CBSE Class 8 Mathematics Fractions in Disguise Long Answer and Case Study Questions with Answers

Long Answer Questions

Teacher's Note:
1. Percentage ÷ 100 gives the decimal.
2. A fraction in simplest form needs the HCF to be cancelled.
3. Percentages above 100 are common in growth questions.

Question 2. A survey of 400 households in a village recorded the main cooking fuel used in each home: LPG 220 households, firewood 100 households, biogas 50 households and electricity 30 households.
(i) Express the number of households using each fuel as a percentage of all the households surveyed.
(ii) Verify that the four percentages add up to 100% and explain why they must.
(iii) Which fuels are used by more than one quarter of the households?
Answer:
(i) Each count is divided by 400, the whole, and multiplied by 100.
LPG = \( \frac{220}{400} \) × 100 = 55%; Firewood = \( \frac{100}{400} \) × 100 = 25%; Biogas = \( \frac{50}{400} \) × 100 = 12.5%; Electricity = \( \frac{30}{400} \) × 100 = 7.5%.
(ii) 55 + 25 + 12.5 + 7.5 = 100. They must add to 100% because every household uses exactly one of the four fuels, so the four groups together make up all 400 households, and the whole is always 100%.
(iii) One quarter is 25%. LPG at 55% is more than a quarter. Firewood is exactly 25%, which is equal to one quarter, not more. Biogas and electricity are far less. So only LPG is used by more than one quarter of the households.
In simple words: LPG 55%, firewood 25%, biogas 12.5% and electricity 7.5% make 100%, and only LPG is above one quarter.

Teacher's Note:
1. The whole (400 households) is the base for every fuel.
2. ‘More than’ excludes the value that is exactly equal.
3. Percentages of non-overlapping groups add to 100%.

Question 3. Kiran runs a cycle shop. He buys a bicycle from the wholesaler for ₹4500 and marks its price as ₹5400. A customer bargains and Kiran finally sells the bicycle after allowing a discount of 5% on the marked price.
(i) Find the selling price of the bicycle.
(ii) Find the profit Kiran makes in rupees.
(iii) Find his profit percentage, stating clearly which price you have taken as 100% and why.
Answer:
(i) Discount = 5% of ₹5400 = 0.05 × 5400 = ₹270.
Selling price = 5400 − 270 = ₹5130. (Check: 95% of 5400 = 0.95 × 5400 = 5130.)
(ii) Profit = selling price − cost price = 5130 − 4500 = ₹630.
(iii) Profit % = \( \frac{\text{profit}}{\text{cost price}} \) × 100 = \( \frac{630}{4500} \) × 100 = 14%.
Here the cost price ₹4500 is taken as 100%, because profit percentage answers the question “how much did I gain compared with what I spent on buying the goods?” Kiran spent ₹4500, so that is the base. Using the marked price or the selling price as the base would give a different, misleading figure.
In simple words: Kiran sells the cycle for ₹5130, gaining ₹630, which is 14% of his cost of ₹4500.

Teacher's Note:
1. Discount is on the marked price; profit is on the cost price.
2. Keep the two bases separate.
3. Check the answer by the 95% route.

Question 4. Sunita wants to deposit ₹15,000 in a fixed deposit for 2 years at 8% per annum. The bank gives her two options: in Option 1 the interest is paid to her every year, and in Option 2 the interest is added back to the deposit each year.
(i) Find the total amount she receives after 2 years under Option 1.
(ii) Find the total amount she receives after 2 years under Option 2.
(iii) Find the difference between the two amounts and explain where this extra money comes from.
Answer:
(i) Option 1 (interest paid out every year): the principal stays ₹15,000 for both years.
Interest each year = 0.08 × 15,000 = ₹1200, so for 2 years = ₹2400.
Amount = p(1 + rt) = 15,000 + 2400 = ₹17,400.
(ii) Option 2 (interest added back, i.e. compounding):
End of Year 1: 15,000 × 1.08 = ₹16,200 (interest ₹1200).
End of Year 2: 16,200 × 1.08 = ₹17,496 (interest ₹1296).
Amount = p(1 + r)^t = 15,000 × (1.08)² = ₹17,496.
(iii) Difference = 17,496 − 17,400 = ₹96. This extra money is the interest earned on the first year’s interest. In Option 2 the ₹1200 earned in Year 1 stays in the deposit, so in Year 2 interest is calculated on ₹16,200 instead of ₹15,000; and 8% of ₹1200 is exactly ₹96.
In simple words: Option 1 gives ₹17,400 and Option 2 gives ₹17,496; the extra ₹96 is interest on the first year’s interest.

