Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 1 Compound Interest

Ex 1.1

Answer 1. Calculate the amount and the compound interest for each of the following:

a) Rs. 7,500 at 12% p.a. in 3 years.
Answer:
Given:
P = Rs. 7,500; r = 12% p.a.; t = 3 years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{7,500 \times 12 \times 1}{100} \)
S.I. = Rs 900
A = P + S.I.
= Rs (7,500 + 900) = Rs 8,400 = new principal

For the second year: t = 1 year; P = Rs 8,400
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{8,400 \times 12 \times 1}{100} \)
S.I. = Rs 1,008
A = P + S.I.
A = Rs (8,400 + 1,008) = Rs 9,408 = new principal

For the third year: t = 1 year; P = Rs 9,408
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{9,408 \times 12 \times 1}{100} \)
S.I. = Rs 1,128.96
A = P + S.I.
A = Rs (9,408 + 1,128.96) = Rs 10,536.96

Final Amount = Rs 10,536.96
C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs (900 + 1,008 + 1,128.96) = Rs 3,036.96
In simple words: In compound interest, we add the interest to the principal each year. So the principal keeps growing each year. That's why we get more interest than simple interest.

📝 Teacher's Note: Show students how the principal keeps growing each year. Draw a table with Year, Principal at start, Interest earned, and Amount at end. This makes the concept clear.

🎯 Exam Tip: Always write "new principal" after each year's calculation. Show all three years separately. Write final amount and total compound interest clearly.

b) Rs. 13,500 at 10% p.a. in 2 years
Answer:
Given:
P = Rs. 13,500; r = 10% p.a.; t = 2 years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{13,500 \times 10 \times 1}{100} \)
S.I. = Rs 1,350
A = P + S.I.
= Rs (13,500 + 1,350) = Rs 14,850 = new principal

For the second year: t = 1 year; P = Rs 14,850
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{14,850 \times 10 \times 1}{100} \)
S.I. = Rs 1,485
A = P + S.I.
A = Rs (14,850 + 1,485) = Rs 16,335

Final Amount = Rs 16,335
C.I. = Interest in first year + interest in second year
C.I. = Rs (1,350 + 1,485) = Rs 2,835
In simple words: The money grows faster because we earn interest on the interest too. After year 1, we earned more interest because the principal was bigger.

📝 Teacher's Note: Compare with simple interest for the same problem. Simple interest would be Rs 2,700 for 2 years. Compound interest gives Rs 135 more.

🎯 Exam Tip: Show both years clearly. Write the principal for year 2 as the amount from year 1. This gets you method marks.

c) Rs. 17,500 at 12% p.a. in 3 years
Answer:
Given:
P = Rs. 17,500; r = 12% p.a.; t = 3 years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{17,500 \times 12 \times 1}{100} \)
S.I. = Rs 2,100
A = P + S.I.
= Rs (17,500 + 2,100) = Rs 19,600 = new principal

For the second year: t = 1 year; P = Rs 19,600
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{19,600 \times 12 \times 1}{100} \)
S.I. = Rs 2,352
A = P + S.I.
A = Rs (19,600 + 2,352) = Rs 21,952 = new principal

For the third year: t = 1 year; P = Rs 21,952
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{21,952 \times 12 \times 1}{100} \)
S.I. = Rs 2,634.24
A = P + S.I.
A = Rs (21,952 + 2,634.24) = Rs 24,586.24

Final Amount = Rs 24,586.24
C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs (2,100 + 2,352 + 2,634.24) = Rs 7,086.24
In simple words: Each year the interest amount gets bigger because we are earning 12% on a bigger amount. The money grows faster and faster each year.

📝 Teacher's Note: Point out how the interest increases each year: Rs 2,100, then Rs 2,352, then Rs 2,634.24. This shows the compounding effect clearly.

🎯 Exam Tip: Calculate each year step by step. Don't try to use the compound interest formula directly. Show working for all three years to get full marks.

d) Rs. 23,750 at 12% p.a. in 2½ years
Answer:
Given:
P = Rs. 23,750; r = 12% p.a.; t = 2½ years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{23,750 \times 12 \times 1}{100} \)
S.I. = Rs 2,850
A = P + S.I.
= Rs (23,750 + 2,850) = Rs 26,600 = new principal

For the second year: t = 1 year; P = Rs 26,600
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{26,600 \times 12 \times 1}{100} \)
S.I. = Rs 3,192
A = P + S.I.
A = Rs (26,600 + 3,192) = Rs 29,792 = new principal

For the third year: t = ½ year; P = Rs 29,792
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{29,792 \times 12 \times 1}{100 \times 2} \)
S.I. = Rs 1,787.52
A = P + S.I.
A = Rs (29,792 + 1,787.52) = Rs 31,579.52

Final Amount = Rs 31,579.52
C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs (2,850 + 3,192 + 1,787.52) = Rs 7,829.52
In simple words: For half a year, we divide the time by 2. So the interest for 6 months is half of what we would get for a full year.

📝 Teacher's Note: Explain that ½ year means 6 months. Show students how to handle fractions in the time. For ½ year, multiply by 1 and divide by 2.

🎯 Exam Tip: Write "t = ½ year" clearly in the third step. Show the division by 2 in your working. This shows you understand fractional time periods.

e) Rs. 30,000 at 8% p.a. in 2½ years
Answer:
Given:
P = Rs. 30,000; r = 8% p.a.; t = 2½ years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{30,000 \times 8 \times 1}{100} \)
S.I. = Rs 2,400
A = P + S.I.
= Rs (30,000 + 2,400) = Rs 32,400 = new principal

For the second year: t = 1 year; P = Rs 32,400
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{32,400 \times 8 \times 1}{100} \)
S.I. = Rs 2,592
A = P + S.I.
A = Rs (32,400 + 2,592) = Rs 34,992 = new principal

For the third year: t = ½ year; P = Rs 34,992
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{34,992 \times 8 \times 1}{100 \times 2} \)
S.I. = Rs 1,399.68
A = P + S.I.
A = Rs (34,992 + 1,399.68) = Rs 36,391.68

Final Amount = Rs 36,391.68
C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs (2,400 + 2,592 + 1,399.68) = Rs 6,391.68
In simple words: Even at 8% rate (which is lower), the compound interest is quite good because the principal keeps growing each year.

📝 Teacher's Note: Compare this 8% rate with the 12% examples above. Students can see how the interest rate affects the final amount.

🎯 Exam Tip: Be careful with decimal calculations in the half year. Round to 2 decimal places for money. Show all working clearly.

f) Rs. 10,000 at 8% p.a. in 2¼ years
Answer:
Given:
P = Rs. 10,000; r = 8% p.a.; t = 2¼ years

For the first year: t = 1 year
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{10,000 \times 8 \times 1}{100} \)
S.I. = Rs 800
A = P + S.I.
= Rs (10,000 + 800) = Rs 10,800 = new principal

For the second year: t = 1 year; P = Rs 10,800
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{10,800 \times 8 \times 1}{100} \)
S.I. = Rs 864
A = P + S.I.
A = Rs (10,800 + 864) = Rs 11,664 = new principal

For the third year: t = ¼ year; P = Rs 11,664
S.I. = \( \frac{P \times r \times t}{100} \)
S.I. = Rs \( \frac{11,664 \times 8 \times 1}{100 \times 4} \)
S.I. = Rs 233.28
A = P + S.I.
A = Rs (11,664 + 233.28) = Rs 11,897.28

Final Amount = Rs 11,897.28
C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs (800 + 864 + 233.28) = Rs 1,897.28
In simple words: ¼ year means 3 months. So we get interest for just 3 months in the last part. That's why the last interest is small.

📝 Teacher's Note: Explain that ¼ year is 3 months. Show students that we divide by 4 instead of 2. Use a calendar to show 3 months if needed.

🎯 Exam Tip: Write "t = ¼ year" clearly. Show "divide by 4" in your fraction. This shows you understand how to handle quarter years correctly.

Question (g). Rs.20,000 at 9% p.a. in 2\(\frac{1}{3}\) years
Answer:
Given:
P = Rs.20,000; r = 9% p.a.; t = 2\(\frac{1}{3}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{20,000 \times 9 \times 1}{100}\)
S.I. = Rs.1,800

A = P + S.I.
= Rs.(20,000 + 1,800) = Rs.21,800 = new principal

For the second year: t = 1 year; P = Rs.21,800
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{21,800 \times 9 \times 1}{100}\)
S.I. = Rs.1,962

A = P + S.I.
A = Rs.(21,800 + 1,962) = Rs.23,762 = new principal

For the third year: t = 1/3 year; P = Rs.23,762
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{23,762 \times 9 \times 1}{100 \times 3}\)
S.I. = Rs.712.86

A = P + S.I.
A = Rs.(23,762 + 712.86) = Rs.24,474.86

C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs.(1,800 + 1,962 + 712.86) = Rs.4,474.86
In simple words: We calculate interest for each year separately. Each year the principal increases because we add the previous year's interest.

📝 Teacher's Note: Show students that compound interest means adding interest to the principal each year. The interest grows bigger each year because the principal gets bigger.

🎯 Exam Tip: Always calculate year by year for compound interest. Show each step clearly. Write the new principal for each year.

Question (h). Rs.25,000 at 8\(\frac{2}{5}\)% p.a. in 1\(\frac{1}{3}\) years
Answer:
Given:
P = Rs.25,000; r = 8\(\frac{2}{5}\)% p.a. = \(\frac{42}{5}\)%; t = 1\(\frac{1}{3}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{25,000 \times 42 \times 1}{100 \times 5}\)
S.I. = Rs.2,100

A = P + S.I.
= Rs.(25,000 + 2,100) = Rs.27,100 = new principal

For the second year: t = 1/3 year; P = Rs.27,100
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{27,100 \times 42 \times 1}{100 \times 5 \times 3}\)
S.I. = Rs.758.80

A = P + S.I.
A = Rs.(27,100 + 758.80) = Rs.27,858.80

C.I. = Interest in first year + interest in second year
C.I. = Rs.(2,100 + 758.80) = Rs.2,858.80
In simple words: First we find interest for 1 full year. Then we find interest for the remaining 1/3 year using the new bigger principal.

📝 Teacher's Note: When time has fractions, break it into full years first, then calculate for the remaining fraction. Students often forget to change the principal.

🎯 Exam Tip: Convert mixed fractions to improper fractions first. Always show the conversion clearly. Calculate interest step by step for each time period.

Question (i). Rs.40,000 at 5\(\frac{1}{4}\)% p.a. in 1\(\frac{1}{3}\) years
Answer:
Given:
P = Rs.40,000; r = 5\(\frac{1}{4}\)% p.a. = \(\frac{21}{4}\)%; t = 1\(\frac{1}{3}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{40,000 \times 21 \times 1}{100 \times 4}\)
S.I. = Rs.2,100

A = P + S.I.
= Rs.(40,000 + 2,100) = Rs.42,100 = new principal

For the second year: t = 1/3 year; P = Rs.42,100
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{42,100 \times 21 \times 1}{100 \times 4 \times 3}\)
S.I. = Rs.736.75

A = P + S.I.
A = Rs.(42,100 + 736.75) = Rs.42,836.75

C.I. = Interest in first year + interest in second year
C.I. = Rs.(2,100 + 736.75) = Rs.2,836.75
In simple words: We find interest for 1 year first, then for the remaining 4 months (which is 1/3 of a year). The principal grows after the first year.

📝 Teacher's Note: Help students understand that 1/3 year equals 4 months. Show them how to convert mixed numbers to improper fractions before calculating.

🎯 Exam Tip: When rate is a mixed number, convert to improper fraction. Show all steps clearly. Never forget to add interest to principal for the next period.

Question (j). Rs.76,000 at 10% p.a. in 2\(\frac{1}{2}\) years
Answer:
Given:
P = Rs.76,000; r = 10% p.a.; t = 2\(\frac{1}{2}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{76,000 \times 10 \times 1}{100}\)
S.I. = Rs.7,600

A = P + S.I.
= Rs.(76,000 + 7,600) = Rs.83,600 = new principal

For the second year: t = 1 year; P = Rs.83,600
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{83,600 \times 10 \times 1}{100}\)
S.I. = Rs.8,360

A = P + S.I.
A = Rs.(83,600 + 8,360) = Rs.91,960 = new principal

For the third year: t = 1/2 year; P = Rs.91,960
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{91,960 \times 10 \times 1}{100 \times 2}\)
S.I. = Rs.4,598

A = P + S.I.
A = Rs.(91,960 + 4,598) = Rs.96,558

C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs.(7,600 + 8,360 + 4,598) = Rs.20,558
In simple words: We calculate interest for 2 full years and then 6 months (half year). Each time we add the interest to make a new principal for the next calculation.

📝 Teacher's Note: Show students that 1/2 year is 6 months. The interest keeps growing each year because the principal gets bigger each time we add interest.

🎯 Exam Tip: Break down the time into complete years first, then calculate for the remaining fraction. Show each year's calculation separately for full marks.

Question (k). Rs.22,500 at 12% p.a. in 1\(\frac{3}{4}\) years
Answer:
Given:
P = Rs.22,500; r = 12% p.a.; t = 1\(\frac{3}{4}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{22,500 \times 12 \times 1}{100}\)
S.I. = Rs.2,700

A = P + S.I.
= Rs.(22,500 + 2,700) = Rs.25,200 = new principal

For the second year: t = 3/4 year; P = Rs.25,200
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{25,200 \times 12 \times 3}{100 \times 4}\)
S.I. = Rs.2,268

A = P + S.I.
A = Rs.(25,200 + 2,268) = Rs.27,468

C.I. = Interest in first year + interest in second year
C.I. = Rs.(2,700 + 2,268) = Rs.4,968
In simple words: We find interest for 1 full year first. Then for 3/4 year (which is 9 months) using the new principal. The total interest is the sum of both.

📝 Teacher's Note: Explain that 3/4 year equals 9 months. Draw a timeline to show students how the principal changes after each period.

🎯 Exam Tip: When time has fractions like 3/4, calculate carefully. Show the fraction multiplication step by step. Always add interests from all periods for total C.I.

Question (l). Rs.16,000 at 15% p.a. in 2\(\frac{2}{3}\) years
Answer:
Given:
P = Rs.16,000; r = 15% p.a.; t = 2\(\frac{2}{3}\) years

For the first year: t = 1 year
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{16,000 \times 15 \times 1}{100}\)
S.I. = Rs.2,400

A = P + S.I.
= Rs.(16,000 + 2,400) = Rs.18,400 = new principal

For the second year: t = 1 year; P = Rs.18,400
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{18,400 \times 15 \times 1}{100}\)
S.I. = Rs.2,760

A = P + S.I.
A = Rs.(18,400 + 2,760) = Rs.21,160 = new principal

For the third year: t = 2/3 year; P = Rs.21,160
S.I. = \(\frac{P \times r \times t}{100}\)
S.I. = Rs.\(\frac{21,160 \times 15 \times 2}{100 \times 3}\)
S.I. = Rs.2,116

A = P + S.I.
A = Rs.(21,160 + 2,116) = Rs.23,276

C.I. = Interest in first year + interest in second year + interest in third year
C.I. = Rs.(2,400 + 2,760 + 2,116) = Rs.7,276
In simple words: We calculate for 2 full years, then 2/3 year (which is 8 months). The interest grows each period because the principal keeps getting bigger.

📝 Teacher's Note: Show students that 2/3 year equals 8 months. Use coins or blocks to demonstrate how the principal grows by adding interest each time.

🎯 Exam Tip: Break complex time periods into years and fractions. Calculate step by step. Always show that principal increases after each period. This gets full marks.

Answer 2.
Answer:
Solution:
(i) \( C_1 = \frac{P \times R \times T}{100} = \frac{65,000 \times 8 \times 1}{100} = \text{Rs. } 5200 \)
\( P_1 = 5200 + 65000 = \text{Rs. } 70200 \)

(ii) \( C_2 = \frac{P \times R \times T}{100} = \text{Rs. } 5616 \)
\( P_2 = \text{Rs. } 75,816 \)

(iii) \( C_1 + C_2 = 5200 + 5616 = 10,816 \)

(iv) \( C_3 = \frac{75816 \times 8 \times 1}{100} = 6065.28 \)

In simple words: We calculate simple interest for each year. The interest earned in the first year gets added to the principal. Then we calculate interest on the new amount.

📝 Teacher's Note: Show students that in compound interest, the principal keeps growing each year. Use a simple example with Rs. 100 and 10% rate to make it clear.

🎯 Exam Tip: Always write the formula first. Show each step clearly. Calculate compound interest by adding previous year's interest to principal before finding next year's interest.

Answer 3.
Answer:
Solution:
(i) \( C_1 = \frac{P \times R \times T}{100} = \frac{75000 \times 1 \times 8}{100} = 6000 \)
\( P_1 = 81000 \)

\( C_2 = \frac{P \times R \times T}{100} = \frac{81000 \times 8 \times 1}{100} = 6480 \)
\( P_2 = 87480 \)

(ii) \( C_3 = \frac{87480 \times 8}{100} = 6998.4 \)
\( P_3 = 94478.4 \)

(iii) \( C_3 = 6998.4 \)

(iv) \( C_4 = \frac{94478.4 \times 1 \times 8}{100} = 7558.272 \)

In simple words: Each year, we add the interest to the principal amount. Then we calculate next year's interest on this new bigger amount.

📝 Teacher's Note: Use a piggy bank example. Every year, the money in the piggy bank grows. Next year's interest is calculated on the bigger amount.

🎯 Exam Tip: Write each year's calculation separately. Show P₁, P₂, P₃ clearly. Don't skip any steps in compound interest problems.

Answer 4.
Answer:
Solution:
(i) \( C_1 = \frac{36000 \times 1 \times 10}{100} = 3600 \)
\( P_1 = 39600 \)

(ii) \( C_2 = \frac{39600 \times 1 \times 10}{200} = 1980 \)
\( P_2 = 41580 \)

In simple words: In the second year, the rate becomes half (10% ÷ 2 = 5%). So we calculate interest at 5% rate on the new principal.

📝 Teacher's Note: Explain that when rate changes, we use the new rate for calculating interest. The rate can go up or down in different years.

🎯 Exam Tip: Read the question carefully for rate changes. Use the correct rate for each year. Show which rate you are using in each step.

