 \)
\( = 50000 \left(\frac{104}{100}\right)^3 = \text{Rs. } 56243.20 \)

Hence, the final Amount = Rs. 56243.20.
Also, the compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 56243.20 - \text{Rs. } 50000 = \text{Rs. } 6243.20 \)
In simple words: Since the interest compounds every six months, 1.5 years means 3 interest periods. We use a half-year rate of 4% for 3 periods to get the final amount.

Exam Tip: Remember that "compounded half-yearly" means the rate is divided by 2 and the time is multiplied by 2. Always write out these modified values before plugging them into the formula.

Question 5. Find the amount on Rs 25,000 for 2 years when the rates of interest for successive years are 4% and 5% respectively.
Answer:
For this case, \( P = \text{Rs. } 25000 \), \( t = 2\text{ years} \), with successive interest rates of \( r_1 = 4\% \) and \( r_2 = 5\% \).
The total accumulated amount is calculated as follows:
\( \text{Amount} = P \left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right) \)
\( = 25000 \left(1 + \frac{4}{100}\right)\left(1 + \frac{5}{100}\right) \)
\( = 25000 \left(\frac{104}{100}\right)\left(\frac{105}{100}\right) = \text{Rs. } 27300 \)

Hence, the final Amount = Rs. 27300.
In simple words: Since the interest rates are different each year, we multiply the principal by the growth factor of the first year, and then by the growth factor of the second year.

Exam Tip: When given successive rates, use the multi-bracket formula rather than calculating year-by-year, as it is much faster and reduces intermediate arithmetic steps.

Question 6. Find the amount on Rs 31,250 for 3 years when the rates of interest for successive years are 8%, 10%, and 12% respectively.
Answer:
For this problem, \( P = \text{Rs. } 31250 \), \( t = 3\text{ years} \), with successive interest rates \( r_1 = 8\% \), \( r_2 = 10\% \), and \( r_3 = 12\% \).
The total accumulated amount is calculated as follows:
\( \text{Amount} = P \left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right)\left(1 + \frac{r_3}{100}\right) \)
\( = 31250 \left(1 + \frac{8}{100}\right)\left(1 + \frac{10}{100}\right)\left(1 + \frac{12}{100}\right) \)
\( = 31250 \left(\frac{108}{100}\right)\left(\frac{110}{100}\right)\left(\frac{112}{100}\right) = \text{Rs. } 41580 \)

Hence, the final Amount = Rs. 41580.
In simple words: When the rate changes every year, we calculate the growth for each year one after the other by multiplying the three separate yearly growth factors with the starting principal.

Exam Tip: Be careful with canceling out zeros in the fraction to simplify the product before doing long multiplication.

Question 7. At what rate percent per annum will Rs 28,000 yield Rs 30,870 in 2 years, interest being compounded annually?
Answer:
Here, \( P = \text{Rs. } 28000 \), \( A = \text{Rs. } 30870 \), and \( t = 2\text{ years} \).
Using the formula for compound interest:
\( A = P \left(1 + \frac{r}{100}\right)^t \)

\( \implies 30870 = 28000 \left(1 + \frac{r}{100}\right)^2 \)

\( \implies \left(1 + \frac{r}{100}\right)^2 = \frac{30870}{28000} = \frac{441}{400} = \left(\frac{21}{20}\right)^2 \)
Taking the square root on both sides:

\( \implies 1 + \frac{r}{100} = \frac{21}{20} \)

\( \implies \frac{r}{100} = \frac{21}{20} - 1 = \frac{1}{20} \)

\( \implies R = \frac{100}{20} = 5 \)

Thus, the rate of interest is 5%.
In simple words: We set up the compound interest formula with our known values and solve for the interest rate. Simplifying the fraction allows us to take the square root of both sides easily to find that the rate is 5%.

Exam Tip: When solving for the rate in a 2-year compounding problem, always try to reduce the fraction on the other side to a perfect square (like 441/400 = (21/20)^2) so you can take the square root easily.

Question 8. In how many years will Rs 15,625 amount to Rs 17,576 at 4% per annum compounded annually?
Answer:
We are given: \( P = \text{Rs. } 15625 \), \( A = \text{Rs. } 17576 \), and \( r = 4\% \).
Using the compound interest formula:
\( A = P \left(1 + \frac{r}{100}\right)^t \)

\( \implies 17576 = 15625 \left(1 + \frac{4}{100}\right)^t \)

\( \implies \left(\frac{26}{25}\right)^t = \frac{17576}{15625} \)
Simplifying the terms, we see that:
\( \frac{17576}{15625} = \left(\frac{26}{25}\right)^3 \)
Thus, we have:
\( \left(\frac{26}{25}\right)^t = \left(\frac{26}{25}\right)^3 \)
Equating the exponents on both sides:

\( \implies t = 3 \)

Therefore, the required time is 3 years.
In simple words: By writing down the formula, we find that the ratio of the final amount to the starting principal is equal to 26/25 raised to the power of time. Since 17,576/15,625 is exactly (26/25) cubed, the time must be 3 years.