Teacher's Note:
1. Compounding earns interest on interest.
2. (1.08)² = 1.1664.
3. Check the extra amount: 8% of 1200 = 96.

Question 5. In the annual examination, Amrita scored 63 marks out of 75 in Mathematics. In Science, the paper was of 125 marks and she scored 80% of them.
(i) Express her Mathematics score as a percentage.
(ii) Find her Science score in marks.
(iii) Find her overall percentage in the two subjects taken together, and state in which subject she did better.
Answer:
(i) Mathematics = \( \frac{63}{75} \) × 100 = 84%.
(ii) 80% of 125 = \( \frac{80}{100} \) × 125 = 100 marks.
(iii) Total marks scored = 63 + 100 = 163. Total maximum marks = 75 + 125 = 200.
Overall = \( \frac{163}{200} \) × 100 = 81.5%.
Mathematics is 84% and Science is 80%, so Amrita did better in Mathematics, even though her Science score (100 marks) is a bigger number than her Mathematics score (63 marks). The raw marks cannot be compared because the papers had different maximum marks; percentages put both on the same base of 100.
Note that 81.5% is not the average of 84% and 80%, because the Science paper carried more marks and pulls the overall figure towards itself.
In simple words: Maths is 84%, Science is 100 marks (80%), overall is 81.5%, and she did better in Maths.

Teacher's Note:
1. Overall percentage = total scored ÷ total maximum × 100.
2. Do not simply average two percentages with different bases.
3. Compare subjects by percentages, not raw marks.

Question 6. The population of a village was 8000 at the beginning of 2023. During 2023 it rose by 5%, and during 2024 it fell by 10%.
(i) Find the population at the end of 2023.
(ii) Find the population at the end of 2024.
(iii) Find the overall percentage change in the population over the two years, and explain why the answer is not simply 5% − 10% = −5%.
Answer:
(i) A rise of 5% means the population becomes 105% of 8000: 1.05 × 8000 = 8400.
(ii) A fall of 10% means it becomes 90% of the 2023 figure: 0.9 × 8400 = 7560.
(iii) Change = 7560 − 8000 = −440. Percentage change = \( \frac{440}{8000} \) × 100 = 5.5% decrease.
It is not 5% − 10% = −5% because the two percentages are taken on different bases. The rise of 5% was on 8000 and came to 400, but the fall of 10% was on the larger number 8400 and came to 840. The correct method is to multiply the factors: 8000 × 1.05 × 0.90 = 8000 × 0.945, and 0.945 means 94.5% of the original, a drop of 5.5%.
In simple words: The population goes 8000 → 8400 → 7560, an overall fall of 5.5%, because the 10% fall is on a bigger number.

Teacher's Note:
1. Successive percentage changes multiply; they do not add.
2. 1.05 × 0.9 = 0.945.
3. Always measure the overall change against the starting value.

Question 7. A provision store offers its customers a choice on every bill above ₹200: either a discount of 20% or a flat discount of ₹60.
(i) Which offer is better for a bill of ₹250? Show the saving in each case.
(ii) Which offer is better for a bill of ₹400? Show the saving in each case.
(iii) Find the bill amount for which the two offers give exactly the same saving, and state what happens for bills above that amount.
Answer:
(i) 20% of 250 = ₹50 saved; the flat offer saves ₹60. Since ₹60 > ₹50, the flat ₹60 discount is better (pay ₹190 instead of ₹200).
(ii) 20% of 400 = ₹80 saved; the flat offer saves ₹60. Since ₹80 > ₹60, the 20% discount is better (pay ₹320 instead of ₹340).
(iii) Let the bill be ₹x. The savings are equal when \( \frac{20}{100} \) × x = 60, so x = \( \frac{60 × 100}{20} \) = ₹300.
For a bill of ₹300 both offers save ₹60. For bills above ₹300 the 20% offer is better (and below ₹300 the flat ₹60 is better), because a percentage discount grows with the bill while a flat discount stays the same.
In simple words: The flat ₹60 is better for ₹250, 20% is better for ₹400, and they are equal at ₹300; above ₹300 the 20% offer wins.

