Unlock the Math Olympiad Champion in Your Child: 100+ Brain-Boosting Questions for Class 6!
Is your child staring at a math problem, looking completely lost? Do you want to transform that confusion into confident problem-solving skills that will not only top exams but also build a powerful logical mind?
For students in Class 6, the world of competitive exams like Math Olympiads can seem daunting. It's not just about memorizing formulas; it's about logic, analysis, and a sharp IQ. The journey from "I can't" to "I solved it!" is one of the most rewarding experiences a young student can have.
This blog post is your ultimate toolkit. We've gone beyond simple practice problems to bring you a massive collection of over 100 questions designed to challenge, engage, and enlighten. We cover every crucial topic for the Bangladesh curriculum, injecting a heavy dose of the tricky, logical, and fundamental thinking that Olympiad champions are made of.
Why This Practice Set is a Game-Changer:
>100 Questions & Answers: More than just volume, it's about variety and depth.
Tricky & IQ-Based: We include the head-scratchers that separate the good from the great.
Step-by-Step Clues: Every problem comes with a strategic clue to guide independent thinking.
Fundamental Focus: Builds a rock-solid understanding of core concepts.
Fully Printable Format: Perfect for study sessions, coaching classes, or last-minute revision.
Printable Format: Grade 6 Math Olympiad Practice Questions
1. Number System
Q1. The product of two numbers is 72,096. If one number is 304, what is the other number?
Clue: Think about the inverse operation of multiplication.
Answer: 72,096 ÷ 304 = 237
Q2. Find the sum of the greatest and the smallest 5-digit numbers formed by the digits 4, 0, 8, 7, 3 (using each digit only once).
Clue: Remember, a number cannot start with 0. Arrange digits in descending order for the greatest and ascending (carefully) for the smallest.
Answer: Greatest: 87,430. Smallest: 30,478. Sum = 87,430 + 30,478 = 117,908
Q3. Which is greater: or ?
Clue: Calculate the actual values.
Answer: and . So, is greater.
Q4. If a number is divided by 7, the quotient is 125 and the remainder is 3. What is the number?
Clue: Use the formula: Dividend = (Divisor × Quotient) + Remainder.
Answer: (7 × 125) + 3 = 875 + 3 = 878
Q5. What is the difference between the place value and the face value of 7 in the number 87,654?
Clue: Place value is 7,000. Face value is just 7.
Answer: 7,000 - 7 = 6,993
2. Average
Q6. The average of five numbers is 26. If one number is excluded, the average becomes 23. What is the excluded number?
Clue: Find the total of five numbers and the total of four numbers. The difference is the excluded number.
Answer: Total of 5 numbers = 26 × 5 = 130. Total of 4 numbers = 23 × 4 = 92. Excluded number = 130 - 92 = 38
Q7. The average weight of 5 boys is 40 kg. When a new boy joins, the average weight becomes 42 kg. What is the weight of the new boy?
Clue: Calculate the total weight before and after. The difference is the new boy's weight.
Answer: Total weight of 5 boys = 40 × 5 = 200 kg. Total weight of 6 boys = 42 × 6 = 252 kg. New boy's weight = 252 - 200 = 52 kg
Q8. The average of 7 consecutive odd numbers is 15. What is the largest of these numbers?
Clue: Consecutive odd numbers are like an arithmetic series. The middle (4th) number will be the average.
Answer: If the 4th number is 15, the numbers are 9, 11, 13, 15, 17, 19, 21. The largest is 21.
3. Fraction & Decimal
Q9. If of a number is 45, what is of the same number?
Clue: First, find the original number.
Answer: Let the number be x. , so x = 45 × = 75. Now, of 75 =
Q10. Simplify:
Clue: Convert all to improper fractions.
Answer: . LCM of 2,3,6 is 6. =
Q11. What is ?
Clue: Remember that .
Answer: , . So,
Q12. By what should be multiplied to get 100?
Clue: Convert to an improper fraction or decimal.
Answer: . Let the multiplier be x. , so x = 100 × = 8
Q13. Simplify:
Clue: Follow the order of operations (division first).
Answer: . Then, 60 × 0.3 = 18
4. Unitary Method, Ratio & Proportion
Q14. If 15 books cost ₹90, what is the cost of 25 such books?
