Simple Interest and Compound Interest

Q1: Find the simple interest and amount on:

i. ₹4500 for \(2\frac{1}{2}\) years at \(7\frac{2}{3}\%\) per annum.

Step 1: Convert mixed fractions to improper fractions
Rate (R) = \(7\frac{2}{3} = \frac{23}{3}\%\)
Time (T) = \(2\frac{1}{2} = \frac{5}{2}\) years
Step 2: Use the formula:
\(SI = \frac{P \times R \times T}{100}\)
\(SI = \frac{4500 \times \frac{23}{3} \times \frac{5}{2}}{100}\)
Step 3: Simplify step-by-step:
\(SI = \frac{4500 \times 23 \times 5}{3 \times 2 \times 100}\)
\(SI = \frac{517500}{600} = ₹862.50\)
Step 4: Find Amount
Amount = Principal + SI = ₹4500 + ₹862.50 = ₹5362.50
Answer: Simple Interest = ₹862.50, Amount = ₹5362.50

ii. ₹6360 for 6 years 3 months at 8% per annum

Step 1: Convert time to years
6 years 3 months = \(6\frac{1}{4} = \frac{25}{4}\) years
Step 2: Apply formula:
\(SI = \frac{6360 \times 8 \times \frac{25}{4}}{100}\)
\(SI = \frac{6360 \times 8 \times 25}{4 \times 100}\)
\(SI = \frac{1272000}{400} = ₹3180\)
Step 3: Find Amount
Amount = ₹6360 + ₹3180 = ₹9540
Answer: Simple Interest = ₹3180, Amount = ₹9540

iii. ₹19200 for 11 months at \(9\frac{3}{4}\%\) per annum

Step 1: Convert time and rate
11 months = \(\frac{11}{12}\) years
\(9\frac{3}{4} = \frac{39}{4}\%\)
Step 2: Apply formula:
\(SI = \frac{19200 \times \frac{39}{4} \times \frac{11}{12}}{100}\)
= \(\frac{19200 \times 39 \times 11}{4 \times 12 \times 100}\)
= \(\frac{8236800}{4800} = ₹1716\)
Step 3: Find Amount
Amount = ₹19200 + ₹1716 = ₹20916
Answer: Simple Interest = ₹1716, Amount = ₹20916

iv. ₹58400 for 75 days at \(6\frac{1}{2}\%\) per annum

Step 1: Convert time and rate
Time = \(\frac{75}{365}\) years
\(6\frac{1}{2} = \frac{13}{2}\%\)
Step 2: Apply formula:
\(SI = \frac{58400 \times \frac{13}{2} \times \frac{75}{365}}{100}\)
= \(\frac{58400 \times 13 \times 75}{2 \times 365 \times 100}\)
= \(\frac{56940000}{73000} = ₹780\)
Step 3: Find Amount
Amount = ₹58400 + ₹780 = ₹59180
Answer: Simple Interest = ₹780, Amount = ₹59180

Q2: Find the simple interest on ₹8600 from 18th October, 2006 to 13th March, 2007 at 8% per annum. Also find the amount.

Step 1: Find the number of days between 18th October, 2006 and 13th March, 2007
➡ Days remaining in October = 31 − 18 = 13
➡ November = 30 days
➡ December = 31 days
➡ January = 31 days
➡ February 2007 = 28 days (not a leap year)
➡ March = 13 days
Total time = 13 + 30 + 31 + 31 + 28 + 13 = 146 days
Step 2: Convert time into years \[ T = \frac{146}{365} = \frac{2}{5} \text{ year} \]Step 3: Use the formula: \[ SI = \frac{P \times R \times T}{100} \\ SI = \frac{8600 \times 8 \times \frac{2}{5}}{100} \\ SI = \frac{137600}{500} = ₹275.20 \]Step 4: Find the Amount
Amount = Principal + SI = ₹8600 + ₹275.20 = ₹8875.20
Answer: Simple Interest = ₹275.20, Amount = ₹8875.20

Q3: Ashish lent ₹10500 to Sunidhi at 7% per annum simple interest. After 5 years, Sunidhi discharged the debt by giving a watch and ₹13000 in cash. What is the value of the watch?