Teacher's Note:
1. Find the break-even point by equating the two savings.
2. Percentage discounts grow with the bill.
3. Flat discounts are better only for small bills.

Question 8. Two shops sell the same water bottle at ₹80 each. Shop A announces “Buy 1 and get 1 free” and Shop B announces “Buy 3 and get 1 free”.
(i) Find the effective price paid per bottle in each shop.
(ii) Find the percentage discount offered by each shop, comparing the free bottles with the total bottles received.
(iii) Sunil needs exactly 4 bottles. Find what he pays in each shop and say which shop he should choose.
Answer:
(i) Shop A: pay for 1, receive 2. ₹80 for 2 bottles, so ₹80 ÷ 2 = ₹40 per bottle.
Shop B: pay for 3, receive 4. 3 × 80 = ₹240 for 4 bottles, so ₹240 ÷ 4 = ₹60 per bottle.
(ii) Shop A: 1 free out of 2 received = \( \frac{1}{2} \) × 100 = 50% discount.
Shop B: 1 free out of 4 received = \( \frac{1}{4} \) × 100 = 25% discount.
The prices give the same figures: \( \frac{40}{80} \) × 100 = 50% saved and \( \frac{20}{80} \) × 100 = 25% saved.
(iii) Shop A: two deals, pay for 2 bottles = ₹160. Shop B: one deal, pay for 3 bottles = ₹240.
Sunil should choose Shop A, saving ₹80 compared with Shop B and ₹160 compared with the full price of ₹320.
In simple words: Shop A works out to ₹40 a bottle (50% off) and Shop B to ₹60 (25% off), so 4 bottles cost ₹160 at A and ₹240 at B.

Teacher's Note:
1. Find the price per bottle to compare offers.
2. Discount % = free items ÷ total items received × 100.
3. ‘Buy 3 get 1’ is only 25% off, not 33%.

Question 9. Surbhi buys cookware from a wholesaler and fixes the price of every item at 40% above her cost price. Later, to clear the stock, she advertises a discount of 50% on the fixed price, believing that she will at least get back the money she spent.
(i) Taking her cost price as x, write expressions for the fixed price and for the price after the discount.
(ii) Does she make a profit or a loss? Find the percentage, and find her loss in rupees if her cost price was ₹16,000.
(iii) What percentage discount should she have offered so that she neither gained nor lost?
Answer:
(i) Let the cost price be x. Fixed (marked) price = x + 40% of x = 1.4x.
Price after a 50% discount = 1.4x × 0.5 = 0.7x.
(ii) She gets back 0.7x, which is less than x, so she makes a loss. Loss = x − 0.7x = 0.3x, and loss % = \( \frac{0.3x}{x} \) × 100 = 30%.
If her cost price was ₹16,000, she receives 0.7 × 16,000 = ₹11,200, so her loss is 16,000 − 11,200 = ₹4800, which is indeed 30% of ₹16,000.
(iii) Let the discount be d (as a decimal) on the fixed price 1.4x. For no profit and no loss, 1.4x(1 − d) = x, so 1 − d = \( \frac{1}{1.4} \) = \( \frac{5}{7} \), giving d = \( \frac{2}{7} \) ≈ 0.2857.
So she should have offered a discount of about 28.57%. Check: 1.4 × (1 − 0.2857) = 1.4 × 0.7143 ≈ 1.0. Her mistake was to think that a 50% discount is safe with a 40% margin; the margin is on the cost price while the discount is on the larger marked price.
In simple words: She sells at 0.7 times her cost, a 30% loss (₹4800 on ₹16,000); a discount of about 28.57% would have let her break even.

Teacher's Note:
1. Write the selling price as a multiple of the cost price.
2. Break-even means SP = CP.
3. Even a discount equal to the markup (40%) would cause a loss.