Answer 5.
Answer:
Solution:
(i) \( C_1 = \frac{P \times R \times T}{100} = \frac{24000 \times 1 \times 10}{100} = 2400 \)
\( P_1 = 26400 \)

\( C_2 = \frac{26400 \times 1 \times 10}{100} = 2640 \)
\( P_2 = 29040 \)

\( C_3 = \frac{29040 \times 1 \times 10}{200} = 2904 \)
\( P_3 = 31944 \)

(ii) Total Interest = 2400 + 2640 + 2904 = 7944

In simple words: We calculate interest for each year separately. Then we add all the interest amounts to get total interest earned.

📝 Teacher's Note: Help students see that compound interest grows faster each year because the principal keeps increasing. Draw a simple chart showing this growth.

🎯 Exam Tip: Calculate each year's interest separately first. Then add them for total interest. Don't forget to mention which part is asking for total interest.

Answer 6.
Answer:
Solution:
(i) \( C_1 = \frac{27500 \times 12 \times 1}{100} = 3300 \)
\( P_1 = 30800 \)

\( C_2 = \frac{30800 \times 12 \times 1}{100} = 3696 \)

Soln: \( P_2 = 34496 \)

\( C_3 = \frac{34496 \times 12 \times 1}{100} = 4139.52 \)
\( P_3 = 38636 \)

(ii) \( C_{total} = 11136 \)

(iii) \( P_3 = 34496 \)

In simple words: Each year the money grows by 12%. After 3 years, the original Rs. 27,500 becomes Rs. 38,636.

📝 Teacher's Note: Show students that 12% means Rs. 12 extra for every Rs. 100. This makes the percentage concept clearer.

🎯 Exam Tip: When the question asks for amount after specific years, calculate step by step until that year. Don't use the compound interest formula shortcut unless you're very sure.

Answer 7.
Answer:
Solution:
(i) \( C_1 = \frac{60000 \times 15 \times 1}{100} = 9000 \)
\( P_1 = 69000 \)

\( C_2 = \frac{69000 \times 15 \times 1}{100} = 10350 \)
\( P_2 = 79350 \)

\( C_3 = \frac{79350 \times 1 \times 15}{100} = 11902.5 \)
\( P_3 = 91252.5 \)

(ii) \( C_{total} = 20541 \)

In simple words: With 15% interest rate, the money grows very fast. In 3 years, Rs. 60,000 becomes Rs. 91,252.

📝 Teacher's Note: Compare this with a lower interest rate example. Show how higher rates make money grow much faster in compound interest.

🎯 Exam Tip: High interest rates mean big numbers. Double-check your calculations. Make sure you add interest to principal correctly each time.

Answer 8.
Answer:
Solution:
\( C_1 = \frac{25000 \times 1 \times 10}{100} = 2500 \)
\( P_1 = 27500 \)

\( C_2 = \frac{27500 \times 10}{100} = 2750 \)
\( P_2 = 30250 \)

\( C_3 = \frac{30250 \times 1 \times 10}{100} = 3025 \)
\( P_3 = 33275 \)

\( C_4 = \frac{33275 \times 10 \times 1}{100} = 1663.75 \)
\( P_4 = 34940 \)

In simple words: We calculate compound interest for 4 years. Each year, the principal grows and earns more interest than the previous year.

📝 Teacher's Note: Point out that in the 4th calculation, there seems to be an error in the original solution. The interest should be 3327.5, not 1663.75.

🎯 Exam Tip: For longer periods, be very careful with calculations. Write each step clearly. Check if your answer makes sense - interest should increase each year.

Answer 9.
Answer:
Solution:
\( C_1 = \frac{16000 \times 15 \times 1}{100} = 2400 \)
\( P_1 = 18400 \)

\( C_2 = \frac{18400 \times 15 \times 1}{100} = 2760 \)
\( P_2 = 21160 \)

\( C_3 = \frac{21160 \times 15 \times 1}{400} = 7935 \)
\( P_3 = 29095 \)

In simple words: We calculate compound interest for 3 years. In the third year calculation, there appears to be a different rate or time period used.

📝 Teacher's Note: Check the third year calculation with students. The denominator changes to 400, which might indicate a different rate or compounding frequency.

🎯 Exam Tip: If the rate or compounding changes in different years, read the question very carefully. Make sure you understand what changes and when.

Answer 10.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{100}\right)^t \)

\( Amount = 24000\left(1 + \frac{10}{100}\right)^3 = 31944 \)

Therefore, Shekhar received Rs. 31944 at the time of maturity.

In simple words: We use the compound interest formula. This gives us the total amount after 3 years directly without calculating year by year.

📝 Teacher's Note: Teach students both methods - year by year calculation and the direct formula. The formula is faster but students should understand the step-by-step method first.

🎯 Exam Tip: Use the compound interest formula when the question asks for final amount only. But if it asks for interest each year, do step-by-step calculations.

Answer 11.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{100}\right)^t \)

\( Amount = 27500\left(1 + \frac{8}{100}\right)^{1.75} = Rs. 3,982 \)

In simple words: When time is not a whole number (like 1.75 years), we use the compound interest formula. This gives us the exact amount.

📝 Teacher's Note: Explain that 1.75 years means 1 year and 9 months (0.75 = 9/12). Use a calculator for fractional powers.

🎯 Exam Tip: For fractional time periods, always use the formula method. Year-by-year calculation becomes very difficult with fractions.

Answer 12.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{200}\right)^{2t} \)

\( Amount = 35000\left(1 + \frac{12}{200}\right)^3 = Rs. 41685.56 \)

In simple words: When interest is compounded half-yearly, we divide the rate by 2 and multiply time by 2. So 12% yearly becomes 6% half-yearly.

📝 Teacher's Note: Explain that half-yearly compounding means interest is calculated and added twice per year. This makes the money grow slightly faster.

🎯 Exam Tip: For half-yearly compounding: rate ÷ 2, time × 2. For quarterly compounding: rate ÷ 4, time × 4. Remember this rule.

Answer 13.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{200}\right)^{2t} \)

\( Amount = 40000\left(1 + \frac{10}{200}\right)^4 = Rs. 48620.25 \)

In simple words: This is half-yearly compounding for 2 years. So we calculate as if it's 4 periods with 5% rate each period.

📝 Teacher's Note: Show students that more frequent compounding (yearly vs half-yearly) gives slightly more money. But the difference is small for normal rates.

🎯 Exam Tip: Read carefully whether it says "per annum" or "half-yearly". Adjust the formula accordingly. Don't mix up yearly and half-yearly rates.

Answer 14.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{200}\right)^{2t} \)

\( Amount = 16000\left(1 + \frac{15}{200}\right)^3 = Rs. 19876.75 \)

\( C = 19876.75 - 16000 = 3876.75 \)

In simple words: First we find the total amount using the formula. Then we subtract the original principal to get only the interest earned.

📝 Teacher's Note: Always clarify whether the question wants amount or compound interest. Amount = Principal + Interest. Interest = Amount - Principal.

🎯 Exam Tip: If question asks for compound interest only, subtract principal from amount. If it asks for amount, give the total value including principal.

Answer 15.
Answer:
Solution:
\( Amount = P\left(1 + \frac{r}{100}\right)^t \)

\( Amount = 12500\left(1 + \frac{16}{100}\right)^{15} = Rs. 15660 \)

In simple words: We use the compound interest formula with 16% rate for 15 years. This gives us the final amount after 15 years.

📝 Teacher's Note: For long periods like 15 years, compound interest makes a huge difference compared to simple interest. Show students this comparison.

🎯 Exam Tip: For long time periods, use a calculator for the power calculation. Make sure you write the formula first, then substitute values clearly.

Exercise 1.2

Answer 1.
Answer:
(a) Rs 12,500 for 2 years at 8% for the first year and 10% for the second year.

Sol: P = Rs.12, 500;

(i) Interest for the first year
T = 1 year, R = 8 % for first year

= Rs \( \frac{12,500 \times 8 \times 1}{100} \)
= Rs.1,000

(ii) Principal for the second year
= Amount after one year

= Rs.12,500 + Rs.1,000
= Rs.13,500

(iii) Interest for the second year
T = 1 year, R = 10 % for second year

= Rs \( \frac{13,500 \times 10 \times 1}{100} \)
= Rs.1,350

Therefore, Amount at the end of 2nd year
= Rs.1,3500 + Rs.1,350

Amount = Rs 14,850

C.I. = A-P
= Rs. (14,850 – 12,500)
C.I. = Rs. 2,350

(b) Rs 15,000 for 2 years at 6% for the first year and 7% for the second year.
Sol: P = Rs.15, 000;

(i) Interest for the first year
T = 1 year, R = 6 % for first year

= Rs \( \frac{15,000 \times 6 \times 1}{100} \)
= Rs.900

(ii) Principal for the second year
= Amount after one year

= Rs.15,000 + Rs.900
= Rs.15,900

(iii) Interest for the second year
T = 1 year, R = 7 % for second year

= Rs \( \frac{15,900 \times 7 \times 1}{100} \)
= Rs.1,113

Therefore Amount at the end of 2nd year
= Rs.15,900 + Rs.1,113
= Rs.17,013

Amount = Rs 17,013
C.I. = A-P
= Rs. (17,013 – 15,000)
C.I. = Rs. 2,013

(c) Rs 12,500 for 3 years at 12% for the first year, 15% for the second year and 17% for the third year.
Sol: P = Rs.12, 500;

(i) Interest for the first year
T = 1 year, R = 12 % for first year

= Rs \( \frac{12,500 \times 12 \times 1}{100} \)
= Rs.1,500

(ii) Principal for the second year
= Amount after one year
= Rs.12,500 + Rs.1,500
= Rs.14,000

(iii) Interest for the second year
T = 1 year, R = 15 % for second year

= Rs \( \frac{14,000 \times 15 \times 1}{100} \)
= Rs.2,100

(iv) Principal for the third year
= Amount after second year

= Rs.14,000 + Rs.2,100
= Rs.16,100

(v) Interest for the third year
T = 1 year, R = 17 % for second year

= Rs \( \frac{16,100 \times 17 \times 1}{100} \)
= Rs.2,737

Therefore Amount at the end of 3rd year
= Rs 16,100 + Rs 2,737
= Rs 18,837

Amount = Rs 18,837
C.I. = A-P
= Rs. (18,837 – 12,500)
C.I.= Rs. 6,337

(d) Rs 20,000 for 3 years at 7\( \frac{1}{2} \)% for the first year, 8% for the second year and 10% for the third year.
Sol: P = Rs.20, 000;

(i) Interest for the first year
T = 1 year, R = 7 \( \frac{1}{2} \) % for first year = \( \frac{15}{2} \) %

= Rs \( \frac{20,000 \times \frac{15}{2} \times 1}{100} \)
= Rs \( \frac{20,000 \times 15 \times 1}{2 \times 100} \)
= Rs.1,500

(ii) Principal for the second year
= Amount after one year

= Rs.20,000 + Rs.1,500
= Rs.21,500

(iii) Interest for the second year
T = 1 year, R = 18 % for second year

= Rs \( \frac{21,500 \times 8 \times 1}{100} \)
= Rs.1,720

(iv) Principal for the third year
= Amount after second year

= Rs.21,500 + Rs.1,720
= Rs.23,220

(v) Interest for the third year
T = 1 year, R = 10 % for second year

= Rs \( \frac{23,220 \times 10 \times 1}{100} \)
= Rs.2,322

Therefore Amount at the end of 3rd year
= Rs.23,220 + Rs.2,322
= Rs.25,542

Amount = Rs 25,542
C.I. = A-P
= Rs. (25,542 – 20,000)
C.I.= Rs. 5,542

In simple words: We calculate interest for each year separately. Each year's interest is added to the principal to get new principal for next year. This is called compound interest.

📝 Teacher's Note: Show students that compound interest means "interest on interest." Each year, the money grows bigger. Next year's interest is calculated on the bigger amount.

🎯 Exam Tip: Always show each year separately. Write "Principal for year 2 = Amount after year 1." This step-by-step method gets full marks.

Answer 2.
Answer:
P = Rs. 25,000, R = 10% p.a.

Interest for first year
= \( \frac{Rs25,000 \times 10 \times 1}{100} \)
= Rs2,500

Amount due after 1st year
= Rs. 25,000 + Rs. 2,500
= Rs 27,500

Amount paid after 1st year = Rs. 7,500

Balance amount = Rs. 27,500 – Rs. 7,500
= Rs. 20,000

Interest for second year
= \( \frac{Rs20,000 \times 10 \times 1}{100} \)
= Rs2,000

Amount due after 2nd year
= Rs. 20,000 + Rs. 2,000
= Rs 22,000

Amount paid after 2nd year = Rs. 7,500

Balance amount = Rs. 22,000 – Rs. 7,500
= Rs. 14,500

Interest for third year
= \( \frac{Rs14,500 \times 10 \times 1}{100} \)
= Rs1,450

Amount due after 3rd year
= Rs. 14,500 + Rs. 1,450
= Rs 15,950

Amount paid after 3rd year = Rs. 7,500

Balance amount = Rs. 15,950 – Rs. 7,500
= Rs. 8,450

Loan outstanding at the beginning of the fourth year = Rs 8,450.

In simple words: When you pay some money each year, the remaining debt becomes smaller. Interest for next year is calculated on the smaller remaining amount.

📝 Teacher's Note: Explain that "balance amount" means "money still owed." Each payment reduces the debt. Next year's interest is only on what is still owed.

🎯 Exam Tip: Always write "Balance amount = Previous amount due – Payment made." Show this calculation clearly for each year to get full marks.

Answer 3.
Answer:
P = Rs. 90,000, R = 15% p.a.
Interest for first year

Step 1: Calculate interest for first year
\[ = \frac{Rs90,000 \times 15 \times 1}{100} \]
\[ = Rs13,500 \]

Step 2: Calculate amount due after 1st year
Amount due after 1st year
= Rs. 90,000 + Rs. 13,500
= Rs 103,500

Step 3: Calculate balance after payment
Amount paid after 1st year = Rs. 35,000
Balance amount = Rs. 103,500 - Rs. 35,000
= Rs. 68,500

Step 4: Calculate interest for second year
Interest for second year
\[ = \frac{Rs68,500 \times 15 \times 1}{100} \]
\[ = Rs10,275 \]

Step 5: Calculate balance after second year payment
Amount due after 2nd year
= Rs. 68,500 + Rs. 10,275
= Rs 78,775
Amount paid after 2nd year = Rs. 35,000
Balance amount = Rs. 78,775 - Rs. 35,000
= Rs. 43,775

Step 6: Calculate interest for third year
Interest for third year
\[ = \frac{Rs43,775 \times 15 \times 1}{100} \]
\[ = Rs6,566.25 \]

Step 7: Calculate final balance
Amount due after 3rd year
= Rs. 43,775 + Rs. 6,566.25
= Rs 50,341.25
Amount paid after 3rd year = Rs.35, 000
Balance amount = Rs. 50,341.25 - Rs.35, 000
= Rs. 15,341.25

Loan outstanding at the beginning of the fourth year = Rs 15,341.25
In simple words: We find how much money is still left to pay after three years. Each year we pay Rs. 35,000 but the remaining money keeps growing because of interest.

📝 Teacher's Note: Show students that each year the interest is calculated on the remaining balance, not the original amount. This is how real loans work - you pay interest only on what you still owe.

🎯 Exam Tip: Always show each year's calculation separately. Write "Given", "Step 1", "Step 2" clearly. Don't forget to subtract the payment from the amount due each year.

Answer 4.
Answer:
P = Rs. 15,000, R = 11% p.a.
Interest for first year

Step 1: Calculate interest for first year
\[ = \frac{Rs15,000 \times 11 \times 1}{100} \]
\[ = Rs1,650 \]

Step 2: Calculate balance after first year payment
Amount due after 1st year
= Rs. 15,000 + Rs. 1,650
= Rs 16,650
Amount paid after 1st year = Rs. 7,550
Balance amount = Rs. 16,650 - Rs. 7,550
= Rs. 9,100

Step 3: Calculate interest for second year
Interest for second year
\[ = \frac{Rs9,100 \times 11 \times 1}{100} \]
\[ = Rs1,001 \]
Amount paid after 2nd year = Rs. 6,101
Balance amount = Rs. 10,101 - Rs. 6,101
= Rs. 4,000

Step 4: Calculate final payment needed
Interest for third year
\[ = \frac{Rs4,000 \times 11 \times 1}{100} \]
\[ = Rs440 \]
Amount due after 3rd year
= Rs. 4,000 + Rs. 440
= Rs 4,440

Pooja needs to pay Rs 4,440 to Sonali at the end of third year to clear her debt.
In simple words: Pooja borrowed Rs. 15,000. She pays some money each year but the leftover money keeps growing with interest. After 3 years she needs Rs. 4,440 to finish paying.

📝 Teacher's Note: Explain that each payment reduces the principal amount. The interest for next year is calculated only on the remaining balance, not the original loan.

🎯 Exam Tip: Always calculate interest on the balance amount from previous year. Show the final answer clearly with "needs to pay" statement.

Answer 5.
Answer:
P = Rs. 18,000, R = 12% p.a.
Interest for first year

Step 1: Calculate interest for first year
\[ = \frac{Rs18,000 \times 12 \times 1}{100} \]
\[ = Rs2,160 \]

Step 2: Calculate balance after first year payment
Amount due after 1st year
= Rs. 18,000 + Rs. 2,160
= Rs 20,160
Amount paid after 1st year = Rs. 5,250
Balance amount = Rs. 20,160 - Rs. 5,250
= Rs. 14,910

Step 3: Calculate interest for second year
Interest for second year
\[ = \frac{Rs14,910 \times 12 \times 1}{100} \]
\[ = Rs1,789.20 \]

Step 4: Calculate balance after second year payment
Amount due after 2nd year
= Rs. 14,910 + Rs. 1,789.20
= Rs 16,699.20
Amount paid after 2nd year = Rs. 5,875
Balance amount = Rs. 16,699.20- Rs. 5,875
= Rs. 10,824.20

Step 5: Calculate interest for third year
Interest for third year
\[ = \frac{Rs10,824.20 \times 12 \times 1}{100} \]
\[ = Rs1,298.904 \]

Step 6: Calculate balance after third year payment
Amount due after 3rd year
= Rs. 10,824.20 + Rs. 1,298.904
= Rs 12,123.10
Amount paid after 3rd year = Rs. 6,875
Balance amount = Rs. 12,123.10- Rs. 6,875
= Rs. 5,248.104

Step 7: Calculate final payment needed
Interest for fourth year
\[ = \frac{Rs5,248.104 \times 12 \times 1}{100} \]
\[ = Rs629.7725 \]
Amount due after 4th year
= Rs. 5,248.104 + Rs. 629.7725
= Rs 5877.876
\[ = Rs5877.87 \]

Archana needs to pay Rs 5877.87 to Ritu at the end of 4th year to clear her debt.
In simple words: Archana borrowed Rs. 18,000. She makes payments each year but the leftover amount keeps growing with 12% interest. After 4 years she needs Rs. 5877.87 to finish paying everything.

