SBI Clerk – Quantitative Aptitude
Topic: Percentages
Percentages form a core part of the SBI Clerk exam.
They test your ability to convert between fractions, decimals and percentages, and to apply those conversions in real‑world problems such as discounts, profit/loss, and interest.
1. Simple Explanation
| Concept | What it means | Example |
|---|---|---|
| Percentage | “Per hundred.” It tells how many parts out of 100 a quantity represents. | 25% of 200 = 0.25 × 200 = 50 |
| Percent Change | Difference between new & old value expressed as a percentage of the old value. | If price rises from ₹80 to ₹100, increase = ((100-80)/80 × 100 = 25%) |
| Discount | Reduction from the list price, expressed as a percentage of the list price. | 20% off on ₹500 → ₹500 – (0.20 × 500) = ₹400 |
| Profit/Loss | Difference between selling & cost price, expressed as a percentage of the cost price. | Selling at ₹120 for a cost of ₹100 → Profit % = ((120-100)/100 × 100 = 20%) |
2. Key Points & Formulae
| Situation | Formula | Notes |
|---|---|---|
| Convert Percent → Decimal | ( \text{Decimal} = \dfrac{\text{Percent}}{100} ) | e.g., 75% = 0.75 |
| Convert Decimal → Percent | ( \text{Percent} = \text{Decimal} × 100 ) | e.g., 0.18 = 18% |
| Percent of a Number | ( \text{Result} = \text{Number} × \dfrac{\text{Percent}}{100} ) | |
| Percent Increase/Decrease | ( \text{New} = \text{Old} × \left(1 ± \dfrac{\text{Percent}}{100}\right) ) | "+" for increase, "−" for decrease |
| Discounted Price | ( \text{Price after discount} = \text{List Price} × \left(1 - \dfrac{\text{Discount%}}{100}\right) ) | |
| Profit/Loss | ( \text{Profit%} = \dfrac{\text{Selling Price} - \text{Cost Price}}{\text{Cost Price}} × 100 ) ( \text{Loss%} = \dfrac{\text{Cost Price} - \text{Selling Price}}{\text{Cost Price}} × 100 ) |
|
| Reverse Percentage (Given Result & Percent) | ( \text{Base} = \dfrac{\text{Result}}{\text{Percent}/100} ) | e.g., 30% of a number is 15 → number = (15 / 0.30 = 50) |
| Compound Interest (Optional) | ( A = P (1 + r/100)^n ) | (r) = annual rate, (n) = years |
3. Worked Example (Mixed Problem)
Question:
A shopkeeper lists a jacket at ₹2,400. He offers a 15% discount to a customer. The customer pays ₹2,040.
The shopkeeper then sells the jacket to a second customer for ₹2,520.a) Confirm the price after discount.
b) Calculate the shopkeeper’s profit percentage for the second sale.
c) If the shopkeeper had bought the jacket for ₹2,000, what would be his profit percentage on the second sale?
a) Price after discount
[ \text{Price after discount} = 2400 \times \left(1 - \dfrac{15}{100}\right) = 2400 \times 0.85 = 2040 ]
✓ Matches the amount paid → discount is correctly applied.
b) Profit % on second sale (Cost Price = ₹2,400)
[ \text{Profit} = 2520 - 2400 = 120 ]
[ \text{Profit %} = \frac{120}{2400} \times 100 = 5% ]
c) Profit % if Cost Price = ₹2,000
[ \text{Profit} = 2520 - 2000 = 520 ]
[ \text{Profit %} = \frac{520}{2000} \times 100 = 26% ]
4. Quick Revision Summary
- Percent → Decimal: divide by 100.
- Decimal → Percent: multiply by 100.
- X% of Y: ( Y × X/100 ).
- Increase/Decrease: ( \text{New} = \text{Old} × (1 ± X/100) ).
- Discounted price: ( \text{Price} = \text{List} × (1 - d/100) ).
- Profit %: ( \frac{\text{SP} - \text{CP}}{\text{CP}} × 100 ).
- Loss %: ( \frac{\text{CP} - \text{SP}}{\text{CP}} × 100 ).
- Reverse: If (X%) of a number = (Y), then number = (Y / (X/100)).
Quick Practice Questions
- What is 12% of 350?
- A price is increased by 8% and then decreased by 5%. What is the net change?
- If a shop offers a 10% discount on ₹1,200, what is the sale price?
- A product is sold for ₹1,500, costing ₹1,200. What is the profit %?
Answer them mentally before checking the solutions to test your speed.
Good luck with your preparation! Focus