Question 10. A rectangular banner is being redesigned. Its length is increased by 25%, but the area of the banner must stay exactly the same as before.
(i) Taking the original length as l and breadth as b, write the new length and form an equation for the new breadth.
(ii) Find the new breadth as a percentage of the original breadth, and hence the exact percentage by which the breadth decreases.
(iii) A student says, “The length went up by 25%, so the breadth must come down by 25%.” Show by substituting actual numbers that this is wrong, and explain the reason.
Answer:
(i) Original area = lb. New length = l + 25% of l = 1.25l.
Let the new breadth be b′. Since the area is unchanged, 1.25l × b′ = l × b.
(ii) Dividing both sides by 1.25l: b′ = \( \frac{b}{1.25} \) = 0.8b.
So the new breadth is 80% of the old breadth, and the decrease is 100% − 80% = 20% exactly.
(iii) Take l = 20 cm and b = 10 cm, so the area is 200 cm². The new length is 25 cm.
If the breadth were cut by 25% it would become 7.5 cm, giving an area of 25 × 7.5 = 187.5 cm², which is less than 200 cm². So the student is wrong.
The correct new breadth is 0.8 × 10 = 8 cm, and 25 × 8 = 200 cm². Correct.
The reason is that the 25% increase is taken on the length, but the decrease is taken on the breadth, and these are different bases. To keep the product fixed, the breadth must be multiplied by \( \frac{1}{1.25} \) = 0.8, which is a 20% cut, not 25%.
In simple words: The new breadth must be 0.8 of the old, a 20% cut; cutting 25% would make the area smaller.

Teacher's Note:
1. Keep the product of the multipliers equal to 1.
2. Test a claim with simple numbers.
3. 1.25 × 0.8 = 1, but 1.25 × 0.75 = 0.9375.

Case Study Based Questions

Question 11. A festive sale at a clothing store. A clothing store put up a big board for its festival sale. Every pair of jeans on the shelf carries a marked price of ₹1600 and is sold at 25% off, while every T-shirt carries a marked price of ₹800 and is sold at 10% off. Nisha walks in with her brother and decides to buy exactly one pair of jeans and one T-shirt. At the billing counter the cashier first works out the discount on each item separately, then subtracts it from the marked price, and finally adds the two selling prices to make the bill. There is no tax to be added on these items.
(i) Find the discount amount on the pair of jeans.
(ii) Find the selling price of the pair of jeans.
(iii) Find the selling price of the T-shirt.
(iv) Find the total amount Nisha pays at the counter.
Answer:
(i) Discount on the jeans = 25% of ₹1600 = \( \frac{25}{100} \) × 1600 = ₹400.
(ii) Selling price of the jeans = 1600 − 400 = ₹1200. (Check: 75% of 1600 = 0.75 × 1600 = 1200.)
(iii) Discount on the T-shirt = 10% of ₹800 = ₹80, so its selling price = 800 − 80 = ₹720.
(iv) Total bill = 1200 + 720 = ₹1920. Nisha saves 400 + 80 = ₹480 on a marked total of ₹2400.
In simple words: The jeans cost ₹1200 and the T-shirt ₹720, so Nisha pays ₹1920.

Teacher's Note:
1. Apply each discount to its own marked price.
2. Overall saving is 480/2400 = 20%, not 25% + 10%.

Question 12. A school checks its attendance. A middle school has 800 students on its rolls. On Monday the class teachers reported that 736 students were present. The head teacher likes to record attendance as a percentage rather than as a count, because the strength of the school changes a little every year and percentages make different days easy to compare. She writes the attendance for the day in the register and also notes how many students were absent. On Tuesday the attendance improved and the register showed that 95% of the students were present.
(i) Find the percentage of students present on Monday.
(ii) Find the percentage of students absent on Monday.
(iii) Find the number of students absent on Monday.
(iv) Find the number of students present on Tuesday.
Answer:
(i) Attendance on Monday = \( \frac{736}{800} \) × 100 = 92%.
(ii) Absent = 100% − 92% = 8%.
(iii) Number absent = 800 − 736 = 64. Check: 8% of 800 = 0.08 × 800 = 64.
(iv) Present on Tuesday = 95% of 800 = 0.95 × 800 = 760, that is, 24 more students than on Monday.
In simple words: Monday’s attendance was 92% with 64 absent, and 760 students came on Tuesday.