📝 Teacher's Note: This is a 4-year problem so students need to be careful with calculations. Remind them that interest is always calculated on the balance from previous year.

🎯 Exam Tip: Keep track of balance amounts carefully. Round final answer to 2 decimal places when dealing with money. Show each year's calculation step by step.

Answer 6.
Answer:
Here, P = Rs 15,000; r = 12% p.a.; t = 2 years

Step 1: Use compound interest formula
\[ A = P\left(1 + \frac{r}{100}\right)^n \]

Step 2: Substitute values
\[ = Rs.15,000\left(1 + \frac{12}{100}\right)^2 \]
\[ = Rs.15,000\left(\frac{112}{100}\right)^2 \]
\[ = Rs.15,000\left(\frac{28}{25}\right)^2 \]
\[ = Rs.15,000 \times \frac{28}{25} \times \frac{28}{25} \]
\[ A = Rs.18,816 \]

Step 3: Calculate remaining balance
Hence, amount due after 2 years = Rs 18,816
Amount paid after 2 years = Rs 7,500
Balance amount = Amount due after 2 years - amount paid after 2 years = cost of the scooter = Rs (18,816 - 7,500)

Cost of the scooter = Rs 11,316
In simple words: The person borrowed Rs. 15,000 to buy a scooter. After 2 years with compound interest, they owed Rs. 18,816. They paid Rs. 7,500, so the scooter cost Rs. 11,316.

📝 Teacher's Note: This uses compound interest formula, not simple interest. Show students how compound interest makes money grow faster because interest earns interest too.

🎯 Exam Tip: Use the compound interest formula \( A = P(1 + \frac{r}{100})^n \). The final answer is what they still owe, which equals the cost of the scooter.

Answer 7.
Answer:
Here, P = Rs 25,000; r = 8.4% p.a.; t = 2 years

Step 1: Use compound interest formula
\[ A = P\left(1 + \frac{r}{100}\right)^n \]

Step 2: Substitute values
\[ = Rs.25,000\left(1 + \frac{8.4}{100}\right)^2 \]
\[ = Rs.25,000\left(1 + \frac{84}{100 \times 10}\right)^2 \]
\[ = Rs.25,000\left(\frac{271}{250}\right)^2 \]
\[ = Rs.25,000 \times \frac{271}{250} \times \frac{271}{250} \]
\[ A = Rs29,376.40 \]

Step 3: Calculate remaining balance
Hence, amount due after 2 years = Rs 29,376.40
Amount paid after 2 years = Rs 17,500
Balance amount = Amount due after 2 years - amount paid after 2 years = cost of the motorcycle = Rs (29,376.40 - 17,500)

Cost of the motorcycle = Rs 11,876.40
In simple words: Someone borrowed Rs. 25,000 to buy a motorcycle. After 2 years with 8.4% compound interest, they owed Rs. 29,376.40. They paid Rs. 17,500, so the motorcycle cost Rs. 11,876.40.

📝 Teacher's Note: The decimal interest rate (8.4%) needs careful calculation. Show students how to convert it to fraction form for easier calculation.

🎯 Exam Tip: When interest rate has decimal like 8.4%, convert to fraction: 8.4/100 = 84/1000. Be careful with decimal calculations and show final answer clearly.

Answer 8.
Answer:
Given:
P = Rs. 10,000, R = 6% p.a.

Step 1: Find interest for first year.
\( \text{Interest} = \frac{Rs10,000 \times 6 \times 1}{100} = Rs600 \)

Step 2: Find amount due after 1st year.
Amount due = Rs. 10,000 + Rs. 600 = Rs 10,600

Step 3: Find balance after partial payment.
Amount paid after 1st year = Rs. 5,600
Balance amount = Rs. 10,600 - Rs. 5,600 = Rs. 5,000

Step 4: Find interest for second year (r = 8% p.a.).
\( \text{Interest} = \frac{Rs5,000 \times 8 \times 1}{100} = Rs400 \)

Step 5: Find amount due after 2nd year.
Amount due = Rs. 5,000 + Rs. 400 = Rs 5,400

Final Answer: Prakash has to return Rs 5,400 to Rajesh at the end of second year.
In simple words: First we find interest for year 1. Then we subtract what was paid and find interest on the remaining amount for year 2.

📝 Teacher's Note: Show students that when someone pays part of the loan, we calculate interest only on the remaining balance. The rate can change each year too.

🎯 Exam Tip: Always show each year separately. Write "Given", "Step 1", "Step 2" clearly. Calculate interest only on the balance amount for the second year.

Answer 9.
Answer:
Given:
P = Rs. 12,500, R = 8% p.a.

Step 1: Find interest for first year.
\( \text{Interest} = \frac{Rs12,500 \times 8 \times 1}{100} = Rs1,000 \)

Step 2: Find amount due after 1st year.
Amount due = Rs. 12,500 + Rs. 1,000 = Rs 13,500

Step 3: Find balance after partial payment.
Amount paid after 1st year = Rs. 7,500
Balance amount = Rs. 13,500 - Rs. 7,500 = Rs. 6,000

Step 4: Find interest for second year (r = 10% p.a.).
\( \text{Interest} = \frac{Rs6,000 \times 10 \times 1}{100} = Rs600 \)

Step 5: Find amount due after 2nd year.
Amount due = Rs. 6,000 + Rs. 600 = Rs 6,600

Final Answer: Meera has to return Rs 6,600 to Rajeev at the end of second year.
In simple words: After paying some money in year 1, the remaining balance earns interest at the new rate in year 2.

📝 Teacher's Note: Emphasize that interest rates can change between years. Students must read the problem carefully to see if rates are different.

🎯 Exam Tip: Write the new interest rate clearly when it changes. Show all calculations step by step. Don't forget to subtract the partial payment amount.

Answer 10.
Answer:
Given:
P = Rs. 50,000, R = \( 7\frac{1}{2} \)% p.a. = \( \frac{15}{2} \)% p.a.

Step 1: Find interest for first year.
\( \text{Interest} = \frac{Rs50,000 \times \frac{15}{2} \times 1}{100} = \frac{Rs50,000 \times 15 \times 1}{2 \times 100} = Rs3,750 \)

Step 2: Find amount due after 1st year.
Amount due = Rs. 50,000 + Rs. 3,750 = Rs 53,750

Step 3: Find balance after partial payment.
Amount paid after 1st year = Rs. 27,750
Balance amount = Rs. 53,750 - Rs. 27,750 = Rs. 26,000

Step 4: Find interest for second year (r = \( 9\frac{1}{4} \)% p.a. = \( \frac{37}{4} \)% p.a.).
\( \text{Interest} = \frac{Rs26,000 \times \frac{37}{4} \times 1}{100} = \frac{Rs26,000 \times 37 \times 1}{4 \times 100} = Rs2,405 \)

Step 5: Find amount due after 2nd year.
Amount due = Rs. 26,000 + Rs. 2,405 = Rs 28,405

Final Answer: Mr. Chatterjee has to return Rs 28,405 to Mr. Patel at the end of second year to clear his loan.
In simple words: When interest rates are mixed numbers (like \( 7\frac{1}{2} \)), convert them to improper fractions first. Then calculate normally.

📝 Teacher's Note: Teach students to convert mixed number percentages to improper fractions. Show \( 7\frac{1}{2} = \frac{15}{2} \) step by step on the board.

🎯 Exam Tip: Always convert mixed number rates to improper fractions first. Write this conversion clearly. Show the fraction calculation step to avoid mistakes.

Exercise 1.3

Answer 1.
Answer:
Given:
Here P = x; A = Rs 9,447.84; t = 3 years; r = 8% p.a.

Using compound interest formula:
\( A = P\left(1 + \frac{r}{100}\right)^t \)

\( Rs9,447.84 = x \left(1 + \frac{8}{100}\right)^3 \)

\( Rs9,447.84 = x \left(\frac{108}{100}\right)^3 \)

\( Rs9,447.84 = x \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25} \)

\( Rs9,447.84 = x \times \frac{19,683}{15,625} \)

\( x = Rs \frac{9,447.84 \times 15,625}{19,683} \)

\( x = Rs7,500 \)

The sum of money will be Rs 7,500.
In simple words: We work backwards from the final amount to find the starting amount. We use the compound interest formula in reverse.

📝 Teacher's Note: Show students how to rearrange the compound interest formula to find P when A is given. Practice cube calculations step by step.

🎯 Exam Tip: Write the formula first. Substitute all known values. Show the cube calculation clearly. Always state what you are finding.

Answer 2.
Answer:
Given:
Here P = x; A = Rs 16,637.50; t = 3 years; r = 10% p.a.

Using compound interest formula:
\( A = P\left(1 + \frac{r}{100}\right)^t \)

\( Rs16,637.50 = x \left(1 + \frac{10}{100}\right)^3 \)

\( Rs16,637.50 = x \left(\frac{11}{10}\right)^3 \)

\( Rs16,637.50 = x \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \)

\( Rs16,637.50 = x \times \frac{1,331}{1,000} \)

\( x = Rs \frac{16,637.50 \times 1,000}{1,331} \)

\( x = Rs12,500 \)

The sum of money will be Rs 12,500.
In simple words: Like the previous problem, we find the starting amount when we know the final amount after compound interest.

📝 Teacher's Note: Emphasize that \( \left(\frac{11}{10}\right)^3 = \frac{11^3}{10^3} = \frac{1331}{1000} \). Students often make errors in cube calculations.

🎯 Exam Tip: Calculate \( 11^3 = 1331 \) carefully. Check your arithmetic. Write the final answer with "Rs" and proper formatting.

Answer 3.
Answer:
For the second year:
Here P = x; A = Rs 7,128; t = 1 year; r = 10% p.a.

\( A = P\left(1 + \frac{r}{100}\right)^t \)

\( Rs7,128 = x \left(1 + \frac{10}{100}\right)^1 \)

\( Rs7,128 = x \left(\frac{11}{10}\right) \)

\( x = Rs \frac{7,128 \times 10}{11} = Rs6,480 \)

The sum of money will be Rs 6,480 at the end of the first year or beginning of the second year.

For the first year:
Here P = x; A = Rs 6,480; t = 1 year; r = 8% p.a.

\( A = P\left(1 + \frac{r}{100}\right)^t \)

\( Rs6,480 = x \left(1 + \frac{8}{100}\right)^1 \)

\( Rs6,480 = x \left(\frac{108}{100}\right) \)

\( x = Rs \frac{6,480 \times 100}{108} = Rs6,000 \)

The sum of money will be Rs 6,000 at the beginning of the first year.
In simple words: We work backwards year by year. First find the amount at end of year 1, then find the starting amount.

📝 Teacher's Note: This is a multi-step backward calculation. Show students to work one year at a time when rates are different each year.

🎯 Exam Tip: Label each year clearly. Show "For second year" and "For first year" as separate calculations. Don't try to do it all in one step.

Answer 4.
Answer:
For the third year:
Here P = x; A = Rs 3,326.40; t = 1 year; r = 12% p.a.

\( A = P\left(1 + \frac{r}{100}\right)^t \)

\( Rs3,326.40 = x \left(1 + \frac{12}{100}\right)^1 \)

\( Rs3,326.40 = x \left(\frac{112}{100}\right) \)
In simple words: This follows the same pattern as Answer 3 - work backwards one year at a time when interest rates change each year.

📝 Teacher's Note: The content appears to be incomplete in the PDF. Remind students that multi-year problems with different rates need step-by-step backward calculation.

🎯 Exam Tip: Always work backwards one year at a time. Label each calculation clearly with the year and rate being used.

Answer 5.
Answer:
Given: P = Rs. 3,000; R = 12% p.a.; T = 3 years

Step 1: Calculate for the third year
\( \Rightarrow x = Rs \frac{3,326.40 \times 100}{112} \)
\( \Rightarrow x = Rs2,970 \)

The sum of money will be Rs 2,970 at the end of the second year or beginning of the third year.

For the second year:
Here P = x; A = Rs 2,970; t = 1 year; r = 10% p.a.
\( \therefore A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow Rs2,970 = x\left(1 + \frac{10}{100}\right)^1 \)
\( \Rightarrow Rs2,970 = x\left(\frac{11}{10}\right) \)
\( \Rightarrow x = Rs \frac{2,970 \times 10}{11} \)
\( \Rightarrow x = Rs2700 \)

The sum of money will be Rs 2,700 at the end of the first year or beginning of the second year.

For the first year:
Here P = x; A = Rs 2,700; t = 1 year; r = 8% p.a.
\( \therefore A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow Rs2,700 = x\left(1 + \frac{8}{100}\right)^1 \)
\( \Rightarrow Rs2,700 = x\left(\frac{108}{100}\right) \)
\( \Rightarrow x = Rs \frac{2,700 \times 100}{108} \)
\( \Rightarrow x = Rs2500 \)

The sum of money will be Rs 2,500 at the beginning of the first year.

For the third year:
Here P = x; A = Rs 13,675.20; t = 1 year; r = 12% p.a.
\( \therefore A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow Rs13,675.20 = x\left(1 + \frac{12}{100}\right)^1 \)
\( \Rightarrow Rs13,675.20 = x\left(\frac{112}{100}\right) \)
\( \Rightarrow x = Rs \frac{13,675.20 \times 100}{112} \)
\( \Rightarrow x = Rs12,210 \)

The sum of money will be Rs 12,210 at the end of the second year or beginning of the third year.

For the second year:
Here P = x; A = Rs 12,210; t = 1 year; r = 11% p.a.
\( \therefore A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow Rs12,210 = x\left(1 + \frac{11}{100}\right)^1 \)
\( \Rightarrow Rs12,210 = x\left(\frac{111}{100}\right) \)
\( \Rightarrow x = Rs \frac{12,210 \times 100}{111} \)
\( \Rightarrow x = Rs11,000 \)

The sum of money will be Rs 11,000 at the end of the first year or beginning of the second year.

For the first year:
Here P = x; A = Rs 11,000; t = 1 year; r = 10% p.a.
\( \therefore A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow Rs11,000 = x\left(1 + \frac{10}{100}\right)^1 \)
\( \Rightarrow Rs11,000 = x\left(\frac{11}{10}\right) \)
\( \Rightarrow x = Rs \frac{11,000 \times 10}{11} \)
\( \Rightarrow x = Rs10,000 \)

The sum of money will be Rs 10,000 at the beginning of the first year.

In simple words: We worked backwards from the final amount to find the starting amount. We used compound interest formula for each year with different rates.

📝 Teacher's Note: Teach students to work backwards step by step. Show them that each year's ending amount becomes the next year's starting amount. Draw a timeline to make it clear.

🎯 Exam Tip: Always write "Given" first. Show each step clearly. Work backwards from the final year to the first year. Write the formula before substituting values.

Answer 6.
Answer:
P = Rs. 4,000; R = 10% p.a.; T = 3 years

Interest for the 1st year:
\( = Rs \frac{4,000 \times 10 \times 1}{100} = Rs400 \)

Principal for the second year:
= Amount at the end of one year + his new savings
= Rs. 4,000 + Rs. 400 + Rs. 4,000
= Rs. 8,400

Interest for the second year:
\( = Rs \frac{8,400 \times 10 \times 1}{100} = Rs840 \)

Compound interest for second year = Rs. 840

Principal for the third year:
= Amount at the end of two years + his new savings
= Rs. 8400 + Rs. 840 + Rs. 4000
= Rs. 13,240

Interest for the third year:
\( = Rs \frac{13,240 \times 10 \times 1}{100} = Rs1,324 \)

Sum due at the end of third year = his savings at the end of third year = Rs. 13,240 + Rs. 1,324 = Rs 14,564

In simple words: Each year he adds Rs. 4,000 to his savings. The bank also adds 10% interest. So his money grows bigger each year because he saves more and gets interest too.

📝 Teacher's Note: Show students that this is different from normal compound interest. Here new money is added every year. Make a table showing principal, interest, and total for each year.

🎯 Exam Tip: Remember to add new savings each year to the principal. Calculate interest on the total amount (old money + new savings + previous interest). Show all steps clearly.

Answer 7.
Answer:
P = Rs. 5,000; R = 12% p.a.; T = 3 years

Interest for the 1st year:
\( = Rs \frac{5,000 \times 12 \times 1}{100} = Rs600 \)

Principal for the second year:
= Amount at the end of one year + his new savings
= Rs. 5,000 + Rs. 600 + Rs. 5,000
= Rs. 10,600

Interest for the second year:
\( = Rs \frac{10,600 \times 12 \times 1}{100} = Rs1,272 \)

Compound interest for second year = Rs. 1,272

Principal for the third year:
= Amount at the end of two years + his new savings
= Rs. 10,600 + Rs. 1,272 + Rs. 5,000
= Rs. 16,872

Interest for the third year:
\( = Rs \frac{16,872 \times 12 \times 1}{100} = Rs2,024.64 \)

Sum due at the end of third year = his savings at the end of third year = Rs. 16,872 + Rs. 2,024.64 = Rs 18,896.64

In simple words: Each year he saves Rs. 5,000 more. The bank gives 12% interest on all his money. After 3 years, his total becomes Rs. 18,896.64.

📝 Teacher's Note: This is the same type as previous problem but with 12% rate instead of 10%. Emphasize that the method stays the same - add new savings each year before calculating interest.

🎯 Exam Tip: Be careful with decimal calculations. Write Rs. 2,024.64 correctly, not Rs. 2,024 or Rs. 2,025. Show the final answer clearly with proper addition.

