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About Comparing Quantities — Class 8 Olympiad

Solve advanced problems on percentages, compound interest, profit-loss, discount, and tax at competition level. This topic is part of the Olympiad Class 8 mathematics syllabus (chapter: Module 8). On this page you can practice 58 questions across three difficulty levels — 20 easy, 20 medium, and 18 hard — each with a visual step-by-step solution, plus a timed 34-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Comparing Quantities — solved examples for Class 8 Olympiad

Example 1easy

If the price of an article is first increased by 20% and then decreased by 20%, what is the net percentage change in the price of the article?
  1. A)0%
  2. B)4% increase
  3. C)4% decrease
  4. D)10% decrease

Step-by-step solution

  1. Let the original price of the article be ₹100.
  2. After a 20% increase, the new price becomes: 100 + (20/100) × 100 = 100 + 20 = ₹120.
  3. Now, the price is decreased by 20%. This decrease is applied to the new price (₹120): 120 - (20/100) × 120 = 120 - 24 = ₹96.
  4. The net change in price is 96 - 100 = -₹4. So, the net percentage change is (4/100) × 100% = 4% decrease.

Answer: 4% decrease

Example 2medium

A's salary is 20% more than B's. C's salary is 25% less than B's. By what percentage is C's salary less than A's salary?
  1. A)37.5%
  2. B)30%
  3. C)25%
  4. D)40%

Step-by-step solution

  1. Let B's salary be ₹100.
  2. A's salary is 20% more than B's, so A's salary = 100 + (20/100) × 100 = ₹120.
  3. C's salary is 25% less than B's, so C's salary = 100 - (25/100) × 100 = ₹75.
  4. Difference between A's and C's salary = 120 - 75 = ₹45. Percentage C's salary is less than A's = (45/120) × 100 = 37.5%.

Answer: 37.5%

Example 3hard

A number is increased by x%, then decreased by (x-10)%. If the final number is 99% of the original number, what is the value of x? Assume x > 10.
  1. A)10
  2. B)20
  3. C)25
  4. D)30

Step-by-step solution

  1. Let the original number be N. After increasing by x%, the number becomes N(1 + x/100).
  2. Then, it is decreased by (x-10)%. So, the final number is N(1 + x/100)(1 - (x-10)/100).
  3. We are given that the final number is 99% of the original, so N(1 + x/100)(1 - (x-10)/100) = 0.99N. Divide by N (assuming N≠0):
  4. (100 + x)/100 × (100 - x + 10)/100 = 99/100
  5. (100 + x)(110 - x) = 9900
  6. 11000 - 100x + 110x - x² = 9900
  7. 11000 + 10x - x² = 9900
  8. x² - 10x - 11000 + 9900 = 0
  9. x² - 10x - 1100 = 0
  10. Using the quadratic formula x = [-b ± sqrt(b² - 4ac)] / 2a, we get:
  11. x = [10 ± sqrt((-10)² - 4 × 1 × (-1100))] / 2
  12. x = [10 ± sqrt(100 + 4400)] / 2
  13. x = [10 ± sqrt(4500)] / 2
  14. x = [10 ± 30√5] / 2
  15. However, since x must be an integer for typical Olympiad problems unless specified, let's recheck the equation or approach. There must be a simpler factorization or integer solution. Let's re-examine (100 + x)(110 - x) = 9900.
  16. Let's test the options. If x=20:
  17. (100+20)(110-20) = 120 × 90 = 10800. This is not 9900. My initial quadratic formulation was off. Let's re-expand:
  18. (1 + x/100)(1 - (x-10)/100) = 0.99
  19. ( (100+x)/100 ) ( (100-(x-10))/100 ) = 99/100
  20. ( (100+x)/100 ) ( (110-x)/100 ) = 99/100
  21. (100+x)(110-x) / 10000 = 99/100
  22. (100+x)(110-x) = 9900
  23. 11000 - 100x + 110x - x² = 9900
  24. 11000 + 10x - x² = 9900
  25. x² - 10x - 1100 = 0. Still this quadratic equation. Let's check my options again. Options are 10, 20, 25, 30. None of these give 0 for x² - 10x - 1100. Let's re-evaluate problem statement.
  26. There must be a simpler factorization. Let's recheck options and calculation. My quadratic equation is correct. The options do not yield integer solutions for this specific equation. Let's adjust the problem slightly to ensure an integer solution from the options or recalculate.
  27. Let N=100. Final number = 99. So 100(1+x/100)(1-(x-10)/100) = 99.
  28. (100+x)(110-x) = 9900. This is correct.
  29. If x = 20: (100+20)(110-20) = 120 × 90 = 10800. This is not 9900.
  30. My problem setup or options must be slightly off for an integer answer. Let's adjust the problem to ensure one of the options works, or reformulate. Let's target x=10 or x=20 for simplicity. If x=10, final = (1+0.1)(1-0) = 1.1N. Not 0.99N.
  31. If x=20: (1 + 20/100)(1 - (20-10)/100) = (1.2)(1 - 10/100) = 1.2 × 0.9 = 1.08. This is 108% of original. Not 99%.
  32. My options or the question itself needs adjustment. The question asks 'final number is 99% of the original number'. So (1+x/100)(1-(x-10)/100) = 0.99. This is correct.
  33. This implies that the value of x must be such that the overall change is a decrease. The product (1+x/100)(1-(x-10)/100) needs to be 0.99. Since (x-10) needs to be positive for a decrease, x>10.
  34. Let's re-think the quadratic: x² - 10x - 1100 = 0. This is the correct algebraic derivation from (100+x)(110-x) = 9900. If I need an integer solution from the options, the options are incompatible with 99%.
  35. Let's craft a question that gives one of the options as a correct answer. How about if the final number is 96% of the original number? Then (100+x)(110-x) = 9600. x² - 10x - 1400 = 0. Still no easy integer solution.
  36. Let's make the percentage change simpler. If a number is increased by x%, then decreased by x%, the final is (1-x²/10000)N. If this is 99% N, then x²/10000 = 0.01, x²=100, x=10. This is too simple.
  37. Revised question: A number is increased by x%, then decreased by (x-10)%. If the final number is 108% of the original number, what is the value of x? Assume x > 10.
  38. Let's re-do the calculation with 108%.
  39. N(1 + x/100)(1 - (x-10)/100) = 1.08N
  40. (100 + x)(110 - x) = 10800
  41. 11000 + 10x - x² = 10800
  42. x² - 10x - 200 = 0
  43. (x - 20)(x + 10) = 0
  44. Since x > 10, x = 20.
  45. This works perfectly with x=20 as an option. I will use this revised problem statement.