Clue: First, find the cost of one book (the unit).
Answer: Cost of 1 book = 90 ÷ 15 = ₹6. Cost of 25 books = 25 × 6 = ₹150
Q15. The ratio of boys to girls in a school is 8:5. If there are 160 boys, how many girls are there?
Clue: Set up a proportion. 8 parts represent 160.
Answer: 8 parts = 160, so 1 part = 160 ÷ 8 = 20. Girls = 5 parts = 5 × 20 = 100
Q16. Two numbers are in the ratio 7:9. If the sum of the numbers is 112, find the numbers.
Clue: Let the numbers be 7x and 9x.
Answer: 7x + 9x = 112 → 16x = 112 → x = 7. The numbers are 7×7=49 and 9×7=63.
Q17. A car travels 180 km in 3 hours. How long will it take to travel 300 km at the same speed?
Clue: First, find the speed (km per hour).
Answer: Speed = 180 km / 3 h = 60 km/h. Time for 300 km = 300 km / 60 km/h = 5 hours
Q18. If A:B = 3:4 and B:C = 5:6, find A:C.
Clue: Make the value of B the same in both ratios.
Answer: LCM of 4 and 5 is 20. A:B = (3×5):(4×5)=15:20. B:C=(5×4):(6×4)=20:24. So, A:B:C=15:20:24. Therefore, A:C = 15:24 or 5:8.
5. Profit, Loss & Percentage
Q19. A man bought a cycle for ₹1200 and sold it for ₹1500. Find his profit percentage.
Clue: Profit % = (Profit / Cost Price) × 100%.
Answer: Profit = 1500 - 1200 = ₹300. Profit % = (300/1200) × 100% = 25%
Q20. A shopkeeper sells a pen at a 12% loss. If the selling price is ₹176, what was the cost price?
Clue: Selling Price = Cost Price - Loss. If loss is 12%, then SP is 88% of CP.
Answer: Let CP be x. Then, 88% of x = 176 → → x = 176 × = ₹200
Q21. What is 25% of 25% of 400?
Clue: Break it down step by step.
Answer: 25% of 400 = . 25% of 100 = 25
Q22. If the price of a commodity is increased by 20%, by what percentage must the new price be reduced to get back the original price?
Clue: This is a tricky one. Use a simple number for the original price, like 100.
Answer: Let original price be ₹100. Increased price = 100 + 20% = ₹120. To get back to ₹100, we need a reduction of ₹20 on ₹120. Reduction % = (20/120) × 100% = % or 16.67%
Q23. Express as a percentage.
Clue: Multiply the fraction by 100%.
Answer:
6. LCM & HCF
Q24. Find the least number which when divided by 12, 15, and 18 leaves a remainder of 5 in each case.
Clue: First, find the LCM of 12, 15, 18. Then add the remainder.
Answer: LCM of 12, 15, 18 is 180. Required number = 180 + 5 = 185
Q25. Two bells ring at intervals of 18 and 24 minutes respectively. They rang together at 12 noon. At what time will they ring together again?
Clue: Find the LCM of the time intervals.
Answer: LCM of 18 and 24 is 72 minutes. So, they will ring together again after 1 hour and 12 minutes, i.e., at 1:12 PM.
Q26. The HCF of two numbers is 11 and their LCM is 7700. If one of the numbers is 275, find the other.
Clue: Use the formula: Product of two numbers = HCF × LCM.
Answer: Let the other number be x. 275 × x = 11 × 7700 → x = (11 × 7700) / 275 = 308
Q27. Find the greatest number that will divide 43, 91, and 183 leaving the same remainder in each case.
Clue: The required number is the HCF of the differences between the numbers.
Answer: Differences: (91-43)=48, (183-91)=92, (183-43)=140. Find HCF of 48, 92, 140. HCF is 4. So, the number is 4.
Q28. What is the smallest number which when increased by 5 is divisible by 12, 18, and 24?
Clue: Find the LCM of 12, 18, 24. The required number is 5 less than that LCM.
Answer: LCM of 12, 18, 24 is 72. Required number = 72 - 5 = 67
7. Measurement
Q29. How many cubes of side 2 cm can be packed in a box of dimensions 12 cm × 10 cm × 8 cm?
Clue: Find the volume of the big box and the volume of one small cube.
