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

CBSE Class 8 Maths Ganita Prakash Part 2 Chapter 1 Fractions in Disguise Practice Paper 27 is a 40-mark test of 20 questions in Sections A to E, given here with step-by-step answers. It checks converting fractions and ratios into percentages, percentage increase and decrease, discount on the marked price, profit and loss percentages, depreciation, simple versus compound interest and why successive percentage changes do not simply add up.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. During a school mathematics exhibition, 240 visitors entered the exhibition on the first day. \( \frac{3}{8} \) of the visitors participated in the mathematics activity zone.
Find: (i) the number of visitors who participated in the activity zone, and
(ii) the percentage of all visitors represented by these participants.
Answer:
(i) Participants = \( \frac{3}{8} \) × 240 = 3 × 30 = 90 visitors.
(ii) Percentage = \( \frac{3}{8} \) × 100 = \( \frac{300}{8} \) = 37.5%.
Check: \( \frac{90}{240} \) × 100 = 37.5%.
In simple words: 3 out of every 8 visitors joined, which is 90 people, or 37.5 out of every 100.

Teacher's Note:
1. To find a fraction of a number, divide by the denominator and multiply by the numerator.
2. Fraction × 100 gives the percentage; 1/8 = 12.5% is worth remembering.

Question 12. A mixture contains millet and water in the ratio 3 : 7. A student claims that millet constitutes 30% of the mixture. Examine the claim mathematically and determine the exact percentage of millet in the mixture. Also state whether the student's claim is correct.
Answer:
Total parts = 3 + 7 = 10.
Fraction of millet = \( \frac{3}{10} \).
Percentage of millet = \( \frac{3}{10} \) × 100 = 30%.
So the student's claim is correct.
In simple words: Millet is 3 parts out of 10 parts, which is 30 out of 100, so 30% is right.

Teacher's Note:
1. For a ratio a : b, the fraction of the first part is a ÷ (a + b), not a ÷ b.
2. Here 3/7 would be wrong; the denominator is the total 10 parts.

Question 13. The price of a school bag was ₹1,600. During a clearance sale, its price was reduced to ₹1,360. Calculate the percentage decrease in the price. Show clearly which quantity is taken as the base while calculating the percentage decrease.
Answer:
Decrease = 1600 − 1360 = ₹240.
Base: the original price ₹1,600 (the price before the change).
Percentage decrease = \( \frac{240}{1600} \) × 100 = 15%.
In simple words: The price fell by ₹240 out of ₹1,600, which is 15 out of every 100.

Teacher's Note:
1. For any percentage change, divide the change by the value before the change.
2. Using ₹1,360 as base would give about 17.6%, which is wrong.

Question 14. A shopkeeper purchases a study table for ₹2,500 and fixes a profit margin of 24% on the cost price. During a sale, he gives a discount of 10% on the resulting selling price. Determine whether he finally makes a profit or a loss, and find the corresponding amount.
Answer:
Profit margin = 24% of 2500 = \( \frac{24}{100} \) × 2500 = ₹600, so the price before discount = 2500 + 600 = ₹3,100.
Discount = 10% of 3100 = ₹310, so final selling price = 3100 − 310 = ₹2,790.
Since ₹2,790 > ₹2,500 (cost price), he makes a profit.
Profit = 2790 − 2500 = ₹290.
In simple words: He first raises the price to ₹3,100, sells at ₹2,790 after discount, and still gains ₹290.

Teacher's Note:
1. The 10% discount is on ₹3,100, not on the cost price.
2. A 24% markup followed by a 10% discount is not a 14% profit: 1.24 × 0.9 = 1.116, an 11.6% profit.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A student scored 54 marks out of 75 in Mathematics and 68 marks out of 90 in Science.
(i) Convert both scores into percentages.
(ii) Identify in which subject the student obtained the higher percentage.
(iii) Explain why comparing only the marks obtained would not give a fair comparison.
Answer:
(i) Mathematics: \( \frac{54}{75} \) × 100 = 72%.
Science: \( \frac{68}{90} \) × 100 = \( \frac{680}{9} \) ≈ 75.56%.
(ii) 75.56% > 72%, so the student did better in Science.
(iii) The two tests have different maximum marks (75 and 90). 68 is more than 54 partly because the Science paper had more marks. Percentages turn both scores into 'marks out of 100', so they can be compared fairly.
In simple words: Change both scores to 'out of 100': Maths is 72 and Science about 75.6, so Science is better.

Teacher's Note:
1. Percentages put different totals on the same scale of 100.
2. Raw marks can only be compared directly when the maximum marks are equal.

Question 16. A shopkeeper buys a jacket for ₹2,400 and marks it at ₹3,000. He offers a discount of 12% on the marked price.
Find: (i) the selling price,
(ii) the profit earned, and
(iii) the percentage profit with respect to the cost price.
Answer:
(i) Discount = 12% of 3000 = \( \frac{12}{100} \) × 3000 = ₹360. Selling price = 3000 − 360 = ₹2,640.
(ii) Profit = SP − CP = 2640 − 2400 = ₹240.
(iii) Profit % = \( \frac{240}{2400} \) × 100 = 10%.
In simple words: He sells at ₹2,640 after the discount, gains ₹240, and that is 10% of what he paid.

Teacher's Note:
1. Discount is on the marked price; profit percentage is on the cost price.
2. Keep the three prices (CP, MP, SP) clearly labelled.

