Simple Interest Concepts

1. Simple Interest — Basic Calculation

• Formula:

$$SI=\frac{P\times R\times T}{100}$$ $$A=P+SI=P\left(1+\frac{RT}{100}\right)$$

where (P) = principal, (R) = rate %, (T) = time in years.

• Example: Find the simple interest on ₹8,000 at 12% per annum for 3 years. Solution:

$$SI=\frac{8000\times12\times3}{100}$$ $$=\boxed{₹2,880}$$

Amount:

$$A=8000+2880=\boxed{₹10,880}$$

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2. Finding Principal, Rate or Time

• Formula:

$$P=\frac{SI\times100}{RT}$$ $$R=\frac{SI\times100}{PT}$$ $$T=\frac{SI\times100}{PR}$$

• Example: The simple interest on a sum at 8% per annum for 5 years is ₹2,000. Find the principal. Solution:

$$P=\frac{2000\times100}{8\times5}$$ $$=\boxed{₹5,000}$$

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3. Amount Under Simple Interest

• Formula:

$$A=P\left(1+\frac{RT}{100}\right)$$

Therefore:

$$A-P=SI$$

• Example: Find the amount on ₹15,000 at 10% SI for 2 years. Solution:

$$A=15000\left(1+\frac{10\times2}{100}\right)$$ $$=15000(1.2) =\boxed{₹18,000}$$

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4. Time in Months or Days

• Formula: Convert time into years:

$$T=\frac{\text{months}}{12}$$

For standard aptitude questions using a 365-day year:

$$T=\frac{\text{days}}{365}$$

• Example: Find the SI on ₹12,000 at 10% per annum for 9 months. Solution:

$$T=\frac9{12}=\frac34\text{ year}$$ $$SI=\frac{12000\times10\times3/4}{100}$$ $$=\boxed{₹900}$$

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5. Difference Between Simple Interests

• Formula: For the same principal and rate:

$$SI_2-SI_1 =\frac{P\times R\times(T_2-T_1)}{100}$$

• Example: The SI on a sum for 5 years is ₹4,000 and for 3 years is ₹2,400. Find the SI for 2 years. Solution: Difference in time:

$$5-3=2\text{ years}$$

Difference in SI:

$$4000-2400=\boxed{₹1,600}$$

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6. Principal Becomes a Given Amount

• Formula: Under SI:

$$A=P\left(1+\frac{RT}{100}\right)$$

If the amount is (kP):

$$k=1+\frac{RT}{100}$$

• Example: At 10% SI, in how many years will a sum become 1.5 times its principal? Solution:

$$1.5=1+\frac{10T}{100}$$ $$0.5=0.1T$$ $$\boxed{T=5\text{ years}}$$

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7. Principal Doubles Under Simple Interest

• Formula: If the amount becomes (2P):

$$2P=P\left(1+\frac{RT}{100}\right)$$

Hence:

$$T=\frac{100}{R}$$

• Example: At 8% SI, how long will it take for a principal to double? Solution:

$$T=\frac{100}{8} =\boxed{12.5\text{ years}}$$

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8. Difference Between Amounts at Different Rates

• Formula: For the same (P,T):

$$A_1-A_2 =\frac{P(R_1-R_2)T}{100}$$

• Example: A sum earns ₹1,200 more SI in 4 years at 12% than at 9%. Find the principal. Solution:

$$1200=\frac{P(12-9)4}{100}$$ $$1200=\frac{12P}{100}$$ $$P=\boxed{₹10,000}$$

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9. Comparing Simple Interest on Two Principals

• Formula:

$$SI_1:SI_2=P_1R_1T_1:P_2R_2T_2$$

• Example: A invests ₹5,000 at 8% for 3 years, while B invests ₹6,000 at 5% for 2 years. Find the ratio of their SIs. Solution:

$$SI_A:SI_B =(5000\times8\times3):(6000\times5\times2)$$ $$=120000:60000 =\boxed{2:1}$$

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10. Splitting Capital Between Two SI Schemes

• Formula:

$$SI_{\text{total}} =\frac{P_1R_1T_1+P_2R_2T_2}{100}$$

If total principal is (P), then:

$$P_2=P-P_1$$

• Example: ₹20,000 is divided between 8% and 10% SI for one year. The total interest is ₹1,800. Find the amount invested at each rate. Solution: Let ₹(x) be invested at 8%.

$$\frac{8x+10(20000-x)}{100}=1800$$ $$8x+200000-10x=180000$$ $$2x=20000$$ $$x=\boxed{₹10,000}$$

Therefore the other part is:

$$\boxed{₹10,000}$$

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11. Weighted Average Rate of Simple Interest

• Formula:

$$R_{\text{effective}} =\frac{P_1R_1+P_2R_2+\cdots}{P_1+P_2+\cdots}$$

when the investment periods are equal.

