Competitive Exams / CUET UG / Simple Interest & Compound Interest

Simple Interest & Compound Interest for CUET UG

Master SI, CI, periodic compounding, effective rate, present worth, depreciation, and population growth with concept-first notes, solved examples, and a timed practice route in the same competitive-exams UI flow.

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Overview

Why This Chapter Matters in CUET

This chapter looks formula-driven on the surface, but most questions really test whether you can identify the correct growth model quickly. SI is flat growth. CI is repeated growth. Present worth, depreciation, and population are just extensions of the same idea.

Once you train that recognition step, the chapter becomes a reliable scoring block because the arithmetic is usually cleaner than ratio-heavy or geometry-heavy topics.

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Section A

Notes & Concept Builder

Identify flat growth vs compounding first

1. Simple Interest Basics 2. Rate, Time, and Principal from SI 3. Doubling, Tripling, and Multiplier Logic in SI 4. Compound Interest Core Idea 5. Half-Yearly and Quarterly Compounding 6. Difference Between SI and CI 7. Effective Annual Rate 8. Mixed Yearly Rates and Successive Growth 9. Present Worth, Depreciation, and Population 10. Exam Strategy for SI and CI

1. Simple Interest Basics

Simple interest is the flat-growth model of the chapter. Interest is always calculated on the original principal, so the interest added every year stays constant.

SI=\frac{PRT}{100} \qquad A=P+SI=P\left(1+\frac{RT}{100}\right)

Because the model is linear, doubling the time doubles the interest and halving the rate halves the interest.

10-second shortcut: In SI, the yearly increase is fixed. If you know one year's interest, multiply directly for the total years.

2. Rate, Time, and Principal from SI

Many CUET questions are reverse questions: amount is given and you must recover the missing rate, time, or principal.

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

Keep time in years. Convert months to fractional years before substitution.

3. Doubling, Tripling, and Multiplier Logic in SI

If a sum doubles under SI, then the total interest equals the principal. If it becomes three times, the interest is twice the principal, and so on.

\text{If amount becomes } mP, \text{ then interest earned }=(m-1)P

This makes doubling and tripling questions extremely fast because principal cancels from both sides.

10-second shortcut: If a sum doubles in 8 years under SI, then it needs 24 years to become four times because required interest is 3P, three times larger.

4. Compound Interest Core Idea

Compound interest is the repeated-growth model. Each period's interest is added back to the principal, so the next interest is calculated on a bigger amount.

A=P\left(1+\frac{R}{100}\right)^n \qquad CI=A-P

This is why CI exceeds SI whenever the same principal, rate, and time are used for more than one period.

5. Half-Yearly and Quarterly Compounding

Periodic compounding changes both the rate per period and the number of periods. This is one of the most tested traps in the chapter.

A=P\left(1+\frac{R}{200}\right)^{2n} \text{ for half-yearly compounding}

A=P\left(1+\frac{R}{400}\right)^{4n} \text{ for quarterly compounding}

10-second shortcut: If compounding happens k times per year, divide the rate by k and multiply the time by k.

6. Difference Between SI and CI

For two years, the difference between compound interest and simple interest comes from the extra interest earned on the first year's interest.

CI-SI=P\left(\frac{R}{100}\right)^2 \qquad \text{for 2 years}

This is one of the highest-value shortcuts in the chapter because it lets you recover principal or rate directly without expanding the full CI formula.

7. Effective Annual Rate

Half-yearly and quarterly compounding create an effective annual rate that is slightly higher than the stated nominal rate.

\text{Effective annual rate} = \left(1+\frac{r}{k}\right)^k - 1

At 10% nominal compounded half-yearly, the effective annual rate is $(1.05)^2 - 1 = 10.25\%$ .

8. Mixed Yearly Rates and Successive Growth

If the yearly rate changes, do not average it blindly. Under CI, simply multiply the yearly growth factors one after another.

A=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)\cdots

This same logic appears in population growth, repeated discounts, and appreciation-depreciation models.

9. Present Worth, Depreciation, and Population

Present worth means reversing compounding. Depreciation means repeated percentage decrease. Population growth means repeated percentage increase.

\text{Present worth} = \frac{\text{Future value}}{\left(1+\frac{R}{100}\right)^n}

\text{Depreciated value} = V\left(1-\frac{r}{100}\right)^n

As soon as you see “grows every year” or “depreciates annually,” shift into compound-percentage thinking.

10. Exam Strategy for SI and CI

First identify the model: flat growth or compounding. Then identify whether the question is forward, reverse, or comparison-based. That three-step check solves most confusion instantly.

Common mistake: Students often forget to adjust rate and periods for half-yearly or quarterly CI, or they use the CI formula for a plain SI question.

