Arithmetic

Percentage, Ratio-Proportion, Profit-Loss and Interest

---

Overview

Arithmetic forms the backbone of the Mathematics section in TS TET Paper I and Paper II. This topic tests your ability to apply fundamental operations to real-world problems involving money, comparisons, and growth over time. Questions typically appear as word problems requiring quick mental math and formula application.

For TS TET, you must master the interconversion between percentages, fractions, and decimals, understand how ratios scale quantities, calculate profit/loss in trade scenarios, and compute simple and compound interest. These concepts also form the foundation for teaching primary and upper primary students, so expect pedagogy-linked questions on how to explain these ideas using real-life examples.

Speed and accuracy matter. Memorise key fraction-percentage equivalents, practise mental shortcuts, and always verify units in word problems.

---

Key Concepts

• **Percentage** means "per hundred." It standardises comparisons by expressing a part as a fraction of 100. Converting between fraction, decimal, and percentage is essential.

• **Ratio** compares two quantities of the same kind (a:b), while **proportion** states that two ratios are equal (a:b = c:d). Cross-multiplication solves most proportion problems.

• **Profit and Loss** measure gain or loss relative to the Cost Price (CP). Profit = SP − CP; Loss = CP − SP. Percentages are always calculated on CP unless stated otherwise.

• **Simple Interest (SI)** grows linearly—interest is calculated only on the principal. **Compound Interest (CI)** grows exponentially—interest is added to principal each period.

• **Marked Price (MP)** and **Discount** are common in retail problems. Discount is calculated on MP, not CP.

• In ratio problems, the **constant of proportionality** (k) helps find actual quantities when only the ratio and sum/difference are given.

• **Successive percentages** (like two discounts or two increases) cannot simply be added; use the net effect formula.

• **Population and depreciation** problems use compound interest logic with growth rate (+r) or decay rate (−r).

---

Formulas / Key Facts

| Concept | Formula | |---------|---------| | Percentage of a number | (Percentage × Number) / 100 | | Fraction to Percentage | (Fraction) × 100% | | Percentage Change | [(New − Old) / Old] × 100% | | Ratio a:b in fractions | a/(a+b) and b/(a+b) of total | | Proportion (cross-multiply) | If a:b = c:d, then a×d = b×c | | Profit | SP − CP | | Loss | CP − SP | | Profit % | (Profit / CP) × 100 | | Loss % | (Loss / CP) × 100 | | SP when Profit % given | CP × (100 + Profit%) / 100 | | SP when Loss % given | CP × (100 − Loss%) / 100 | | Discount | MP − SP | | Discount % | (Discount / MP) × 100 | | Simple Interest | SI = (P × R × T) / 100 | | Amount (SI) | A = P + SI = P(1 + RT/100) | | Compound Interest | A = P(1 + R/100)^T | | CI | A − P | | Successive % change | Net = a + b + (ab/100) for two changes a% and b% |

**Key fraction-percentage equivalents:**

• 1/2 = 50%, 1/3 ≈ 33.33%, 1/4 = 25%, 1/5 = 20%, 1/6 ≈ 16.67%, 1/8 = 12.5%, 1/10 = 10%

---

Worked Examples

### Example 1: Percentage Increase **Problem:** A teacher's salary increased from ₹25,000 to ₹28,000. Find the percentage increase.

**Solution:**

• Increase = 28,000 − 25,000 = ₹3,000
• Percentage Increase = (3,000 / 25,000) × 100 = 12%

**Answer:** 12%

---

### Example 2: Ratio and Proportion **Problem:** Two numbers are in the ratio 3:5. If their sum is 96, find the numbers.

**Solution:**

• Total parts = 3 + 5 = 8
• Value of one part = 96 / 8 = 12
• First number = 3 × 12 = 36
• Second number = 5 × 12 = 60

**Answer:** 36 and 60

---

### Example 3: Profit and Loss **Problem:** A shopkeeper buys an article for ₹400 and sells it at a profit of 15%. Find the selling price.

**Solution:**

• SP = CP × (100 + Profit%) / 100
• SP = 400 × (100 + 15) / 100 = 400 × 115 / 100 = ₹460

**Answer:** ₹460

---

### Example 4: Simple Interest **Problem:** Find the simple interest on ₹8,000 at 5% per annum for 3 years.

**Solution:**

• SI = (P × R × T) / 100
• SI = (8,000 × 5 × 3) / 100 = 1,20,000 / 100 = ₹1,200

**Answer:** ₹1,200

---

### Example 5: Compound Interest **Problem:** Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.

**Solution:**

• A = P(1 + R/100)^T
• A = 10,000 × (1 + 10/100)² = 10,000 × (1.1)² = 10,000 × 1.21 = ₹12,100
• CI = A − P = 12,100 − 10,000 = ₹2,100

**Answer:** ₹2,100

---

Common Mistakes

• **Calculating profit/loss percentage on SP instead of CP** → Always use CP as the base for profit% and loss% unless the question specifies otherwise.

• **Adding successive percentages directly** → Two successive discounts of 10% each ≠ 20% discount. Use: Net = 10 + 10 + (10×10)/100 = 19% discount, or calculate step by step.

• **Confusing ratio with actual values** → A ratio 2:3 does not mean the numbers are 2 and 3; multiply by the constant k to get actual values.

• **Forgetting to convert time units in interest problems** → If rate is per annum but time is in months, convert months to years (e.g., 6 months = 0.5 years).

• **Applying SI formula for CI problems** → SI grows linearly; CI grows on accumulated amount. For CI, always use the power formula.

• **Ignoring "on MP" vs "on CP"** → Discount is always on Marked Price; profit/loss is on Cost Price. Mixing these leads to wrong answers.

---

Quick Reference

1. Percentage = (Part / Whole) × 100

2. In ratio a:b, first quantity = a/(a+b) × Total

3. Profit% and Loss% are always on CP

4. SI = PRT/100; CI uses A = P(1 + R/100)^T

5. Successive changes: Net = a + b + ab/100

6. Discount% is calculated on Marked Price, not Cost Price