Answer 8.
Answer:
P = Rs. 500; R = 10% p.a.; T = 3 years

Interest for the 1st year:
\( = Rs \frac{500 \times 10 \times 1}{100} = Rs50 \)

Principal for the second year:
= Amount at the end of one year + his new savings
= Rs. 500 + Rs. 50 + Rs. 550
= Rs. 1,100

Interest for the second year:
\( = Rs \frac{1,100 \times 10 \times 1}{100} = Rs110 \)

Compound interest for second year = Rs. 110

Principal for the third year:
= Amount at the end of two years + his new savings
= Rs. 1,100 + Rs. 110 + Rs. 600
= Rs. 1,810

Interest for the third year:
\( = Rs \frac{1,810 \times 10 \times 1}{100} = Rs181 \)

Sum due at the end of third year = his savings at the end of third year = Rs 1,810 + Rs. 181 = Rs 1,991

In simple words: He saves different amounts each year - Rs. 500, then Rs. 550, then Rs. 600. Each time the bank adds 10% interest on all his money. His final amount is Rs. 1,991.

📝 Teacher's Note: Point out that in this problem, the yearly savings amounts are different (Rs. 500, Rs. 550, Rs. 600). Students often miss this detail and assume equal savings each year.

🎯 Exam Tip: Read the problem carefully to see how much is saved each year. The amounts may be different. Write down the savings amount for each year before starting calculations.

Answer 9.
Answer:
P = Rs. 4,000; R = 15% p.a.; T = 3 years

Interest for the 1st year
= Rs \( \frac{4,000 \times 15 \times 1}{100} \)
= Rs 600

Principal for the second year
= Amount at the end of one year + her new savings
= Rs. 4,000 + Rs. 600 + Rs. 5,000
= Rs. 9,600

Interest for the second year
= Rs \( \frac{9,600 \times 15 \times 1}{100} \)
= Rs 1,440

Compound interest for second year
= Rs. 1,440

Principal for the third year
= Amount at the end of two years + her new savings
= Rs. 9,600 + Rs. 1,440 + Rs. 6000
= Rs. 17,040

Interest for the third year
= Rs \( \frac{17,040 \times 15 \times 1}{100} \)
= Rs 2,556

Sum due at the end of third year = her savings at the end of third year = Rs. 17,040 + Rs. 2,556 = Rs 19,596
In simple words: Each year she adds money to her savings. The interest is calculated on the new total amount. After 3 years, she has Rs. 19,596.

📝 Teacher's Note: Show students that each year the principal changes because new money is added. This makes it different from normal compound interest. Draw a table to make it clear.

🎯 Exam Tip: Always calculate year by year when new money is added. Write "new principal" clearly for each year. Show all working steps to get full marks.

Answer 10.
Answer:
Let value of car be Rs x.
V₀ = Rs x; n = 3; r = 10% for first 2 years and 8% for 3rd year.

\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^T \)

\( V_t = Rs.x \times \left(1 - \frac{10}{100}\right)^2 \left(1 - \frac{8}{100}\right) \)

\( V_t = Rs.x \times \frac{9}{10} \times \frac{9}{10} \times \frac{23}{25} \)

\( V_t = Rs.x \times \frac{1863}{2500} \)

\( V_t = Rs.0.7452x \)

Depreciation in the value of car = Rs (x-0.7452x) = Rs 0.2548x

Percentage change in depreciation
= \( \frac{0.2548x}{x} \times 100 \)
= 25.48%

Percentage change = 25.48%
In simple words: The car loses value each year. After 3 years, its value drops by about 25.48%. This means the car is worth about 74% of its original price.

📝 Teacher's Note: Explain that depreciation means the value goes down, not up. Use a real car example - a car that costs Rs. 10 lakh new will be worth about Rs. 7.45 lakh after 3 years.

🎯 Exam Tip: Remember to multiply all the depreciation rates together. When rates change each year, calculate step by step. Always convert to percentage at the end.

Answer 11.
Answer:
Let value of machine be Rs x.
V₀ = Rs x; n = 3; r = 10% for first year, 12% for 2nd year and 15% for 3rd year.

\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( V_t = Rs.x \times \left(1 - \frac{10}{100}\right)\left(1 - \frac{12}{100}\right)\left(1 - \frac{15}{100}\right) \)

\( V_t = Rs.x \times \frac{9}{10} \times \frac{22}{25} \times \frac{17}{20} \)

\( V_t = Rs.x \times \frac{3366}{5000} \)

\( V_t = Rs.0.6732x \)

Depreciation in the value of car = Rs (x-0.6732x) = Rs 0.3268x

Percentage change in depreciation
= \( \frac{0.3268x}{x} \times 100 \)
= 32.68%

Percentage change = 32.68%
In simple words: The machine loses about 33% of its value in 3 years. If it cost Rs. 100, it will be worth only Rs. 67 after 3 years.

📝 Teacher's Note: Point out that different depreciation rates each year is common for machines. They lose value faster as they get older and need more repairs.

🎯 Exam Tip: When depreciation rates are different each year, multiply all the remaining value fractions. Write each year's calculation clearly to avoid mistakes.

Answer 12.
Answer:
Let value of the scooter be Rs x.
V₀ = Rs x; n = 2; r = 12%

Depreciation in the first year =
\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( V_t = Rs.x \times \left(1 - \frac{12}{100}\right) \)

\( V_t = Rs.x \times \frac{22}{25} \)

\( V_t = Rs.0.88x \)

Depreciation in the second year =
\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( V_t = Rs.0.88x \times \left(1 - \frac{12}{100}\right) \)

\( V_t = Rs.0.88x \times \frac{22}{25} \)

\( V_t = Rs.0.7744x \)

Depreciation in the value of scooter in the second year
= Rs (0.88x-0.7744x) = Rs 2,640
\( 0.1056x = Rs 2,640 \)
\( x = Rs 25,000 \)

The original value of the scooter was Rs 25,000.
In simple words: We worked backwards from the second year's depreciation amount to find the original price. The scooter originally cost Rs. 25,000.

📝 Teacher's Note: This is a reverse calculation. Start with what you know (Rs. 2,640 loss in year 2) and work backwards. Show students how the depreciation in year 2 is calculated on the already reduced value.

🎯 Exam Tip: In reverse problems, set up the equation carefully. The depreciation in year 2 is calculated on the value at start of year 2, not the original value. Double-check your answer.

Answer 13.
Answer:
Let value of the refrigerator be Rs x.
V₀ = Rs x; n = 2; r = 8%

Depreciation in the first year =
\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( V_t = Rs.x \times \left(1 - \frac{8}{100}\right) \)

\( V_t = Rs.x \times \frac{23}{25} \)

\( V_t = Rs.0.92x \)

Depreciation in the second year =
\( V_t = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( V_t = Rs.0.92x \times \left(1 - \frac{8}{100}\right) \)

\( V_t = Rs.0.92x \times \frac{23}{25} \)

\( V_t = Rs.0.8464x \)

Depreciation in the value of refrigerator in the second year
= Rs (0.92x-0.8464x) = Rs 2,392
\( 0.0736x = Rs 2,392 \)
\( x = Rs 32,500 \)

The original value of the refrigerator was Rs 32,500.
In simple words: The refrigerator originally cost Rs. 32,500. After 2 years of 8% depreciation each year, it lost Rs. 2,392 in the second year alone.

📝 Teacher's Note: Refrigerators depreciate slower than vehicles because they last longer. Show students that 8% per year is a reasonable rate for appliances. Compare with cars that depreciate faster.

🎯 Exam Tip: Always check if the depreciation asked is for a specific year or total depreciation. Here they asked for "second year" only, so calculate carefully. Show the value at start and end of year 2.

Answer 14.
Let value of the machine be Rs x.

Given:
\( V_0 = \text{Rs } x \); n = 2; r = 15%

Step 1: Depreciation in the first year
\( V_e = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( \Rightarrow V_e = \text{Rs.} x \times \left(1 - \frac{15}{100}\right) \)

\( \Rightarrow V_e = \text{Rs.} x \times \frac{17}{20} \)

\( \Rightarrow V_e = \text{Rs} 0.85x \)

Step 2: Depreciation in the second year when r is 12%
\( V_e = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( \Rightarrow V_e = \text{Rs} 0.85x \times \left(1 - \frac{12}{100}\right) \)

\( \Rightarrow V_e = \text{Rs} 0.85x \times \frac{22}{25} \)

\( \Rightarrow V_e = \text{Rs} 0.748x \)

Step 3: Calculate depreciation amount
Depreciation in the value of machine in the second year
= Rs (0.85x-0.748x) = Rs 1,632

\( \Rightarrow 0.102x = \text{Rs } 1,632 \)

\( \Rightarrow x = \text{Rs } 16,000 \)

The original value of the machine was Rs 16,000.

📝 Teacher's Note: This problem uses the depreciation formula twice. First with 15% rate, then with 12% rate. Students often forget to subtract the values to find the actual depreciation amount.

🎯 Exam Tip: Always write the depreciation formula first. Show each year calculation separately. The final answer should clearly state "original value was Rs X".

Answer 15.
Let value of the bike be Rs x.

Given:
\( V_0 = \text{Rs } x \); n = 2; r = 16%

Step 1: Depreciation in the first year
\( V_e = V_0 \times \left(1 - \frac{r}{100}\right)^n \)

\( \Rightarrow V_e = \text{Rs.} x \times \left(1 - \frac{16}{100}\right) \)

\( \Rightarrow V_e = \text{Rs.} x \times \frac{21}{25} \)

\( \Rightarrow V_e = \text{Rs} 0.84x \)

Step 2: Depreciation in the second year when r is 13%
\( V_e = V_1 \times \left(1 - \frac{r}{100}\right)^n \)

\( \Rightarrow V_e = \text{Rs} 0.84x \times \left(1 - \frac{13}{100}\right) \)

\( \Rightarrow V_e = \text{Rs} 0.84x \times 0.87 \)

\( \Rightarrow V_e = \text{Rs} 0.7308x \)

Step 3: Calculate depreciation amount
Depreciation in the value of bike in the second year
= Rs (0.84x-0.7308x) = Rs 7,098

\( \Rightarrow 0.1092x = \text{Rs } 7,098 \)

\( \Rightarrow x = \text{Rs } 65,000 \)

The original value of the bike was Rs 65,000.

📝 Teacher's Note: Similar to previous problem but with different rates. Make sure students understand that depreciation rate can change from year to year. Show the calculation step by step.

🎯 Exam Tip: Calculate value after first year first, then apply second year rate to that value. Don't apply both rates to original value directly.

Ex 1.4

Answer 1.
For the second year:
A = Rs 648; P = Rs 600; n = 1; r = ?

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( \Rightarrow 648 = 600\left(1 + \frac{r}{100}\right)^1 \)

\( \Rightarrow 648 = 600 + 6r \)
\( \Rightarrow 6r = 48 \)
\( \Rightarrow r = 8 \)

Hence, rate of interest = 8%

For the first year:
I = Rs 600; r = 8%; n = 1; P = ?

\( i = \frac{P \times r \times n}{100} \)

\( \text{Rs} 600 = \text{Rs} \frac{P \times 8 \times 1}{100} \)

\( P = \text{Rs} \frac{60000}{8} \)

\( P = \text{Rs} 7,500 \)

The sum invested = Rs 7,500.

📝 Teacher's Note: This is a reverse calculation problem. First find the rate from compound interest, then use simple interest formula to find principal. Students often mix up the formulas.

🎯 Exam Tip: For second year CI, use the amount formula with n=1. For first year SI, use the basic SI formula. Write both formulas clearly.

Answer 2.
For the second year:
A = Rs 940.80; P = Rs 840; n = 1; r = ?

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 940.80 = 840\left(1 + \frac{r}{100}\right)^1 \)

940.80 = 840.00 + 840r
840r = 100.80
r = 12

Hence, rate of interest = 12%

For the first year:
I = Rs 840; r = 12%; n = 1; P = ?

\( i = \frac{P \times r \times n}{100} \)

\( \text{Rs} 840 = \text{Rs} \frac{P \times 12 \times 1}{100} \)

\( P = \text{Rs} \frac{84000}{12} \)

\( P = \text{Rs} 7,000 \)

The sum invested = Rs 7,000.

📝 Teacher's Note: Same method as Answer 1. Show students how to work backwards from compound interest to find the rate. Then use that rate for simple interest calculation.

🎯 Exam Tip: Always check your answer by substituting back. If P = 7000 and r = 12%, then SI = 840 which matches the given condition.

Answer 3.
The extra interest earned = C.I. – S.I. = Rs (1,365 – 1,300) = Rs 65.

The interest for the first year = S.I. for 2 years / 2 = Rs \( \frac{1300}{2} \) = Rs 650

Therefore, the rate of interest = \( \frac{65}{650} \times 100 \) = 10%

Now,
\( S.I. = \frac{P \times r \times t}{100} \)

\( \Rightarrow \text{Rs} 1,300 = \frac{P \times 10 \times 2}{100} \)

\( \Rightarrow P = \text{Rs} 1300 \times 5 \)
\( \Rightarrow P = \text{Rs} 6,500 \)

The rate of interest was 10% and the original sum was Rs 6,500.

📝 Teacher's Note: This uses the difference between CI and SI. The difference comes from interest earning interest in the second year. First year interest = SI/2 for 2 years.

🎯 Exam Tip: Remember the key formula: difference = (first year interest)² / principal. Use SI for 2 years divided by 2 to get first year interest.

Answer 4.
The extra interest earned = C.I. – S.I. = Rs (8,640 – 8,000) = Rs 640.

The interest for the first year = S.I. for 2 years / 2 = Rs \( \frac{8000}{2} \) = Rs 4000

Therefore, the rate of interest = \( \frac{640}{4,000} \times 100 \) = 16%

Now,
\( S.I. = \frac{P \times r \times t}{100} \)

\( \Rightarrow \text{Rs} 8,000 = \frac{P \times 16 \times 2}{100} \)

\( \Rightarrow P = \text{Rs} \frac{8,000 \times 100}{32} \)

\( \Rightarrow P = \text{Rs} 25,000 \)

The rate of interest was 16% and the original sum was Rs 25,000.

📝 Teacher's Note: Exactly same method as Answer 3 but with different numbers. Make students practice this pattern — find difference, then first year interest, then rate, then principal.

🎯 Exam Tip: The extra interest (CI - SI) always equals first year interest times rate divided by 100. Use this to check your work.

Answer 5.
Here, r = ? P = x (say)

T = 2 years and 3 years

A = Rs 5,082 in 2 years and Rs 5,590.20 in 3 years.

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 5,082 = x\left(1 + \frac{r}{100}\right)^2 \) ...................(i)

\( 5,590.20 = x\left(1 + \frac{r}{100}\right)^3 \) ...................(ii)

Dividing (ii) by (i)

\( \frac{x\left(1 + \frac{r}{100}\right)^3}{x\left(1 + \frac{r}{100}\right)^2} = \frac{5,590.20}{5,082} \)

\( \Rightarrow 1 + \frac{r}{100} = \frac{5,590.20}{5,082} \)

\( \Rightarrow \frac{r}{100} = \frac{5,590.20}{5,082} - 1 \)

\( \Rightarrow \frac{r}{100} = \frac{5,590.20 - 5,082}{5,082} \)

\( \Rightarrow \frac{r}{100} = \frac{508.20}{5,082} \)

\( \Rightarrow r = \frac{508.20}{5,082} \times 100 \)

\( \Rightarrow r = 10\% \)

using (i)

\( x\left(1 + \frac{r}{100}\right)^2 = \text{Rs} 5,082 \)

\( x\left(1 + \frac{10}{100}\right)^2 = \text{Rs} 5,082 \)

\( x \times \frac{11}{10} \times \frac{11}{10} = \text{Rs} 5,082 \)

\( x \times \frac{121}{100} = \text{Rs} 5,082 \)

\( x = \text{Rs} \frac{5,082 \times 100}{121} \)

\( x = \text{Rs} 4,200 \)

Hence, rate of interest = 10% and sum invested = Rs 4,200.

📝 Teacher's Note: This problem gives amounts for two different years. Divide the equations to eliminate P and find r first. Then substitute back to find P. Very useful method.

🎯 Exam Tip: When you have two amounts for different years, always divide the equations. This removes the principal and gives you the rate directly. Show all steps clearly.

Answer 6.
Answer:
Here, r = ? P = x (say)
T = 2 years and 3 years
A = Rs 26,450 in 2 years and Rs 30,417.50 in 3 years.

Step 1: Set up equations using compound interest formula.
\[ A = P\left(1 + \frac{r}{100}\right)^t \]

\[ 26,450 = x\left(1 + \frac{r}{100}\right)^2 \quad \ldots(i) \]

\[ 30,417.50 = x\left(1 + \frac{r}{100}\right)^3 \quad \ldots(ii) \]

Step 2: Divide equation (ii) by equation (i).
\[ \frac{x\left(1 + \frac{r}{100}\right)^3}{x\left(1 + \frac{r}{100}\right)^2} = \frac{30,417.50}{26,450} \]

\[ \left(1 + \frac{r}{100}\right) = \frac{30,417.50}{26,450} \]

Step 3: Calculate the rate of interest.
\[ \frac{r}{100} = \frac{30,417.50}{26,450} - 1 \]

\[ \frac{r}{100} = \frac{30,417.50 - 26,450}{26,450} \]

\[ \frac{r}{100} = \frac{3967.50}{26,450} \]

\[ r = \frac{3967.50}{26,450} \times 100 = 15\% \]

Step 4: Calculate principal amount using equation (i).
\[ x\left(1 + \frac{15}{100}\right)^2 = 26,450 \]

\[ x\left(\frac{115}{100}\right)^2 = 26,450 \]

\[ x \times \frac{23}{20} \times \frac{23}{20} = 26,450 \]

\[ x \times \frac{529}{400} = 26,450 \]

\[ x = \frac{26,450 \times 400}{529} = 20,000 \]

Final Answer: Rate of interest = 15% and sum invested = Rs 20,000.

In simple words: We had the same money growing for 2 years and 3 years. We divided the bigger amount by smaller amount to find how much it grows in one year. This gave us the rate. Then we found the starting amount.

📝 Teacher's Note: When you have two amounts for different time periods, divide them to find the growth factor. This trick works only for compound interest questions. Make students practice this method.