Exam Tip: When solving for time, simplify the base fraction first. Then, express the ratio of Amount to Principal as a power of that base fraction to find the exponent directly.

Question 9. In how many years will Rs 2,000 amount to Rs 2,662 at 10% per annum compounded annually?
Answer:
Here we have: \( P = \text{Rs. } 2000 \), \( A = \text{Rs. } 2662 \), and \( r = 10\% \).
Using the compound interest formula:
\( A = P \left(1 + \frac{r}{100}\right)^t \)

\( \implies 2662 = 2000 \left(1 + \frac{10}{100}\right)^t \)

\( \implies \left(\frac{11}{10}\right)^t = \frac{2662}{2000} \)
Simplifying the fraction on the right gives:
\( \frac{2662}{2000} = \frac{1331}{1000} = \left(\frac{11}{10}\right)^3 \)
Thus, we have:
\( \left(\frac{11}{10}\right)^t = \left(\frac{11}{10}\right)^3 \)
By comparing the powers on both sides:

\( \implies t = 3 \)

Therefore, the required time is 3 years.
In simple words: The ratio of our amount to our principal simplifies to 1,331/1,000, which is exactly the cube of 11/10. Therefore, it takes 3 years to reach this amount.

Exam Tip: Always reduce the fraction representing the ratio of A/P to its lowest terms; this makes it much easier to identify it as a perfect square, cube, or higher power.

Question 10. The simple interest on a certain sum of money for 3 years at 4% per annum is Rs 600. Find the compound interest on the same sum for the same period at the same rate.
Answer:
We know that Simple Interest is given by the formula:
\( \text{Simple Interest} = \frac{P \times r \times t}{100} \)

\( \implies 600 = \frac{P \times 4 \times 3}{100} \)

\( \implies P = \frac{60000}{12} = \text{Rs. } 5000 \)

Now, we find the compound interest for this principal \( P = \text{Rs. } 5000 \) at \( r = 4\% \) for \( t = 3\text{ years} \):
\( \text{Amount} = P \left(1 + \frac{r}{100}\right)^t = 5000 \left(1 + \frac{4}{100}\right)^3 \)
\( = 5000 \left(\frac{26}{25}\right)^3 = \text{Rs. } 5624.32 \)

Hence, the final Amount = Rs. 5624.32.
The compound interest is:
\( \text{C.I.} = A - P = \text{Rs. } 5624.32 - \text{Rs. } 5000 = \text{Rs. } 624.32 \)
In simple words: First, we use the simple interest of Rs. 600 to find that the starting principal is Rs. 5,000. Then, we use the compound interest formula on this Rs. 5,000 to find the total interest, which comes out to Rs. 624.32.

Exam Tip: Do this problem in two distinct steps. First find the principal using the Simple Interest formula, and only then proceed to calculate the Compound Interest.

Question 11. The compound interest on a certain sum of money for 2 years is Rs 40.80 and the simple interest on the same sum for the same period is Rs 40. Find the sum and the rate of interest.
Answer:
Let the principal sum be Rs \( P \) and the annual interest rate be \( r\% \).
We are given \( t = 2\text{ years} \), \( \text{C.I.} = \text{Rs. } 40.80 \), and \( \text{S.I.} = \text{Rs. } 40 \).
Since Simple Interest is calculated as:
\( \text{Simple Interest} = \frac{P \times r \times t}{100} \)

\( \implies 40 = \frac{P \times r \times 2}{100} \)

\( \implies Pr = \frac{4000}{2} = 2000 \)

Now, using the formula for Compound Interest:
\( \text{C.I.} = A - P = P \left[\left(1 + \frac{r}{100}\right)^t - 1\right] \)

\( \implies 40.80 = P \left[\left(1 + \frac{r}{100}\right)^2 - 1\right] \)
Expanding the binomial term inside the brackets:

\( \implies 40.80 = P \left[1 + \frac{r^2}{10000} + \frac{2r}{100} - 1\right] \)

\( \implies 40.80 = P \left[\frac{r^2}{10000} + \frac{2r}{100}\right] \)
Factoring out \( r \) to form the term \( Pr \):

\( \implies 40.80 = Pr \left[\frac{r}{10000} + \frac{2}{100}\right] \)