Step 1: Use the Simple Interest formula \[ SI = \frac{P \times R \times T}{100} \\ SI = \frac{10500 \times 7 \times 5}{100} \\ SI = \frac{367500}{100} = ₹3675 \]Step 2: Find total amount due after 5 years
Amount = Principal + Interest
= ₹10500 + ₹3675 = ₹14175
Step 3: Cash paid = ₹13000
So, value of the watch = Total amount due − Cash paid
= ₹14175 − ₹13000 = ₹1175
Answer: The value of the watch is ₹1175

Q4: In what time will the simple interest on ₹7560 be ₹1102.50 at \(6\frac{1}{4}\)% per annum?

Step 1: Use the formula for Simple Interest \[ SI = \frac{P \times R \times T}{100} \] We are given:
SI = ₹1102.50, P = ₹7560, R = \(6\frac{1}{4} = \frac{25}{4}\)%
Step 2: Substitute into the formula and solve for T \[ 1102.50 = \frac{7560 \times \frac{25}{4} \times T}{100} \]Step 3: Simplify step-by-step \[ 1102.50 = \frac{7560 \times 25 \times T}{400} \\ 1102.50 = \frac{189000 \times T}{400} \]Multiply both sides by 400: \[ 1102.50 \times 400 = 189000 \times T \\ 441000 = 189000 \times T \]Now divide both sides by 189000: \[ T = \frac{441000}{189000} = \frac{49}{21} = \frac{7}{3} \text{ years} \\ T = 2\frac{1}{3} \text{ years} \]Answer: The required time is \(2\frac{1}{3}\) years or 2 years 4 months

Q5: In how much time will ₹25600 amount to ₹35664, when money is worth \(9\frac{1}{4}\)% per annum simple interest?

Step 1: Use formula: \[ SI = A – P \] Where A = ₹35664, P = ₹25600 \[ SI = 35664 – 25600 = ₹10064 \]Step 2: Convert rate to improper fraction \[ R = 9\frac{1}{4} = \frac{37}{4}\% \]Step 3: Use SI formula: \[ SI = \frac{P \times R \times T}{100} \] Substitute values: \[ 10064 = \frac{25600 \times \frac{37}{4} \times T}{100} \]Step 4: Simplify the equation: \[ 10064 = \frac{25600 \times 37 \times T}{400} \\ 10064 = 64 \times 37 \times T \\ 10064 = 2368 \times T \]Step 5: Solve for T \[ T = \frac{10064}{2368} = \frac{10064 \div 16}{2368 \div 16} = \frac{629}{148} = 4.25 \text{ years } \\ T = 4 \frac{1}{4} \text{ years} \]Answer: The required time is approximately \(4\frac{1}{4}\) years or 4 years 3 months

Q6: At what rate percent per annum will ₹1625 amount to ₹2080 in \(3\frac{1}{2}\) years?

Step 1: Use formula: \[ SI = A – P \] Where A = ₹2080 and P = ₹1625 \[ SI = 2080 – 1625 = ₹455 \]Step 2: Convert time to improper fraction \[ T = 3\frac{1}{2} = \frac{7}{2} \text{ years} \]Step 3: Use the formula: \[ SI = \frac{P \times R \times T}{100} \] Substitute values: \[ 455 = \frac{1625 \times R \times \frac{7}{2}}{100} \\ 455 = \frac{1625 \times 7 \times R}{200} \]Step 4: Multiply both sides by 200 \[ 455 \times 200 = 11375 \times R \\ 91000 = 11375 \times R \\ R = \frac{91000}{11375} = 8 \]Answer: The required rate of interest is 8% per annum

Q7: At what rate percent per annum will the simple interest on ₹6720 br ₹1911 in 3 years 3 months?

Step 1: Use the simple interest formula: \[ SI = \frac{P \times R \times T}{100} \] Where:
SI = ₹1911, P = ₹6720, T = 3 years 3 months = \(3\frac{1}{4} = \frac{13}{4}\) years
Step 2: Substitute into the formula: \[ 1911 = \frac{6720 \times R \times \frac{13}{4}}{100} \\ 1911 = \frac{6720 \times 13 \times R}{400} \]Step 3: Multiply both sides by 400: \[ 1911 \times 400 = 6720 \times 13 \times R \\ 764400 = 87360 \times R \]Step 4: Solve for R: \[ R = \frac{764400}{87360} = 8.75 \]Answer: The required rate of interest is 8.75% per annum

Q8: At what rate percent of simple interest will a sum of money double itself in 12 years?

