Mastering percentages goes far beyond calculating basic fractions—it is the hidden key to cracking some of the trickiest questions in competitive exams like SSC CGL, Banking, and Railways. If you want to conquer quantitative aptitude, percentages form the core foundation that ties seamlessly into topics like Ratio and Proportion and Profit and Loss.
Welcome to Part 8 of our Percentage Master Series (Questions 71 to 80). In this post, we trade lengthy, time-consuming traditional formulas for lightning-fast shortcut methods. Whether you are dealing with complex election vote counts, successive population changes, income-savings ratios, or alligation-based alloy mixtures, these exam-tested techniques will help you arrive at accurate answers in seconds.
Let’s dive right into these 10 high-yield questions and level up your math preparation!
Table of Contents
Percentage Practice Set 8: Advanced Questions (Q71-80) with Shortcuts
71. Election – Registered Voters
📝 Question
In an election between two candidates, 20% of the registered voters did not cast their votes, and 200 of the cast votes were declared invalid. The winning candidate secured 55% of the valid votes and defeated the opponent by 800 votes. If the winning candidate received 4,400 valid votes, find the total number of registered voters.
⚡ Shortcut Method
Winning candidate = 55% of valid votes = 4,400
Therefore, 100% valid votes = 4,400 × 100/55 = 8,000
Opponent’s votes = 8,000 − 4,400 = 3,600
Cast votes = 8,000 + 200 = 8,200
Since 20% did not vote, 80% of registered voters = 8,200
Registered voters = 8,200 × 100/80 = 10,250
✅ Answer
10,250
72. Income and Savings
📝 Question
The income of A is 25% more than that of B. B’s expenditure is 20% less than A’s expenditure. If both A and B save 30% of their respective incomes, and the difference between their savings is Rs. 3,000, find the income of A in Rupees.
⚡ Shortcut Method
Let B’s income = 100
A’s income = 125
Both save 30%, so their savings ratio is also:
A : B = 125 : 100 = 5 : 4
Difference = 1 part = Rs. 3,000
Therefore, B’s savings = Rs. 12,000
Since B saves 30%:
B’s income = 12,000 × 100/30 = Rs. 40,000
A’s income = 40,000 × 125/100 = Rs. 50,000
Consistency check:
A’s expenditure = 70% of 50,000 = Rs. 35,000
B’s expenditure = 70% of 40,000 = Rs. 28,000
Rs. 28,000 is exactly 20% less than Rs. 35,000. ✔️
✅ Answer
Rs. 50,000
73. Milk-Water Mixture
📝 Question
A vessel contains 500 litres of a milk-water mixture with 30% water. ‘x’ litres of pure milk is added, reducing the water concentration to 20%. Then, ‘y’ litres of pure water is added to the resulting mixture, increasing the water concentration to 25%. Find the value of (x − y).
⚡ Shortcut Method
Initial water = 30% of 500 = 150 litres
After adding x litres milk:
150/(500 + x) = 20%
Therefore:
500 + x = 750
x = 250 litres
New mixture = 750 litres
After adding y litres water:
(150 + y)/(750 + y) = 25%
Solving gives:
y = 50 litres
Therefore:
x − y = 250 − 50 = 200
✅ Answer
200
74. Increase in Sugar Price
📝 Question
The price of sugar increases by 30%. Due to this, a family decreases its monthly consumption by 20% and consequently ends up spending Rs. 2,400 more on sugar per month than before. Find their initial monthly expenditure on sugar in Rupees.
⚡ Shortcut Method
Price multiplier = 130%
Consumption multiplier = 80%
New expenditure multiplier:
130% × 80% = 104%
Thus, expenditure increases by 4%.
Given increase = Rs. 2,400
Therefore:
4% of initial expenditure = 2,400
Initial expenditure = 2,400 × 100/4
= Rs. 60,000
✅ Answer
Rs. 60,000
75. Salesman’s Commission
📝 Question
A salesman receives a commission of 9% on total sales and an additional bonus of 1% on sales exceeding Rs. 20,000. If his total earnings are Rs. 6,800, find his total sales in Rupees.
