📚 SSC JE – Quantitative Aptitude
Topic: Percentages
Purpose – Percentages are one of the most frequently asked quantitative topics in the SSC JE exam.
They help you solve problems involving proportional change, profit/loss, discounts, interest, and various real‑life scenarios in a quick and systematic way.
1️⃣ Simple Explanation
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Percentage is a way of expressing a number as a fraction of 100.
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In equations, % is equivalent to the fraction /100.
[ \text{Value} = \frac{\text{Percentage} \times \text{Base}}{100} ]
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Example:
25% of 200 = (\frac{25 \times 200}{100} = 50).
2️⃣ Key Points (Cheat‑Sheet)
| Concept | Formula / Shortcut | Notes |
|---|---|---|
| Percent of a number | (\frac{p}{100} \times N) | (p) = percentage, (N) = number |
| Percentage change (increase/decrease) | (\frac{\text{New} - \text{Old}}{\text{Old}} \times 100% ) | Positive = increase, negative = decrease |
| Profit / Loss | Profit % = (\frac{P}{C} \times 100%) Loss % = (\frac{L}{C} \times 100%) |
(C) = cost price, (P) = profit, (L) = loss |
| Discount | Discount % = (\frac{D}{MRP} \times 100%) | (D) = discount, (MRP) = marked price |
| Compound Interest (C.I.) | (\text{A} = P(1 + r/n)^{nt}) (r) = annual rate, (n) = compounding per year, (t) = years |
Convert (r) to decimal before using |
| Simple Interest | (SI = \frac{P \times r \times t}{100}) | (r) in % |
| Percentage of two numbers | (\frac{a}{b} \times 100%) | “a” is part, “b” is whole |
| Percentage of sum | (\frac{a}{a+b} \times 100%) | Commonly used in “x% of (a+b)” |
| Percentage difference | (\frac{ | a-b |
| Work/Speed | (\text{Speed} = \frac{\text{Distance}}{\text{Time}}) → then apply percent | Often “X% faster” -> scale speed accordingly |
| Ratio to Percentage | Ratio (m:n) → ( \frac{m}{m+n} \times 100%) and ( \frac{n}{m+n} \times 100%) | Convert any ratio to percent |
3️⃣ Worked Example (All‑in‑One)
Problem
A shopkeeper purchases a batch of goods for ₹12 000.
He sells them at a 20 % profit.
During the sale, he offers a 10 % discount on the selling price.
If the goods are sold in 12 days at a constant rate, find:
- Selling price (SP) before discount.
- Discount amount.
- Net amount received.
- Daily earning.
Step‑by‑Step Solution
| # | Calculation | Result |
|---|---|---|
| 1. Selling Price before discount | (SP = CP \times (1 + \frac{20}{100})) | (SP = 12,000 \times 1.20 = ₹14,400) |
| 2. Discount amount | (D = SP \times \frac{10}{100}) | (D = 14,400 \times 0.10 = ₹1,440) |
| 3. Net amount received | (Net = SP - D) | (Net = 14,400 - 1,440 = ₹12,960) |
| 4. Daily earning | (Daily = \frac{Net}{12}) | (Daily = \frac{12,960}{12} = ₹1,080) |
Answer
- SP before discount: ₹14 400
- Discount: ₹1 440
- Net received: ₹12 960
- Daily earning: ₹1 080
4️⃣ Quick Revision Summary
- Percent of a number → multiply by that number/100.
- Change percent → (new‑old)/old × 100.
- Profit/Loss % → (profit/cost) × 100 or (loss/cost) × 100.
- Discount % → (discount/marked price) × 100.
- Simple Interest → (P \times r \times t / 100).
- Compound Interest → (P(1 + r