Strong performance in this chapter comes from recognizing the right model in the first five seconds.

Solved Practice

Solved Examples

Try first, then open the reasoning

Exam Trap: Most mistakes happen when students forget to convert months into years, or they fail to adjust both rate and periods for half-yearly and quarterly CI.

Example 1: What is the simple interest on Rs 18,000 at 12% for 2 years?

Use $SI=\frac{PRT}{100}$ .

$SI=\frac{18000 \times 12 \times 2}{100}=4320$ .

Example 2: A sum earns Rs 2,700 as SI in 3 years at 9%. Find the principal.

$P=\frac{SI\times100}{RT}=\frac{2700\times100}{9\times3}=10000$ .

Example 3: At what rate will Rs 12,000 earn Rs 3,600 in 2.5 years under SI?

$R=\frac{SI\times100}{PT}=\frac{3600\times100}{12000\times2.5}=12\%$ .

Example 4: A sum doubles in 8 years under SI. Find the rate.

If a sum doubles, interest equals principal.

So $RT=100$ . With $T=8$ , $R=12.5\%$ .

Example 5: A sum becomes three times in 10 years under SI. Find the rate.

Tripling means interest is $2P$ .

So $RT=200$ , giving $R=20\%$ .

Example 6: Find the amount on Rs 8,000 for 2 years at 10% compounded annually.

$A=8000(1.1)^2=9680$ .

Example 7: What is the CI on Rs 15,000 for 2 years at 8% compounded annually?

$A=15000(1.08)^2=17496$ .

So $CI=17496-15000=2496$ .

Example 8: A sum becomes Rs 19,845 in 2 years at 5% compounded annually. Find the principal.

$P=\frac{19845}{(1.05)^2}=18000$ .

Example 9: At what rate will Rs 16,000 become Rs 19,360 in 2 years at annual CI?

We need $(1+r)^2=19360/16000=1.21$ .

So $1+r=1.1$ and $r=10\%$ .

Example 10: How many years will Rs 20,000 become Rs 24,200 at 10% annual CI?

$20000(1.1)^2=24200$ , so the time is 2 years.

Example 11: Find the amount on Rs 40,000 for 1 year at 12% compounded half-yearly.

Half-yearly rate is 6% and periods are 2.

$A=40000(1.06)^2=44944$ .

Example 12: What is the CI on Rs 20,000 for 18 months at 8% compounded half-yearly?

Half-yearly rate is 4% and periods are 3.

$A=20000(1.04)^3=22497.28$ , so $CI=2497.28$ .

Example 13: What is the amount on Rs 16,000 for 9 months at 12% compounded quarterly?

Quarterly rate is 3% and periods are 3.

$A=16000(1.03)^3=17482.94$ .

Example 14: Find the CI on Rs 12,500 for 1 year at 16% compounded quarterly.

Quarterly rate is 4% and periods are 4.

$A=12500(1.04)^4=14623.20$ , so $CI=2123.20$ .

Example 15: Difference between CI and SI on Rs 25,000 for 2 years at 12% is:

For 2 years, difference $=P\left(\frac{R}{100}\right)^2$ .

$=25000\times\frac{144}{10000}=360$ .

Example 16: At the same rate for 2 years, CI exceeds SI by Rs 100 on a principal of Rs 10,000. Find the rate.

$100=10000\left(\frac{R}{100}\right)^2$ gives $R^2=100$ .

So $R=10\%$ .

Example 17: A sum of Rs 10,000 is invested at 10%, 12%, and 8% in successive years under CI. Find the amount after 3 years.

Use successive multiplication: $10000\times1.10\times1.12\times1.08=13305.60$ .

Example 18: The population of a town is 50,000 and grows by 4% annually. Find the population after 2 years.

Population growth follows compounding.

$50000(1.04)^2=54080$ .

Example 19: The value of a machine depreciates 10% per year. If present value is Rs 58,320 after 2 years, find original price.

$P(0.9)^2=58320$ .

So $P=\frac{58320}{0.81}=72000$ .

Example 20: Present worth of Rs 14,520 due after 2 years at 10% CI is:

Present worth $=\frac{14520}{(1.1)^2}=12000$ .

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Section B

The Test Zone

This chapter has a separate practice route with 4 sectional sessions of 10 questions each and a mixed 40-question mock. Every question runs on a 60-second timer so we train formula selection, speed, and reverse-reading skill together.

Sectional Tests

4 focused sessions on SI basics, annual CI, periodic compounding, and application-style questions.

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Full Mixed Mock

40 questions blending SI, CI, effective rate, present worth, depreciation, and growth in CUET style.

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