Introduction
Percentages are the quiet backbone of banking maths. They hide inside profit and loss, data interpretation, mixtures, elections and even geometry. A question may never use the word “percentage”, yet one wrong step with a 25% increase or a 20% decrease can cost you the whole answer.
Welcome to Part 10 of the BrainQuro Percentage series. This part brings you 10 percentage tough questions (Q91 to Q100) written at SBI PO and IBPS PO Mains level. Each question uses a different idea: successive change, election votes, spending chains, price and consumption, sectional marks, acid mixtures, discount and profit, comparison chains, geometry and literacy data. Every answer is a clean integer, just like in real exams.
How to use this post: read a question, try it on your own for at least 5 minutes, and only then read the solution. If you get stuck, the step-by-step method will show you the shortcut you missed.
Let us begin.
Percentage Series: All Parts
Table of Contents
Percentage Tough Questions with Solutions
Attempt every question before looking at the solution. The challenge question at the end is for the comments section.
Question 91
The population of a town increases by 20% in the first year, decreases by 10% in the second year and increases by 25% in the third year. If the population at the end of the three years is 13,500, find the original population of the town.
Solution
Step 1: Let the original population be P.
Step 2: Multiplying factors: 20% increase = 1.2, 10% decrease = 0.9, 25% increase = 1.25.
Step 3: Overall factor = 1.2 x 0.9 x 1.25 = 1.35.
Step 4: P x 1.35 = 13,500.
Step 5: P = 13,500 / 1.35 = 10,000.
Answer
10,000
Question 92
In an election between two candidates, 10% of the registered voters did not vote, and 5% of the votes cast were invalid. The winner received 60% of the valid votes and won by 1,368 votes. Find the number of registered voters.
Solution
Step 1: Let the number of registered voters be V.
Step 2: Votes cast = 90% of V = 0.9V.
Step 3: Valid votes = 95% of 0.9V = 0.855V.
Step 4: Winner gets 60% and loser gets 40% of the valid votes, so the winning margin = 20% of valid votes.
Step 5: 20% of valid votes = 1,368, so valid votes = 1,368 / 0.2 = 6,840.
Step 6: 0.855V = 6,840, so V = 6,840 / 0.855 = 8,000.
Answer
8,000
Question 93
A person spends 30% of his income on food. He spends 25% of the remaining amount on rent, and then 20% of what is left on transport. After these expenses he is left with Rs 8,400. Find the total amount spent on rent and transport together.
Solution
Step 1: Let the income be I.
Step 2: After food, remaining = 70% of I = 0.7I.
Step 3: Rent = 25% of 0.7I = 0.175I. Remaining = 0.7I – 0.175I = 0.525I.
Step 4: Transport = 20% of 0.525I = 0.105I. Remaining = 0.525I – 0.105I = 0.42I.
Step 5: 0.42I = 8,400, so I = 20,000.
Step 6: Rent + transport = 0.175I + 0.105I = 0.28I = 0.28 x 20,000 = 5,600.
Answer
Rs 5,600
Question 94
A family spends Rs 9,000 per month on rice and wheat in the ratio 2 : 1. The price of rice increases by 25% and the price of wheat decreases by 20%. As a result, the family reduces its rice consumption by 20% and its wheat consumption by 10%. Find the new monthly expenditure on rice and wheat together.
Solution
Step 1: Original spending on rice = 9,000 x 2/3 = Rs 6,000. On wheat = 9,000 x 1/3 = Rs 3,000.
Step 2: Expenditure = price x quantity, so the new expenditure is the old one multiplied by the price factor and the quantity factor.
Step 3: New spending on rice = 6,000 x 1.25 x 0.8 = 6,000 x 1 = Rs 6,000.
Step 4: New spending on wheat = 3,000 x 0.8 x 0.9 = 3,000 x 0.72 = Rs 2,160.
Step 5: New total expenditure = 6,000 + 2,160 = 8,160.
Answer
Rs 8,160
Question 95
In an exam, the maximum marks of Section A and Section B are in the ratio 3 : 2. A student scores 60% in Section A and 40% in Section B. The pass mark is 50% of the total marks, and the student passes by 12 marks. Find the marks scored by the student in Section B.
Solution
Step 1: Let the maximum marks of Section A and Section B be 3k and 2k. Total = 5k.
Step 2: Marks scored in A = 60% of 3k = 1.8k. Marks scored in B = 40% of 2k = 0.8k.