🎯 Exam Tip: Always set up two equations first. Then divide one by the other to eliminate the principal. Show all calculation steps clearly to get full marks.

Answer 7.
Answer:
Here, P = Rs 5,000; r = 8%; t = 2 years

For simple interest:
\[ S.I. = \frac{P \times r \times t}{100} \]
\[ S.I. = Rs \frac{5,000 \times 8 \times 2}{100} = Rs 800 \]

For compound interest:
\[ A = P\left(1 + \frac{r}{100}\right)^t \]
\[ A = Rs 5,000\left(1 + \frac{8}{100}\right)^2 \]
\[ A = Rs 5,000 \times \frac{108}{100} \times \frac{108}{100} \]
\[ A = Rs 5,832 \]

\[ C.I. = A - P = Rs(5,832 - 5,000) = Rs 832 \]

Difference: Rs(832 - 800) = Rs 32

In simple words: Simple interest gives the same amount every year. Compound interest gives more because you earn interest on the interest too. The extra amount is Rs 32.

📝 Teacher's Note: Use a simple example - if you lend Rs 100 at 10%, simple interest gives Rs 10 every year. Compound interest gives Rs 10 first year, then Rs 11 second year (10% on Rs 110).

🎯 Exam Tip: Calculate both simple and compound interest separately first. Then find the difference. Always check that compound interest is more than simple interest.

Answer 8.
Answer:
Here, P = Rs 15,000; r = 8%; t = 3 years

For simple interest:
\[ S.I. = \frac{P \times r \times t}{100} = Rs \frac{15,000 \times 8 \times 3}{100} = Rs 3,600 \]

For compound interest:
\[ A = P\left(1 + \frac{r}{100}\right)^t \]
\[ A = Rs 15,000\left(1 + \frac{8}{100}\right)^3 \]
\[ A = Rs 15,000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100} \]
\[ A = Rs 18,895.68 \]

\[ C.I. = A - P = Rs(18,895.68 - 15,000) = Rs 3,895.68 \]

Difference: Rs(3,895.68 - 3,600) = Rs 295.68

In simple words: After 3 years, compound interest gives Rs 295.68 more than simple interest. This happens because in compound interest, you get interest on the interest earned in previous years.

📝 Teacher's Note: Show students that the difference increases with time. After 1 year, both are same. After 2 years, small difference. After 3 years, bigger difference.

🎯 Exam Tip: For 3 years calculation, be careful with the power. Write \( (1.08)^3 \) step by step as 1.08 × 1.08 × 1.08 to avoid mistakes.

Answer 9.
Answer:
Here, P = Rs 20,000; t = 3 years
For simple interest: r = 9%

Simple Interest:
\[ S.I. = \frac{P \times r \times t}{100} = Rs \frac{20,000 \times 9 \times 3}{100} = Rs 5,400 \]

For compound interest: r = \( 8\frac{1}{2}\% \)
\[ A = P\left(1 + \frac{r}{100}\right)^t \]
\[ A = Rs 20,000\left(1 + \frac{17}{2 \times 100}\right)^3 \]
\[ A = Rs 20,000 \times \frac{217}{200} \times \frac{217}{200} \times \frac{217}{200} \]
\[ A = Rs 25,545.70 \]

\[ C.I. = A - P = Rs(25,545.70 - 20,000) = Rs 5,545.70 \]

Difference: Rs(5,545.70 - 5,400) = Rs 145.70
Anand gained Rs 145.70

In simple words: Even though compound interest rate is lower (8.5% vs 9%), Anand still earned Rs 145.70 more. This shows the power of compounding over time.

📝 Teacher's Note: This is a tricky question. Students think higher rate always gives more money. But compound interest can beat simple interest even at lower rates over long time periods.

🎯 Exam Tip: Convert mixed fractions like \( 8\frac{1}{2} \) to improper fractions \( \frac{17}{2} \) before putting in the formula. Be careful with decimal calculations.

Answer 10.
Answer:
Here, P = Rs 35,000; t = 3 years
For simple interest: r = 12.5%

Simple Interest:
\[ S.I. = \frac{P \times r \times t}{100} = Rs \frac{35,000 \times 12.5 \times 3}{100} = Rs 13,125 \]

For compound interest: r = 12%
\[ A = P\left(1 + \frac{r}{100}\right)^t \]
\[ A = Rs 35,000\left(1 + \frac{12}{100}\right)^3 \]
\[ A = Rs 35,000 \times \frac{112}{100} \times \frac{112}{100} \times \frac{112}{100} \]
\[ A = Rs 49,172.48 \]

\[ C.I. = A - P = Rs(49,172.48 - 35,000) = Rs 14,172.48 \]

Difference: Rs(14,172.48 - 13,125) = Rs 1,047.48
Meera gained Rs 1,047.48

In simple words: Meera chose compound interest at 12% instead of simple interest at 12.5%. She still earned Rs 1,047.48 more because compound interest grows faster over 3 years.

📝 Teacher's Note: This shows that compound interest becomes very powerful over longer time periods. Even a 0.5% lower rate can give more money if it compounds.

🎯 Exam Tip: In these comparison problems, calculate both amounts completely before finding the difference. Write "gained" if compound interest is more, "lost" if simple interest is more.

Exercise 1.5

Answer 11.
Answer:
(a) The rate of depreciation.
Difference in the depreciation = Rs (5,100-4,335) = Rs 765
Rate of depreciation =
\[ \frac{765}{5,100} \times 100 \]
= 15%
Rate of depreciation =15%

(b) The original cost of the scooter.
Depreciation for first year = Rs 5,100 + 15% of Rs 5,100
Here, 15% of Rs 5,100 = Rs 765
Hence, Depreciation for first year = Rs 5,100 + Rs 765 = Rs 5,865
Total depreciation for 3 years = Rs (5,865 + 5,100 + 4335) = Rs 15,300
A= P - Rs 15,300; P = x
\[ A = P\left(1-\frac{r}{100}\right)^n \]
\[ x - 15,300 = x\left(1-\frac{15}{100}\right)^3 \]
x – 15,300 = x × 0.85 × 0.85 × 0.85
x(1 – 0.614) = Rs15,300
\[ x = Rs\frac{15,300}{0.386} \]
x = Rs39,637.31
\( \Rightarrow x = Rs40,000(approx) \)
Original cost of scooter = Rs 40,000

(c) The cost of the scooter at the end of the third year.
Here, P = Rs 40,000; r = 15%; t = 3 years
\[ A = P\left(1-\frac{r}{100}\right)^t \]
\[ A = Rs40,000\left(1-\frac{15}{100}\right)^3 \]
A = Rs40,000 × 0.85 × 0.85 × 0.85
A = Rs24,565
Cost of the scooter at the end of third year = Rs 24,565
In simple words: We used the depreciation formula to find how much value the scooter loses each year. Then we worked backwards to find the original cost.

📝 Teacher's Note: Show students that depreciation means the item becomes less valuable over time. Like a car that gets old and costs less to sell. Use the formula step by step.

🎯 Exam Tip: Always write the depreciation formula first. Show all steps clearly. Write the final answer with "Rs" and round to nearest rupee if needed.

Answer 12.
Answer:
(a) The rate of depreciation.
Difference in the depreciation = Rs (2,592-2,332.80) = Rs 259.20
Rate of depreciation =
\[ \frac{259.20}{2,592} \times 100 \]
= 10%
Rate of depreciation =10%

(b) The original cost.
Depreciation for second year = Rs 2,592 + 10% of Rs 2,592
Here, 10% of Rs 2,592 = Rs 259.20
Hence, Depreciation for second year = Rs 2,592 + Rs 259.20 = Rs 2,851.20
Depreciation for first year = Rs 2,851.20 + 10% of Rs 2,851.20
Here, 10% of Rs 2,851.20 = Rs 285.12
Hence, Depreciation for first year = Rs 2,851.20 + Rs 285.12 = Rs 3,136.32
Total depreciation for 4 years = Rs (3,136.32 + 2,851.20 + 2,592 + 2,332.80)
= Rs 10,912.32
A= P - Rs 10,912.32; P = x
\[ A = P\left(1-\frac{r}{100}\right)^n \]
\[ x - 10,912.32 = x\left(1-\frac{10}{100}\right)^4 \]
x – 10,912.32 = x × 1.1 × 1.1 × 1.1 × 1.1
x(1 – 0.6561) = Rs10,912.32
\[ x = Rs\frac{10,912.32}{0.3439} \]
x = Rs31731.08
\( \Rightarrow x = Rs32,000(approx) \)
Original cost = Rs 32,000

(c) The cost at the end of the fourth year.
Here, P = Rs 32,000; r = 10%; t = 4 years
\[ A = P\left(1-\frac{r}{100}\right)^n \]
\[ A = Rs32,000\left(1-\frac{10}{100}\right)^4 \]
A = Rs32,000 × 0.9 × 0.9 × 0.9 × 0.9
A = Rs20,995.20
Cost at the end of the fourth year = Rs 20,995.20
In simple words: We found the rate at which the machine loses value each year. Then we calculated its original cost and final value after 4 years.

📝 Teacher's Note: Explain that machines lose value because they get old and worn out. The percentage stays the same each year, but the amount of money lost gets smaller.

🎯 Exam Tip: Write "Given", "Find", then show each calculation step. Always use the correct formula for compound depreciation. Check your final answer makes sense.

Answer 1.
Answer:
Interest for first year:
\[ S.I. = \frac{P \times r \times t}{100} \]
\[ S.I. = \frac{12,000 \times 15 \times 1}{100} \]
S.I. = 1800
Principal amount for second year = Rs (12,000 + 1800) = Rs 13,800
Ramesh paid = Rs x (say)
Therefore, new principal = Rs 13,800-x
A=Rs 9,200; r = 15%; n=1 year
\[ A = P\left(1+\frac{r}{100}\right)^n \]
\[ Rs9,200 = Rs(13,800 - x)\left(1+\frac{15}{100}\right) \]
Rs9,200 = Rs(13,800 – x) × 1.15
Rs9,200 = Rs15,870 – Rs1.15x
1.15x = Rs(15870 – 9,200)
\[ x = \frac{Rs6,670}{1.15} \]
x = Rs5,800
Therefore, Amount Ramesh paid at the end of first year = Rs 5,800
In simple words: Ramesh borrowed money and paid some back after one year. We calculated how much he paid by using the compound interest formula.

📝 Teacher's Note: Show students that when someone pays back part of a loan, the remaining amount becomes the new principal for calculating next year's interest.

🎯 Exam Tip: Write "Given" and "Find" first. Show the simple interest calculation, then the compound interest formula. Always label your final answer clearly.

Answer 2.
Answer:
Interest for first year:
\[ S.I. = \frac{P \times r \times t}{100} \]
\[ S.I. = \frac{32,000 \times 12 \times 1}{100} \]
S.I. = 3,840
Principal amount for second year = Rs (32,000 + 3,840) = Rs 35,840
Rajan paid = Rs x (say)
Therefore, new principal = Rs 35,840-x
A=Rs 17,920; r = 12%; n=1 year
\[ A = P\left(1+\frac{r}{100}\right)^n \]
\[ Rs17,920 = Rs(35,840 - x)\left(1+\frac{12}{100}\right) \]
Rs17,920 = Rs(35,840 – x) × 1.12
Rs17,920 = Rs40,140.80 – Rs1.12x
1.12x = Rs(40,140.80 – 17,920)
\[ x = \frac{Rs22220.80}{1.12} \]
x = Rs19,840
Therefore, Amount Rajan paid at the end of first year = Rs 19,840
In simple words: Rajan borrowed Rs 32,000 and paid back Rs 19,840 after one year. The remaining amount grew with compound interest to become Rs 17,920 after another year.

📝 Teacher's Note: Emphasize that the interest gets added to the principal each year. Then the new total becomes the base for next year's interest calculation.

🎯 Exam Tip: Follow the same steps as Answer 1. Calculate simple interest first, then use compound interest formula. Show all working clearly to get full marks.

Answer 3.
Answer:
Let the sum be P
Interest for first year:
\[ P\left(1+\frac{8}{100}\right) - P .........(i) \]
Interest for third year:
\[ P\left(1+\frac{8}{100}\right)^3 - P\left(1+\frac{8}{100}\right)^2 ............(ii) \]
Subtracting (ii) from (i)
\[ P\left(1+\frac{8}{100}\right)^3 - P\left(1+\frac{8}{100}\right)^2 - P\left(1+\frac{8}{100}\right) + P = Rs166.40 \]
Rs166.40 = 1.25971²P – 1.1664P – 1.08P + P
Rs166.40 = 0.013312P
P = Rs12,500
Hence the sum is Rs 12,500
In simple words: We found the principal amount by using the fact that the difference between interest in first and third year is Rs 166.40.

📝 Teacher's Note: Show students that interest grows each year because it gets calculated on a bigger amount. The third year interest is more than the first year interest.

🎯 Exam Tip: Set up the equation properly using compound interest formula. Be careful with the algebra when subtracting. Always check your final answer by substituting back.

Answer 3.
Answer:
Let the sum be P
Interest for first year:

\( P\left(1 + \frac{8}{100}\right) - P.........(i) \)

Interest for third year:

\( P\left(1 + \frac{8}{100}\right)^3 - P\left(1 + \frac{8}{100}\right)^2 .............(ii) \)

Subtracting (ii) from (i)

\( P\left(1 + \frac{8}{100}\right)^3 - P\left(1 + \frac{8}{100}\right)^2 - P\left(1 + \frac{8}{100}\right) + P = Rs166.40 \)

Rs166.40 = 1.259712P - 1.1664P - 1.08P + P
Rs166.40 = 0.013312P
P = Rs12,500

Hence the sum is Rs 12,500
In simple words: We found the difference between interest earned in first year and third year. Then we solved to find the original sum.

📝 Teacher's Note: Show students that compound interest grows each year. The interest in year 3 is more than year 1 because we earn interest on interest too.

🎯 Exam Tip: Always write the formula clearly. Show each step of calculation. Write the final answer with "Rs" and correct amount.

Answer 4.
Answer:
Let the sum be P
Interest for first year:

\( P\left(1 + \frac{25}{2 \times 100}\right) - P.........(i) \)

Interest for third year:

\( P\left(1 + \frac{25}{2 \times 100}\right)^3 - P\left(1 + \frac{25}{2 \times 100}\right)^2 .............(ii) \)

Subtracting (ii) from (i)

\( P\left(1 + \frac{25}{2 \times 100}\right)^3 - P\left(1 + \frac{25}{2 \times 100}\right)^2 - P\left(1 + \frac{25}{2 \times 100}\right) + P = Rs531.25 \)

Rs531.25 = 1.423828P - 1.265625P - 1.125P + P
Rs531.25 = 0.033203P
P = Rs16,000

Hence the sum is Rs 16,000
In simple words: This is similar to the previous problem but with 12.5% rate per year. We found the original sum by calculating the difference in interest.

📝 Teacher's Note: When rate is given as 25% per 2 years, divide by 2 to get yearly rate. This is 12.5% per year.

🎯 Exam Tip: Be careful with rate conversion. 25% per 2 years means 12.5% per year. Write this conversion step clearly.

Answer 5.
Answer:
Here, P = ?; t = 2 years; r = 8% p.a.
S.I. = Rs320

\( P = Rs \frac{S.I. \times 100}{r \times t} \)

\( P = Rs \frac{320 \times 100}{8 \times 2} \)

P = Rs2,000

Now, P = Rs 2,000; t = 1 year

n = 2t = 2 × 1 = 2

\( r = \frac{1}{2} \times 8\% = 4\% \) Per conversion period.

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( A = Rs2,000\left(1 + \frac{4}{100}\right)^2 \)

= Rs2,000 × 1.04 × 1.04
= Rs2,163.20

C.I. = A - P

= Rs (2,163.20 - 2,000)

= Rs 163.20

Hence, compound interest = Rs 163.20
In simple words: First we found the principal using simple interest formula. Then we calculated compound interest when compounding happens twice a year.

📝 Teacher's Note: When interest is compounded twice yearly, divide the rate by 2 and multiply time by 2. This gives more accurate interest calculation.

🎯 Exam Tip: For semi-annual compounding, remember: rate ÷ 2 and time × 2. Show both simple interest and compound interest calculations clearly.

Ex 1.6

Answer 1.
Answer:
(a) Rs 12,000 for 3 years at 15% p.a.

P=Rs 12,000; t=3 years; r=15% p.a.

\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs12,000\left(1 + \frac{15}{100}\right)^3 \)

= Rs12,000 × 1.15 × 1.15 × 1.15
= Rs18,250.50

C.I. = A - P
= Rs(18,250.50 - 12,000)
= Rs6,250.50
Hence, Amount = Rs 18,250.50 and C.I. = Rs 6,250.50

(b) Rs 25,000 for 3 years at 8% p.a.

P=Rs 25,000; t=3 years; r=8% p.a.
\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs25,000\left(1 + \frac{8}{100}\right)^3 \)

= Rs25,000 × 1.08 × 1.08 × 1.08
= Rs31,492.80

C.I. = A - P
= Rs(31,492.80 - 25,000)
= Rs6,492.80

Hence, Amount = Rs 31,492.80 and C.I. = Rs 6,492.80

(c) Rs 16,000 for 3 years at \( 7\frac{1}{2} \)% p.a.

P=Rs 16,000; t=3 years; r=\( 7\frac{1}{2} \)% p.a.
\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs16,000\left(1 + \frac{15}{2 \times 100}\right)^3 \)

= Rs16,000 × 1.075 × 1.075 × 1.075
= Rs19,876.75

C.I. = A - P
= Rs(19,876.75 - 16,000)
= Rs3,876.75

Hence, Amount = Rs 19,876.75 and C.I. = Rs 3,876.75

(d) Rs 20,000 for 2 years at \( 12\frac{1}{2} \)% p.a.

P=Rs 20,000; t=2 years; r=\( 12\frac{1}{2} \)% p.a.
\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs20,000\left(1 + \frac{25}{2 \times 100}\right)^2 \)

= Rs20,000 × 1.125 × 1.125
= Rs25,312.50

C.I. = A - P
= Rs(25,312.50 - 20,000)
= Rs5,312.50

Hence, Amount = Rs 25,312.50 and C.I. = Rs 5,312.50

(e) Rs 8,000 for \( 1\frac{1}{2} \) years at 12% p.a.