\( \implies 40.80 = Pr \left[\frac{r + 200}{10000}\right] \)
Substituting \( Pr = 2000 \):

\( \implies 40.80 = 2000 \left[\frac{r + 200}{10000}\right] = \frac{r + 200}{5} \)

\( \implies r + 200 = 40.80 \times 5 = 204 \)

\( \implies r = 204 - 200 = 4\% \)

Now, using \( Pr = 2000 \), we can find the principal:

\( \implies P = \frac{2000}{r} = \frac{2000}{4} = \text{Rs. } 500 \)

Thus, the sum is Rs. 500 and the rate of interest is 4%.
In simple words: The difference between compound interest and simple interest for 2 years is the interest on the first year's interest. By finding that this difference is Rs. 0.80, we find the rate is 4% and the original sum is Rs. 500.

Exam Tip: Expressing the compound interest equation in terms of \( Pr \) allows you to easily substitute the simple interest value and avoid solving a complex quadratic equation.

Question 12. The difference between the compound interest and the simple interest on a certain sum of money for 2 years at 8% per annum is Rs 448. Find the sum.
Answer:
Let the principal sum be Rs \( P \).
Since \( \text{C.I.} = A - P \):
\( \text{C.I.} = P \left(1 + \frac{8}{100}\right)^2 - P = P \left(\frac{108}{100}\right)^2 - P \)
\( = \frac{11664P}{10000} - P = \frac{1664P}{10000} \)

We also calculate the Simple Interest:
\( \text{S.I.} = \frac{P \times 8 \times 2}{100} = \frac{16P}{100} \)

We are given that the difference between C.I. and S.I. is Rs. 448:
\( \text{C.I.} - \text{S.I.} = \text{Rs. } 448 \)

\( \implies \frac{1664P}{10000} - \frac{16P}{100} = 448 \)

\( \implies \frac{1664P - 1600P}{10000} = 448 \)

\( \implies \frac{64P}{10000} = 448 \)

\( \implies 64P = 4480000 \)

\( \implies P = \text{Rs. } 70000 \)

Hence, the sum is Rs. 70,000.
In simple words: We write algebraic expressions for both the compound interest and simple interest. Setting their difference equal to Rs. 448 allows us to solve for the original sum of Rs. 70,000.

Exam Tip: Alternatively, for a 2-year difference, you can use the direct formula: \( \text{C.I.} - \text{S.I.} = P \left(\frac{r}{100}\right)^2 \). This saves significant computation time.

Question 13. The difference between the compound interest and the simple interest on Rs 50,000 for 2 years is Rs 125. Find the rate of interest per annum.
Answer:
Let the annual interest rate be \( r\% \).
The Simple Interest for 2 years on Rs. 50,000 is:
\( \text{S.I. in 2 years} = \text{Rs. } \frac{50000 \times r \times 2}{100} = \text{Rs. } 1000r \)

The Compound Interest for 2 years on Rs. 50,000 is:
\( \text{C.I. in 2 years} = A - P = 50000 \left(1 + \frac{r}{100}\right)^2 - 50000 \)

We are given that the difference between C.I. and S.I. is Rs. 125:
\( \text{C.I.} - \text{S.I.} = \text{Rs. } 125 \)

\( \implies 50000 \left(1 + \frac{r}{100}\right)^2 - 50000 - 1000r = 125 \)

\( \implies 50000 \left(1 + \frac{r^2}{10000} + \frac{2r}{100}\right) - 50000 - 1000r = 125 \)

\( \implies 50000 + 5r^2 + 1000r - 50000 - 1000r = 125 \)

\( \implies 5r^2 = 125 \)

\( \implies r^2 = 25 \)

\( \implies r = \pm 5 \)
Since the rate of interest cannot be negative:

\( \implies r = 5\% \)

Therefore, the rate of interest is 5%.
In simple words: By expanding the compound interest equation and subtracting the simple interest, the linear terms cancel out beautifully, leaving a simple quadratic equation that yields an interest rate of 5%.

Exam Tip: Expanding the squared term \( \left(1 + \frac{r}{100}\right)^2 \) first is the key to simplifying the algebraic expression and canceling out the \( 1000r \) term.

