RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest

Exercise 12A

Question 1. Find the simple interest and amount when principal is Rs. 6400, rate is 6% per annum, and time is 2 years.
Answer: Given: P = Rs. 6400, R = 6%, T = 2 years
Using the formula: S.I. = \( \frac{P \times R \times T}{100} = \frac{6400 \times 6 \times 2}{100} = \text{Rs. } 768 \)
Amount = P + S.I. = 6400 + 768 = Rs. 7168
In simple words: Calculate the interest earned using P times R times T divided by 100, then add it to the principal to get the total amount.

Exam Tip: Always identify which values you have and which you need to find. The formula S.I. = PRT/100 is your foundation - memorise it.

Question 2. Find the amount when principal is Rs. 2650, rate is 8% per annum, and time is 2.5 years.
Answer: Given: P = Rs. 2650, R = 8%, T = 2\( \frac{1}{2} \) years = \( \frac{5}{2} \) years
S.I. = \( \frac{P \times R \times T}{100} = \frac{2650 \times 8 \times 5}{100 \times 2} = \text{Rs. } 530 \)
Amount = P + S.I. = 2650 + 530 = Rs. 3180
In simple words: Convert the mixed time into an improper fraction first, then apply the simple interest formula and add the interest to the principal.

Exam Tip: When time is given as a mixed number or in months/days, convert it to years first before substituting into the formula.

Question 3. Find the amount when principal is Rs. 1500, rate is 12% per annum, and time is 3\( \frac{3}{12} \) years.
Answer: Given: P = Rs. 1500, R = 12%, T = 3 + \( \frac{3}{12} = \frac{13}{4} \) years
S.I. = \( \frac{P \times R \times T}{100} = \frac{1500 \times 12 \times 13}{100 \times 4} = \text{Rs. } 585 \)
Amount = P + S.I. = 1500 + 585 = Rs. 2085
In simple words: Simplify the fractional time first, calculate the interest using the formula, then find the total amount.

Exam Tip: Verify your conversion of time - use a common denominator when simplifying fractions of years.

Question 4. Find the amount when principal is Rs. 9600, rate is 7.5% per annum, and time is 5 months.
Answer: Given: P = Rs. 9600, R = 7\( \frac{1}{2} \)%, T = 5 months = \( \frac{5}{12} \) years
S.I. = \( \frac{P \times R \times T}{100} = \frac{9600 \times 15 \times 5}{100 \times 2 \times 12} = \text{Rs. } 300 \)
Amount = P + S.I. = 9600 + 300 = Rs. 9900
In simple words: Convert months into years by dividing by 12, then use the standard interest formula.

Exam Tip: Always convert time to years - if given in months, divide by 12; if in days, divide by 365.

Question 5. Find the amount when principal is Rs. 5000, rate is 9% per annum, and time is 146 days.
Answer: Given: P = Rs. 5000, R = 9%, T = 146 days = \( \frac{146}{365} \) years
S.I. = \( \frac{P \times R \times T}{100} = \frac{5000 \times 9 \times 146}{100 \times 365} = \text{Rs. } 180 \)
Amount = P + S.I. = 5000 + 180 = Rs. 5180
In simple words: Express the number of days as a fraction of 365, then substitute into the interest formula.

Exam Tip: When days are given, always use 365 as the divisor to convert to years (unless a leap year is specifically mentioned).

Question 6. Find the time when principal is Rs. 6400, simple interest is Rs. 1152, and rate is 6% per annum.
Answer: Given: P = Rs. 6400, S.I. = Rs. 1152, R = 6%
Using T = \( \frac{S.I. \times 100}{P \times R} = \frac{1152 \times 100}{6400 \times 6} = \frac{1152}{384} = 3 \) years
In simple words: Rearrange the simple interest formula to isolate time, then substitute the known values and solve.

Exam Tip: When finding time, remember to use the rearranged formula T = (S.I. × 100)/(P × R) - this is essential for inverse problems.

Question 7. Find the time when principal is Rs. 9540, simple interest is Rs. 1908, and rate is 8% per annum.
Answer: Given: P = Rs. 9540, S.I. = Rs. 1908, R = 8%
T = \( \frac{S.I. \times 100}{P \times R} = \frac{1908 \times 100}{9540 \times 8} = \frac{10}{8} = 2\frac{1}{4} \) years
In simple words: Apply the rearranged time formula, substitute the values, and simplify to get the answer in years.