Teacher's Note:
1. Present % + absent % = 100%.
2. Use the total roll of 800 as the base each day.

Question 13. The level in an overhead tank. The overhead tank of an apartment block holds 5000 L of water when completely full. A float gauge on the side shows how full the tank is. At 6 a.m. one morning the caretaker noted that the tank was 80% full. Through the day the families in the block drew water for cooking, washing and cleaning, and by 8 p.m. a total of 1200 L had been used. No fresh water was pumped into the tank during the day. The caretaker records the use of the day and the water left, both as a percentage of the full capacity of the tank.
(i) Find the volume of water in the tank at 6 a.m.
(ii) Express the water used during the day as a percentage of the full capacity.
(iii) Find the volume of water left in the tank at 8 p.m.
(iv) Express the water left at 8 p.m. as a percentage of the full capacity.
Answer:
(i) Water at 6 a.m. = 80% of 5000 = 0.8 × 5000 = 4000 L.
(ii) Water used = \( \frac{1200}{5000} \) × 100 = 24% of the full capacity.
(iii) Water left = 4000 − 1200 = 2800 L.
(iv) Water left = \( \frac{2800}{5000} \) × 100 = 56%. Check: 80% − 24% = 56%, since both percentages are taken on the same base of 5000 L.
In simple words: The tank had 4000 L, 24% of its capacity was used, and 2800 L (56%) was left.

Teacher's Note:
1. Percentages on the same base can be subtracted directly.
2. The base here is always the full capacity 5000 L.

Question 14. The strike rate of a batter. In a one-day match, Meera opened the batting for her state team and scored 48 runs off 40 balls before she was out. The strike rate of a batter is the number of runs scored for every 100 balls faced, so it is really a percentage in disguise: it is found as \( \frac{runs}{balls} \) × 100. The scorer noted that 36 of the 48 runs of Meera came from boundaries. The team finished with a total of 240 runs, and the whole innings lasted 120 balls.
(i) Find the strike rate of Meera.
(ii) What percentage of her own runs came from boundaries?
(iii) What percentage of the total runs of the team did Meera score?
(iv) What percentage of the balls of the team did Meera face? (Round to two decimal places.)
Answer:
(i) Strike rate = \( \frac{48}{40} \) × 100 = 120. (It is above 100 because she scored more runs than the balls she faced.)
(ii) Runs from boundaries = \( \frac{36}{48} \) × 100 = 75% of her runs.
(iii) Share of the team total = \( \frac{48}{240} \) × 100 = 20%.
(iv) Balls faced = \( \frac{40}{120} \) × 100 ≈ 33.33% of the balls of the innings. So Meera faced about a third of the balls but scored only a fifth of the runs.
In simple words: Meera’s strike rate was 120; 75% of her runs were boundaries, and she made 20% of the runs from 33.33% of the balls.

Teacher's Note:
1. Strike rate is runs per 100 balls.
2. Each part has a different whole; identify it first.

Question 15. A raise in salary. Ramesh works in a printing press and earned ₹24,000 a month last year. At the yearly review his employer gave him a raise of 12.5%, which took effect from April. Ramesh plans his spending carefully. Out of his new salary he puts aside 20% as savings every month, and he pays a house rent of ₹9450 a month. His employer has also promised a further raise of 10% on the new salary from next April, provided the press meets its order targets.
(i) Find the new monthly salary of Ramesh after the raise.
(ii) Find the amount he saves each month from the new salary.
(iii) Express his monthly rent as a percentage of his new salary.
(iv) Find the monthly salary he will get from next April if the promised raise is given.
Answer:
(i) New salary = 24,000 × 1.125 = ₹27,000. (The raise is 12.5% of 24,000 = ₹3000.)
(ii) Savings = 20% of 27,000 = 0.2 × 27,000 = ₹5400 per month.
(iii) Rent as a percentage of salary = \( \frac{9450}{27000} \) × 100 = 35%.
(iv) Next April: 27,000 × 1.10 = ₹29,700. This 10% raise is worth ₹2700 because it is taken on the new, larger salary.
In simple words: His salary becomes ₹27,000; he saves ₹5400, rent is 35% of salary, and next year he will get ₹29,700.

