Simple & Compound Interestसाधारण ब्याज एवं चक्रवृद्धि ब्याज
Master Simple and Compound Interest calculations, rate-time conversions, difference formulas for 2 and 3 years, and population growth models specifically tailored for the Super TET Junior Mathematics exam. This high-yielding module carries 3-4 questions (9-12 marks) requiring high computational speed and exact formula application under negative marking constraints.
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Syllabus Focus & High-Yield Strategy: Master Simple and Compound Interest calculations, rate-time conversions, difference formulas for 2 and 3 years, and population growth models specifically tailored for the Super TET Junior Mathematics exam. This high-yielding module carries 3-4 questions (9-12 marks) requiring high computational speed and exact formula application under negative marking constraints.
1. Fundamental Axioms, Definitions & Foundational Properties
PRT — Principal $ imes$ Rate $ imes$ Time divided by $100$ gives Simple Interest!2. Formulas, Standard Theorems & Algorithmic Procedures
| Category / Case | Formula / Mathematical Rule | Topper Calculation Shortcut |
|---|---|---|
| Simple Interest (SI) | $SI = \frac{P \times R \times T}{100}$ | $SI = P \times (R \times T\%)$ |
| Compound Interest (CI) | $A = P \left(1 + \frac{R}{100}\right)^n$, $CI = A - P$ | Effective Rate via Successive Percentage Formula: $x + y + \frac{xy}{100}$ |
3. Word Problem Modeling & Practical Applications
4. Mathematical Edge Cases, Sign Conventions & Common Traps
PAR — Principal, Amount, Rate; always identify these three before selecting your equation!Comprehensive Mathematical Matrix & Method Comparison
| Parameter / Dimension | Method / Concept A | Method / Concept B | Exam Speed & Accuracy Takeaway |
|---|---|---|---|
| Growth Nature | Linear (Arithmetic Progression) | Exponential (Geometric Progression) | SI is constant per year; CI accelerates yearly. |
| Formula Base | $SI = \frac{P \times R \times T}{100}$ | $A = P \left(1 + \frac{R}{100}\right)^n$ | Use formula fractions for speed during exam calculations. |
| 2-Year Difference | $SI_2 = \frac{2PR}{100}$ | $CI_2 - SI_2 = P \left(\frac{R}{100}\right)^2$ | Memorize the 2-year and 3-year difference formulas to save 90 seconds. |
| Compounding Shift | Annual Flat Rate | Half-Yearly / Quarterly Adjustments | Halve rate and double time for half-yearly compounding. |
1. If a sum of money becomes 3 times itself in 5 years at simple interest, in how many years will it become 7 times itself at the same rate?
At simple interest, if a sum becomes $n$ times in $t_1$ years, the time $t_2$ to become $m$ times is given by $t_2 = t_1 \times \frac{m - 1}{n - 1}$. Substituting values: $t_2 = 5 \times \frac{7 - 1}{3 - 1} = 5 \times \frac{6}{2} = 15 \text{ years}$. Distractor 10 years comes from incorrect proportional scaling ($5 \times \frac{7}{3}$). Option 2 is correct.
2. What is the difference between compound interest and simple interest on ₹ $8000$ for $2$ years at $5\%$ per annum compounded annually?
Using the direct difference formula for 2 years: $\text{Difference} = P \left(\frac{R}{100}\right)^2$. Here $P = 8000$, $R = 5$. Difference $= 8000 \times \left(\frac{5}{100}\right)^2 = 8000 \times \frac{25}{10000} = \frac{8000 \times 1}{400} = \text{₹ } 20$. Option 1 is correct.
3. At what rate percent per annum compound interest will ₹ $1000$ amount to ₹ $1331$ in $3$ years?
Using the compound amount formula: $A = P \left(1 + \frac{R}{100}\right)^n$. $1331 = 1000 \left(1 + \frac{R}{100}\right)^3$. $\frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3$. $\left(\frac{11}{10}\right)^3 = \left(1 + \frac{R}{100}\right)^3$. Taking cube root on both sides: $\frac{11}{10} = 1 + \frac{R}{100} \implies \frac{11}{10} - 1 = \frac{R}{100} \implies \frac{1}{10} = \frac{R}{100} \implies R = 10\%$. Option 2 is correct.