Exam Tip: Express the final answer as a mixed number if it comes out as an improper fraction - this shows clear understanding of the concept.

Question 8. Find the time when principal is Rs. 5000, amount is Rs. 6450, and rate is 12% per annum.
Answer: Given: P = Rs. 5000, A = Rs. 6450, R = 12%
S.I. = A - P = 6450 - 5000 = Rs. 1450
T = \( \frac{S.I. \times 100}{P \times R} = \frac{1450 \times 100}{5000 \times 12} = \frac{20}{12} = 2\frac{5}{12} \) years = 2 years 5 months
In simple words: First find the simple interest by subtracting principal from amount, then use the time formula to calculate how long the money was invested.

Exam Tip: When amount is given instead of simple interest, always subtract principal first. Convert fractional years back to years and months for the final answer.

Question 9. Find the rate when principal is Rs. 8250, simple interest is Rs. 1100, and time is 2 years.
Answer: Given: P = Rs. 8250, S.I. = Rs. 1100, T = 2 years
R = \( \frac{S.I. \times 100}{P \times T} = \frac{1100 \times 100}{8250 \times 2} = \frac{1100}{165} = 6.67\% \)
In simple words: Rearrange the simple interest formula to solve for the rate by dividing simple interest times 100 by principal times time.

Exam Tip: Use the formula R = (S.I. × 100)/(P × T) when finding rate. Round the final answer appropriately, showing at least two decimal places for accuracy.

Question 10. Find the rate when principal is Rs. 5200, simple interest is Rs. 975, and time is 2.5 years.
Answer: Given: P = Rs. 5200, S.I. Rs. 975, T = 2\( \frac{1}{2} \) years = \( \frac{5}{2} \) years
R = \( \frac{S.I. \times 100}{P \times T} = \frac{975 \times 100 \times 2}{5200 \times 5} = \frac{195}{26} = 7.5\% \)
In simple words: Apply the rate formula, substitute all values including the converted time fraction, then simplify to find the percentage.

Exam Tip: When time is fractional, ensure you substitute it correctly into the denominator. Double-check your calculation by working backwards - verify that P × R × T / 100 equals the given S.I.

Question 11. Find the principal when simple interest is Rs. 900, rate is 9% per annum, and time is 3 years.
Answer: Given: S.I. = Rs. 900, R = 9%, T = 3 years
P = \( \frac{S.I. \times 100}{R \times T} = \frac{900 \times 100}{9 \times 3} = \frac{90000}{27} = \text{Rs. } 3333.33 \)
In simple words: Rearrange the simple interest formula to isolate principal, substitute the known values, and perform the division.

Exam Tip: When finding principal, use P = (S.I. × 100)/(R × T). If the answer is not a whole number, express it as a decimal or fraction, and verify by substituting back into the original formula.

Question 12. Find the principal when amount is Rs. 4521.20, rate is 9% per annum, and time is 3 years.
Answer: Given: A = Rs. 4521.20, R = 9%, T = 3 years
S.I. = A - P = 4521.20 - P
Substituting into the formula: 4521.20 - P = \( \frac{P \times 9 \times 3}{100} \)
4521.20 = P + \( \frac{27P}{100} \) = P\( (1 + 0.27) \) = 1.27P
P = \( \frac{4521.20}{1.27} = \text{Rs. } 3560 \)
R = \( \frac{S.I. \times 100}{P \times T} = \frac{961.20 \times 100}{3560 \times 3} = 9\% \)
In simple words: Express simple interest as the difference between amount and principal, then substitute this relationship into the formula and solve for principal.

Exam Tip: When amount is given, set up the equation A = P + (PRT/100) and solve for P algebraically - this avoids circular reasoning.

Question 13. Find the principal and amount when simple interest is Rs. 2640, rate is 12% per annum, and time is 3 years 8 months.
Answer: Given: S.I. = Rs. 2640, R = 12%, T = 3 years 8 months = 3\( \frac{8}{12} = \frac{44}{12} \) years
P = \( \frac{S.I. \times 100}{R \times T} = \frac{2640 \times 100}{12 \times 44} = \text{Rs. } 6000 \)
A = P + S.I. = 6000 + 2640 = Rs. 8640
In simple words: Convert the time to a fraction of years, use the rearranged formula to find principal, then add the simple interest to get the amount.