Answer: Volume of box = 12×10×8 = 960 cm³. Volume of one cube = 2×2×2=8 cm³. Number of cubes = 960 ÷ 8 = 120
Q30. A rectangular field is 40 m long and 30 m wide. A path 2 m wide is built outside it. Find the area of the path.
Clue: The path is outside. Find the area of the outer rectangle (field+path) and subtract the area of the field.
Answer: Outer length = 40+2+2=44 m. Outer width = 30+2+2=34 m. Area of path = (44×34) - (40×30) = 1496 - 1200 = 296 m²
Q31. Convert 5 square kilometers into square meters.
Clue: 1 km = 1000 m, so 1 km² = 1,000,000 m².
Answer: 5 km² = 5 × 1,000,000 m² = 5,000,000 m²
Q32. A tank is 1.5 m long, 1 m wide, and 0.8 m deep. How many liters of water can it hold? (1 m³ = 1000 liters)
Clue: Find the volume in cubic meters first.
Answer: Volume = 1.5 × 1 × 0.8 = 1.2 m³. Capacity in liters = 1.2 × 1000 = 1200 liters
8. Geometry
Q33. The perimeter of a square is 36 cm. Find its area.
Clue: Find the side length first.
Answer: Side = Perimeter ÷ 4 = 36 ÷ 4 = 9 cm. Area = side² = 9 × 9 = 81 cm²
Q34. The angles of a triangle are in the ratio 2:3:4. Find the largest angle.
Clue: The sum of angles in a triangle is 180°.
Answer: Let angles be 2x, 3x, 4x. 2x+3x+4x=180 → 9x=180 → x=20. Largest angle = 4x = 4×20 = 80°
Q35. How many diagonals does a regular hexagon have?
Clue: Use the formula: Number of diagonals = where n is the number of sides.
Answer: For a hexagon, n=6. Diagonals = 9
Q36. The area of a rectangle is 112 cm². If its length is 14 cm, find its perimeter.
Clue: Find the breadth using the area formula.
Answer: Area = Length × Breadth → 112 = 14 × Breadth → Breadth = 112 ÷ 14 = 8 cm. Perimeter = 2×(Length+Breadth) = 2×(14+8) = 2×22 = 44 cm
Q37. An angle is 30° more than its complement. What is the angle?
Clue: Complementary angles sum to 90°. Let the angle be x.
Answer: x + (x - 30) = 90 → 2x - 30 = 90 → 2x = 120 → x = 60°
9. Algebra: Basics & Equations
Q38. If a = 2 and b = -3, find the value of .
Clue: Substitute the values carefully, minding the negative sign.
Answer: (2)² - (2)(-3) + (-3)² = 4 - (-6) + 9 = 4 + 6 + 9 = 19
Q39. Add: and
Clue: Combine like terms (x with x, y with y, z with z).
Answer: (3x+2x) + (2y-3y) + (-5z+4z) = 5x - y - z
Q40. Solve for x:
Clue: Get all x terms on one side and constants on the other.
Answer: 5x - 3x = 9 + 7 → 2x = 16 → x = 8
Q41. Simplify:
Clue: Multiply the coefficients and the variables separately.
Answer: (2×3×-4) × (a×b×c) = -24abc
Q42. If , what is the value of 2x + 3?
Clue: First, find the value of x from the equation.
Answer: → x = 7×5 = 35. Then, 2x+3 = 2(35)+3 = 70+3 = 73
Q43. The sum of three consecutive integers is 48. Find the numbers.
Clue: Let the numbers be x, x+1, x+2.
Answer: x + (x+1) + (x+2) = 48 → 3x + 3 = 48 → 3x = 45 → x=15. The numbers are 15, 16, 17.
More Challenging & Logical/IQ-Based Problems
Q44. If 10 cats can catch 10 mice in 10 minutes, how long will it take for 100 cats to catch 100 mice?
Clue: This is a logical rate problem. Think about the rate per cat.
Answer: 10 cats catch 10 mice in 10 minutes, so 1 cat catches 1 mouse in 10 minutes. Therefore, 100 cats will catch 100 mice in the same time it takes 1 cat to catch 1 mouse, which is 10 minutes.
Q45. A bat and a ball together cost $1.10. The bat costs $1.00 more than the ball. How much does the ball cost?
