CBSE Class 8 Mathematics Fractions in Disguise Practice Paper Assignment Set 02 with Answers

This is CBSE Class 8 Mathematics Ganita Prakash Part 2 Chapter 1 Fractions in Disguise Practice Paper 29: 40 marks, 20 questions in Sections A to E, with a step-by-step answer under every question. It covers turning fractions such as 7/16 and decimals such as 0.875 into percentages, finding a number from its percentage, marked price from selling price, gross and net profit on sales, overall loss on two articles, discount followed by GST, and comparing food labels with different packet sizes.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A water tank is initially \( \frac{3}{8} \) full. Then 20% of the remaining empty capacity is filled. What fraction of the tank is full now? Express the answer as a percentage.
Answer:
Empty part = 1 − \( \frac{3}{8} \) = \( \frac{5}{8} \).
Water added = 20% of \( \frac{5}{8} \) = \( \frac{1}{5} \) × \( \frac{5}{8} \) = \( \frac{1}{8} \).
Now full = \( \frac{3}{8} \) + \( \frac{1}{8} \) = \( \frac{4}{8} \) = \( \frac{1}{2} \).
As a percentage: \( \frac{1}{2} \) × 100 = 50%.
In simple words: Half the tank is now full.

Teacher's Note:
1. The 20% is of the empty part, not of the whole tank.
2. Draw the tank in 8 equal parts to see it: 3 full, then 1 more of the 5 empty parts.

Question 12. A recipe contains rava and sugar in the ratio 3 : 2. If the total mixture weighs 2.5 kg, find the quantity and percentage of each ingredient.
Answer:
Total parts = 3 + 2 = 5.
Rava = \( \frac{3}{5} \) × 2.5 kg = 1.5 kg; percentage = \( \frac{3}{5} \) × 100 = 60%.
Sugar = \( \frac{2}{5} \) × 2.5 kg = 1 kg; percentage = \( \frac{2}{5} \) × 100 = 40%.
Check: 1.5 + 1 = 2.5 kg and 60% + 40% = 100%.
In simple words: Rava is 1.5 kg (60%) and sugar is 1 kg (40%).

Teacher's Note:
1. For a ratio, divide by the total number of parts.
2. Always check that the parts add back to the whole.

Question 13. An article is sold for ₹2040 after allowing a discount of 15%. Find its marked price.
Answer:
After a 15% discount, selling price = 85% of the marked price.
85% of MP = 2040, so MP = 2040 × \( \frac{100}{85} \) = ₹2400.
Check: 15% of 2400 = 360 and 2400 − 360 = 2040.
In simple words: ₹2,040 is 85% of the marked price, so the marked price is ₹2,400.

Teacher's Note:
1. Work backwards from the selling price using the remaining percentage (100 − 15 = 85).
2. Adding 15% of 2040 to 2040 is wrong; the 15% was of the marked price.

Question 14. A business has sales of ₹2,40,000. Its cost of purchasing the goods is ₹1,60,000 and its other expenses are ₹12,000. Find the gross profit, net profit and net profit percentage on sales.
Answer:
Gross profit = sales − cost of goods = 2,40,000 − 1,60,000 = ₹80,000.
Net profit = gross profit − other expenses = 80,000 − 12,000 = ₹68,000.
Net profit % on sales = \( \frac{68000}{240000} \) × 100 = \( \frac{85}{3} \) ≈ 28.33%.
In simple words: After paying for the goods and other costs, ₹68,000 is left, which is about 28.33% of the sales.

Teacher's Note:
1. Gross profit ignores other expenses; net profit subtracts them.
2. Here the percentage is on sales, as the question says, not on cost.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Two students score \( \frac{7}{12} \) and \( \frac{5}{8} \) of the maximum marks respectively. Who has the higher percentage score and by how many percentage points?
Answer:
First student: \( \frac{7}{12} \) × 100 = \( \frac{175}{3} \) ≈ 58.33%.
Second student: \( \frac{5}{8} \) × 100 = 62.5%.
62.5% > 58.33%, so the second student scored higher.
Difference = 62.5 − 58.33 ≈ 4.17 percentage points (exactly \( \frac{25}{6} \) = 4⅙).
In simple words: The second student got 62.5% and the first about 58.3%, so the second is ahead by about 4.17 points.

Teacher's Note:
1. Change both fractions to percentages before comparing.
2. The gap between two percentages is in percentage points, not percent.

Question 16. A sum of ₹16,000 is invested at 10% per annum for 2 years. Find the amount when the growth is compounded annually. Also find the difference between this amount and the amount obtained by calculating 10% on the original principal each year.
Answer:
Compounded: Year 1 interest = 10% of 16,000 = ₹1,600, amount = ₹17,600.
Year 2 interest = 10% of 17,600 = ₹1,760, amount = 17,600 + 1,760 = ₹19,360.
On original principal each year: interest = 1,600 × 2 = ₹3,200, amount = ₹19,200.
Difference = 19,360 − 19,200 = ₹160.
In simple words: Compounding gives ₹19,360, which is ₹160 more than ₹19,200, because the second year also earns interest on the first year's ₹1,600.