Exam Tip: Always convert time given in years and months into a single fractional form before calculation. Check that your principal value, when multiplied by rate and time then divided by 100, gives exactly the stated simple interest.

Question 14. What principal will yield a simple interest of Rs. 829.50 in 3 years at 10% per annum?
Answer: Given: S.I. = Rs. 829.50, T = 3 years, R = 10%
Using the rearranged formula: P = \( \frac{S.I. \times 100}{R \times T} = \frac{829.50 \times 100}{10 \times 3} = \frac{82950}{30} = \text{Rs. } 2765 \)
Hence, the sum is Rs. 2765.
In simple words: Apply the principal formula P = (S.I. × 100)/(R × T) with the given values to find the starting amount that produces the specified interest.

Exam Tip: Verify your answer by calculating S.I. = (2765 × 10 × 3)/100 to confirm you get Rs. 829.50.

Question 15. What principal will amount to Rs. 3920 in 3 years at 7.5% per annum?
Answer: Given: A = Rs. 3920, R = 7\( \frac{1}{2} \)%, T = 3 years
Using A = P\( (1 + \frac{R \times T}{100}) \), we get: 3920 = P\( (1 + \frac{7.5 \times 3}{100}) = P(1 + 0.225) = 1.225P \)
P = \( \frac{3920}{1.225} = \text{Rs. } 3200 \)
Therefore, the required sum is Rs. 3200.
In simple words: Use the amount formula rearranged to find principal by dividing the amount by the factor (1 plus the rate-time product divided by 100).

Exam Tip: When solving for principal from amount, use the direct formula P = A / (1 + RT/100) rather than finding S.I. separately - it's faster and reduces error.

Question 16. A man borrowed some money at 11% per annum for 2 years 3 months. He paid Rs. 4491 as amount. Find the principal and simple interest.
Answer: Given: R = 11%, T = 2 years 3 months = 2\( \frac{3}{12} = \frac{27}{12} \) years, A = Rs. 4491
Using A = P\( (1 + \frac{R \times T}{100}) \): 4491 = P\( (1 + \frac{11 \times 27}{100 \times 12}) = P(1 + \frac{99}{400}) = P(\frac{499}{400}) \)
P = \( \frac{4491 \times 400}{499} = \text{Rs. } 3600 \)
S.I. = A - P = 4491 - 3600 = Rs. 1188
In simple words: Convert time to years as a fraction, substitute into the amount formula, then solve for principal algebraically and find the interest by subtraction.

Exam Tip: Always verify: check that P + (P × R × T / 100) equals the given amount, confirming your principal value is correct.

Question 17. At what rate will Rs. x amount to Rs. 12,122 in 3 years 3 months?
Answer: Given: P = Rs. x, A = Rs. 12,122, T = 3 years 3 months = 3\( \frac{1}{4} = \frac{13}{4} \) years
Using A = P\( (1 + \frac{R \times T}{100}) \): 12,122 = x\( (1 + \frac{R \times 13}{100 \times 4}) = x(1 + \frac{13R}{400}) \)
12,122 = x + \( \frac{13Rx}{400} \)
Dividing by x: \( \frac{12,122}{x} = 1 + \frac{13R}{400} \)
S.I. = A - P = 12,122 - x
Since S.I. = \( \frac{P \times R \times T}{100} \): 12,122 - x = \( \frac{x \times R \times 13}{400} \)
Using the given amount and principal, we can solve: 10,450 = \( \frac{13Rx}{400} \)
Therefore: R = \( \frac{10,450 \times 400}{13x} \) = 2.5%, given that x = 10,450
Now, S.I. = \( \frac{10,450 \times R \times 13}{400} = \text{Rs. } 2508 \)
∴ A = P + S.I. = Rs. 10,450 + Rs. 2508 = Rs. 12,958
In simple words: Set up the amount equation with the unknown principal and rate, then use the relationship between amount, principal, and simple interest to solve for the rate algebraically.

Exam Tip: When the principal is not explicitly given, express your rate answer in terms of the principal, or note that additional information would be needed to find a numerical rate.

Question 18. Find the rate of interest if Rs. 3600 amounts to Rs. 4734 in 3.5 years.
Answer: Given: P = Rs. 3600, A = Rs. 4734, T = 3\( \frac{1}{2} = \frac{7}{2} \) years
S.I. = A - P = 4734 - 3600 = Rs. 1134
R = \( \frac{S.I. \times 100}{P \times T} = \frac{1134 \times 100 \times 2}{3600 \times 7} = \frac{226,800}{25,200} = 9\% \)
In simple words: Find simple interest by subtracting principal from amount, then apply the rate formula with the converted time in years.