Step 1: Let the principal be ₹100
Since the money doubles, the amount becomes ₹200
So, Simple Interest = ₹200 − ₹100 = ₹100
Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \] Substitute values: \[ 100 = \frac{100 \times R \times 12}{100} \]Step 3: Simplify and solve: \[ 100 = 12R \\ R = \frac{100}{12} = 8.33\% \]Answer: The required rate of interest is \(8\frac{1}{3}\)% per annum

Q9: Simple interest on a certain sum is \(\frac{9}{16}\) of the sum. Find the rate per cent and the time if both are numerically equal.

Step 1: Let the principal = ₹P
Given: SI = \(\frac{9}{16} \times P\)
Also given: Rate and Time are equal, let each be \(x\)
Step 2: Use SI formula: \[ SI = \frac{P \times R \times T}{100} \\ \frac{9}{16} \times P = \frac{P \times x \times x}{100} \]Step 3: Cancel out P (non-zero) \[ \frac{9}{16} = \frac{x^2}{100} \\ x^2 = \frac{9 \times 100}{16} = \frac{900}{16} \\ x^2 = 56.25 \\ \Rightarrow x = \sqrt{56.25} = 7.5 \]Answer: Rate = 7.5%, Time = 7.5 years

Q10: What sum will yield ₹406 as simple interest in 1 year 2 months at \(6\frac{1}{4}\)% per annum?

Step 1: Use the formula: \[ SI = \frac{P \times R \times T}{100} \] We are given:
SI = ₹406, R = \(6\frac{1}{4} = \frac{25}{4}\%\), T = 1 year 2 months = \(\frac{14}{12} = \frac{7}{6}\) years
Step 2: Substitute into formula: \[ 406 = \frac{P \times \frac{25}{4} \times \frac{7}{6}}{100} \\ 406 = \frac{175P}{24 \times 100} = \frac{175P}{2400} \]Step 3: Cross multiply: \[ 406 \times 2400 = 175P \\ 974400 = 175P \\ \Rightarrow P = \frac{974400}{175} = ₹5568 \]Answer: The required sum is ₹5568

Q11: What sum will amount to ₹1748 in \(2\frac{1}{2}\) years at \(7\frac{1}{2}\)% per annum simple interest?

Step 1: Let the required sum be ₹P
Given: Amount (A) = ₹1748
Rate (R) = \(7\frac{1}{2} = \frac{15}{2}\%\)
Time (T) = \(2\frac{1}{2} = \frac{5}{2}\) years
Step 2: Use the formula: \[ A = P + SI \quad \text{and} \quad \\ SI = \frac{P \times R \times T}{100} \\ 1748 = P + \frac{P \times \frac{15}{2} \times \frac{5}{2}}{100} \]Step 3: Simplify the interest expression: \[ SI = \frac{75P}{400} = \frac{3P}{16} \\ 1748 = P + \frac{3P}{16} = \frac{16P + 3P}{16} = \frac{19P}{16} \]Step 4: Solve for P: \[ P = \frac{1748 \times 16}{19} = \frac{27968}{19} = ₹1472 \]Answer: The required sum is ₹1472

Q12: A sum of money becomes \(\frac{8}{5}\) of itself in 5 years at a certain rate of simple interest. Find the rate of interest.

Step 1: Let the principal = ₹P
Then, amount = \(\frac{8}{5} \times P = \frac{8P}{5}\)
Step 2: Use formula: \[ SI = A – P = \frac{8P}{5} – P = \frac{3P}{5} \]Step 3: Use simple interest formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow \quad \frac{3P}{5} = \frac{P \times R \times 5}{100} \]Step 4: Cancel P on both sides (non-zero) and simplify: \[ \frac{3}{5} = \frac{5R}{100} \\ \frac{3}{5} = \frac{R}{20} \\ \Rightarrow \quad R = \frac{3 \times 20}{5} = 12 \]Answer: The required rate of interest is 12% per annum

Q13: What sum of money lent at \(12\frac{1}{2}\)% per annum will produce the same interest in 4 years as ₹8560 produces in 5 years at 12% per annum?

Step 1: Use simple interest formula: \[ SI = \frac{P \times R \times T}{100} \] First, calculate interest produced by ₹8560 at 12% for 5 years: \[ SI_1 = \frac{8560 \times 12 \times 5}{100} = \frac{513600}{100} = ₹5136 \]Step 2: Let the required principal be ₹P
This amount at \(12\frac{1}{2} = \frac{25}{2}\)% for 4 years must also yield ₹5136 interest \[ \frac{P \times \frac{25}{2} \times 4}{100} = 5136 \\ \frac{100P}{2} = 5136 \times 100 \\ \Rightarrow 50P = 513600 \\ \Rightarrow P = \frac{513600}{50} = ₹10272 \]Answer: The required sum is ₹10272

Q14: If ₹1250 amount to ₹1550 in 3 years at simple interest, what will ₹3200 amount to in 4 years at the same rate?