⚡ Shortcut Method
Let total sales = S
Commission = 9% of S
Additional bonus = 1% of (S − 20,000)
Therefore:
9% of S + 1% of (S − 20,000) = 6,800
This gives:
10% of S − 200 = 6,800
10% of S = 7,000
S = Rs. 70,000
✅ Answer
Rs. 70,000
76. Population Change
📝 Question
The population of a town increased by 20% in the first year, decreased by 25% in the second year, increased by 40% in the third year, and decreased by 10% in the fourth year. If the final population at the end of the fourth year is 56,700, find the initial population of the town.
⚡ Shortcut Method
Overall multiplier:
1.20 × 0.75 × 1.40 × 0.90
= 1.134
Therefore:
Initial population × 1.134 = 56,700
Initial population = 56,700/1.134
= 50,000
✅ Answer
50,000
77. Examination – Two Subjects
📝 Question
In an examination, 60% of the candidates passed in English, 50% passed in Mathematics, and 15% passed in both subjects. If 350 candidates failed in both subjects, find the total number of candidates who appeared for the exam.
⚡ Shortcut Method
Candidates passing at least one subject:
60% + 50% − 15% = 95%
Therefore, candidates failing both:
100% − 95% = 5%
Given 5% = 350
Total candidates:
350 × 100/5 = 7,000
✅ Answer
7,000
78. Alloy Mixture
📝 Question
40 kg of Alloy A containing 30% copper is mixed with ‘m’ kg of Alloy B containing 60% copper to form a resulting mixture containing 45% copper. Find the value of m in kg.
⚡ Shortcut Method
Using alligation:
Alloy A = 30% copper
Alloy B = 60% copper
Required mixture = 45% copper
Ratio of A : B:
(60 − 45) : (45 − 30)
= 15 : 15
= 1 : 1
Therefore, equal quantities of both alloys are required.
Since Alloy A = 40 kg,
m = 40 kg
✅ Answer
40 kg
79. Income and Expenditure
📝 Question
A man spends 30% of his monthly income on rent, 25% of the remainder on food, and 20% of the further remainder on clothing. If he deposits the remaining amount into his savings account, which totals Rs. 10,080 per month, find his monthly income in Rupees.
⚡ Shortcut Method
After rent:
100% − 30% = 70%
After food:
75% of 70% = 52.5%
After clothing:
80% of 52.5% = 42%
Thus, savings = 42% of income.
Given:
42% of income = Rs. 10,080
Income = 10,080 × 100/42
= Rs. 24,000
✅ Answer
Rs. 24,000
80. Depreciation of Machine
📝 Question
The value of a commercial machine depreciates by 10% each year. If its value two years ago was Rs. 2,00,000, find its present value in Rupees.
⚡ Shortcut Method
The machine depreciates by 10% every year, so its value becomes 90% each year.
After 2 years:
Present value = 2,00,000 × 90% × 90%
= 2,00,000 × 0.81
= Rs. 1,62,000
✅ Answer
Rs. 1,62,000
📚 Recommended Reading
- Advanced percentage problems
- Percentage questions for banking exams
- Previous Percentage Practice Set
- Next Percentage Practice Set (Coming Soon)
- Percentage increase and decrease practice questions
Challenge Question (Try this and drop your answer in the comments!)
📝 Question
In a certain town, the population of men, women, and children is in the ratio of $5 : 4 : 3$. If the population of men, women, and children increases by 20%, 25%, and 40% respectively in the first year, and the total population becomes 24,600 at the end of the year, find the initial total population of the town.
👇 Solve this using shortcut concepts and drop your answer in the comments below! Let’s see who gets it right first.
Frequently Asked Questions (FAQs)
Q1. How can I solve percentage questions faster in competitive exams like SSC CGL and Banking?
Answer: To solve percentage questions quickly, avoid traditional algebraic formulas and master multiplier concepts and fraction-to-percentage conversions. Breaking down multi-step problems like successive percentage changes or income-expenditure ratios into direct sequential multipliers can save crucial seconds in exam halls.
Q2. Are these hard-level percentage questions relevant for Banking Mains examinations?
Answer: Yes. Questions involving variable mixtures (like milk-water or alloy alligations), multi-layered income-savings deductions, and invalid votes in elections are specifically tailored for the Mains level of exams like IBPS PO, SBI PO, and RBI Assistant.
Q3. Why is mastering percentages important for other quantitative aptitude chapters?
Answer: Percentages form the backbone of arithmetic. A strong grip on percentage concepts directly helps you solve problems in Profit and Loss, Simple and Compound Interest, Ratio and Proportion, and Data Interpretation (DI) much more efficiently.