Clue: This is a classic trick. Don't jump to 10 cents.
Answer: Let the ball cost x dollars. Then bat costs x+1.00. x + (x+1.00) = 1.10 → 2x + 1.00 = 1.10 → 2x = 0.10 → x = 0.05. The ball costs 5 cents.
Q46. What is the next number in the sequence: 2, 6, 12, 20, 30, ...?
Clue: Look at the differences between consecutive numbers: 4, 6, 8, 10, ...
Answer: The differences are increasing by 2. So next difference is 12. 30 + 12 = 42. (Also note: 1×2, 2×3, 3×4, 4×5, 5×6, 6×7=42)
Q47. If today is Monday, what day will it be after days?
Clue: . Find the remainder when 32 is divided by 7 (days in a week).
Answer: 32 ÷ 7 gives a remainder of 4 (since 28 is divisible by 7). So, Monday + 4 days = Friday
Q48. A clock shows 3:15. What is the angle between the hour and minute hands?
Clue: The hour hand also moves as the minutes pass. At 3:15, the hour hand is 1/4 of the way between 3 and 4.
Answer: Minute hand at 15 minutes is at 90°. Hour hand at 3:00 is at 90°. In 15 minutes, the hour hand moves 0.5°×15=7.5°. So, hour hand is at 97.5°. The difference is 97.5° - 90° = 7.5°.
Q49. If you write all the numbers from 1 to 100, how many times do you write the digit '5'?
Clue: Count in the units place and the tens place separately.
Answer: Units place: 5,15,25,...,95 → 10 times. Tens place: 50,51,...,59 → 10 times. Total = 20 times.
Q50. The cost of 4 pens and 3 pencils is ₹90. The cost of 7 pens and 2 pencils is ₹130. What is the cost of a pen?
Clue: This is a pair of linear equations. Let the cost of a pen be x and a pencil be y.
Answer:
4x + 3y = 90 ...(1)
7x + 2y = 130 ...(2)
Multiply (1) by 2 and (2) by 3:
8x + 6y = 180
21x + 6y = 390
Subtract the first from the second: (21x-8x) + (6y-6y) = 390-180 → 13x = 210 → x = 210/13 = ₹16.15 (approx.) (For a class 6 level, numbers are usually chosen to give whole number answers. Let's adjust the problem: If 4pens+3pencils=90 and 7pens+2pencils=125, then solving gives x=15, y=10. So, a pen costs ₹15).
*(This list provides 50 high-quality questions. To reach over 100, you can create variations of these problems by changing the numbers and contexts. For example, change the numbers in the average, ratio, or profit-loss questions. The core concepts and logical structure will remain the same, providing excellent practice.)*
Tips for Preparation:
Practice regularly and time yourself.
Review mistakes to understand the underlying concept.
Learn shortcuts and tricks for calculation.
Focus on understanding word problems and translating them into mathematical equations.
Good luck with your preparations
Part 2: More Challenging & Logical/IQ-Based Problems (Continued)
Q51. A number is as much greater than 36 as it is less than 86. Find the number.
Clue: The number is exactly in the middle of 36 and 86.
Answer: The number is the average of 36 and 86. (36 + 86) ÷ 2 = 122 ÷ 2 = 61
Q52. If a number is multiplied by 7 and then 15 is added, the result is 50. What is the number?
Clue: Work backwards. Start from 50 and reverse the operations.
Answer: (50 - 15) ÷ 7 = 35 ÷ 7 = 5
Q53. The sum of the digits of a two-digit number is 9. If 27 is added to the number, its digits get reversed. Find the number.
Clue: Let the number be 10x + y, where x is the tens digit and y is the units digit.
Answer: x + y = 9. Also, (10x + y) + 27 = 10y + x → 9x - 9y = -27 → x - y = -3. Solving x+y=9 and x-y=-3, we get 2x=6, x=3, y=6. The number is 36.
Q54. A farmer has ducks and cows. If there are 35 heads and 110 legs, how many ducks does he have?
Clue: Ducks have 1 head and 2 legs. Cows have 1 head and 4 legs. You can set up equations.
Answer: Let d be ducks, c be cows. d + c = 35. 2d + 4c = 110. Multiply the first equation by 2: 2d + 2c = 70. Subtract from the second: (2d+4c) - (2d+2c) = 110-70 → 2c=40 → c=20. Then d=35-20=15 ducks.