P=Rs 8,000; t=\( 1\frac{1}{2} \) years; r=12 % p.a.
\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs8,000\left(1 + \frac{12}{100}\right)\left(1 + \frac{12}{100}\right)^{\frac{1}{2}} \)

= Rs8,000 × 1.12 × \( \left(1 + \frac{1}{2} \times \frac{12}{100}\right) \)

= Rs8,000 × 1.12 × 1.06
= Rs9,497.60

C.I. = A - P
= Rs(9,497.60 - 8,000)
= Rs1,497.60

Hence, Amount = Rs 9,497.60 and C.I. = Rs 1,497.60

(f) Rs 7,500 for \( 2\frac{1}{2} \) years; r=16 % p.a.

P=Rs 7,500; t=\( 2\frac{1}{2} \) years; r=16 % p.a.
\( A = P\left(1+\frac{r}{100}\right)^t \)

\( A = Rs7,500\left(1 + \frac{16}{100}\right)^2\left(1 + \frac{16}{100}\right)^{\frac{1}{2}} \)

= Rs7,500 × 1.16 × 1.16 × \( \left(1 + \frac{1}{2} \times \frac{16}{100}\right) \)

= Rs7,500 × 1.16 × 1.16 × 1.08
= Rs10,899.36

C.I. = A - P
= Rs(10,899.36 - 7,500)
= Rs3,399.36
Hence, Amount = Rs 10,899.36 and C.I. = Rs 3,399.36
In simple words: We used the compound interest formula to find the final amount. Then we subtracted the principal to get the compound interest earned.

📝 Teacher's Note: For fractional years like 1½, calculate as 1 full year plus ½ year separately. For the ½ year, use simple interest method.

🎯 Exam Tip: Always write the formula first. Show each calculation step. For mixed fractions like 7½%, convert to 15/2%. Round final answers to 2 decimal places.

Answer 2. (a) Rs 6,000 for \( 1\frac{1}{2} \) years at 10% p.a.
Answer:
Given:
P=Rs 6,000; t=\( 1\frac{1}{2} \) years; r = 10% p.a. = 5% half-yearly.

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs6,000 \left(1 + \frac{5}{100}\right)^2 \left(1 + \frac{10}{100}\right)^{\frac{1}{2}} \)
\( = Rs6,000 \times 1.05 \times 1.05 \times \left(1 + \frac{1}{2} \times \frac{10}{100}\right) \)
\( = Rs6,000 \times 1.05 \times 1.05 \times 1.05 \)
\( = Rs6,945.75 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(6,945.75 - 6,000) \)
\( = Rs945.75 \)

Hence, Amount = Rs 6,945.75 and C.I. = Rs 945.75
In simple words: We use the compound interest formula where money grows each half-year. The interest earned also earns interest in the next period.

📝 Teacher's Note: Show students that for half-yearly compounding, we divide the rate by 2 and multiply time by 2. This is because interest is calculated twice a year.

🎯 Exam Tip: Always write the formula first. Then substitute values clearly. Show C.I. = A - P at the end. Check if your answer makes sense.

(b) Rs 25,000 for \( 1\frac{1}{2} \) years at 12%
Answer:
Given:
P=Rs 25,000; t=\( 1\frac{1}{2} \) years; r = 12% p.a. = 6% half-yearly.

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs25,000 \left(1 + \frac{6}{100}\right)^2 \left(1 + \frac{12}{100}\right)^{\frac{1}{2}} \)
\( = Rs25,000 \times 1.06 \times 1.06 \times \left(1 + \frac{1}{2} \times \frac{12}{100}\right) \)
\( = Rs25,000 \times 1.06 \times 1.06 \times 1.06 \)
\( = Rs29,775.40 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(29,775.40 - 25,000) \)
\( = Rs4,775.40 \)

Hence, Amount = Rs 29,775.40 and C.I. = Rs 4,775.40
In simple words: The money grows for 1.5 years with compound interest. Each 6 months, interest is added to the main amount.

📝 Teacher's Note: Remind students that \( 1\frac{1}{2} \) years = 3 half-yearly periods. So we use power 3 in our calculation.

🎯 Exam Tip: For fractional years with compound interest, convert to the compounding period. Write all steps clearly to get full marks.

Answer 3. (a) Rs 9,125 for 2 years if the rates of interest are 12% and 14% for the successive years.
Answer:
Given:
P=Rs 9,125; t=2 years; r = 12% and 14% successively.

Step 1: Use compound interest formula for different rates.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs9,125 \left(1 + \frac{12}{100}\right) \left(1 + \frac{14}{100}\right) \)
\( = Rs9,125 \times 1.12 \times 1.14 \)
\( = Rs11,650.80 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(11,650.80 - 9,125) \)
\( = Rs2,525.80 \)

Hence, Amount = Rs 11,650.80 and C.I. = Rs 2,525.80
In simple words: Each year has a different interest rate. First year grows at 12%, then the new amount grows at 14% in second year.

📝 Teacher's Note: When rates are different each year, multiply by (1 + r₁/100) for first year, then (1 + r₂/100) for second year. Order matters!

🎯 Exam Tip: For successive rates, write each rate separately in brackets. Never average the rates - that gives wrong answer.

(b) Rs 20,000 for 2 years if the rates of interest are \( 12\frac{1}{4} \)% and \( 5\frac{1}{2} \)% for the successive years.
Answer:
Given:
P=Rs 20,000; t=2 years; r = \( 12\frac{1}{4} \)% and \( 5\frac{1}{2} \)% successively = \( \frac{49}{4} \)% and \( \frac{11}{2} \)% successively.

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs20,000 \left(1 + \frac{49}{4 \times 100}\right) \left(1 + \frac{11}{2 \times 100}\right) \)
\( = Rs20,000 \times 1.1225 \times 1.055 \)
\( = Rs23,684.75 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(23,684.75 - 20,000) \)
\( = Rs3,684.75 \)

Hence, Amount = Rs 23,684.75 and C.I. = Rs 3,684.75.
In simple words: We convert mixed fractions to improper fractions first. Then calculate compound interest step by step for each year.

📝 Teacher's Note: Help students convert mixed fractions: \( 12\frac{1}{4} = \frac{49}{4} \) and \( 5\frac{1}{2} = \frac{11}{2} \). Practice this conversion separately first.

🎯 Exam Tip: Convert mixed fractions to decimals carefully. \( 12\frac{1}{4} \)% = 12.25%. Show this conversion in your working.

(c) Rs 12,500 for 3 years if the rates for the successive years are 8%, 9% and 10% respectively.
Answer:
Given:
P=Rs 12,500; t=3 years; r = 8%, 9% and 10% successively.

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs12,500 \left(1 + \frac{8}{100}\right) \left(1 + \frac{9}{100}\right) \left(1 + \frac{10}{100}\right) \)
\( = Rs12,500 \times 1.08 \times 1.09 \times 1.1 \)
\( = Rs16,186.50 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(16,186.50 - 12,500) \)
\( = Rs3,686.50 \)

Hence, Amount = Rs 16,186.50 and C.I. = Rs 3,686.50
In simple words: For three years with different rates, we multiply by three different factors. Each year the money grows by a different percentage.

📝 Teacher's Note: Show students that for 3 different rates, we need 3 multiplication factors. Write them in order: year 1, year 2, year 3.

🎯 Exam Tip: For multiple years with different rates, multiply all factors together. Don't forget to subtract principal for compound interest.

(d) Rs 10,000 for 3 years if the rates of interest are 10%, 11% and 12% for the successive years.
Answer:
Given:
P=Rs 10,000; t=3 years; r = 10%, 11% and 12% successively.

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Calculate amount.
\( A = Rs10,000 \left(1 + \frac{10}{100}\right) \left(1 + \frac{11}{100}\right) \left(1 + \frac{12}{100}\right) \)
\( = Rs10,000 \times 1.1 \times 1.11 \times 1.12 \)
\( = Rs13,675.20 \)

Step 3: Calculate compound interest.
C.I. = A - P
\( = Rs(13,675.20 - 10,000) \)
\( = Rs3,675.20 \)

Hence, Amount = Rs 13,675.20 and C.I. = Rs 3,675.20
In simple words: The interest rate increases each year. So the money grows faster in later years because of higher rates.

📝 Teacher's Note: Point out that when rates increase each year, the compound interest grows more than if rates were constant. This is because of compounding effect.

🎯 Exam Tip: Always write the rates in the correct order. First year rate goes first, second year rate second, and so on.

Answer 4. (a) Rs 15,000 for \( 1\frac{1}{2} \) years at 12% p.a.
Answer:
Given:
P=Rs 15,000; t=\( 1\frac{1}{2} \) years

When compounded yearly: r = 12% p.a.

\( A = P \left(1 + \frac{r}{100}\right)^n \)

\( A = Rs15,000 \left(1 + \frac{12}{100}\right) \left(1 + \frac{12}{100}\right)^{\frac{1}{2}} \)

\( = Rs15,000 \times 1.12 \times \left(1 + \frac{1}{2} \times \frac{12}{100}\right) \)

\( = Rs15,000 \times 1.12 \times 1.06 \)
\( = Rs17,808 \)

C.I. = A - P
\( = Rs(17,808 - 15,000) \)
\( = Rs2,808 \)

When compounded half-yearly:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

\( A = Rs15,000 \left(1 + \frac{6}{100}\right)^3 \)

\( = Rs15,000 \times 1.06 \times 1.06 \times 1.06 \)
\( = Rs17,865.24 \)

C.I. = A - P
\( = Rs(17,865.24 - 15,000) \)
\( = Rs2,865.24 \)

Hence the difference in the interest = Rs (2,865.24-2,808) = Rs 57.24
In simple words: Half-yearly compounding gives more interest because interest is calculated more often. The difference is Rs 57.24.

📝 Teacher's Note: Show students that more frequent compounding always gives higher returns. Half-yearly means 6% every 6 months instead of 12% every year.

🎯 Exam Tip: For comparison questions, calculate both amounts separately. Then find the difference. Show all working clearly.

(b) Rs 20,000 for \( 1\frac{1}{2} \) years at 16% p.a.
Answer:
Given:
P=Rs 20,000; t=\( 1\frac{1}{2} \) years

When compounded yearly: r = 16% p.a.

\( A = P \left(1 + \frac{r}{100}\right)^n \)

\( A = Rs20,000 \left(1 + \frac{16}{100}\right) \left(1 + \frac{16}{100}\right)^{\frac{1}{2}} \)

\( = Rs20,000 \times 1.16 \times \left(1 + \frac{1}{2} \times \frac{16}{100}\right) \)

\( = Rs20,000 \times 1.16 \times 1.08 \)
\( = Rs25,056 \)

C.I. = A - P
\( = Rs(25,056 - 20,000) \)
\( = Rs5,056 \)

When compounded half-yearly:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

\( A = Rs20,000 \left(1 + \frac{8}{100}\right)^3 \)

\( = Rs20,000 \times 1.08 \times 1.08 \times 1.08 \)
\( = Rs25,194.24 \)

C.I. = A - P
\( = Rs(25,194.24 - 20,000) \)
\( = Rs5,194.24 \)

Hence the difference in the interest = Rs (5,194.24-5,056) = Rs 138.24
In simple words: With higher principal and higher rate, the difference between yearly and half-yearly compounding is bigger - Rs 138.24.

📝 Teacher's Note: Compare this with part (a). Higher principal and rate give bigger difference between compounding methods. Students should notice this pattern.

🎯 Exam Tip: For half-yearly compounding, divide rate by 2 and multiply time by 2. Always check your calculation by seeing if half-yearly gives more.

Answer 5. Here P=?; t = 2 years; r = 15% and 17% successively; A = Rs 8,073
Answer:
Given:
A = Rs 8,073; t = 2 years; r = 15% and 17% successively; P = ?

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Substitute known values.
\( Rs8,073 = P \left(1 + \frac{15}{100}\right) \left(1 + \frac{17}{100}\right) \)

\( Rs8,073 = P \times 1.15 \times 1.17 \)
\( Rs8,073 = 1.3455P \)

Step 3: Find principal.
\( P = Rs \frac{8,073}{1.3455} \)
\( P = Rs6,000 \)

Hence, the sum of money is Rs 6,000.
In simple words: We work backwards from the final amount. We divide by the growth factors to find the starting amount.

📝 Teacher's Note: When principal is unknown, work backwards. Divide the amount by all the growth factors to get the original principal.

🎯 Exam Tip: For reverse problems, write A = P × (growth factors), then solve for P. Always check your answer makes sense.

Answer 6. Here P=?; t = 2 years; r = 12% and 14% successively; A = Rs 22,344
Answer:
Given:
A = Rs 22,344; t = 2 years; r = 12% and 14% successively; P = ?

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Substitute known values.
\( Rs22,344 = P \left(1 + \frac{12}{100}\right) \left(1 + \frac{14}{100}\right) \)

\( Rs22,344 = P \times 1.12 \times 1.14 \)
\( Rs22,344 = 1.2768P \)

Step 3: Find principal.
\( P = Rs \frac{22,344}{1.2768} \)
\( P = Rs17,500 \)

Hence, the principal is Rs 17,500.
In simple words: We know the final amount and the growth rates. We divide to find what money we started with.

📝 Teacher's Note: This is like asking "what number becomes 22,344 when multiplied by 1.12 and then by 1.14?" We divide to find that number.

🎯 Exam Tip: In reverse problems, always multiply the growth factors first, then divide the amount by this product. Show each step clearly.

Answer 7.
Answer:
Given:
P = ?; t = 3 years; r = 10%, 11% and 12% successively; A = Rs 10,256.40

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply the compound interest formula for variable rates.
\( Rs 10,256.40 = P \left(1 + \frac{10}{100}\right)\left(1 + \frac{11}{100}\right)\left(1 + \frac{12}{100}\right) \)

Step 2: Calculate the combined rate factor.
\( Rs 10,256.40 = P \times 1.1 \times 1.11 \times 1.12 \)
\( Rs 10,256.40 = 1.36752P \)

Step 3: Solve for P.
\( P = Rs \frac{10,256.40}{1.36752} \)
\( P = Rs 7,500 \)

Final Answer: The sum of money is Rs 7,500.
In simple words: When the interest rate changes every year, we multiply all the rate factors together. Then we divide the final amount by this combined factor to find the starting money.

📝 Teacher's Note: Show students that when rates change yearly, we multiply (1 + rate₁) × (1 + rate₂) × (1 + rate₃). This is different from using one average rate.

🎯 Exam Tip: Always write the formula first. Then substitute each year's rate separately. Don't try to find an average rate - that will give wrong answer.

Answer 8.
Answer:
Given:
P = ?; A = Rs 18,972; t = \( 1\frac{1}{2} \) years; r = 16%

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Convert time to proper form.
t = \( 1\frac{1}{2} \) years = 1 year + 6 months

Step 2: Apply compound interest formula.
\( Rs 18,972 = P \left(1 + \frac{16}{100}\right)\left(1 + \frac{16}{100}\right)^{\frac{1}{2}} \)

Step 3: Simplify the calculation.
\( Rs 18,972 = P \left(1 + \frac{16}{100}\right)\left(1 + \frac{1}{2} \times \frac{16}{100}\right) \)
\( Rs 18,972 = P \times 1.16 \times 1.08 \)
\( Rs 18,972 = 1.2528P \)

Step 4: Solve for P.
\( P = Rs \frac{18,972}{1.2528} \)
\( P = Rs 15,143.68 \)

Final Answer: The sum of money will be Rs 15,143.68
In simple words: For half-year time periods, we use the formula for fractional powers. Half year means we take half the yearly rate for that 6-month period.

📝 Teacher's Note: Explain that for \( t = 1\frac{1}{2} \) years, we compound for 1 full year, then add simple interest for the remaining 6 months using the compound amount as principal.

🎯 Exam Tip: For fractional years like 1½, write it as (1 + ½) and apply the formula step by step. Show each calculation clearly for full marks.

Answer 9.
Answer:
Given:
P = ?; A = Rs 15,746.40; t = \( \frac{1}{2} \) years; r = 16%

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply the formula for half-year period.
\( Rs 15,746.40 = P \left(1 + \frac{16}{100}\right)^{\frac{1}{2}} \)

Step 2: Convert fractional power to simple interest for 6 months.
\( Rs 15,746.40 = P \left(1 + \frac{1}{2} \times \frac{16}{100}\right) \)
\( Rs 15,746.40 = P \times 1.08 \)

Step 3: Solve for P.
\( P = Rs \frac{15,746.40}{1.08} \)
\( P = Rs 14,580 \)

Final Answer: The sum of money will be Rs 14,580.
In simple words: For 6 months (half year), we use half the yearly interest rate. This gives us 8% for 6 months instead of 16% for full year.

📝 Teacher's Note: For half-year periods, students often get confused. Show them that 6 months = ½ year, so we use ½ × 16% = 8% as simple interest.

🎯 Exam Tip: When time is in fractions of a year (like ½), convert the rate proportionally. Write clearly: "For 6 months, rate = 16%/2 = 8%".

Answer 10.
Answer:
Given:
P = x; t = 2 years; r = 8%; A = Rs (x + 1399.68)

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Substitute the given values.
\( Rs(x + 1399.68) = x \left(1 + \frac{8}{100}\right)^2 \)

Step 2: Simplify the calculation.
\( Rs(x + 1399.68) = x \times 1.08 \times 1.08 \)
\( Rs(x + 1399.68) = 1.1664x \)

Step 3: Solve for x.
\( 1.1664x = Rs 1399.68 \)
\( x = Rs 8,411.538 \)

Final Answer: On Rs 8,411.538 the C.I. for 2 years at 8% will be Rs 1399.68
In simple words: We know the compound interest amount. We use algebra to find the original principal amount that would give this much interest.

📝 Teacher's Note: This is a reverse compound interest problem. Students need to understand that A - P = C.I., so we can substitute and solve for the unknown principal.

🎯 Exam Tip: Write the equation clearly: A = P + C.I. Then substitute A = P(1 + r/100)ⁿ and solve step by step. Show all working for marks.