Answer: 20

Practice questions on Comparing Quantities

  1. Q1.easy

    A shopkeeper sold two watches for ₹990 each. On one, he gained 10% and on the other, he lost 10%. What was his overall profit or loss percentage in the entire transaction?
    1. A)No profit, no loss
    2. B)1% profit
    3. C)1% loss
    4. D)2% loss
    Show answer

    Answer: 1% loss

    Hint: Calculate the cost price (CP) for each watch separately. Remember that profit/loss percentages are always calculated on the cost price.

  2. Q2.easy

    A manufacturer marks the price of an item 30% above its cost price. He then allows a discount of 15%. What is his profit percentage?
    1. A)10.5%
    2. B)15%
    3. C)12.5%
    4. D)14%
    Show answer

    Answer: 10.5%

    Hint: Assume a convenient cost price (e.g., ₹100) to easily calculate the marked price and then the selling price after the discount.

  3. Q3.easy

    In a town, 60% of the population are males and the rest are females. If 20% of the males are children and 30% of the females are children, what percentage of the total population are children?
    1. A)24%
    2. B)22%
    3. C)26%
    4. D)28%
    Show answer

    Answer: 24%

    Hint: Assume a total population of 100 for easy percentage calculations. Calculate the number of male children and female children separately.

  4. Q4.medium

    A shopkeeper sells two articles for ₹12000 each. On one, he gains 20%, and on the other, he loses 20%. What is his overall profit or loss percentage in the entire transaction?
    1. A)4% loss
    2. B)4% profit
    3. C)No profit, no loss
    4. D)2% loss
    Show answer

    Answer: 4% loss

    Hint: Remember that when two items are sold at the same selling price, one at X% profit and the other at X% loss, there is always a loss. Calculate the cost price for each article individually.

  5. Q5.medium

    A sum of money invested at compound interest doubles itself in 5 years. In how many years will it become eight times itself at the same rate of compound interest?
    1. A)10 years
    2. B)15 years
    3. C)20 years
    4. D)25 years
    Show answer

    Answer: 15 years

    Hint: Compound interest grows exponentially. If it doubles in 't' years, it will become 2^n times in n*t years. Consider how many 'doubling periods' are needed for 8 times the amount.

  6. Q6.medium

    A refrigerator is marked at ₹25000. The shopkeeper offers two successive discounts of 10% and 5%. If he still makes a profit of 12.5%, what was the cost price of the refrigerator?
    1. A)₹18000
    2. B)₹19000
    3. C)₹20000
    4. D)₹21000
    Show answer

    Answer: ₹19000

    Hint: First, calculate the selling price after applying both successive discounts on the marked price. Then, use the profit percentage to find the cost price.

  7. Q7.hard

    A dishonest fruit seller claims to sell fruits at a 10% profit. However, he uses a faulty weighing balance which weighs 800 grams for every kilogram. What is his actual profit percentage?
    1. A)37.5%
    2. B)32.5%
    3. C)25%
    4. D)20%
    Show answer

    Answer: 37.5%

    Hint: Consider the effective cost price and selling price for a transaction. The seller gains profit from both marking up the price and using a false weight.

  8. Q8.hard

    If the length of a cuboid is measured with an error of +10%, its width with an error of -10%, and its height with an error of +5%, what is the approximate percentage error in the calculated volume?
    1. A)+5%
    2. B)+4.5%
    3. C)+3.5%
    4. D)+4.95%
    Show answer

    Answer: +4.95%

    Hint: Let the original dimensions be L, W, H. Calculate the new dimensions and the new volume. Then find the percentage change relative to the original volume.

  9. Q9.hard

    A shopkeeper allows a 20% discount on the marked price of an article and still makes a profit of 25%. If the article costs him ₹P, what is its marked price?
    1. A)1.5P
    2. B)1.5625P
    3. C)1.6P
    4. D)1.75P
    Show answer

    Answer: 1.5625P

    Hint: Relate Cost Price (CP) to Selling Price (SP) using profit, and relate Selling Price (SP) to Marked Price (MP) using discount. Then find the relation between MP and CP.

These are 9 of the 58 questions available for Comparing Quantities. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.