Introduction
Welcome to Part 2 of the BrainQuro Percentage Questions with Answers Series.
This practice set contains 10 completely original percentage questions (Questions 11–20) designed for IBPS PO, SBI PO, RBI Assistant, LIC AAO, RRB, SSC CGL, and other competitive exams.
Unlike traditional solutions, every question is solved using a shortcut approach that helps improve speed and accuracy. The questions cover important concepts such as:
- Successive Percentage Change
- Reverse Percentage
- Markup & Discount
- Population
- Depreciation
- Salary
- Examination Marks
If you haven’t attempted the previous set, solve Percentage Questions with Answers – Part 1 before continuing.
Table of Contents
Question 11
Question
The marked price of a laptop was increased by 25%. During a festive sale, a 12% discount was offered on the increased price. If the customer finally paid ₹55,000, find the original marked price of the laptop.
Information
- Increase = 25%
- Discount = 12%
- Final Price = ₹55,000
Solution
Assume the original marked price = 100 parts
After a 25% increase:
100 → 125 parts
After a 12% discount:
125 × 88% = 110 parts
So,
110 parts = ₹55,000
1 part = ₹500
Original marked price = 100 × ₹500
= ₹50,000
✅ Final Answer
₹50,000
Question 12
Question
A candidate secured 88% marks in an examination and obtained 704 marks. Find the maximum marks of the examination.
Information
- Percentage Obtained = 88%
- Marks Obtained = 704
Solution
88% = 22/25
22 parts = 704
1 part = 32
Maximum marks = 25 × 32
= 800
✅ Final Answer
800 Marks
Question 13
Question
A dealer marked the price of a refrigerator 60% above its cost price and then allowed a 25% discount. If the customer finally paid ₹48,000, find the cost price of the refrigerator.
Information
- Markup = 60%
- Discount = 25%
- Selling Price = ₹48,000
Solution
Assume the cost price = 100 parts
After 60% markup:
100 → 160 parts
After 25% discount:
160 × 75% = 120 parts
So,
120 parts = ₹48,000
1 part = ₹400
Cost price = 100 × ₹400
= ₹40,000
✅ Final Answer
₹40,000
Question 14
Question
The population of a city increased by 25% in the first year and by 20% in the second year. If the population at the end of the second year became 1,50,000, find the original population.
Information
- First Year Increase = 25%
- Second Year Increase = 20%
- Final Population = 1,50,000
Solution
Assume the original population = 100 parts
After the first year:
100 → 125 parts
After the second year:
125 → 150 parts
So,
150 parts = 1,50,000
1 part = 1,000
Original population = 100 × 1,000
= 1,00,000
✅ Final Answer
1,00,000
Question 15
Question
A shopkeeper marked an air conditioner 50% above its cost price. He then offered successive discounts of 20% and 10%. If the customer finally paid ₹81,000, find the cost price of the air conditioner.
Information
- Markup = 50%
- First Discount = 20%
- Second Discount = 10%
- Selling Price = ₹81,000
Solution
Assume the cost price = 100 parts
After 50% markup:
100 → 150 parts
After 20% discount:
150 → 120 parts
After 10% discount:
120 → 108 parts
So,
108 parts = ₹81,000
1 part = ₹750
Cost price = 100 × ₹750
= ₹75,000
✅ Final Answer
₹75,000
Question 16
Question
A candidate obtained 62.5% marks in an examination and scored 750 marks. Find the maximum marks of the examination.
Information
- Percentage Obtained = 62.5%
- Marks Obtained = 750
Solution
62.5% = 5/8
5 parts = 750
1 part = 150
Maximum marks = 8 × 150
= 1,200
✅ Final Answer
1,200 Marks
Question 17
Question
A wholesaler increased the price of a water purifier by 20%. A retailer purchased it and marked it up by 25%. During a promotional sale, the retailer offered a 20% discount. If the customer finally paid ₹72,000, find the wholesaler’s original price.
Information
- Wholesaler’s Price Increase = 20%
- Retailer’s Markup = 25%
- Retailer’s Discount = 20%
- Final Selling Price = ₹72,000
Solution
Assume the wholesaler’s original price = 100 parts
After 20% increase:
100 → 120 parts
After 25% markup:
120 → 150 parts
After 20% discount:
150 × 80% = 120 parts
So,
120 parts = ₹72,000
1 part = ₹600
Original price = 100 × ₹600
= ₹60,000
✅ Final Answer
₹60,000
Question 18
Question
In an examination, 18% of the students failed. If 1,722 students passed the examination, find the total number of students who appeared.
Information
- Failed Students = 18%
- Passed Students = 1,722
Solution
Passed percentage = 100% − 18% = 82%
82% = 41/50
41 parts = 1,722
1 part = 42
Total students = 50 × 42
= 2,100
✅ Final Answer
2,100 Students
Question 19
Question
The value of a machine depreciated by 20% in the first year and by 10% in the second year. If its value after two years is ₹2,16,000, find its original value.
Information
- First Year Depreciation = 20%
- Second Year Depreciation = 10%
- Present Value = ₹2,16,000
Solution
Assume the original value = 100 parts
After 20% depreciation:
100 → 80 parts
After 10% depreciation:
80 → 72 parts
So,
72 parts = ₹2,16,000
1 part = ₹3,000
Original value = 100 × ₹3,000
= ₹3,00,000
✅ Final Answer
₹3,00,000
Question 20
Question
The monthly salary of an employee was increased by 30%. He then started saving 20% of his new salary. If his monthly savings are ₹23,400, find his monthly salary before the increment.
Information
- Salary Increase = 30%
- Savings = 20% of New Salary
- Monthly Savings = ₹23,400
Solution
Since savings are 20%, the new salary represents 100%.
20% = 1/5
If 1 part = ₹23,400
New salary = 5 parts
= ₹1,17,000
Now,
New salary = 130 parts
130 parts = ₹1,17,000
1 part = ₹900
Original salary = 100 × ₹900
= ₹90,000
✅ Final Answer
₹90,000
After completing this practice set, continue with:
- Percentage Questions with Answers (1–10) – Part 1
- Percentage Questions with Answers (21–30) – Part 3
- Profit and Loss Questions with Answers
⭐ Shortcut Trick of the Day
Instead of remembering dozens of formulas, memorize these important percentage-to-fraction conversions:
| Percentage | Fraction |
|---|---|
| 12.5% | 1/8 |
| 20% | 1/5 |
| 25% | 1/4 |
| 37.5% | 3/8 |
| 40% | 2/5 |
| 50% | 1/2 |
| 62.5% | 5/8 |
| 72% | 18/25 |
| 84% | 21/25 |
| 88% | 22/25 |
These conversions help solve banking exam questions much faster.
⚠️ Common Mistakes to Avoid
- Never add successive percentage increases and decreases directly.
- Always identify whether the given value represents the original amount or the final amount.
- Don’t apply discount and markup on the same base value.
- Convert familiar percentages into fractions whenever possible.
- In reverse percentage questions, first determine what percentage the given amount represents.
- Read words like “after,” “before,” “increased,” and “decreased” carefully before solving.
💡 Banking Exam Speed Tips
- Use the 100 parts method for markup, discount, and reverse percentage questions.
- Memorize common fraction equivalents like 5/8 = 62.5% and 22/25 = 88%.
- Look for percentage changes that simplify calculations, such as +25% followed by −20%.
- Solve mentally whenever the values divide evenly instead of using lengthy calculations.
- In prelims, try to solve each percentage question within 45–60 seconds.