Exam Tip: Always find S.I. first when amount is given - this makes calculating the rate straightforward using the standard formula R = (S.I. × 100)/(P × T).

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Exam Tip: Review your work once complete to ensure all given values have been correctly identified and substituted.

Question 20. A man deposits Rs. 640 and gets Rs. 768 after 2 years 6 months. He then deposits Rs. 850 and gets Rs. 1054 after 3 years. Find the rate of interest for each deposit.
Answer: For the first deposit:
P = Rs. 640, A = Rs. 768, T = 2 years 6 months = \( \frac{5}{2} \) years
S.I. = A - P = 768 - 640 = Rs. 128
R = \( \frac{S.I. \times 100}{P \times T} = \frac{128 \times 100 \times 2}{640 \times 5} = 8\% \)

For the second deposit:
P = Rs. 850, A = Rs. 1054, T = 3 years
S.I. = A - P = 1054 - 850 = Rs. 204
R = \( \frac{S.I. \times 100}{P \times T} = \frac{204 \times 100}{850 \times 3} = 8\% \)
In simple words: For each deposit, calculate the simple interest earned, then use the rate formula to find the percentage for that investment period.

Exam Tip: When dealing with multiple deposits or investments, treat each one separately and calculate the rate for each transaction using the same method.

Question 21. Find the rate of interest if a sum of money becomes 1.2 times itself in 5 years.
Answer: Let the sum be Rs. x.
Amount = 1.2x
∴ S.I. = A - P = 1.2x - x = 0.2x

Let the rate be R%.
S.I. = \( \frac{P \times R \times T}{100} \)

\( \frac{0.2x}{1} = \frac{x \times R \times 5}{100} \)

\( \Rightarrow 3x \times 20 = R \times x \times 5 \)

\( \Rightarrow R = \frac{3x \times 20}{x \times 5} = 12 \)

Hence, the rate of interest is 12%.
In simple words: Express the simple interest in terms of the original principal (the difference between the new amount and principal), then substitute into the rate formula and solve.

Exam Tip: When a multiple of the principal is given (like 1.2 times), the actual sum cancels out in the calculation - this is a powerful shortcut that works for all such problems.

Question 22. A sum of money amounts to Rs. 837 in 3 years and Rs. 783 in 2 years. Find the principal and the rate of interest.
Answer: Amount in 3 years = (Principal + S.I. for 3 years) = Rs. 837
Amount in 2 years = (Principal + S.I. for 2 years) = Rs. 783

On subtracting:
S.I. for 1 year = (837 - 783) = Rs. 54
S.I. for 2 years = \( (54 \times 2) \) = Rs. 108
∴ Sum = Amount for 2 years - S.I. for 2 years = 783 - 108 = Rs. 675

P = Rs. 675, S.I. = Rs. 108 and T = 2 years
R = \( \frac{S.I. \times 100}{P \times T} = \frac{108 \times 100 \times 1}{675 \times 2} = 8\% \)
In simple words: Find the simple interest for one year by subtracting the two amounts, then use that to find principal and rate step by step.

Exam Tip: The difference between amounts at two different times gives you the interest earned in the intervening period - this is the key insight for solving such problems efficiently.

Question 23. A sum of money amounts to Rs. 5475 in 5 years and Rs. 4745 in 3 years. Find the principal and the rate of interest.
Answer: Amount in 5 years = (Principal + S.I. for 5 years) = Rs. 5475
Amount in 3 years = (Principal + S.I. for 3 years) = Rs. 4745

On subtracting:
S.I. for 2 years = (5475 - 4745) = Rs. 730
S.I. for 3 years = \( (\frac{730}{2} \times 3) \) = Rs. 1095
∴ Sum = Amount for 3 years - S.I. for 3 years = 4745 - 1095 = Rs. 3650

P = Rs. 3650, S.I. = Rs. 1095, T = 3 years
R = \( \frac{S.I. \times 100}{P \times T} = \frac{1095 \times 100}{3650 \times 3} = 10\% \)
In simple words: Find the interest earned in a 2-year interval, use that to calculate interest for any other period, then determine principal and rate using standard formulas.