4. A sum is lent at compound interest compounding half-yearly. If the annual rate is $20\%$ p.a. and time is $1$ year, what is the effective annual rate of interest?
For half-yearly compounding, rate per half-year $R' = \frac{20}{2} = 10\%$ and time periods $n = 1 \times 2 = 2$. Amount ratio on Re $1$ = $\left(1 + \frac{10}{100}\right)^2 = (1.1)^2 = 1.21$. Effective interest $= 1.21 - 1 = 0.21$ or $21\%$. Distractor $20\%$ is the nominal rate. Option 1 is correct.
5. The population of a town increases at the rate of $5\%$ per annum. If its current population is $80,000$, what will be its population after $3$ years?
Using population growth formula: $P_3 = P_0 \left(1 + \frac{R}{100}\right)^n = 80000 \left(1 + \frac{5}{100}\right)^3 = 80000 \left(\frac{21}{20}\right)^3 = 80000 \times \frac{9261}{8000} = 10 imes 9261 = 92,610$. Option 0 is correct.
6. If the simple interest on a certain sum for $3$ years at $4\%$ per annum is ₹ $600$, what will be the compound interest on the same sum for the same period and rate?
First find Principal $P$: $SI = \frac{P \times R \times T}{100} \implies 600 = \frac{P \times 4 \times 3}{100} \implies P = \frac{600 \times 100}{12} = \text{₹ } 5000$. Now calculate compound amount for 3 years at $4\%$: $A = 5000 \left(1 + \frac{4}{100}\right)^3 = 5000 \left(\frac{26}{25}\right)^3 = 5000 \times \frac{17576}{15625} = \text{₹ } 5624.32$. Thus $CI = 5624.32 - 5000 = \text{₹ } 624.32$. Option 2 is correct.
7. At a certain rate of simple interest, a principal of ₹ $4000$ amounts to ₹ $5000$ in $4$ years. If the rate of interest is increased by $2\%$ p.a., what will be the new total amount?
Extra simple interest due to $2\%$ increase in rate over $4$ years on ₹ $4000$ is $\Delta SI = \frac{4000 \times 2 \times 4}{100} = \text{₹ } 320$. New Amount = Old Amount + Extra Interest = $5000 + 320 = \text{₹ } 5320$. Option 1 is correct.
8. What sum of money will produce ₹ $140$ as simple interest in $4$ years at $5\%$ per annum?
Using $P = \frac{SI \times 100}{R \times T}$: $P = \frac{140 \times 100}{5 \times 4} = \frac{14000}{20} = \text{₹ } 700$. Option 2 is correct.
9. If the difference between CI and SI on a certain sum of money for $3$ years at $10\%$ per annum is ₹ $31$, find the principal sum.
Using the 3-year difference formula: $\text{Diff} = \frac{PR^2(300+R)}{100^3}$. Here $\text{Diff} = 31, R = 10$. $31 = \frac{P \times 10^2 \times (300+10)}{100^3} = \frac{P \times 100 \times 310}{1000000} = \frac{31000 P}{1000000} = \frac{31P}{1000} \implies P = \frac{31 \times 1000}{31} = \text{₹ } 1000$. Option 2 is correct.
10. A sum of money invested at compound interest doubles itself in $4$ years. In how many years will it become $8$ times itself?
Under compound interest, if a sum becomes $2$ times in $4$ years, it becomes $2^k$ times in $k \times 4$ years. We want it to be $8$ times, which is $2^3$. Therefore, required time $= 3 \times 4 = 12 \text{ years}$. Option 2 is correct.
5-Minute Exam Simulation Mode
Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.
High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)
Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)
Compound vs Simple Interest Growth
SCG
S - Simple is Constant, C - Compound is Cumulative, G - Geometric Growth.
Core Variable Identification
PRT
P - Principal, R - Rate per annum, T - Time in years.
Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)
Negative Marking Trap 1: Forgetting to convert months into years (dividing by 12) before applying interest formulas.
Negative Marking Trap 2: Using the simple interest formula when compound interest is explicitly stated in the word problem.
Negative Marking Trap 3: Confusing the 2-year difference formula with the 3-year difference formula, leading to severe calculation mismatch.
Negative Marking Trap 4: Calculating only the Compound Interest amount $A$ when the question specifically asks for the net Compound Interest $CI = A - P$.