Q55. Find the next letter in the sequence: O, T, T, F, F, S, S, ?
Clue: Think about the first letters of numbers (One, Two, Three, Four, Five, Six, Seven, ...).
Answer: The next number is Eight, so the next letter is E.
Q56. If 6 workers can build a wall in 10 days, how many days will 15 workers take to build the same wall?
Clue: This is an inverse proportion problem. More workers means less time.
Answer: Worker-Days = 6 workers × 10 days = 60 worker-days. Time for 15 workers = 60 worker-days / 15 workers = 4 days.
Q57. A number when divided by 133 gives a remainder of 49. What is the remainder when the number is divided by 7?
Clue: 133 is divisible by 7 (133 ÷ 7 = 19). So, the remainder will be based on the remainder of 49.
Answer: 49 ÷ 7 = 7 exactly. So, the remainder is 0.
Q58. A is older than B. C is younger than A but older than B. D is younger than E but older than A. Who is the youngest?
Clue: Arrange them from youngest to oldest based on the clues.
Answer: From the clues: E > D > A > C > B. The youngest is B.
Q59. If 25% of a number is 75, then 45% of the same number is?
Clue: First, find the number.
Answer: Let the number be x. 0.25x = 75 → x = 75 / 0.25 = 300. 45% of 300 = 0.45 × 300 = 135.
Q60. The cost price of 20 articles is equal to the selling price of 15 articles. Find the profit percentage.
Clue: Let CP of 1 article be ₹1. Find SP of 1 article and then calculate profit %.
Answer: CP of 20 articles = SP of 15 articles = ₹20 (assuming CP=₹1 each). So, SP of 1 article = 20/15 = ₹4/3. Profit on 1 article = (4/3) - 1 = ₹1/3. Profit % = ((1/3) / 1) × 100% = 33.33%.
10. Algebra: Expressions & Equations (Continued)
Q61. Subtract from .
Clue: Be careful with the minus sign. It's (5x²+4x-3) - (3x²-2x+5).
Answer: 5x²+4x-3 -3x² +2x -5 = (5x²-3x²) + (4x+2x) + (-3-5) = 2x² + 6x - 8
Q62. Solve for y:
Clue: First, eliminate the denominator by multiplying both sides by 3.
Answer: 2y + 1 = 5 × 3 → 2y + 1 = 15 → 2y = 14 → y = 7
Q63. If and , find the value of .
Clue: Substitute the values carefully, especially the negative sign for p.
Answer: 2(-2)² - 3(-2)(3) + (3)² = 2(4) - 3(-6) + 9 = 8 + 18 + 9 = 35
Q64. Multiply:
Clue: Use the distributive property (FOIL method).
Answer: (2a)(4a) + (2a)(-5b) + (3b)(4a) + (3b)(-5b) = 8a² -10ab +12ab -15b² = 8a² + 2ab - 15b²
Q65. The length of a rectangle is (2x+1) cm and its width is (x-2) cm. If its perimeter is 40 cm, find the value of x.
Clue: Perimeter of rectangle = 2(Length + Width).
Answer: 2( (2x+1) + (x-2) ) = 40 → 2(3x -1) = 40 → 6x - 2 = 40 → 6x = 42 → x = 7
11. Advanced Number System & Arithmetic
Q66. Find the unit digit of .
Clue: Find the pattern of the units digit of powers of 7: 7¹=7, 7²=49, 7³=343, 7⁴=2401, 7⁵=16807... The pattern is 7, 9, 3, 1 and it repeats every 4 powers.
Answer: Find the remainder when 95 is divided by 4. 95 ÷ 4 gives remainder 3. The 3rd digit in the cycle (7, 9, 3, 1) is 3.
Q67. Simplify:
Clue: Calculate each root separately.
Answer: √169 = 13, ∛216 = 6, √64 = 8. So, 13 + 6 - 8 = 11
Q68. Which of the following is divisible by 3? 1234, 2345, 3456, 4567
Clue: A number is divisible by 3 if the sum of its digits is divisible by 3.
Answer: Check sums: 1+2+3+4=10, 2+3+4+5=14, 3+4+5+6=18 (divisible by 3), 4+5+6+7=22. So, 3456 is divisible by 3.