Exam Tip: When you know the interest for one time period, you can scale it proportionally for any other period since simple interest is linear with time.

Question 24. A sum of Rs. 3000 is divided into two parts. One part is invested at 8% per annum for 4 years and the other at 9% per annum for 4 years. The total simple interest is Rs. 1020. Find the two parts.
Answer: Let the first part be Rs. x.
Second part = (3000 - x)

∴ S.I. on x at 8% per annum for 4 years = \( \frac{x \times 8 \times 4}{100} = \frac{8x}{25} \)

S.I. on (3000 - x) at 9% per annum for 4 years = \( \frac{(3000 - x) \times 9 \times 4}{100} = \frac{27000 - 9x}{50} \)

\( \Rightarrow \frac{8x}{25} = \frac{27000 - 9x}{50} = 1020 \)

\( \Rightarrow 8x + 27000 - 9x = 51000 \)
\( \Rightarrow 16x + 9r = 27000 \)
\( \Rightarrow 16x + 9r = 27000 \)
\( \Rightarrow x = 1080 \)

∴ First part = Rs. 1080
Second part = (3000 - 1080) = Rs. 1920
In simple words: Express both parts as variables, calculate the interest earned on each, set their sum equal to the total interest given, and solve the resulting equation.

Exam Tip: When a sum is divided into parts with different rates and times, setting up an equation with the total interest condition is the most systematic approach.

Question 25. A sum of Rs. 3600 is divided into two parts. One part is invested at 9% per annum for 1 year and the other at 10% per annum for 1 year. The total amount obtained is Rs. 3933. Find the two parts.
Answer: Let the first part be Rs. x.
Second part = (3600 - x)

∴ S.I. on x at 9% per annum for 1 year = \( \frac{x \times 9 \times 1}{100} = \frac{9x}{100} \)

And, S.I. on (3600 - x) at 10% per annum for 1 year = \( \frac{(3600 - x) \times 10 \times 1}{100} = \frac{3600 - x}{10} \)

\( \Rightarrow \frac{9x}{100} + \frac{3600 - x}{10} = 333 \)

\( \Rightarrow x + 36000 = 33300 \)
\( \Rightarrow - x = 33300 - 36000 = 2700 \)
\( \Rightarrow x = 2700 \)

First part = Rs. 2700
Second part = (3600 - 2700) = Rs. 900
In simple words: Set up variables for the two parts, calculate the simple interest on each at their respective rates, add them to get the total interest (which equals the difference between total amount and original sum), then solve.

Exam Tip: Remember that the total amount is principal plus the combined interest from both parts - use this relationship to form your equation rather than trying to work with amounts directly.

Exercise 12B

Question 1. (a) Rs. 125. If principal is Rs. 6250, simple interest is 4% per annum, and time is 6 months, find the simple interest.
Answer: Principal = Rs. 6250
Simple Interest Rate = 4% per annum
Time = 6 months = \( \frac{1}{2} \) years
Simple Interest = \( \frac{P \times R \times T}{100} = \frac{6250 \times 4 \times 1}{100 \times 2} = \frac{250}{2} = \text{Rs. } 125 \)
In simple words: Substitute the principal, rate, and converted time (in years) into the simple interest formula to calculate the interest earned.

Exam Tip: Always ensure time is in years before applying the formula - convert months by dividing by 12 and days by dividing by 365.

Question 2. (b) Rs.3500. If amount is Rs. 3605, time is 219 days, and rate is 5% per annum, find the principal.
Answer: Amount = Rs. 3605
Time = 219 days = \( \frac{219}{365} \) days
Rate = 5% per annum
Amount = Sum \( + \frac{\text{Sum} \times \text{Rate} \times \text{Time}}{100} \)
Amount = Sum \( \left(1 + \frac{\text{Rate} \times \text{Time}}{100}\right) \)
Sum = \( \frac{1 + \frac{219 \times 5}{365 \times 100}}{3605} = \frac{3605 \times 36500}{37500} = \text{Rs. } 3500 \)
In simple words: Use the amount formula rearranged to find principal by dividing the given amount by the factor containing rate and time.

Exam Tip: Convert days to years using 365 as the divisor, then apply the rearranged amount formula to find the principal directly.

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Exam Tip: Ensure you identify all given values before beginning your calculation.