Step 1: Calculate the simple interest on ₹1250 \[ SI = 1550 – 1250 = ₹300 \]Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 300 = \frac{1250 \times R \times 3}{100} \]Step 3: Solve for R: \[ 300 = \frac{3750R}{100} \\ \Rightarrow 300 = 37.5R \\ \Rightarrow R = \frac{300}{37.5} = 8 \]Step 4: Use R = 8% to calculate new interest on ₹3200 for 4 years: \[ SI = \frac{3200 \times 8 \times 4}{100} = \frac{102400}{100} = ₹1024 \]Step 5: Calculate amount: \[ \text{Amount} = ₹3200 + ₹1024 = ₹4224 \]Answer: The amount will be ₹4224

Q15: A sum of money lent at simple interest amounts to ₹3224 in 2 years and ₹4160 in 5 years. Find the sum and the rate of interest.

Step 1: Use the formula: \[ \text{SI} = \text{Amount after 5 years} – \text{Amount after 2 years} \\ SI = 4160 – 3224 = ₹936 \]Step 2: Find the interest for 1 year
Time difference = 5 − 2 = 3 years
So interest for 3 years = ₹936
Interest for 1 year = \(\frac{936}{3} = ₹312\)
Step 3: Interest for 2 years = ₹312 × 2 = ₹624
Now use: \[ \text{Principal} = \text{Amount after 2 years} – \text{SI for 2 years} = 3224 – 624 = ₹2600 \]Step 4: Use formula to find rate: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 312 = \frac{2600 \times R \times 1}{100} \\ \Rightarrow 312 = 26R \\ \Rightarrow R = \frac{312}{26} = 12 \]Answer: Sum = ₹2600, Rate = 12% per annum

Q16: A lends ₹2500 to B and a certain sum to C at the same time at 7% per annum simple interest. If, after 4 years, A altogether receives 1120 as interest from B and C, find the sum lent to C.

Step 1: Use simple interest formula: \[ SI = \frac{P \times R \times T}{100} \] For B: Principal = ₹2500, Rate = 7%, Time = 4 years \[ SI_B = \frac{2500 \times 7 \times 4}{100} = \frac{70000}{100} = ₹700 \]Step 2: Total interest received = ₹1120
Interest from C = ₹1120 − ₹700 = ₹420
Step 3: Let the amount lent to C be ₹P
We use the SI formula again: \[ 420 = \frac{P \times 7 \times 4}{100} \\ \Rightarrow 420 = \frac{28P}{100} \] Multiply both sides by 100: \[ 42000 = 28P \Rightarrow P = \frac{42000}{28} = ₹1500 \]Answer: The sum lent to C is ₹1500

Q17: The simple interest on a certain sum for 3 years at 8% annum is ₹96 more than the simple interest on the same sum for 2 years at 9% per annum. Find the sum.

Step 1: Let the sum be ₹P
Use the formula: \[ SI = \frac{P \times R \times T}{100} \]Step 2: SI for 3 years at 8%: \[ SI_1 = \frac{P \times 8 \times 3}{100} = \frac{24P}{100} \]SI for 2 years at 9%: \[ SI_2 = \frac{P \times 9 \times 2}{100} = \frac{18P}{100} \]Step 3: According to the question: \[ SI_1 – SI_2 = 96 \\ \Rightarrow \frac{24P}{100} – \frac{18P}{100} = 96 \\ \Rightarrow \frac{6P}{100} = 96 \\ \Rightarrow P = \frac{96 \times 100}{6} = ₹1600 \]Answer: The required sum is ₹1600

Q18: Two equal sums of money were lent at simple interest at 11% p.a. for \(3\frac{1}{2}\) years and \(4\frac{1}{2}\) years respectively. If the difference in interests for two periods was ₹412.50, find each sum.