• Example: ₹6,000 is invested at 8% and ₹4,000 at 12% for one year. Find the effective rate. Solution:

$$R=\frac{6000(8)+4000(12)}{10000}$$ =\frac{48000+48000}{10000} =\boxed{9.6%}

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12. Loan Repayment in Equal Installments — Basic

• Formula: If total interest is calculated on the original principal for the entire period and then divided equally:

$$\text{Installment} =\frac{P+SI}{n}$$

Note: This is different from reducing-balance installments.

• Example: A loan of ₹12,000 at 10% SI is repaid in 3 equal installments, with interest calculated on the original principal for 3 years. Find each installment. Solution:

$$SI=\frac{12000\times10\times3}{100}=3600$$

Total repayment:

$$12000+3600=15600$$

Each installment:

$$\frac{15600}{3} =\boxed{₹5,200}$$

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13. Installment Paid at Different Times

• Formula: Equate the value of all installments to the amount due at the appropriate time using SI:

$$A=P\left(1+\frac{RT}{100}\right)$$

• Example: A debt of ₹11,000 is due after 2 years at 10% SI. It is paid in two equal installments at the end of years 1 and 2. Find each installment. Solution: Amount due after 2 years:

$$11000\left(1+\frac{10\times2}{100}\right)=13200$$

Let each installment be (x).

At the end of year 2, the first installment has effectively earned one year’s SI:

$$1.1x+x=13200$$ $$2.1x=13200$$ $$\boxed{x=₹6,285.71}$$

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14. Interest Allocation Between Multiple Investments

• Formula:

$$\text{Total SI} =SI_1+SI_2+\cdots$$

For each investment:

$$SI_i=\frac{P_iR_iT_i}{100}$$

• Example: ₹30,000 is invested partly at 6% and partly at 10% for one year. If the total interest is ₹2,400, find the amount invested at 10%. Solution: Let ₹(x) be invested at 10%.

$$\frac{10x+6(30000-x)}{100}=2400$$ $$10x+180000-6x=240000$$ $$4x=60000$$ $$\boxed{x=₹15,000}$$

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15. Finding Rate from Principal and Amount

• Formula:

$$SI=A-P$$

Then:

$$R=\frac{(A-P)100}{PT}$$

• Example: ₹8,000 becomes ₹9,920 in 4 years under SI. Find the rate. Solution:

$$SI=9920-8000=1920$$ $$R=\frac{1920\times100}{8000\times4}$$ =\boxed{6%}

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Advanced Variants

16. Finding Principal from Difference in Amounts

• Formula:

$$A_1-A_2 =\frac{P(R_1-R_2)T}{100}$$

• Example: A sum earns ₹2,500 more when invested at 10% instead of 8% for 5 years. Find the principal. Solution:

$$2500=\frac{P(10-8)5}{100}$$ $$2500=\frac{10P}{100}$$ $$P=\boxed{₹25,000}$$

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17. Difference Between SI and Principal

• Formula: If:

$$SI=P$$

then:

$$\frac{PRT}{100}=P$$

Hence:

$$RT=100$$

• Example: At what rate will the SI equal the principal in 8 years? Solution:

$$R\times8=100$$ \boxed{R=12.5%}

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18. Principal and Rate in Ratio Form

• Formula: If two investments have equal SI:

$$P_1R_1T_1=P_2R_2T_2$$

Therefore:

$$P_1:P_2=R_2T_2:R_1T_1$$

• Example: Two sums earn equal SI for the same time at 8% and 12%. Find the ratio of their principals. Solution:

$$P_1(8)=P_2(12)$$ $$P_1:P_2=12:8 =\boxed{3:2}$$

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19. Successive Loans / Separate Interest Periods

• Formula: Calculate each period separately:

$$SI_{\text{total}} =\sum\frac{P_iR_iT_i}{100}$$

Do not compound the interest unless the question explicitly specifies it.

• Example: ₹10,000 is borrowed at 8% SI for 2 years and then ₹5,000 is borrowed at 10% SI for 3 years. Find the total interest. Solution: First loan:

$$SI_1=\frac{10000\times8\times2}{100}=1600$$

Second loan:

$$SI_2=\frac{5000\times10\times3}{100}=1500$$

Total:

$$\boxed{₹3,100}$$

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20. Simple Interest with Changing Rates

• Formula: When the rate changes after different periods:

$$SI_{\text{total}} =\frac{P(R_1T_1+R_2T_2+\cdots)}{100}$$

• Example: ₹20,000 is invested at 8% for 2 years and 10% for the next 3 years under SI. Find the total interest. Solution:

$$SI=\frac{20000(8\times2+10\times3)}{100}$$ $$=\frac{20000(46)}{100}$$ $$=\boxed{₹9,200}$$