Question 4. (c) 8%. Let the sum be Rs. x. Rate of interest = r%. Time = 2\( \frac{1}{2} \) years = \( \frac{5}{2} \) years. Amount = \( \frac{6}{5} \) × Sum. Rate = ?. Amount = \( \frac{6}{5} \) × Sum. Principal + \( \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100} = \frac{6}{5} \) × Principal. \( \Rightarrow x \left(1 + \frac{5r}{100 \times 2}\right) = \frac{6}{5} x \). \( \Rightarrow 1 + \frac{r}{40} = \frac{6}{5} \). \( \Rightarrow r = 40 \times \frac{1}{5} = 8 \). So, the rate of interest is 8%.
Answer: Given: Time = 2\( \frac{1}{2} \) years = \( \frac{5}{2} \) years; Amount is \( \frac{6}{5} \) of the principal.
Using the formula A = P\( \left(1 + \frac{R \times T}{100}\right) \):
\( \frac{6}{5}P = P\left(1 + \frac{R \times 5}{100 \times 2}\right) \)
\( \frac{6}{5} = 1 + \frac{5R}{200} \)
\( \frac{6}{5} - 1 = \frac{5R}{200} \)
\( \frac{1}{5} = \frac{5R}{200} \)
\( R = \frac{200}{5 \times 5} = 8 \)
So, the rate of interest is 8%.
In simple words: When amount is expressed as a multiple of principal, substitute into the amount formula and solve for the rate algebraically.

Exam Tip: Problems stating amounts as multiples of principal are simplified because the principal cancels out, leaving only the rate to be found.

Question 5. (b) 9 months. Let the time be t years. Principal = Rs. 8000. Amount = Rs. 8360. Rate = 6% per annum. Amount = Principal \( \left(1 + \frac{\text{Rate} \times \text{Time}}{100}\right) \). \( \frac{8360}{8000} = 1 + \frac{6t}{100} \). \( \Rightarrow \frac{8360}{8000} - 1 = \frac{6t}{100} \). \( \Rightarrow t = \left(\frac{8360 - 8000}{8000}\right) \times \frac{100}{6} \). \( = \frac{360}{8000} \times \frac{100}{6} \). \( = \frac{6}{12} \times 12 \text{ months} = 9 \text{ months} \).
Answer: Given: Principal = Rs. 8000, Amount = Rs. 8360, Rate = 6% per annum
Using A = P\( \left(1 + \frac{R \times T}{100}\right) \):
\( 8360 = 8000\left(1 + \frac{6t}{100}\right) \)
\( \frac{8360}{8000} = 1 + \frac{6t}{100} \)
\( 1.045 = 1 + \frac{6t}{100} \)
\( \frac{6t}{100} = 0.045 \)
\( t = \frac{0.045 \times 100}{6} = 0.75 \text{ years} = 9 \text{ months} \)
In simple words: Rearrange the amount formula to isolate time, substitute all values, then convert the fractional year result into months for the final answer.

Exam Tip: When time is to be found as months, calculate it in years first, then multiply the decimal portion by 12 to get the number of months.

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Answer: [Content appears incomplete in the provided document.]
In simple words: [Awaiting complete question and working]

Exam Tip: Review the complete question and all given data before beginning your solution.

Question 7. (b) 8%. Time = 5 years. Simple Interest = \( \frac{2}{5} \)P. \( \Rightarrow \frac{\text{P} \times \text{Rate} \times \text{Time}}{100} = \frac{2}{5}P \). \( \Rightarrow \frac{\text{Rate} \times 2}{100} = \frac{2}{5} \). \( \Rightarrow \text{Rate} = \frac{2 \times 100}{5 \times 2} = \frac{2 \times 100}{5 \times 2} \). \( \Rightarrow \text{Rate} = 8\% \).
Answer: Given: Time = 5 years, Simple Interest = \( \frac{2}{5} \)P
Using S.I. = \( \frac{P \times R \times T}{100} \):
\( \frac{2}{5}P = \frac{P \times R \times 5}{100} \)
\( \frac{2}{5} = \frac{5R}{100} \)
\( \text{Rate} = \frac{2 \times 100}{5 \times 5} = 8\% \)
In simple words: Express the given simple interest as a fraction of principal, substitute into the formula, and the principal cancels out leaving you to solve for rate.

Exam Tip: When simple interest is given as a fraction of principal, this type of cancellation makes the calculation straightforward - the principal always cancels out.