Teacher's Note:
1. The ₹160 is 10% of the first year's interest ₹1,600.
2. Check: 16000 × 1.1 × 1.1 = 19360.

Question 17. A school purchases a piece of equipment marked at ₹4800. A discount of 25% is offered, and GST of 5% is charged on the discounted price. Find the final amount payable.
Answer:
Discount = 25% of 4800 = \( \frac{1}{4} \) × 4800 = ₹1,200.
Price after discount = 4800 − 1200 = ₹3,600.
GST = 5% of 3600 = \( \frac{5}{100} \) × 3600 = ₹180.
Final amount = 3600 + 180 = ₹3,780.
Check: 4800 × 0.75 × 1.05 = 3780.
In simple words: After 25% off the price is ₹3,600; adding 5% GST makes it ₹3,780.

Teacher's Note:
1. Discount first, then GST on the reduced price.
2. 25% is the same as one quarter.

Section D: Long Answer Question (5 Marks)

Question 18. A trader sells two different articles for ₹54,000 each. On the first article, he gains 8%, while on the second article, he incurs a loss of 10%.
Find:
(i) the cost price of the first article,
(ii) the cost price of the second article,
(iii) the total cost price,
(iv) the overall profit or loss, and
(v) the overall percentage profit or loss.
Answer:
(i) Gain 8%, so SP = 108% of CP. CP = 54000 × \( \frac{100}{108} \) = ₹50,000.
(ii) Loss 10%, so SP = 90% of CP. CP = 54000 × \( \frac{100}{90} \) = ₹60,000.
(iii) Total CP = 50,000 + 60,000 = ₹1,10,000.
(iv) Total SP = 2 × 54,000 = ₹1,08,000. Since SP < CP, there is a loss of ₹2,000.
(v) Loss % = \( \frac{2000}{110000} \) × 100 = \( \frac{20}{11} \) ≈ 1.82%.
In simple words: The two articles cost ₹1,10,000 together but were sold for ₹1,08,000, a loss of ₹2,000, about 1.82%.

Teacher's Note:
1. Find each cost price by working backwards from the selling price (divide by 1.08 and by 0.9).
2. Do not subtract 10% − 8% = 2%; the percentages are on different cost prices.

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

Question 19. Case Study 1 – Comparing Food Labels
A school nutrition club compares the labels of two packaged food products. For Product A, 84 g of the 120 g packet is made up of the main nutritional ingredient. For Product B, 105 g of the 150 g packet is made up of the same type of ingredient. The students first compare these proportions by converting them into percentages. They also notice that Product A contains 18 g of food chemicals, while Product B contains 20 g. The teacher asks them to compare both products carefully because the packet sizes are different. The students must use percentages rather than simply comparing the masses printed on the labels.
(i) Find the percentage of the main ingredient in Product A.
(ii) Find the percentage of the main ingredient in Product B.
(iii) Find the percentage of food chemicals in each product.
(iv) Which product has the smaller proportion of food chemicals?
Answer:
(i) Product A: \( \frac{84}{120} \) × 100 = 70%.
(ii) Product B: \( \frac{105}{150} \) × 100 = 70% (the same as Product A).
(iii) Product A: \( \frac{18}{120} \) × 100 = 15%; Product B: \( \frac{20}{150} \) × 100 = \( \frac{40}{3} \) ≈ 13.33%.
(iv) 13.33% < 15%, so Product B has the smaller proportion of food chemicals, even though it has more grams of them (20 g against 18 g).
In simple words: Both have 70% main ingredient, but B is only about 13.3% chemicals against 15% for A.

Teacher's Note:
1. Compare proportions, not raw masses, when the totals are different.
2. A larger amount can still be a smaller share of a bigger packet.

Question 20. Case Study 2 – A Festival Exhibition Stall
A school mathematics club sets up a stall at an educational exhibition. The club purchases a model for ₹2400 and decides to mark its price 25% above the cost price. To attract visitors, the stall offers a discount of 12% on the marked price. A GST of 5% is then charged on the discounted selling price. The students are asked to distinguish between the selling price before GST and the final amount paid by a customer. They also investigate the actual profit made by the club before GST, because the GST collected from the customer is a tax and is not part of the selling price retained by the seller.
(i) Find the marked price of the model.
(ii) Find the discount and the selling price before GST.
(iii) Find the GST and the final amount paid by the customer.
(iv) Find the profit percentage earned by the club before GST.
Answer:
(i) Mark-up = 25% of 2400 = ₹600, so marked price = ₹3,000.
(ii) Discount = 12% of 3000 = ₹360; selling price before GST = 3000 − 360 = ₹2,640.
(iii) GST = 5% of 2640 = ₹132; final amount paid = 2640 + 132 = ₹2,772.
(iv) Profit = 2640 − 2400 = ₹240; profit % = \( \frac{240}{2400} \) × 100 = 10%. The ₹132 GST goes to the government, so it is not part of the profit.
In simple words: The model is marked ₹3,000, sold for ₹2,640 plus ₹132 GST (₹2,772 in all), and the club's profit is 10%.

Teacher's Note:
1. Profit is worked out on the selling price before GST.
2. Check: 1.25 × 0.88 = 1.10, a 10% profit.