Step 3: Total scored = 1.8k + 0.8k = 2.6k.
Step 4: Pass mark = 50% of 5k = 2.5k.
Step 5: The student passes by 12 marks, so 2.6k – 2.5k = 0.1k = 12, so k = 120.
Step 6: Maximum marks of Section B = 2k = 240. Marks scored in B = 40% of 240 = 96.
Answer
96 marks
Question 96
A vessel contains 80 litres of a solution that has 25% acid. 16 litres of the solution are removed and replaced with pure acid. How many litres of water must now be added to bring the acid concentration back to 25%?
Solution
Step 1: Acid in the original solution = 25% of 80 = 20 litres.
Step 2: The 16 litres removed contain 25% acid = 4 litres. Acid left = 20 – 4 = 16 litres in 64 litres of solution.
Step 3: Adding 16 litres of pure acid gives 16 + 16 = 32 litres of acid in 80 litres of solution (that is 40%).
Step 4: For 25% concentration, the total solution must be 32 / 0.25 = 128 litres.
Step 5: Water to be added = 128 – 80 = 48 litres.
Answer
48 litres
Question 97
A shopkeeper marks an article 60% above its cost price. He gives successive discounts of 20% and 10% on the marked price and sells each article for Rs 4,608. If he sells 25 such articles, find his total profit.
Solution
Step 1: Let the cost price of one article be C. Marked price = 1.6C.
Step 2: Selling price = 1.6C x 0.8 x 0.9 = 1.152C.
Step 3: 1.152C = 4,608, so C = 4,000.
Step 4: Profit on one article = 4,608 – 4,000 = Rs 608.
Step 5: Profit on 25 articles = 608 x 25 = 15,200.
Answer
Rs 15,200
Question 98
A is 25% more than B and C is 40% less than A. If the sum of A, B and C is Rs 3,600, find the difference between A and C.
Solution
Step 1: Take B = 100. Then A = 125 (25% more than B).
Step 2: C = 60% of A = 0.6 x 125 = 75.
Step 3: A : B : C = 125 : 100 : 75 = 5 : 4 : 3.
Step 4: Sum of ratio = 12 parts = 3,600, so 1 part = 300.
Step 5: A = 5 x 300 = 1,500 and C = 3 x 300 = 900.
Step 6: Difference = 1,500 – 900 = 600.
Answer
Rs 600
Question 99
The length and breadth of a rectangle are in the ratio 5 : 3. The length is increased by 40% and the breadth is decreased by 25%, due to which the area increases by 300 sq m. Find the perimeter of the new rectangle.
Solution
Step 1: Let the length = 5k and breadth = 3k. Original area = 15k squared.
Step 2: New length = 5k x 1.4 = 7k. New breadth = 3k x 0.75 = 2.25k.
Step 3: New area = 7k x 2.25k = 15.75k squared.
Step 4: Increase in area = 15.75k squared – 15k squared = 0.75k squared = 300, so k squared = 400 and k = 20.
Step 5: New length = 7 x 20 = 140 m. New breadth = 2.25 x 20 = 45 m.
Step 6: New perimeter = 2 x (140 + 45) = 370 m.
Answer
370 metres
Question 100
In a town, 55% of the population are males and the rest are females. 60% of the males and 40% of the females are literate. If the number of illiterate people in the town is 9,800, find the number of literate females.
Solution
Step 1: Let the total population be P. Males = 0.55P and females = 0.45P.
Step 2: Literate males = 60% of 0.55P = 0.33P. Literate females = 40% of 0.45P = 0.18P.
Step 3: Total literate = 0.33P + 0.18P = 0.51P.
Step 4: Illiterate = P – 0.51P = 0.49P = 9,800, so P = 20,000.
Step 5: Literate females = 0.18 x 20,000 = 3,600.
Answer
3,600
Challenge Question for You
In a college, 72% of the students passed in Maths, 68% passed in Reasoning and 12% failed in both subjects. If 1,560 students passed in both subjects, find the number of students who passed in exactly one subject.
Think you have it? Drop your answer in the comments below, along with your method. We will reveal the correct answer and the shortest solution in the next post.
What’s Next
That wraps up Part 10 of the Percentage series. If you want to build speed, revise the solutions, note which concept tripped you up, and keep practising percentage tough questions like these before the next part.
Stay tuned for Part 11 on BrainQuro, and keep practising. See you in the next one.