Q69. Find the value of:
Clue: Pair them up: (1-2) + (3-4) + (5-6) + ... + (99-100). Each pair equals -1.
Answer: There are 50 such pairs. So, 50 × (-1) = -50
Q70. The product of two numbers is 2070. Their HCF is 15. What is their LCM?
Clue: Use the relationship: HCF × LCM = Product of Numbers.
Answer: LCM = (Product of Numbers) / HCF = 2070 / 15 = 138
12. Geometry (Advanced)
Q71. How many degrees does the minute hand of a clock move in 18 minutes?
Clue: The minute hand moves 360° in 60 minutes, so it moves 6° per minute.
Answer: 18 minutes × 6° per minute = 108°
Q72. The angles of a quadrilateral are in the ratio 1:2:3:4. Find the largest angle.
Clue: The sum of angles in a quadrilateral is 360°.
Answer: Let angles be x, 2x, 3x, 4x. x+2x+3x+4x=360 → 10x=360 → x=36. Largest angle = 4x = 4×36 = 144°
Q73. The area of a circle is 154 cm². Find its circumference. (Take π = 22/7)
Clue: First, find the radius from the area. Area = πr².
Answer: πr² = 154 → (22/7)×r²=154 → r²=(154×7)/22=49 → r=7 cm. Circumference = 2πr = 2×(22/7)×7 = 44 cm.
Q74. A cube has a total surface area of 96 cm². What is its volume?
Clue: A cube has 6 faces. Find the area of one face to get the side length.
Answer: Area of one face = 96 cm² / 6 = 16 cm². Side length = √16 = 4 cm. Volume = side³ = 4×4×4 = 64 cm³.
Q75. In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What is the measure of each base angle?
Clue: The sum of angles in a triangle is 180°. Let each base angle be x.
Answer: x + x + 40 = 180 → 2x = 140 → x = 70°
13. Tricky Word Problems & Logical Puzzles
Q76. A man is 4 years older than his wife, and 5 times as old as his son. If the son is 6 years old, how old is the wife?
Clue: Find the man's age first.
Answer: Son's age = 6. Man's age = 5 × 6 = 30 years. Wife's age = 30 - 4 = 26 years.
Q77. A train 150m long is running at 72 km/h. How long will it take to cross a 250m long platform?
Clue: The total distance to be covered is the length of the train plus the length of the platform. Convert speed to m/s.
Answer: Speed = 72 km/h = 72 × (1000/3600) m/s = 20 m/s. Distance = 150m + 250m = 400m. Time = Distance/Speed = 400 / 20 = 20 seconds.
Q78. In a class, 20 students passed in Mathematics, 25 passed in English, and 15 passed in both. If every student passed at least one subject, how many students are in the class?
Clue: Use the principle: n(A∪B) = n(A) + n(B) - n(A∩B)
Answer: Total students = 20 + 25 - 15 = 30 students.
Q79. If 3 mangoes and 4 oranges cost ₹85, and 5 mangoes and 2 oranges cost ₹95, what is the cost of 1 mango?
Clue: Set up two equations. Let the cost of a mango be 'm' and an orange be 'o'.
Answer:
3m + 4o = 85 ...(1)
5m + 2o = 95 ...(2)
Multiply equation (2) by 2: 10m + 4o = 190 ...(3)
Subtract equation (1) from (3): (10m-3m) + (4o-4o) = 190-85 → 7m = 105 → m = ₹15
Q80. Walking at 4/5th of his usual speed, a man is 12 minutes late. What is his usual time to cover the distance?
Clue: Speed and time are inversely proportional. If new speed is 4/5 of usual speed, new time is 5/4 of usual time.
Answer: Let usual time be T. New time = (5/4)T. The difference is (5/4)T - T = (1/4)T. This equals 12 minutes. So, (1/4)T = 12 → T = 48 minutes.
14. Fractions, Decimals & Percentages (Advanced)
Q81. What is 25% of 50% of 1000?
Clue: Work step by step.
Answer: 50% of 1000 = 500. 25% of 500 = 125.
Q82. If 0.025 ÷ x = 0.25, then x = ?
Clue: 0.025 ÷ x = 0.25 can be rewritten as 0.025 / x = 0.25.
Answer: x = 0.025 / 0.25 = 0.1 or 1/10.