Teacher's Note:
1. 12.5% = 1/8, so the raise is 24,000 ÷ 8.
2. Each new raise is on the latest salary.

Question 16. Counting the votes. A municipal ward has 24,000 registered voters. On polling day the officials reported that 75% of the registered voters actually cast their votes; none of these votes was rejected. When the counting finished, the winning candidate had 9000 votes and the runner-up had 6300 votes, with the remaining votes shared among the other candidates. The newspapers next morning described the result in percentages, and readers noticed that the share of the winner looked quite different depending on whether it was compared with the votes cast or with all the registered voters.
(i) Find the number of votes cast in the ward.
(ii) Find the votes of the winner as a percentage of the votes cast.
(iii) Find the votes of the winner as a percentage of the registered voters.
(iv) Express the margin of victory as a percentage of the votes cast.
Answer:
(i) Votes cast = 75% of 24,000 = 0.75 × 24,000 = 18,000.
(ii) Winner’s share of the votes cast = \( \frac{9000}{18000} \) × 100 = 50%.
(iii) Winner’s share of the registered voters = \( \frac{9000}{24000} \) × 100 = 37.5%. The same 9000 votes give two different percentages because the base is different.
(iv) Margin = 9000 − 6300 = 2700 votes, which is \( \frac{2700}{18000} \) × 100 = 15% of the votes cast.
In simple words: 18,000 votes were cast; the winner got 50% of them (37.5% of all voters) and won by a margin of 15%.

Teacher's Note:
1. A percentage depends on the base chosen.
2. Always say what the percentage is ‘of’.

Question 17. Reading a GST bill. Goods and Services Tax (GST) is charged as a percentage of the price of an item and is collected by the shop on behalf of the government. At an electronics counter, a pair of headphones is priced at ₹2400 before tax, and GST is charged on it at 18%. The cashier adds the tax to the price to make the final bill. The same customer later looks at a printed bill for a music speaker which shows only one figure, the final amount of ₹5900 paid, with a note saying that the price includes GST at 18%.
(i) Find the GST charged on the headphones.
(ii) Find the final amount paid for the headphones.
(iii) Which of these gives the final amount directly: 2400 + 18, 2400 × 1.18, or 2400 × 0.18? Give a reason.
(iv) Find the price of the music speaker before GST was added.
Answer:
(i) GST = 18% of ₹2400 = \( \frac{18}{100} \) × 2400 = ₹432.
(ii) Final amount = 2400 + 432 = ₹2832.
(iii) 2400 × 1.18 gives it directly. Adding 18% means the final price is 100% + 18% = 118% of the original, and 118% = 1.18. (2400 + 18 adds a rate to rupees, and 2400 × 0.18 gives only the tax, not the total.)
(iv) If the price before tax is P, then 1.18P = 5900, so P = \( \frac{5900}{1.18} \) = ₹5000 (GST ₹900).
In simple words: The headphones carry ₹432 GST and cost ₹2832; the speaker was ₹5000 before GST.

Teacher's Note:
1. Price with GST = price × 1.18.
2. To remove GST, divide by 1.18, not subtract 18%.

Question 18. Know your contents. Madhav has learnt in school to read the label on a packet before buying processed food. He picks up a 250 g pack of biscuits and reads the contents printed on the back: wheat flour 140 g, sugar 60 g, fat 35 g and other ingredients 15 g. The label gives the weights in grams only. Madhav wants to compare this pack with another brand that sells a 400 g pack and prints its contents as percentages, so he decides to rewrite the figures on his pack as percentages of the total weight.
(i) Find the percentage of sugar in the pack.
(ii) Find the percentage of wheat flour in the pack.
(iii) Find the percentage of fat and the percentage of the other ingredients.
(iv) Check that the four percentages add up to 100%, and explain why converting to percentages helps Madhav compare the two packs.
Answer:
(i) Sugar = \( \frac{60}{250} \) × 100 = 24%.
(ii) Wheat flour = \( \frac{140}{250} \) × 100 = 56%.
(iii) Fat = \( \frac{35}{250} \) × 100 = 14%; other ingredients = \( \frac{15}{250} \) × 100 = 6%.
(iv) 24 + 56 + 14 + 6 = 100, as the four ingredients make up the whole 250 g pack. Percentages help because the two packs weigh different amounts (250 g and 400 g); turning both into “out of 100” puts them on the same base, so the shares can be compared at a glance.
In simple words: The pack is 24% sugar, 56% flour, 14% fat and 6% other, adding to 100%, which makes comparison with any pack size easy.

