Simple and Compund Interest — BSSC Quantitative Aptitude

Explanation:

• Interest = Amount - Principal = 6525 - 4500 = 2025.

• Rate = $\frac{I \times 100}{P \times T}$.

• Rate = $\frac{2025 \times 100}{4500 \times 5} = 9\%$.

Condition: Simple Interest remains constant every year.

Formula: $SI = \frac{P \times R \times T}{100}$.

Exam Trick:

• 1 year interest = $\frac{2025}{5} = 405$.

• Rate = $\frac{405}{4500} \times 100 = 9\%$.

Explanation:

• 1st year CI = 10% of 1000 = 100.

• 2nd year CI = 100 (on principal) + 10% of 100 (on interest) = 110.

• Total CI = 100 + 110 = 210.

Condition: Interest is calculated on the accumulated interest of previous years.

Formula: $A = P\left(1 + \frac{R}{100}\right)^T$

Exam Trick:

• Effective rate for 2 years at 10% = $10 + 10 + \frac{10 \times 10}{100} = 21\%$.

• CI = 21% of 1000 = 210.

Explanation:

• Let Principal be P.

• Difference = $P\left(\frac{R}{100}\right)^2$.

• $30 = P\left(\frac{5}{100}\right)^2 = P\left(\frac{1}{20}\right)^2$.

• $P = 30 \times 400 = 12000$.

Condition: This formula is valid only for the difference between CI and SI for exactly 2 years.

Formula: $D = P\left(\frac{R}{100}\right)^2$.

Exam Trick:

• Effective CI for 2 years = 10.25%, SI = 10%.

• Difference = 0.25%.

• 0.25% of $P = 30 \Rightarrow 1\% = 120 \Rightarrow 100\% = 12000$.

Explanation:

• Let principal be P. Amount becomes 2P, so Interest = P in 7 years.

• For amount to become 5P, Interest needed = 4P.

• Since SI is constant, time taken = $4 \times 7 = 28$ years.

Condition: The rate of interest is the same throughout the period.

Formula: $T_2 = \frac{N_2 - 1}{N_1 - 1} \times T_1$

Exam Trick:

• Interest for 2 times = 1 unit $\rightarrow$ 7 years.

• Interest for 5 times = 4 units $\rightarrow 4 \times 7 = 28$ years.

Explanation:

• The sum becomes 2 times in 3 years.

• It becomes $2 \times 2 = 4$ times in next 3 years (Total 6 years).

• It becomes $4 \times 2 = 8$ times in next 3 years (Total 9 years).

Condition: The interest is compounded annually.

Formula: If a sum becomes $x$ times in $T$ years, it becomes $x^n$ times in $n \times T$ years.

Exam Trick:

• $2^1 \rightarrow$ 3 years.

• $8 = 2^3 \rightarrow 3 \times 3 = 9$ years.

Explanation:

• Rate per quarter = $\frac{40\%}{4} = 10\%$.

• Time = 1 year = 4 quarters.

• Effective rate for 4 cycles of 10% = 46.41%.

• CI = 46.41% of 80000 = 37128.

Condition: Quarterly compounding means interest is calculated 4 times a year.

Formula: $A = P\left(1 + \frac{R}{400}\right)^{4T}$

Exam Trick:

• Remember Pascal's ratio for 4 cycles: 4, 6, 4, 1.

• CI = $4(8000) + 6(800) + 4(80) + 1(8) = 32000 + 4800 + 320 + 8 = 37128$.

Explanation:

• For 3 years at 10%, SI = 30%.

• For 3 years at 10%, CI = 33.1%.

• Difference = 33.1% - 30% = 3.1%.

• 3.1% of Principal = 155.

• $1\% = 50$, so $100\% = 5000$.

Condition: Rate is compounded annually.

Formula: Difference for 3 years = $P\left(\frac{R}{100}\right)^2 \left(\frac{R}{100} + 3\right)$.

Exam Trick:

• Direct difference percentage for 10% over 3 years is always 3.1%.

• $P = \frac{155}{3.1} \times 100 = 5000$.

Explanation:

• 1st year CI = 20% of 6000 = 1200.

• 2nd year CI = 1st year CI + 20% of 1st year CI = $1200 + 240 = 1440$.

• 3rd year CI = 2nd year CI + 20% of 2nd year CI = $1440 + 288 = 1728$.

• Difference = 1728 - 1440 = 288.

Condition: The CI for any year is the previous year's CI increased by the rate of interest.

Formula: CI of $n^{th}$ year = CI of $(n-1)^{th}$ year $\times \left(1 + \frac{R}{100}\right)$.

Exam Trick:

• The difference between the 2nd and 3rd year CI is simply the interest on the 2nd year's CI.

• Difference = 20% of 1440 = 288.

Explanation:

• 73 days = $\frac{73}{365}$ = $\frac{1}{5}$ year.

• For the first two years, the interest rate is 20% per annum.

• For the remaining $\frac{1}{5}$ year, the effective rate is $\frac{1}{5} \times 20\% = 4\%$.

• We calculate successive percentages: first 20% and 20% gives $20 + 20 + \frac{20 \times 20}{100} = 44\%$.

• Now, applying successive percentage with 44% and 4%: $44 + 4 + \frac{44 \times 4}{100} = 48 + 1.76 = 49.76\%$.

• CI = 49.76% of 5000 = $50 \times 49.76 = 2488$.

Condition: Time is given in fractional years, so the rate for the fraction is adjusted proportionally.

Formula: Successive increase $X + Y + \frac{XY}{100}$.

Exam Trick:

• Do not calculate year by year with large numbers. Find the net effective percentage rate first (49.76%) and multiply directly by the principal.

Explanation:

• Let A's share be A and B's share be B.

• According to the problem: $A \left(1 + \frac{4}{100}\right)^7 = B \left(1 + \frac{4}{100}\right)^9$.

• $\frac{A}{B} = \left(1 + \frac{4}{100}\right)^2 = \left(\frac{26}{25}\right)^2 = \frac{676}{625}$.

• Total parts = $676 + 625 = 1301$.

• $1301 \text{ parts} = 3903 \Rightarrow 1 \text{ part} = 3$.

• A's share = $676 \times 3 = 2028$.

Condition: The compounding rate applies equally to both, differing only by the time invested.

Formula: $A = P\left(1 + \frac{R}{100}\right)^T$.

Exam Trick:

• The ratio of amounts for a 2-year gap at 4% ($\frac{1}{25}$) directly becomes $(25+1)^2 : (25)^2 = 676 : 625$. Multiply ratio parts directly to match the total sum.