Q83. A fraction becomes 4/5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes 1/2. Find the fraction.
Clue: Let the fraction be x/y. Set up two equations.
Answer:
(x+1)/(y+1) = 4/5 -> 5x+5=4y+4 -> 5x-4y=-1 ...(1)
(x-5)/(y-5) = 1/2 -> 2x-10=y-5 -> 2x-y=5 ...(2)
Solve (2) for y: y=2x-5. Substitute in (1): 5x - 4(2x-5) = -1 -> 5x-8x+20=-1 -> -3x=-21 -> x=7. Then y=2(7)-5=9. The fraction is 7/9.
Q84. Express the recurring decimal 0.454545... as a fraction in its simplest form.
Clue: Let x = 0.454545... Then 100x = 45.454545... Subtract the first from the second.
Answer: 100x - x = 45.454545... - 0.454545... -> 99x = 45 -> x = 45/99 = 5/11.
Q85. A's income is 25% more than B's income. By what percent is B's income less than A's income?
Clue: Use a simple number for B's income. Let B's income be 100.
Answer: If B=100, A=125. B's income is 25 less than A's. Percentage less = (25/125) × 100% = 20%.
15. Final Mixed Bag (Ultimate Challenge)
Q86. Find the smallest perfect square number which is divisible by 12, 15, and 20.
Clue: Find the LCM of 12,15,20. Then find the square factors needed to make it a perfect square.
Answer: LCM of 12,15,20 is 60. 60 = 2² × 3 × 5. To make it a perfect square, we need 3² and 5². So, multiply by 3×5=15. The number is 60×15=900.
Q87. If 20% of (x+y) = 50% of (x-y), then what percent of x is y?
Clue: Form an equation and find the ratio of x and y.
Answer: 0.2(x+y) = 0.5(x-y) -> 2(x+y) = 5(x-y) -> 2x+2y=5x-5y -> 7y=3x -> y = (3/7)x. So, y is of x ≈ 42.86% of x.
Q88. A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 27 is subtracted from the number, its digits are reversed. Find the number.
Clue: Let the units digit be x. Then tens digit is 2x. The number is 10(2x)+x=21x.
Answer: The number with digits reversed is 10x+2x=12x. So, 21x - 27 = 12x -> 9x=27 -> x=3. The number is 21×3=63.
Q89. A can do a work in 10 days and B can do it in 15 days. They work together for 2 days and then B leaves. In how many more days will A finish the work?
Clue: Find their combined work rate and how much work is left.
Answer: A's rate=1/10 work/day, B's rate=1/15 work/day. Combined rate=1/10+1/15=1/6. In 2 days, they complete 2×(1/6)=1/3 of the work. Work left = 1 - 1/3 = 2/3. Time for A to finish = (2/3) / (1/10) = (2/3)×10 = 20/3 days or 6.67 days.
Q90. The ratio of the ages of a father and his son is 5:2. If the father was 35 years old when the son was born, what is the father's present age?
Clue: The age difference is constant. Let the father's age be 5x and son's age be 2x.
Answer: Age difference = 5x - 2x = 3x. This equals 35 years. So, 3x=35 -> x=35/3. Father's age = 5×(35/3) = 175/3 = 58.33 years (or 58 years and 4 months). For a cleaner answer, the problem might use numbers like ratio 3:1 and difference 30, giving father=45. But based on these numbers, the answer is 58.33 years.
Q91. Simplify:
Clue: Simplify from the bottom up.
Answer: Start with the denominator: 1 + 1/2 = 3/2. Then, 1 / (3/2) = 2/3. Then, 1 + (2/3) = 5/3. Then, 1 / (5/3) = 3/5. Finally, 1 + (3/5) = 8/5 or 1.6.
Q92. A student multiplied a number by 3/5 instead of 5/3. What is the percentage error?
Clue: Find the difference between the correct and wrong value, and compare it to the correct value.
Answer: Let the number be 15 (LCM of 5 and 3). Wrong result = 15 × (3/5) = 9. Correct result = 15 × (5/3) = 25. Error = 25 - 9 = 16. Percentage Error = (16 / 25) × 100% = 64%.
Q93. If the side of a square is increased by 20%, its area is increased by what percent?
Clue: Let original side be 10. Original area=100. New side=12, new area=144.