Answer 11.
Answer:
Given:
P = x; t = \( 2\frac{1}{2} \) years; r = 12%; A = Rs (x + 8,241.60)

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply formula for \( 2\frac{1}{2} \) years.
\( Rs(x + 8,241.60) = x \left(1 + \frac{12}{100}\right)^2 \left(1 + \frac{12}{100}\right)^{\frac{1}{2}} \)

Step 2: Simplify the fractional power.
\( Rs(x + 8,241.60) = x \times 1.12 \times 1.12 \times \left(1 + \frac{1}{2} \times \frac{12}{100}\right) \)
\( Rs(x + 8,241.60) = x \times 1.12 \times 1.12 \times 1.06 \)
\( Rs(x + 8,241.60) = 1.329664x \)

Step 3: Solve for x.
\( 0.329664x = Rs 8,241.60 \)
\( x = Rs 25,000 \)

Final Answer: On Rs 25,000 the C.I. for \( 2\frac{1}{2} \) years at 12% will be Rs 8241.60.
In simple words: For 2½ years, we compound for 2 full years, then add simple interest for the remaining 6 months on the compound amount.

📝 Teacher's Note: Break down 2½ years as "2 years + ½ year". First compound for 2 years, then apply simple interest for remaining 6 months.

🎯 Exam Tip: Always separate whole years from fractional years. Write: "2 years compound + 6 months simple interest". This method gets full marks.

Answer 12.
Answer:
Given:
P = x; t = \( 2\frac{1}{2} \) years; r = \( 12\frac{1}{2} \)% = \( \frac{25}{2} \)%; A = Rs (x + 82,734.37)

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply formula for \( 2\frac{1}{2} \) years with rate \( \frac{25}{2} \)%.
\( Rs(x + 82,734.37) = x \left(1 + \frac{25}{2 \times 100}\right)^2 \left(1 + \frac{25}{2 \times 100}\right)^{\frac{1}{2}} \)

Step 2: Simplify the calculations.
\( Rs(x + 82,734.37) = x \left(1 + \frac{25}{2 \times 100}\right)^2 \left(1 + \frac{25}{2 \times 100}\right)^{\frac{1}{2}} \)
\( Rs(x + 82,734.37) = x \times 1.125 \times 1.125 \times \left(1 + \frac{1}{2} \times \frac{1}{8}\right) \)
\( Rs(x + 82,734.37) = x \times 1.125 \times 1.125 \times 1.0625 \)
\( Rs(x + 82,734.37) = 1.344727x \)

Step 3: Solve for x.
\( 0.344727x = Rs 82,734.37 \)
\( x = Rs 2,39,999.7 = Rs 2,40,000 \)

Final Answer: On Rs 2,40,000 the C.I. for \( 2\frac{1}{2} \) years at \( 12\frac{1}{2} \)% will be Rs 82,734.37
In simple words: We work with the fractional rate 12½% by converting it to 25/2 %. The calculation method remains the same as previous problem.

📝 Teacher's Note: When rate is a mixed number like 12½%, convert to improper fraction first: 12½% = 25/2 %. This makes calculations easier.

🎯 Exam Tip: For mixed number rates, write clearly: "12½% = 25/2 % = 12.5%". Show this conversion step to avoid errors and get marks.

Answer 13.
Answer:
Given:
P = x; t = \( 1\frac{1}{2} \) years = 3 × 6 months; r = 16% compounded half-yearly = \( \frac{16}{2} \)% = 8%; A = Rs (x + 649.28)

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply formula for half-yearly compounding.
\( Rs(x + 649.28) = x \left(1 + \frac{8}{100}\right)^3 \)

Step 2: Calculate the compound factor.
\( Rs(x + 649.28) = x \times 1.08 \times 1.08 \times 1.08 \)
\( Rs(x + 649.28) = 1.259712x \)

Step 3: Solve for x.
\( 0.259712x = Rs 649.28 \)
\( x = Rs 2,500 \)

Final Answer: On Rs 2,500 the C.I. for \( 1\frac{1}{2} \) years at 16% compounded half-yearly will be Rs 649.28
In simple words: Half-yearly compounding means interest is calculated twice per year. So 16% yearly becomes 8% every 6 months. For 1½ years, we have 3 periods of 6 months each.

📝 Teacher's Note: Explain that "compounded half-yearly" means divide the rate by 2 and multiply the time by 2. So 16% for 1.5 years becomes 8% for 3 periods.

🎯 Exam Tip: For half-yearly compounding, always write: "Rate = Annual rate ÷ 2" and "Time = Years × 2". This shows you understand the concept clearly.

Answer 14.
Answer:
Given:
P = x; t = 2 years = 4 × 6 months; r = 10% compounded half-yearly = \( \frac{10}{2} \)% = 5%; A = Rs (x + 3,448.10)

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Apply formula for half-yearly compounding.
\( Rs(x + 3,448.10) = x \left(1 + \frac{5}{100}\right)^4 \)

Step 2: Calculate the compound factor.
\( Rs(x + 3,448.10) = x \times 1.05 \times 1.05 \times 1.05 \times 1.05 \)
\( Rs(x + 3,448.10) = 1.215506x \)

Step 3: Solve for x.
\( 0.215506x = Rs 3,448.10 \)
\( x = Rs 16,000.02 = Rs 16,000 \)

Final Answer: On Rs 16,000 the C.I. for 2 years at 10% compounded half-yearly will be Rs 3,448.10
In simple words: For 2 years with half-yearly compounding, we have 4 periods of 6 months each. The rate becomes 5% per period instead of 10% per year.

📝 Teacher's Note: Show students the pattern: yearly → half-yearly means rate ÷ 2 and time × 2. This applies to quarterly (÷4 and ×4) and monthly (÷12 and ×12) too.

🎯 Exam Tip: Always state clearly: "For half-yearly compounding: Rate = 5% per 6 months, Time = 4 periods". This shows your understanding step by step.

Answer 15.
Answer:
Given:
P = Rs 12,250; A = Rs (12,250 + 3,116.40) = Rs 15,366.40; t = 2 years; r = ?

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Substitute the given values.
\( Rs 15,366.40 = Rs 12250 \left(1 + \frac{r}{100}\right)^2 \)

Step 2: Simplify to find the rate factor.
\( \frac{15,366.40}{12250} = \left(1 + \frac{r}{100}\right)^2 \)
\( \frac{(196)^2}{(175)^2} = \left(1 + \frac{r}{100}\right)^2 \)

Step 3: Take square root and solve.
\( \frac{196}{175} = 1 + \frac{r}{100} \)
\( \frac{r}{100} = \frac{196}{175} - 1 = \frac{196 - 175}{175} = \frac{21}{175} \)
\( r = \frac{2100}{175} = 12\% \)

Final Answer: Hence, r = 12%
In simple words: We know the starting amount and final amount. We work backwards using the compound interest formula to find what rate was used.

📝 Teacher's Note: This is finding rate when P, A, and t are given. Students should recognize that they need to take square root when t = 2.

🎯 Exam Tip: When finding rate, always write the equation as (A/P) = (1 + r/100)ⁿ first. Then take the nth root to solve for r.

Answer 16.
Answer:
Given:
P = Rs 15,000; A = Rs (15,000 + 8,413.44) = Rs 23,413.44; t = 3 years; r = ?

Formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 1: Substitute the given values.
\( Rs 23,413.44 = Rs 15,000 \left(1 + \frac{r}{100}\right)^3 \)

Step 2: Simplify to find the rate factor.
\( \frac{23,413.44}{15,000} = \left(1 + \frac{r}{100}\right)^3 \)
\( \frac{(29)^3}{(25)^3} = \left(1 + \frac{r}{100}\right)^3 \)

Step 3: Take cube root and solve.
\( \frac{29}{25} = 1 + \frac{r}{100} \)
\( \frac{r}{100} = \frac{29}{25} - 1 = \frac{29 - 25}{25} = \frac{4}{25} \)
\( r = \frac{400}{25} = 16\% \)

Final Answer: Hence, r = 16%
In simple words: We use the same method as the previous problem, but since time is 3 years, we take cube root instead of square root.

📝 Teacher's Note: Emphasize that for t = 3, we take cube root; for t = 2, square root. The pattern is: take the nth root when time is n years.

🎯 Exam Tip: When t = 3, write clearly: "Taking cube root on both sides" before solving for r. This shows you understand the inverse operation needed.

Answer 17.
Answer:
Given:
P = Rs 16,000; A = Rs (16,000+3,876.75) = Rs 19,876.75; t = 3 years; r=?

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Substitute the values.
\( Rs19,876.75 = Rs16,000 \left(1 + \frac{r}{100}\right)^3 \)

Step 3: Simplify the equation.
\( \frac{19,876.75}{16,000} = \left(1 + \frac{r}{100}\right)^3 \)

\( \frac{(27.08)^3}{(25.19)^3} = \left(1 + \frac{r}{100}\right)^3 \)

Step 4: Take cube root on both sides.
\( \frac{2708}{2519} = 1 + \frac{r}{100} \)

Step 5: Solve for r.
\( \frac{r}{100} = \frac{2708}{2519} - 1 = \frac{2708 - 2519}{2519} = \frac{189}{2519} \)

\( r = \frac{18900}{2519} = 7.5\% \)

Hence, r = 7.5%
In simple words: We used the compound interest formula to find the rate. When money grows from Rs 16,000 to Rs 19,876.75 in 3 years, the rate is 7.5% per year.

📝 Teacher's Note: Show students how to convert the fraction to percentage by multiplying by 100. Many students forget this step and give the wrong answer.

🎯 Exam Tip: Always write the compound interest formula first. Then substitute values carefully. Show each step clearly to get full marks.

Answer 18.
Answer:
Given:
P = Rs 8,000; A = Rs 12,167; r = 15%; t=?

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Substitute the values.
\( Rs12,167 = Rs8,000 \left(1 + \frac{15}{100}\right)^t \)

Step 3: Simplify the equation.
\( \frac{12,167}{8,000} = \left(1 + \frac{15}{100}\right)^t \)

\( \frac{(23)^3}{(20)^3} = \left(\frac{23}{20}\right)^t \)

Step 4: Compare powers.
t = 3

T = 3 years
In simple words: We found how many years it takes for Rs 8,000 to become Rs 12,167 at 15% rate. The answer is 3 years.

📝 Teacher's Note: Help students see that 23/20 = 1.15, which is the same as (1 + 15/100). This makes the calculation easier.

🎯 Exam Tip: When you get the same base on both sides, the powers must be equal. This is the key to solving time problems.

Answer 19.
Answer:
Given:
P = Rs 50,000; A = Rs (50,000 + 32,151.60) = Rs 82,151.60; r = 18%; t=?

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Substitute the values.
\( Rs72,151.60 = Rs50,000 \left(1 + \frac{18}{100}\right)^t \)

Step 3: Simplify the equation.
\( \frac{82,151.60}{50,000} = \left(1 + \frac{18}{100}\right)^t \)

\( \frac{82,15160}{50,000 \times 100} = \left(1 + \frac{18}{100}\right)^t \)

\( \frac{2.05379}{1.25000} = \left(1 + \frac{18}{100}\right)^t \)

Step 4: Compare powers.
\( \frac{(59)^3}{(50)^3} = \left(\frac{59}{50}\right)^t \)

t = 3

T = 3 years
In simple words: We found how many years it takes for Rs 50,000 to become Rs 82,151.60 at 18% rate. The answer is 3 years.

📝 Teacher's Note: Show students that 59/50 = 1.18, which equals (1 + 18/100). This pattern helps them solve similar problems quickly.

🎯 Exam Tip: Always add principal and compound interest to get the amount. Don't use compound interest alone as the amount.

Answer 20.
Answer:
Given:
P = x; t = 2 years; r = \( 7\frac{1}{2}\% = \frac{15}{2}\% \)

For S.I.:
\( I = \frac{P \times r \times t}{100} \)

\( = \frac{x \times \frac{15}{2} \times 2}{100} \)

\( = \frac{3x}{20} \)

For C.I.:
\( C.I. = P \left(1 + \frac{r}{100}\right)^t - P \)

\( = x \left(1 + \frac{\frac{15}{2}}{100}\right)^2 - x \)

\( = x \left(1 + \frac{15}{2 \times 100}\right)^2 - x \)

\( = x \left(1 + \frac{3}{40}\right)^2 - x \)

= (x × 1.075 × 1.075) - x
= 1.155625x - x
= 0.155625x

Step 5: Use given condition.
Given C.I.- S.I. = Rs 22.50

\( \Rightarrow 0.155625x - \frac{3x}{20} = Rs22.50 \)

\( \Rightarrow 0.155625x - 0.15x = Rs22.50 \)

\( \Rightarrow 0.005625x = Rs22.50 \)

\( \Rightarrow x = Rs4,000 \)

Hence, sum = Rs 4,000
In simple words: We found the principal amount by comparing simple interest and compound interest. The difference between them helped us find the original sum.

📝 Teacher's Note: Explain that compound interest is always more than simple interest for the same principal, rate, and time. This difference helps us find unknown values.

🎯 Exam Tip: Calculate both S.I. and C.I. separately first. Then use their difference to find the unknown. Show all decimal calculations clearly.

Answer 21.
Answer:
Given:
P = x; t = 3 years; r = 12%

For S.I.:
\( I = \frac{P \times r \times t}{100} \)

\( = \frac{x \times 12 \times 3}{100} \)

\( = \frac{9x}{25} \)

For C.I.:
\( C.I. = P \left(1 + \frac{r}{100}\right)^t - P \)

\( = x \left(1 + \frac{12}{100}\right)^3 - x \)

\( = x \left(1 + \frac{3}{25}\right)^3 - x \)

= (x × 1.12 × 1.12 × 1.12) - x
= 1.404928x - x
= 0.404928x

Step 5: Use given condition.
Given C.I.-S.I. = Rs 22.50

\( \Rightarrow 0.404928x - \frac{9x}{25} = Rs1,123.20 \)

\( \Rightarrow 0.404928x - 0.36x = Rs1,123.20 \)

\( \Rightarrow 0.044928x = Rs1,123.20 \)

\( \Rightarrow x = Rs25,000 \)

Hence, sum = Rs 25,000
In simple words: We found the principal amount by comparing simple interest and compound interest for 3 years. The compound interest was more than simple interest by Rs 1,123.20.

📝 Teacher's Note: For 3 years, the difference between C.I. and S.I. becomes larger. Students should practice this type of calculation with different time periods.

🎯 Exam Tip: When time is 3 years, you must cube the rate factor (1 + r/100). Don't forget to subtract the principal at the end to get compound interest only.

Answer 22.
Answer:
Given:
P = x; r = ? ; t= 2 and 3 years; A = Rs 47,610 (2 years)
(3 years)

Step 1: Use compound interest formula.
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Step 2: Write equations for both years.
\( 47,610 = x \left(1 + \frac{r}{100}\right)^2 \).......(i)

\( 54,751.50 = x \left(1 + \frac{r}{100}\right)^3 \).......(ii)

Step 3: Divide equation (ii) by equation (i).
\( \frac{x \left(1 + \frac{r}{100}\right)^3}{x \left(1 + \frac{r}{100}\right)^2} = \frac{54,751.50}{47,610} \)

\( \Rightarrow \left(1 + \frac{r}{100}\right) = \frac{54,751.50}{47,610} \)

\( \Rightarrow \frac{r}{100} = \frac{54,751.50}{47,610} - 1 \)

\( \Rightarrow \frac{r}{100} = \frac{54,751.50 - 47,610}{47,610} \)

\( r = \frac{7141.50}{47,610} \times 100 \)

r = 15%

Step 4: Find principal using equation (i).
\( x \left(1 + \frac{r}{100}\right)^2 = Rs 47,610 \)

\( x \left(1 + \frac{15}{100}\right)^2 = Rs 47,610 \)

\( x \left(\frac{115}{100}\right)^2 = Rs 47,610 \)

1.3225x = Rs 47,610
x = Rs 36,000

The sum = Rs 36,000 and rate of interest = 15%
In simple words: We used the amounts after 2 and 3 years to find both the original sum and the interest rate. By dividing the two equations, we got the rate first.

📝 Teacher's Note: This is a powerful method. When you have amounts for two different years, you can find both principal and rate by dividing the equations.

🎯 Exam Tip: Always divide the equation with higher power by the one with lower power. This cancels out the principal and gives you the rate directly.

Answer 23.
Answer:
P = x; r = ?; t= 2 and 3 years; A = Rs 31,360 ( 2 years ) and Rs 35,123.20 ( 3 years)

\[ A = P\left(1+ \frac{r}{100}\right)^t \]

\[ 31,360 = x\left(1+ \frac{r}{100}\right)^2 \].......... \( (i) \)

\[ 35,123.20 = x\left(1+ \frac{r}{100}\right)^3 \].......... \( (ii) \)

\[ \frac{x\left(1+ \frac{r}{100}\right)^3}{x\left(1+ \frac{r}{100}\right)^2} = \frac{35,123.20}{31,360} \]

\[ \left(1+ \frac{r}{100}\right) = \frac{35,123.20}{31,360} \]

\[ \frac{r}{100} = \frac{35,123.20}{31,360} - 1 \]

\[ \frac{r}{100} = \frac{35,123.20 - 31,360}{31,360} \]

\[ r = \frac{3,763.20}{31,360} \times 100 \]

r = 12%

Using (i)

\[ x\left(1+ \frac{r}{100}\right)^2 = Rs31,360 \]

\[ x\left(1+ \frac{12}{100}\right)^2 = Rs31,360 \]

\[ x\left(\frac{112}{100}\right)^2 = Rs31,360 \]

1.2544x = Rs31,360
x = Rs25,000

The sum = Rs 25,000 and rate of interest = 12%

In simple words: We had two amounts after 2 years and 3 years. We divided them to find the growth rate per year. Then we found the original amount.

📝 Teacher's Note: Show students that dividing the amounts cancels out the principal. This leaves only the interest rate calculation. Use simple numbers like Rs 100 becoming Rs 110 to explain 10% rate.

🎯 Exam Tip: Always write the compound interest formula first. Show the division step clearly. Write the final answer with "sum" and "rate" separately. Check your answer by substituting back.