Question 8. (c) 22 years. Given: R₁ = 12%, R₂ = 10%, P₁ = Rs. 8000, P₂ = Rs. 9100. Let their amount s be equal in T years. Amount₁ = S.I.₁ + P₁ = \( \frac{P_1 \times R_1 \times T}{100} + P_1 = \frac{8000 \times 12 \times T}{100} + 8000 = 960T + 8000 \). Amount₂ = S.I.₂ + P₂ = \( \frac{P_2 \times R_2 \times T}{100} + P_2 = \frac{9100 \times 10 \times T}{100} + 9100 = 910T + 9100 \). Amount₁ = Amount₂ \( \Rightarrow 960T + 8000 = 910T + 9100 \) \( \Rightarrow 960T - 910T = 9100 - 8000 \) \( \Rightarrow 50T = 1100 \) \( \Rightarrow T = 22 \). Hence, after 22 years their amounts will be equal.
Answer: Given: R₁ = 12%, R₂ = 10%, P₁ = Rs. 8000, P₂ = Rs. 9100
Let the amounts be equal after T years.
Amount₁ = \( \frac{P_1 \times R_1 \times T}{100} + P_1 = \frac{8000 \times 12 \times T}{100} + 8000 = 960T + 8000 \)
Amount₂ = \( \frac{P_2 \times R_2 \times T}{100} + P_2 = \frac{9100 \times 10 \times T}{100} + 9100 = 910T + 9100 \)
Setting them equal:
\( 960T + 8000 = 910T + 9100 \)
\( 50T = 1100 \)
\( T = 22 \)
Hence, after 22 years their amounts will be equal.
In simple words: Express each final amount in terms of T, set them equal to each other, and solve the resulting linear equation to find the number of years required.

Exam Tip: When two investments reach equal amounts, setting up the amount equations and equating them is the standard method - always express amounts as principal plus simple interest in terms of time.

Question 9. [Question text not provided in source]
Answer: [Content appears incomplete in the provided document.]
In simple words: [Awaiting complete question and working]

Exam Tip: Ensure all problem data is available before attempting a solution.

Question 10. (d) y² = zx
Answer: Given that S.I. on x equals the quantity \( \frac{z \times R \times T}{100} \), we have y = S.I. on x ... (i). Similarly, z equals S.I. on y, which gives us z = S.I. on y = \( \frac{y \times R \times T}{100} \) ... (ii).

Dividing equation (i) by equation (ii):

\[ \frac{y}{z} = \frac{\frac{z \times R \times T}{100}}{\frac{y \times R \times T}{100}} \]

\[ \frac{y}{z} = \frac{z}{y} \]

\[ y^2 = zx \]
In simple words: When you set up the ratios of simple interest on two different amounts and divide them, the cross-multiplication produces this relationship between y, z, and x.

Exam Tip: Always express both simple interest formulas clearly before dividing - this step is where marks are awarded.

Question 11. (a) 1¼ years
Answer: The rate is 10% per annum. Using the simple interest formula, we know that Simple Interest equals 0.125 times the Principal. This means:

\[ \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100} = 0.125 \times \text{Principal} \]

\[ \frac{\text{Time}}{10} = 0.125 \]

\[ \text{Time} = 1.25 = 1\frac{1}{4} \text{ years} \]
In simple words: Divide 0.125 by the rate (as a decimal) to get the time needed for the interest to build up to that fraction of the principal.

Exam Tip: Convert all percentages to decimals before solving - it makes the algebra cleaner and reduces errors.

Question 12. (b) Rs 2400
Answer: The rate given is 3⅓% per annum, which equals \( \frac{15}{4} \)% per annum. The time period is 2⅓ years, which is \( \frac{7}{4} \) years. Using the simple interest formula:

\[ S.I. = \frac{P \times R \times T}{100} \]

\[ \Rightarrow P = \frac{210 \times 100}{\frac{15}{4} \times \frac{7}{4}} \]

\[ \Rightarrow P = 600 \times 4 \]

\[ \Rightarrow P = \text{Rs } 2400 \]
In simple words: Rearrange the simple interest formula to solve for the principal by dividing the given interest by the product of the rate and time (both converted to fractions), then multiply by 100.

Exam Tip: Always convert mixed numbers and percentages to improper fractions and their decimal equivalents before substituting into the formula - this prevents calculation mistakes.