Teacher's Note:
1. Divide each part by the total weight 250 g.
2. Percentages compare packs of different sizes fairly.

Question 19. Monthly sales of a shop. A sports shop keeps a record of its monthly sales. In April the sales were ₹4,00,000. The owner noted the change in each of the next two months as a percentage of the sales of the previous month. In May, because of the school season, the sales rose by 15% compared with April. In June, the rains kept customers away and the sales fell by 20% compared with May. At the end of June the assistant of the owner claimed, “We are only 5% below April, because 15% up and 20% down makes 5% down.” The owner disagreed and worked out the figures month by month.
(i) Find the sales in May.
(ii) Find the sales in June.
(iii) Find the overall percentage change in sales from April to June.
(iv) Explain why the reasoning of the assistant is wrong.
Answer:
(i) May sales = 4,00,000 × 1.15 = ₹4,60,000 (a rise of ₹60,000).
(ii) June sales = 4,60,000 × 0.80 = ₹3,68,000 (a fall of ₹92,000).
(iii) Overall change = 3,68,000 − 4,00,000 = −32,000, so the percentage change = \( \frac{32000}{400000} \) × 100 = 8% decrease from April to June.
(iv) The assistant added and subtracted the percentages as if both were on the same base. They are not: the 15% rise was on April’s ₹4,00,000 and gave ₹60,000, while the 20% fall was on May’s larger ₹4,60,000 and took away ₹92,000. The correct way is to multiply the factors: 1.15 × 0.80 = 0.92, that is 92% of April, an 8% fall.
In simple words: Sales went ₹4,00,000 → ₹4,60,000 → ₹3,68,000, an 8% fall, not 5%, because the 20% was on a bigger amount.

Teacher's Note:
1. Successive percentage changes are multiplied.
2. Each change has its own base.
3. Check: 1.15 × 0.8 = 0.92.

Question 20. Who is left in the hall? A three-day science workshop is being held in a hall. On the first day there are 200 participants in the hall, and exactly 95% of them are school students; the rest are teachers who have come as mentors. On the second day the organisers find that the hall is too crowded, so some of the school students are moved to another hall for a separate session. Not a single teacher leaves the hall. After the students have been moved, the organisers announce that school students now form 90% of the people remaining in the hall.
(i) Find the number of school students and the number of teachers on the first day.
(ii) When some students leave the hall, which of the two numbers stays unchanged, and why?
(iii) Find the number of people remaining in the hall when the students form 90% of them.
(iv) Hence find how many school students were moved out, and verify your answer.
Answer:
(i) School students = 95% of 200 = 0.95 × 200 = 190; teachers = 200 − 190 = 10.
(ii) The number of teachers stays unchanged at 10, because only students are moved out. Their count is fixed even though their percentage share changes.
(iii) When students form 90% of the people left, teachers form 100% − 90% = 10% of them. So 10% of the new total is 10 teachers, giving a new total of 10 × \( \frac{100}{10} \) = 100 people.
(iv) Students remaining = 90% of 100 = 90, so students moved out = 190 − 90 = 100 students. Verification: of the 100 people left, 90 are students and 10 are teachers, and \( \frac{90}{100} \) × 100 = 90%. Moving out more than half of the students lowered their share only from 95% to 90%, because the fixed group of teachers becomes a bigger share of a much smaller total.
In simple words: There were 190 students and 10 teachers; for students to be 90%, only 100 people can remain, so 100 students were moved out.

Teacher's Note:
1. Work with the group whose number does not change.
2. If 10 teachers are 10%, the whole is 100.
3. Small percentage changes can hide large changes in numbers.