Question 14. What sum of money will amount to Rs 15,729 in 2 years if the rates of interest for successive years are 5% and 7% respectively?
Answer:
We are given: \( \text{Amount} = \text{Rs. } 15729 \), \( n = 2\text{ years} \), with successive interest rates \( r_1 = 5\% \) and \( r_2 = 7\% \).` `
Using the formula for successive compounding rates:
\( A = P \left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right) \)

\( \implies 15729 = P \left(1 + \frac{5}{100}\right)\left(1 + \frac{7}{100}\right) \)

\( \implies 15729 = P \left(\frac{105}{100}\right)\left(\frac{107}{100}\right) \)
Solving for \( P \):

\( \implies P = \frac{15729 \times 100 \times 100}{105 \times 107} \)

\( \implies P = \text{Rs. } 14000 \)

Hence, the principal sum is Rs. 14,000.
In simple words: We plug the target amount and the two consecutive rates into our formula. Solving for the starting principal tells us that Rs. 14,000 is the sum that grows to Rs. 15,729.

Exam Tip: Reduce the fractions or look for common factors between the numerator and denominator to simplify the final calculation of \( P \).

Question 15. At what rate percent per annum compound interest will Rs 12,000 amount to Rs 13,891.50 in 3 years?
Answer:
We are given: \( A = \text{Rs. } 13891.50 \), \( P = \text{Rs. } 12000 \), and \( n = 3\text{ years} \).
Using the compound interest formula:
\( A = P \left(1 + \frac{r}{100}\right)^3 \)

\( \implies 13891.50 = 12000 \left(1 + \frac{r}{100}\right)^3 \)

\( \implies \frac{13891.50}{12000} = \left(1 + \frac{r}{100}\right)^3 \)

\( \implies \frac{1389150}{1200000} = \left(1 + \frac{r}{100}\right)^3 \)

\( \implies \frac{9261}{8000} = \left(1 + \frac{r}{100}\right)^3 \)
Since \( 9261 = 21^3 \) and \( 8000 = 20^3 \):

\( \implies \left(\frac{21}{20}\right)^3 = \left(1 + \frac{r}{100}\right)^3 \)
Taking the cube root on both sides:

\( \implies \frac{21}{20} = 1 + \frac{r}{100} \)

\( \implies \frac{r}{100} = \frac{21}{20} - 1 = \frac{1}{20} \)

\( \implies r = 5\% \)

Thus, the rate of interest is 5%.
In simple words: We find the ratio of the final amount to the principal. After simplifying, we see it is a perfect cube, allowing us to find the interest rate by taking the cube root of both sides.

Exam Tip: For a 3-year period, the ratio of A/P will always simplify to a perfect cube (like 9261/8000 = (21/20)^3). Memorizing basic cubes up to 25 will help you identify these instantly.

Question 16. Divide Rs 16,820 between A and B, aged 27 years and 25 years respectively, so that when they reach 40 years of age, their shares, with compound interest at 5% per annum, are equal.
Answer:
Let the share of A be Rs \( x \).
This leaves B's share as Rs \( (16820 - x) \).

For A: \( P = \text{Rs. } x \), \( r = 5\% \), and the compounding period is \( n = 40 - 27 = 13\text{ years} \).
The accumulated amount is:
\( A = P \left(1 + \frac{r}{100}\right)^n = x \left(1 + \frac{5}{100}\right)^{13} = x \left(\frac{21}{20}\right)^{13} \)

For B: \( P = \text{Rs. } (16820 - x) \), \( r = 5\% \), and the compounding period is \( n = 40 - 25 = 15\text{ years} \).
The accumulated amount is:
\( A = P \left(1 + \frac{r}{100}\right)^n = (16820 - x) \left(1 + \frac{5}{100}\right)^{15} = (16820 - x) \left(\frac{21}{20}\right)^{15} \)

Since both receive the same final sum when they turn 40:

\( \implies x \left(\frac{21}{20}\right)^{13} = (16820 - x) \left(\frac{21}{20}\right)^{15} \)
Dividing both sides by \( \left(\frac{21}{20}\right)^{13} \):

\( \implies x = (16820 - x) \left(\frac{21}{20}\right)^2 \)

\( \implies x = (16820 - x) \left(\frac{441}{400}\right) \)

\( \implies 400x = 441(16820 - x) \)

\( \implies 400x = 441 \times 16820 - 441x \)

\( \implies 841x = 441 \times 16820 \)

\( \implies x = \frac{441 \times 16820}{841} = 441 \times 20 = \text{Rs. } 8820 \)

Therefore, the shares are as follows:
Share of A = Rs. 8,820
Share of B = Rs. 16,820 - Rs. 8,820 = Rs. 8,000
In simple words: Since A is older, A's money will compound for fewer years than B's. To end up with equal amounts at age 40, the older person (A) must receive a larger initial share (Rs. 8,820) than the younger person (B, who gets Rs. 8,000).

Exam Tip: When dividing an equation with exponents on both sides (like 13 and 15), always divide through by the smaller exponent to leave a simple squared term on one side.