Question 17. A bank offers 8% annual interest on a fixed deposit. Riya deposits ₹25,000 for 2 years. Calculate the final amount in each of the following cases:
(i) interest is paid separately every year without compounding,
(ii) interest is compounded annually.
Hence, find how much more money is received under annual compounding.
Answer:
(i) Interest each year = 8% of 25000 = \( \frac{8}{100} \) × 25000 = ₹2,000. For 2 years = 2 × 2000 = ₹4,000. Amount = 25000 + 4000 = ₹29,000.
(ii) Year 1: interest ₹2,000, so amount = ₹27,000.
Year 2: interest = 8% of 27000 = \( \frac{8}{100} \) × 27000 = ₹2,160, so final amount = 27000 + 2160 = ₹29,160.
Extra money under compounding = 29160 − 29000 = ₹160.
(The ₹160 is 8% of the first year's interest ₹2,000: interest earned on interest.)
In simple words: Compounding gives ₹160 more because in the second year the bank also pays interest on the first year's ₹2,000 interest.

Teacher's Note:
1. Simple interest: same interest every year. Compound interest: interest grows each year.
2. Check: 25000 × 1.08 × 1.08 = 29160.

Section D: Long Answer Question (5 Marks)

Question 18. A community centre is planning a large educational fair. The organisers estimate that 40% of the total expected visitors will be students, 35% will be parents and the remaining visitors will be teachers and other guests. The organisers expect 2,400 visitors in all. On the day of the fair, the actual number of visitors is 15% greater than the expected number.
Answer the following:
(i) Find the expected number of students, parents and teachers/other guests separately.
(ii) Find the actual number of visitors on the day of the fair.
(iii) If 60% of the actual visitors are adults, find the number of adult visitors.
(iv) Express the actual number of visitors as a percentage of the expected number.
Answer:
(i) Students = 40% of 2400 = \( \frac{40}{100} \) × 2400 = 960.
Parents = 35% of 2400 = \( \frac{35}{100} \) × 2400 = 840.
Teachers and other guests = 100% − 40% − 35% = 25%; 25% of 2400 = 600.
Check: 960 + 840 + 600 = 2400.
(ii) Increase = 15% of 2400 = 360, so actual visitors = 2400 + 360 = 2,760.
(iii) Adults = 60% of 2760 = \( \frac{60}{100} \) × 2760 = 1,656.
(iv) \( \frac{2760}{2400} \) × 100 = 115% (100% expected + 15% extra).
In simple words: Split 2,400 as 960, 840 and 600; 15% more gives 2,760 visitors, of whom 1,656 are adults, and 2,760 is 115% of 2,400.

Teacher's Note:
1. The 60% in part (iii) is taken of the actual number 2,760, not of 2,400.
2. A 15% increase means the new value is 115% of the old value.

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

Question 19. Case Study 1 – The School Nutrition Drive
During a school nutrition drive, the students of Class VIII prepared a healthy food mixture for a community exhibition. The mixture contained millet and water in the ratio 2 : 3. The students prepared 750 ml of the mixture. They observed that converting the ratio into a fraction helped them determine the percentage of millet. Later, they increased the quantity of the mixture while maintaining the same proportion. The teacher asked them to estimate the percentage first and then calculate the exact quantity. The students also compared their mixture with another preparation containing 30% millet to decide which had the larger proportion of millet.
Answer the following:
(i) What fraction of the mixture is millet?
(ii) What percentage of the mixture is millet?
(iii) How much millet is present in 750 ml of the mixture?
(iv) Which mixture contains the greater proportion of millet: this mixture or the mixture containing 30% millet?
Answer:
Millet : water = 2 : 3, so total parts = 2 + 3 = 5.
(i) Fraction of millet = \( \frac{2}{5} \).
(ii) Percentage = \( \frac{2}{5} \) × 100 = 40%.
(iii) Millet = 40% of 750 ml = \( \frac{2}{5} \) × 750 = 300 ml (and water = 450 ml).
(iv) 40% > 30%, so this (Class VIII) mixture has the greater proportion of millet.
In simple words: Millet is 2 of every 5 parts, which is 40%, so 750 ml has 300 ml of millet, more than a 30% mixture.

Teacher's Note:
1. Ratio 2 : 3 gives the fraction 2/5, not 2/3.
2. Making the mixture bigger in the same ratio does not change the percentage.

Question 20. Case Study 2 – The School Book Fair
A school organised a book fair to encourage reading among students. A reference book had a marked price of ₹1,200. During the first stage of the sale, the shopkeeper offered a 20% discount. To clear the remaining stock, he then offered an additional 10% discount on the reduced price. A student argued that the total discount was 30% because 20% + 10% = 30%. The teacher asked the students to calculate both prices carefully and compare the actual reduction with the marked price. This activity helped them understand why successive percentage changes must be applied to the changed quantity.
Answer the following:
(i) Find the price after the first 20% discount.
(ii) Find the price after the additional 10% discount.
(iii) Find the actual percentage discount on the marked price.
(iv) Is the student's claim that the total effective discount is 30% correct? Give a mathematical reason.
Answer:
(i) First discount = 20% of 1200 = ₹240, so price = 1200 − 240 = ₹960.
(ii) Second discount = 10% of 960 = ₹96, so price = 960 − 96 = ₹864.
(iii) Total reduction = 1200 − 864 = ₹336; percentage = \( \frac{336}{1200} \) × 100 = 28%.
(iv) No. The second 10% is taken on ₹960, not on ₹1,200, so it is only ₹96 instead of ₹120. Total discount = 240 + 96 = ₹336, which is 28% of ₹1,200, not 30%.
Multiplier check: 0.8 × 0.9 = 0.72, so the buyer pays 72%, a discount of 28%.
In simple words: The second discount is on a smaller price, so together the discounts make 28%, not 30%.

Teacher's Note:
1. Successive discounts multiply; they do not add.
2. Two successive discounts of a% and b% give a total of a + b − (ab ÷ 100) percent: 20 + 10 − 2 = 28.