Step 1: Let the sum lent in each case be ₹P
Rate (R) = 11% p.a.
Time 1 = \(3\frac{1}{2} = \frac{7}{2}\) years
Time 2 = \(4\frac{1}{2} = \frac{9}{2}\) years
Step 2: Use the simple interest formula: \[ SI = \frac{P \times R \times T}{100} \\ SI_1 = \frac{P \times 11 \times \frac{7}{2}}{100} = \frac{77P}{200} \\ SI_2 = \frac{P \times 11 \times \frac{9}{2}}{100} = \frac{99P}{200} \]Step 3: Difference in interests is ₹412.50 \[ SI_2 – SI_1 = \frac{99P}{200} – \frac{77P}{200} = \frac{22P}{200} \\ \Rightarrow \frac{22P}{200} = 412.50 \\ \frac{11P}{100} = 412.50 \\ \Rightarrow P = \frac{412.50 \times 100}{11} = ₹3750 \]Answer: Each sum is ₹3750

Q19: Divide ₹6000 into two parts so that the simple interest on the first part for 9 months at 12% per annum is to the simple interest on for \(1\frac{1}{2}\) years at 10% per annum.

Step 1: Let the first part be ₹x
Then the second part = ₹(6000 − x)
Step 2: Use S.I. formula: \[ SI = \frac{P \times R \times T}{100} \] Time 1 = 9 months = \(\frac{3}{4}\) years, Rate 1 = 12%
Time 2 = \(1\frac{1}{2} = \frac{3}{2}\) years, Rate 2 = 10%
Step 3: Set S.I. from both parts equal: \[ \frac{x \times 12 \times \frac{3}{4}}{100} = \frac{(6000 – x) \times 10 \times \frac{3}{2}}{100} \] Simplify both sides: \[ \frac{36x}{400} = \frac{180000 – 30x}{200} \] Multiply both sides by 400 to eliminate denominator: \[ 36x = 2 \times (180000 – 30x) \\ \Rightarrow 36x = 360000 – 60x \\ \Rightarrow 96x = 360000 \\ \Rightarrow x = \frac{360000}{96} = ₹3750 \]Step 4: Second part = ₹6000 − ₹3750 = ₹2250
Answer: First part = ₹3750, Second part = ₹2250

Q20: Divide a sum of ₹13500 into two parts such that if one part be lent at \(8\frac{1}{3}\)% per annum for 2years 9 months and the other at \(7\frac{1}{2}\)% per annum for 1 year 8 months, the total interest received is ₹2375.

Step 1: Let the first part be ₹x
Then the second part = ₹(13500 − x)
Step 2: Convert rates and times:
Rate 1 = \(8\frac{1}{3} = \frac{25}{3}\)%
Time 1 = 2 years 9 months = \(2\frac{3}{4} = \frac{11}{4}\) years
Rate 2 = \(7\frac{1}{2} = \frac{15}{2}\)%
Time 2 = 1 year 8 months = \(1\frac{2}{3} = \frac{5}{3}\) years
Step 3: Use SI formula: \[ \text{SI}_1 = \frac{x \times \frac{25}{3} \times \frac{11}{4}}{100} = \frac{275x}{12 \times 100} = \frac{275x}{1200} = \frac{275x \div 25}{1200 \div 25} = \frac{11x}{48} \\ \text{SI}_2 = \frac{(13500 – x) \times \frac{15}{2} \times \frac{5}{3}}{100} = \frac{(13500 – x) \times 75}{600} \]Step 4: Total interest is ₹2375 \[ \frac{11x}{48} + \frac{75(13500 – x)}{600} = 2375 \] Simplify each term: \[ \frac{11x}{48} + \frac{(13500 – x)}{8} = 2375 \] Take LCM = 48: \[ \frac{11x + 6(13500 – x)}{48} = 2375 \\ 11x + 81000 – 6x = 2375 \times 48 \\ \Rightarrow 5x + 81000 = 114000 \\ 5x = 114000 – 81000 = 33000 \\ \Rightarrow x = \frac{33000}{5} = ₹6600 \]Step 5: Second part = ₹13500 − ₹6600 = ₹6900
Answer: First part = ₹6600, Second part = ₹6900

Q21: In what time will a sum of money lent at \(8\frac{1}{3}\)% simple interest become 4 times of itself.

Step 1: Let the sum be ₹P
Amount becomes 4 times of itself, so: \[ A = 4P \Rightarrow SI = A – P = 4P – P = 3P \]Step 2: Use the S.I. formula: \[ SI = \frac{P \times R \times T}{100} \] Given: SI = 3P and Rate \(R = 8\frac{1}{3} = \frac{25}{3}\%\)
Step 3: Substitute in the formula: \[ 3P = \frac{P \times \frac{25}{3} \times T}{100} \] Cancel P (non-zero): \[ 3 = \frac{25T}{300} \\ \Rightarrow 3 \times 300 = 25T \\ \Rightarrow 900 = 25T \\ \Rightarrow T = \frac{900}{25} = 36 \text{ years} \]Answer: The required time is 36 years

Q22: A certain sum of money lent out at \(6\frac{2}{3}\)% p.a. produces the same simple interest in 6 years as ₹3200 lent out at \(8\frac{2}{5}\)% p.a. for 7 years. Find the sum.

