CBSE Class 8 Mathematics Fractions in Disguise Practice Paper Worksheet Set 01 with Answers

Practice Paper 28 for CBSE Class 8 Mathematics Ganita Prakash Part 2 Chapter 1 Fractions in Disguise has 20 questions for 40 marks across Sections A to E, each followed by a worked solution. The questions practise comparing fractions, decimals and percentages, finding the original number after a percentage increase, mark-up with discount, GST on a discounted price, 'Buy One Get One' offers and the small net loss from an equal rise and fall in a population.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A school garden uses \( \frac{7}{20} \) of its area for vegetables and \( \frac{3}{10} \) for flowers. What percentage of the garden is left for other purposes?
Answer:
Area used = \( \frac{7}{20} \) + \( \frac{3}{10} \) = \( \frac{7}{20} \) + \( \frac{6}{20} \) = \( \frac{13}{20} \).
Area left = 1 − \( \frac{13}{20} \) = \( \frac{7}{20} \).
Percentage left = \( \frac{7}{20} \) × 100 = 35%.
(Another way: \( \frac{7}{20} \) = 35% and \( \frac{3}{10} \) = 30%, so 100% − 35% − 30% = 35%.)
In simple words: Vegetables take 35% and flowers 30%, so 35% of the garden is left.

Teacher's Note:
1. Make the denominators equal before adding fractions.
2. Changing each fraction to a percent first is often quicker.

Question 12. Riya scores 54 marks out of 75 in Mathematics, while Ananya scores 68 marks out of 90 in Science. Who has the higher percentage score?
Answer:
Riya: \( \frac{54}{75} \) × 100 = 72%.
Ananya: \( \frac{68}{90} \) × 100 = \( \frac{680}{9} \) ≈ 75.56%.
75.56% > 72%, so Ananya has the higher percentage score (by about 3.56 percentage points).
In simple words: Out of 100, Riya gets 72 and Ananya about 75.6, so Ananya did better.

Teacher's Note:
1. Different maximum marks must be turned into percentages before comparing.
2. The difference of two percentages is measured in 'percentage points'.

Question 13. The price of a school bag is increased from ₹1600 to ₹1840. Find the percentage increase.
Answer:
Increase = 1840 − 1600 = ₹240.
Percentage increase = \( \frac{240}{1600} \) × 100 = 15% (the original price ₹1600 is the base).
In simple words: The price went up by ₹240 on ₹1,600, which is 15%.

Teacher's Note:
1. Divide the change by the original value.
2. Check: 115% of 1600 = 1840.

Question 14. A shopkeeper buys an article for ₹900 and marks it 30% above the cost price. He then offers a discount of 10% on the marked price. Find the marked price and the selling price.
Answer:
He marks the article 30% above cost: 30% of 900 = \( \frac{30}{100} \) × 900 = ₹270.
Marked price = 900 + 270 = ₹1170.
Discount = 10% of 1170 = ₹117.
Selling price = 1170 − 117 = ₹1053.
(His actual profit is then 1053 − 900 = ₹153, i.e. 17%, not 30%.)
In simple words: He marks the article at ₹1,170 and, after the 10% discount, sells it for ₹1,053.

Teacher's Note:
1. The marked price is built on the cost price; the discount is taken on the marked price.
2. Because of the discount, the real profit (17%) is less than the planned 30%.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A laptop has a marked price of ₹50,000. A discount of 12% is offered on the marked price. GST at 18% is then charged on the discounted price. Find the final amount paid by the customer.
Answer:
Discount = 12% of 50,000 = \( \frac{12}{100} \) × 50000 = ₹6,000.
Price after discount = 50,000 − 6,000 = ₹44,000.
GST = 18% of 44,000 = \( \frac{18}{100} \) × 44000 = ₹7,920.
Final amount = 44,000 + 7,920 = ₹51,920.
Check: 50000 × 0.88 × 1.18 = 51920.
In simple words: Take off 12% to get ₹44,000, then add 18% tax to pay ₹51,920.

Teacher's Note:
1. Order matters: the discount comes first, and GST is charged on the reduced price.
2. Use multipliers: 0.88 for a 12% discount, 1.18 for 18% GST.

Question 16. A shopkeeper buys a set of articles for ₹3200. He marks the price 25% above the cost price and then gives a discount of 20% on the marked price. Find his selling price and determine whether he makes a profit or loss.
Answer:
Mark-up = 25% of 3200 = ₹800, so marked price = 3200 + 800 = ₹4000.
Discount = 20% of 4000 = ₹800.
Selling price = 4000 − 800 = ₹3200.
SP = CP = ₹3200, so there is neither profit nor loss.
Multiplier check: 1.25 × 0.8 = 1.
In simple words: Marking up by 25% and then cutting by 20% brings the price back to exactly ₹3,200.

Teacher's Note:
1. The 25% is on ₹3200, but the 20% is on the larger ₹4000, so both amounts are ₹800.
2. A 25% increase is exactly undone by a 20% decrease (5/4 × 4/5 = 1).