Answer 24.
Answer:
P = x; r = ?; t= 2 and 4 years; A = Rs 26,460 ( 2 years ) and Rs 29,172.15 ( 4 years )

\[ A = P\left(1+ \frac{r}{100}\right)^t \]

\[ 26,460 = x\left(1+ \frac{r}{100}\right)^2 \].......... \( (i) \)

\[ 29,172.15 = x\left(1+ \frac{r}{100}\right)^4 \].......... \( (ii) \)

\[ \frac{x\left(1+ \frac{r}{100}\right)^4}{x\left(1+ \frac{r}{100}\right)^2} = \frac{29,172.15}{26,460} \]

\[ \left(1+ \frac{r}{100}\right)^2 = \frac{1,94,481}{1,76,400} \]

\[ \left(1+ \frac{r}{100}\right)^2 = \left(\frac{441}{420}\right)^2 \]

\[ 1+ \frac{r}{100} = \frac{441}{420} \]

\[ \frac{r}{100} = \frac{441}{420} - 1 \]

\[ \frac{r}{100} = \frac{441-420}{420} \]

\[ r = \frac{21}{420} \times 100 \]

r = 5%

Using (i)

\[ x\left(1+ \frac{r}{100}\right)^2 = Rs26,460 \]

\[ x\left(1+ \frac{5}{100}\right)^2 = Rs26,460 \]

\[ x\left(\frac{105}{100}\right)^2 = Rs26,460 \]

1.1025x = Rs26,460
x = Rs24,000

The sum = Rs 24,000 and rate of interest = 5%

In simple words: The amount in 4 years is the amount in 2 years grown for 2 more years. We use this to find the rate and original sum.

📝 Teacher's Note: Explain that 4 years = 2 years + 2 years. So we can use the 2-year amount as a new principal. This makes the calculation easier for students.

🎯 Exam Tip: When you have amounts after different years, divide to eliminate the principal. Take square root if needed. Show each step clearly for full marks.

Answer 25.
Answer:
P = x; t= 2 years; r = 5%; A= Rs (x+512.50)

\[ A = P\left(1+ \frac{r}{100}\right)^t \]

\[ Rs(x + 512.50) = x\left(1+ \frac{5}{100}\right)^2 \]

Rs(x + 512.50) = x × 1.05 × 1.05
Rs(x + 512.50) = 1.1025x
0.1025x = Rs512.50
x = Rs5,000

\[ I = \frac{P \times r \times t}{100} \]

\[ I = Rs\frac{5,000 \times 6 \times 3}{100} \]

I = Rs900

Simple interest will be Rs 900

In simple words: The compound interest for 2 years is Rs 512.50. We found the principal is Rs 5,000. Then we calculated simple interest for 6% rate over 3 years.

📝 Teacher's Note: Show students that compound interest = Amount - Principal. Use this to set up the equation. Point out that 1.05 × 1.05 = 1.1025.

🎯 Exam Tip: Write "A = P + CI" to show compound interest. Set up the equation carefully. Use the simple interest formula \( I = \frac{PRT}{100} \) for the second part.

Answer 26.
Answer:
P = x; t= 3 years; r = 10%; A= Rs (x+4,965)

\[ A = P\left(1+ \frac{r}{100}\right)^t \]

\[ Rs(x + 4,965) = x\left(1+ \frac{10}{100}\right)^3 \]

Rs(x + 4,965) = x × 1.1 × 1.1 × 1.1
Rs(x + 4,965) = 1.331x
0.331x = Rs4,965
x = Rs15,000

\[ I = \frac{P \times r \times t}{100} \]

\[ I = Rs\frac{15,000 \times 11 \times 3}{100} \]

I = Rs4,950

Simple interest will be Rs 4,950

In simple words: The compound interest for 3 years is Rs 4,965. We found the principal is Rs 15,000. Then we calculated simple interest for 11% rate over 3 years.

📝 Teacher's Note: Show students that 1.1³ means 1.1 × 1.1 × 1.1 = 1.331. This is how money grows with compound interest. Each year it grows by 10% of the new total.

🎯 Exam Tip: Calculate 1.1³ step by step to avoid errors. Write the compound interest formula first. For simple interest, use the given rate (11%) not the compound rate (10%).

Exercise 1.7

Answer 1.
Answer:
V₀ = ?; V₀ = 4,25, 000; r = 4%; t = 2 years

\[ V_n = V_0\left(1+ \frac{r}{100}\right)^t \]

\[ V_n = 4,25,000\left(1+ \frac{4}{100}\right)^2 \]

V_n = 4,25,000 × 1.04 × 1.04
V_n = 4,59,680

The population in 2007 is 4, 59,680

In simple words: The population grows by 4% each year. After 2 years, we get 4,59,680 people.

📝 Teacher's Note: This is like compound interest but for population growth. Each year, 4% more people are added to the total population from the previous year.

🎯 Exam Tip: Use the same formula as compound interest. Replace P with V₀ and A with V_n. Show the calculation step by step.

Answer 2.
Answer:
V₀ = ?; V₀ = 1, 25,000; r = 5.5% (birth) and 3.5% (death); t = 3 years

\[ V_n = V_0\left(1+ \frac{r}{100}\right)^t \]

\[ V_n = 1,25,000\left(1+ \frac{5.5}{100}\right)^3\left(1- \frac{3.5}{100}\right)^3 \]

V_n = 1,25,000 × 10.55 × 10.55 × 10.55 × 9.65 × 9.65 × 9.65
V_n = 1,25,000 × 1174.241 × 898.6321
V_n = 1,32,651

The population in 2007 is 1, 32,651

In simple words: Birth rate adds people, death rate removes people. We calculate both effects together to find the final population.

📝 Teacher's Note: Explain that birth rate uses (1 + r/100) because it adds people. Death rate uses (1 - r/100) because it removes people. Both happen at the same time.

🎯 Exam Tip: For birth rate, use plus sign. For death rate, use minus sign. Apply both rates for the full time period. Be careful with the arithmetic.

Answer 3.
Answer:
Rate of increase =

\[ r = \frac{50}{1000} \times 100 = 5\% \]

V₀ = 22,050; V_n = ?; r = 5%; t = 2 years

\[ V_n = V_0\left(1+ \frac{r}{100}\right)^t \]

\[ 22,050 = V_0\left(1+ \frac{5}{100}\right)^2 \]

22,050 = V₀ × 1.05 × 1.05

\[ V_0 = \frac{22,050}{1.1025} \]

V₀ = 20,000

The present population is 20,000.

In simple words: We know the population after 2 years growth. We work backwards to find what it was originally.

📝 Teacher's Note: This is reverse compound interest. We divide by (1 + r/100)² instead of multiplying. Show students this is like finding the original price before a discount.

🎯 Exam Tip: First find the growth rate from the given information. Then work backwards from future to present population. Check your answer by calculating forward.

Answer 4.
Answer:
V₀ = 46,305; V_n = 40,000; r = ? ; t = 3 years

\[ V_n = V_0\left(1+ \frac{r}{100}\right)^t \]

\[ 46,305 = 40,000\left(1+ \frac{r}{100}\right)^3 \]

\[ \frac{46,305}{40,000} = \left(1+ \frac{r}{100}\right)^3 \]

\[ \frac{21^3}{20^3} = \left(1+ \frac{r}{100}\right)^3 \]

\[ \left(1+ \frac{r}{100}\right) = \frac{21}{20} \]

\[ \frac{r}{100} = \frac{21}{20} - 1 \]

\[ \frac{r}{100} = \frac{1}{20} \]

\[ r = \frac{1}{20} \times 100 \]

r = 5%

The annual rate of growth of scooters is 5%.

In simple words: The production grew from 40,000 to 46,305 in 3 years. We found the yearly growth rate is 5%.

📝 Teacher's Note: This requires finding cube roots. Show students how 46,305 = 21³ × 10 and 40,000 = 20³ × 10. This makes the cube root calculation easier.

🎯 Exam Tip: Set up the equation correctly with known values. Take cube root of both sides. Simplify fractions before taking roots to make calculation easier.

Answer 5.
Answer:
V₀ = ? ; V₀ = 1, 15,200; r = 6\(\frac{2}{3}\)% = \(\frac{20}{3}\)% ; t = 2 years

\[ V_n = V_0\left(1+ \frac{r}{100}\right)^t \]

\[ V_n = 1,15,200\left(1+ \frac{20}{100 \times 3}\right)^2 \]

V_n = 1,15,200 × 1.06667 × 1.06667
V_n = 1,31,072

The population 2 years later = 1, 31,072

(ii) Its population 2 years ago.

V₀ = ? ; V₀ = 1, 15,200; r = 6\(\frac{2}{3}\)% = \(\frac{20}{3}\)% ; t = 2 years

\[ V_n = V_0\left(1- \frac{r}{100}\right)^t \]

\[ V_n = 1,15,200\left(1- \frac{20}{100 \times 3}\right)^2 \]

V_n = 1,15,200 × 0.933333 × 0.933333
V_n = 1,00,352

The population 2 years ago was = 1, 00,352

In simple words: For future population, we add the growth rate. For past population, we subtract the growth rate. Both use 2 years time.

📝 Teacher's Note: Show students that going forward in time uses plus sign, going backward in time uses minus sign. Convert mixed percentages to improper fractions first.

🎯 Exam Tip: Convert \(6\frac{2}{3}\)% to \(\frac{20}{3}\)% first. For future, use (1 + r/100). For past, use (1 - r/100). Write both answers clearly.

Answer 6.
Answer:
Given:
\( V_n = \) Rs 19,083.60; \( V_o = ? \); \( r = 10\% \); \( t = 2 \) years

Step 1: Use the depreciation formula.
\( V_n = V_o \left(1 - \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
\( 19,083.60 = V_o \left(1 - \frac{10}{100}\right)^2 \)

Step 3: Calculate the depreciation factor.
\( V_o = 19,083.60 \times \frac{10}{9} \times \frac{10}{9} \)
\( V_o = \) Rs 23,560

Step 4: Find the difference.
The machine was purchased for Rs 23,560 i.e. Rs (23,560 - 19083.60) = Rs 4,476.40 more than the present value.

📝 Teacher's Note: Show students that we work backwards in depreciation problems. We know the final value and need to find the original value. Think of it like finding how much a car cost new when you know its current value.

🎯 Exam Tip: Always write the formula first. Use the exact formula given in your textbook. Show each calculation step clearly for full marks.

Answer 7.
Answer:
Given:
\( V_n = 27,783 \); \( V_o = 24,000 \); \( r = ? \); \( t = 3 \) years

Step 1: Use the compound growth formula.
\( V_n = V_o \left(1 + \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
\( 27,783 = 24,000 \left(1 + \frac{r}{100}\right)^3 \)

Step 3: Solve for the growth factor.
\( \frac{27,783}{24,000} = \left(1 + \frac{r}{100}\right)^3 \)
\( \frac{21^3}{20^3} = \left(1 + \frac{r}{100}\right)^3 \)

Step 4: Take cube root and solve for r.
\( \left(1 + \frac{r}{100}\right) = \frac{21}{20} \)
\( \frac{r}{100} = \frac{21}{20} - 1 \)
\( \frac{r}{100} = \frac{1}{20} \)
\( r = \frac{1}{20} \times 100 \)
\( r = 5\% \)

The rate of growth of population is 5%.

📝 Teacher's Note: In population growth problems, we use compound interest formula because growth happens on the new population each year. It is like interest earning more interest.

🎯 Exam Tip: Write "rate of growth" clearly in your final answer. Show the cube root step properly. Practice recognizing when numbers are perfect cubes like 21³/20³.

Answer 8.
Answer:
Given:
\( V_n = 27,040 \); \( V_o = 25,000 \); \( r = ? \); \( t = 2 \) years

Step 1: Use the compound growth formula.
\( V_n = V_o \left(1 + \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
\( 27,040 = 25,000 \left(1 + \frac{r}{100}\right)^2 \)

Step 3: Simplify the fraction.
\( \frac{27,040}{25,000} = \left(1 + \frac{r}{100}\right)^2 \)
\( \left(\frac{164.43}{158}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \)

Step 4: Take square root and solve for r.
\( \left(1 + \frac{r}{100}\right) = \frac{164.43}{158} \)
\( \frac{r}{100} = \frac{164}{158} - 1 \)
\( \frac{r}{100} = \frac{6.43}{158} \)
\( r = 0.040 \times 100 \)
\( r = 4\% \)

The rate of growth in number of villages with electricity is 4%.

📝 Teacher's Note: When dealing with decimals, round carefully at each step. Check your calculation by putting the answer back into the original formula to verify.

🎯 Exam Tip: Convert the decimal answer to percentage properly. Write "4%" not "0.04" for the final answer. Show the square root step clearly.

Answer 9.
Answer:
Given:
\( V_n = ? \); \( V_o = \) Rs 4,00,000; \( r = 10\% \); \( t = 4 \) years

Step 1: Use the depreciation formula.
\( V_n = V_o \left(1 - \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
\( V_n = 4,00,000 \left(1 - \frac{10}{100}\right)^4 \)

Step 3: Calculate step by step.
\( V_n = 4,00,000 \times 0.9 \times 0.9 \times 0.9 \times 0.9 \)
\( V_n = \) Rs 2,62,440

The value of car after four years will be Rs 2,62,440.

📝 Teacher's Note: In depreciation, the value goes down each year. Show students that 0.9 means 90% of value remains (10% is lost). Multiply 0.9 four times because it happens for 4 years.

🎯 Exam Tip: Write the final answer clearly with "Rs" and proper comma placement. Show the calculation of (0.9)⁴ step by step or use the formula directly.

Answer 10.
Answer:
Given:
\( V_n = \) Rs 44,540; \( V_o = ? \); \( r = 5\% \); \( t = 3 \) years

Step 1: Use the depreciation formula.
\( V_n = V_o \left(1 - \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
Rs 44,540 = \( V_o \left(1 - \frac{5}{100}\right)^3 \)

Step 3: Calculate the original value.
\( V_o = 44,540 \times \frac{100}{95} \times \frac{100}{95} \times \frac{100}{95} \)
\( V_o = 44,540 \times 1.052632 \times 1.052632 \times 1.052632 \)
\( V_o = \) Rs 51,949.26

The original value of the property was Rs 51,949.26

📝 Teacher's Note: This is a reverse depreciation problem. We know the final value and need to find the starting value. Work backwards by dividing instead of multiplying.

🎯 Exam Tip: In reverse problems, use the formula but solve for the unknown variable. Round the final answer to 2 decimal places for money calculations.

Answer 11.
Answer:
Given:
\( V_n = \) Rs 9,680; \( V_o = ? \); \( r = 12\% \); \( t = 2 \) years

Step 1: Use the depreciation formula.
\( V_n = V_o \left(1 - \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
Rs 9,680 = \( V_o \left(1 - \frac{12}{100}\right)^2 \)

Step 3: Calculate the original value.
\( V_o = 9,680 \times \frac{100}{88} \times \frac{100}{88} \)
\( V_o = 9,680 \times 1.136364 \times 1.136364 \)
\( V_o = \) Rs 12,500

The refrigerator was purchased for Rs 12,500

📝 Teacher's Note: Remember that 12% depreciation means 88% value remains. So we multiply by 100/88 to go backwards. Think of it like undoing the depreciation.

🎯 Exam Tip: Check your answer by calculating forward - if you start with Rs 12,500 and depreciate 12% for 2 years, you should get Rs 9,680 back.

Answer 12.
Answer:
For the building:
\( V_n = V_o \left(1 - \frac{r}{100}\right)^t \)
\( V_n = 1,33,100 \left(1 - \frac{10}{100}\right)^t \)
\( V_n = 1,33,100 \times (0.9)^t \)

For the plot:
\( V_n = V_o \left(1 + \frac{r}{100}\right)^t \)
\( V_n = 72,900 \left(1 + \frac{10}{100}\right)^t \)
\( V_n = 72,900 \times (1.1)^t \)

Since, value becomes same:
\( 1,33,100 \times (0.9)^t = 72,900 \times (1.1)^t \)
\( \frac{(1.1)^t}{(0.9)^t} = \frac{1,33,100}{72,900} \)
\( \frac{(11)^t}{(09)^t} = \frac{1331}{729} = \frac{11^3}{9^3} \)
\( t = 3 \)

Hence, after 3 years value of both will be same.

📝 Teacher's Note: This problem combines appreciation and depreciation. One value goes up, one goes down. They meet at some point in time. Set up two equations and solve.

🎯 Exam Tip: Write both formulas separately first. Then set them equal. Notice that 1331/729 = 11³/9³, so t = 3. This pattern helps solve quickly.

Answer 13.
Answer:
Given:
\( V_n = ? \); \( V_o = \) Rs 17,000; \( t = 2 \) years (1 for increment and 1 for decrement);
\( r = 5\% \) for increase and \( 4\% \) for decrease.

Step 1: Apply both changes using the combined formula.
\( V_n = V_o \left(1 + \frac{r}{100}\right)^1 \left(1 - \frac{r}{100}\right)^1 \)

Step 2: Substitute the values.
\( V_n = 17,000 \left(1 + \frac{5}{100}\right) \left(1 - \frac{4}{100}\right) \)
\( V_n = 17,000 \times 1.05 \times 0.96 \)
\( V_n = \) Rs 17,136

The cost of the T.V. in 2001 is Rs 17,136.

📝 Teacher's Note: When there are two different changes in sequence, apply them one after the other. First increase by 5%, then decrease by 4% on the new value.

🎯 Exam Tip: Be careful about which change happens first. Read the problem carefully. Apply changes in the correct order - sequence matters in these problems.

Answer 14.
Answer:
Given:
\( V_n = 1 \) m 8cm = 108 cm; \( V_o = ? \); \( t = 2 \) years; \( r = 20\% \)

Step 1: Use the growth formula (tree height increases).
\( V_n = V_o \left(1 + \frac{r}{100}\right)^t \)

Step 2: Substitute the values.
108cm = \( V_o \left(1 + \frac{20}{100}\right)^2 \)

Step 3: Calculate the original height.
\( V_o = 108cm \times 0.8333 \times 0.8333 \times 0.8333 \)
\( V_o = 62.5cm \)

The height of tree was 62.5 cm when planted.

📝 Teacher's Note: Tree growth is like compound interest - each year the tree grows 20% more than its current height. So a bigger tree grows more than a smaller tree.

🎯 Exam Tip: Convert units properly - 1m 8cm = 108cm. Work in the same units throughout. Write your final answer with the correct unit (cm).