Step 1: Let the required sum be ₹P
Rate 1 = \(6\frac{2}{3} = \frac{20}{3}\)%
Time 1 = 6 years
Given: ₹3200 lent at \(8\frac{2}{5} = \frac{42}{5}\)% for 7 years
Step 2: Use the formula for simple interest: \[ \text{SI}_1 = \frac{P \times \frac{20}{3} \times 6}{100} = \frac{120P}{300} = \frac{2P}{5} \\ \text{SI}_2 = \frac{3200 \times \frac{42}{5} \times 7}{100} = \frac{3200 \times 294}{500} = \frac{940800}{500} = ₹1881.60 \]Step 3: Set the two SIs equal: \[ \frac{2P}{5} = 1881.60 \\ \Rightarrow P = \frac{1881.60 \times 5}{2} = ₹4704 \]Answer: The required sum is ₹4704

Q23: Naveen and Praveen borrowed ₹42000 and ₹55000 respectively for \(3\frac{1}{2}\) years at the same rate of interest. If Praveen has to pay 3640 more than Naveen, find the rate of interest.

Step 1: Let the rate of interest be R% per annum.
Time for both = \(3\frac{1}{2} = \frac{7}{2}\) years
Step 2: Use the simple interest formula: \[ SI = \frac{P \times R \times T}{100} \] Interest paid by Naveen: \[ SI_1 = \frac{42000 \times R \times \frac{7}{2}}{100} = \frac{147000R}{100} \] Interest paid by Praveen: \[ SI_2 = \frac{55000 \times R \times \frac{7}{2}}{100} = \frac{192500R}{100} \]Step 3: Difference in interest = ₹3640 \[ \frac{192500R}{100} – \frac{147000R}{100} = 3640 \\ \Rightarrow \frac{(192500 – 147000)R}{100} = 3640 \\ \Rightarrow \frac{45500R}{100} = 3640 \]Multiply both sides by 100: \[ 45500R = 364000 \\ \Rightarrow R = \frac{364000}{45500} = 8 \]Answer: The rate of interest is 8% per annum

Q24: A sum of money was put at simple interest at a certain rate for 2 years. If this sum had been put at 3% higher rate, it would have earned ₹720 more as interest. Find the sum.

Step 1: Let the principal (sum) be ₹P and the original rate be R% per annum.
Step 2: Interest at original rate for 2 years: \[ SI_1 = \frac{P \times R \times 2}{100} \]Interest at (R + 3)% for 2 years: \[ SI_2 = \frac{P \times (R + 3) \times 2}{100} \]Step 3: Difference in interest = ₹720 \[ \frac{P \times (R + 3) \times 2}{100} – \frac{P \times R \times 2}{100} = 720 \] Factor out common terms: \[ \frac{2P}{100}[(R + 3) – R] = 720 \\ \Rightarrow \frac{2P}{100} \times 3 = 720 \\ \frac{6P}{100} = 720 \\ \Rightarrow P = \frac{720 \times 100}{6} = ₹12000 \]Answer: The sum is ₹12000

Q25: A sum of money invested at 6% p.a. simple interest for a certain period of time yields ₹960 as interest. If this sum had been invested for 5 years more, it would have yielded ₹2160 as interest. Find the sum.

Step 1: Let the sum be ₹P and the time be T years.
Rate = 6% p.a.
Simple Interest (SI) for T years = ₹960 \[ \text{Using } SI = \frac{P \times R \times T}{100} \\ \Rightarrow \frac{P \times 6 \times T}{100} = 960 \tag{1} \]Simple Interest for (T + 5) years = ₹2160 \[ \Rightarrow \frac{P \times 6 \times (T + 5)}{100} = 2160 \tag{2} \]Step 2: Subtract Equation (1) from (2): \[ \frac{P \times 6 (T + 5 – T)}{100} = 2160 – 960 \\ \Rightarrow \frac{P \times 6 \times 5}{100} = 1200 \\ \Rightarrow \frac{30P}{100} = 1200 \\ \Rightarrow P = \frac{1200 \times 100}{30} = ₹4000 \]Answer: The required sum is ₹4000