Question 17. The population of a town is 80,000. It increases by 5% in one year and then decreases by 5% in the following year. Find the population after two years and the overall percentage change.
Answer:
Year 1: increase = 5% of 80,000 = 4,000, so population = 84,000.
Year 2: decrease = 5% of 84,000 = 4,200, so population = 84,000 − 4,200 = 79,800.
Overall change = 80,000 − 79,800 = 200 (a decrease); percentage = \( \frac{200}{80000} \) × 100 = 0.25% decrease.
Check: 1.05 × 0.95 = 0.9975, i.e. 0.25% less.
In simple words: The population ends at 79,800, which is 200 fewer people, a 0.25% fall.

Teacher's Note:
1. The second year's 5% is on 84,000, not on 80,000.
2. Equal % rise then fall always gives a small net fall of (r²/100)% = 25/100 = 0.25%.

Section D: Long Answer Question (5 Marks)

Question 18. A stationery dealer purchases 50 identical school bags at ₹640 each. He marks each bag 35% above its cost price and offers a discount of 20% on the marked price. Find:
(i) the marked price of each bag,
(ii) the selling price of each bag,
(iii) the profit on each bag,
(iv) the total profit on all 50 bags, and
(v) the percentage profit.
Answer:
(i) Mark-up = 35% of 640 = \( \frac{35}{100} \) × 640 = ₹224. Marked price = 640 + 224 = ₹864.
(ii) Discount = 20% of 864 = \( \frac{20}{100} \) × 864 = ₹172.80. Selling price = 864 − 172.80 = ₹691.20.
(iii) Profit per bag = 691.20 − 640 = ₹51.20.
(iv) Total profit = 50 × 51.20 = ₹2,560.
(v) Profit % = \( \frac{51.20}{640} \) × 100 = 8%.
Check: 1.35 × 0.8 = 1.08, an 8% profit.
In simple words: Each bag is marked ₹864, sold for ₹691.20, gives ₹51.20 profit, which is ₹2,560 on 50 bags, an 8% profit.

Teacher's Note:
1. The percentage profit is the same for one bag and for all 50 bags.
2. Combine mark-up and discount with multipliers to check the final percentage quickly.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1 – Planning a School Excursion
A school is organising an educational excursion for 200 students. The students are divided between Classes VIII and IX in the ratio 11 : 9. The teachers conducting the trip observe that 40% of the Class VIII students have opted for a particular science activity. The activity involves studying proportions and percentages through measurements collected during the visit. The students are asked to determine what fraction of the total group has selected the activity and express this fraction as a percentage. They then compare this percentage with another group in which 30% of all students have selected a different activity.
(i) Find the number of Class VIII students.
(ii) Find the number of Class VIII students who selected the science activity.
(iii) What percentage of the total students selected the science activity?
(iv) Which activity has the greater participation percentage: the science activity or the other activity selected by 30% of all students?
Answer:
(i) Total parts = 11 + 9 = 20. Class VIII = \( \frac{11}{20} \) × 200 = 110 students.
(ii) 40% of 110 = \( \frac{40}{100} \) × 110 = 44 students.
(iii) \( \frac{44}{200} \) = \( \frac{11}{50} \) of the whole group; \( \frac{44}{200} \) × 100 = 22%.
(iv) 30% > 22%, so the other activity has the greater participation percentage.
In simple words: 110 students are in Class VIII, 44 of them chose science, which is only 22% of all 200, less than 30%.

Teacher's Note:
1. 40% of Class VIII is not 40% of everyone: always check which whole the percentage refers to.
2. Shortcut for (iii): 40% of 11/20 = 0.4 × 55% = 22%.

Question 20. Case Study 2 – A School Book Fair
A school conducts a book fair to encourage students to read mathematics and science books. A particular reference book has a marked price of ₹1200. To attract more students, the bookseller first offers a discount of 20% on the marked price. During the final hour of the fair, an additional discount of 10% is offered on the already discounted price. A student argues that the total discount is therefore 30%, because 20% + 10% = 30%. The teacher asks the students to calculate the actual final price and the actual overall percentage discount to check the statement.
(i) Find the price after the first discount.
(ii) Find the price after the additional 10% discount.
(iii) Find the total discount in rupees.
(iv) Find the actual overall percentage discount. Is it 30%?
Answer:
(i) 20% of 1200 = ₹240, so price = 1200 − 240 = ₹960.
(ii) 10% of 960 = ₹96, so price = 960 − 96 = ₹864.
(iii) Total discount = 1200 − 864 = ₹336.
(iv) Overall discount = \( \frac{336}{1200} \) × 100 = 28%. It is not 30%, because the extra 10% is taken on ₹960, not on ₹1200 (₹96 instead of ₹120).
In simple words: The two discounts together save ₹336, which is 28% of ₹1,200, not 30%.

Teacher's Note:
1. Successive discounts are multiplied: 0.8 × 0.9 = 0.72, so 28% off.
2. Adding the percentages only works when both are taken on the same base.