Answer: Increase in area = 44. Percentage increase = (44/100)×100% = 44%.
Q94. Find the area of the shaded region if the larger square has a side of 10cm and the smaller square inside it has a side of 6cm.
Clue: The shaded region is the area between the two squares. Subtract the area of the small square from the area of the large square.
Answer: Area of large square = 10×10=100 cm². Area of small square=6×6=36 cm². Shaded area = 100 - 36 = 64 cm².
Q95. A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?
Clue: If the principal is P, the amount is 2P after 10 years, so the Simple Interest is P.
Answer: SI = (P × R × T)/100 -> P = (P × R × 10)/100 -> 1 = (10R)/100 -> R = 10% per annum.
Q96. The volume of a cube is 512 cm³. What is the length of its diagonal?
Clue: Find the side of the cube first. The space diagonal of a cube is .
Answer: Side = ∛512 = 8 cm. Length of diagonal = cm ≈ 13.86 cm.
Q97. If 5 men or 7 women can earn ₹875 per day, how much would 10 men and 5 women earn per day?
Clue: Find the earning rate of one man and one woman.
Answer: 5 men = ₹875, so 1 man = ₹175 per day. 7 women = ₹875, so 1 woman = ₹125 per day. 10 men + 5 women = (10×175) + (5×125) = 1750 + 625 = ₹2375.
Q98. A number when divided by 5 leaves a remainder of 3. What is the remainder when the square of that number is divided by 5?
Clue: Let the number be 5k+3. Find its square.
Answer: (5k+3)² = 25k² + 30k + 9 = 5(5k²+6k+1) + 4. So, the remainder is 4.
Q99. The sum of three consecutive multiples of 7 is 126. Find the smallest multiple.
Clue: Let the multiples be 7x, 7(x+1), 7(x+2).
Answer: 7x + 7(x+1) + 7(x+2) = 126 -> 7(3x+3)=126 -> 21(x+1)=126 -> x+1=6 -> x=5. Smallest multiple = 7×5 = 35.
Q100. A person spends 75% of his income. If his income increases by 20% and his expenditure increases by 10%, by what percent do his savings increase?
Clue: Use an initial income of 100 for easy calculation.
Answer: Initial: Income=100, Expenditure=75, Savings=25.
New: Income=120, Expenditure=75×1.10=82.5, Savings=120-82.5=37.5.
Increase in savings = 37.5 - 25 = 12.5. Percentage increase = (12.5/25)×100% = 50%.
Q101. A train passes a pole in 15 seconds and a 200m long platform in 35 seconds. What is its length?
Clue: The time to pass the pole is the time to cover its own length. The time to pass the platform is the time to cover its own length plus the platform's length.
Answer: Let the length be L meters. Speed = L/15 m/s.
To pass the platform: (L+200) / (L/15) = 35 -> 15(L+200)/L = 35 -> 15L+3000=35L -> 20L=3000 -> L=150 meters.
Q102. If and , find the value of .
Clue: Use the identity .
Answer: -> -> 100 = a²+b²+42 -> a²+b² = 58.
This comprehensive list of 102 questions covers all the required areas with a strong emphasis on logical, analytical, and IQ-based thinking, perfectly tailored for a Class 6 student preparing for a competitive Math Olympiad in Bangladesh.
Beyond the Numbers: How to Use This Guide for Maximum Effect
The Daily Dose: Solve 5-10 questions daily to build a consistent habit.
The Clue is Key: When stuck, read the clue carefully. It’s designed to spark the right thought process without giving away the answer.
Learn from Mistakes: Review the solution for every incorrect answer. Understanding why you went wrong is more important than getting it right.
Time Yourself: As confidence grows, time the sessions to simulate exam pressure.
Make it Fun: Turn it into a game! Challenge siblings or parents to see who can solve a problem first.
Ready to Transform Your Child's Math Journey?
The path to becoming a math champion is paved with consistent practice, curiosity, and the right resources. This collection is more than just homework; it's a training ground for future engineers, scientists, and innovators.
Don't let this opportunity slide! Scroll down, dive into the questions, and watch your child's confidence and logical ability soar.
Let the problem-solving begin! Share your child's success story or a tricky problem in the comments below. #MathOlympiad #Class6Math #BangladeshEducation
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