CBSE Class 8 Mathematics Proportional Reasoning 1 Long Answer and Case Study Questions with Answers

Long Answer Questions

Teacher's Note:
1. Reduce every ratio and group the equal ones.
2. Show the factors to explain why one frame is the odd one out.

Question 2. Equivalent ratios are the heart of proportional reasoning.
(i) Write any three ratios that are proportional to 6 : 11, and show the factor used each time.
(ii) Fill in the missing terms of these ratios, each of which is proportional to 20 : 36: 5 : ____, ____ : 72, 15 : ____
(iii) Explain why multiplying (or dividing) both terms of a ratio by the same non-zero number keeps the new ratio proportional to the original one.
Answer:
(i) Factor 2: 12 : 22. Factor 3: 18 : 33. Factor 5: 30 : 55. Each reduces back to 6 : 11.
(ii) 20 : 36 = 5 : 9 (HCF 4).
5 : 9, since 20 ÷ 4 = 5, so 36 ÷ 4 = 9.
40 : 72, since 72 = 36 × 2, so the first term is 20 × 2 = 40.
15 : 27, since 15 = 20 × \( \frac{3}{4} \), so the second term is 36 × \( \frac{3}{4} \) = 27.
(iii) If both terms of a : b are multiplied by k, the new ratio is ka : kb. Dividing both its terms by k gives back a : b, so the two ratios have the same simplest form.
By cross multiplication, a × kb = b × ka, which confirms a : b :: ka : kb.
In simple words: Multiplying both terms by the same number gives a ratio with the same simplest form.

Teacher's Note:
1. Find the factor from the term you know, then apply it to the other.
2. The proof in (iii) uses both tests of proportion.

Question 3. A car uses 6 L of petrol to travel 93 km on a highway, and this rate stays the same for the whole trip.
(i) Find the distance the car travels on 1 L of petrol.
(ii) Find the petrol needed for a journey of 279 km.
(iii) Find the distance the car can cover on 14 L of petrol, and write the ratio of distance to petrol in its simplest form.
Answer:
(i) Distance and petrol are in direct proportion. Distance per litre = \( \frac{93}{6} \) = 15.5 km.
(ii) Petrol needed = 279 ÷ 15.5 = 18 L. Check: 6 × 279 = 1674 = 93 × 18.
(iii) Distance on 14 L = 15.5 × 14 = 217 km.
Distance : petrol = 93 : 6. HCF is 3, so the simplest form is 31 : 2, that is, 31 km for every 2 L (15.5 km per litre).
In simple words: The car does 15.5 km per litre: 279 km needs 18 L and 14 L gives 217 km.

Teacher's Note:
1. Find the rate for one litre first.
2. A simplest form with whole numbers can hide a decimal rate like 15.5.

Question 4. A box of 120 laddus is to be shared between two hostels, Anand and Bela, in the ratio 7 : 5.
(i) Find the total number of equal groups the 120 laddus must be split into, and the size of one group.
(ii) Find the number of laddus each hostel receives.
(iii) Check your answer by writing the ratio of the two shares in its simplest form, and then write the general rule for dividing a quantity x in the ratio m : n.
Answer:
(i) Groups = 7 + 5 = 12; size of one group = \( \frac{120}{12} \) = 10 laddus.
(ii) Anand = 7 × 10 = 70 laddus; Bela = 5 × 10 = 50 laddus.
(iii) 70 + 50 = 120, so all laddus are used. 70 : 50 = 7 : 5 (HCF 10), the required ratio.
General rule: make m + n equal groups, each of size \( \frac{x}{m+n} \). The two parts are \( m \times \frac{x}{m+n} \) and \( n \times \frac{x}{m+n} \), and they are always in the ratio m : n.
In simple words: 12 groups of 10 laddus give Anand 70 and Bela 50.

Teacher's Note:
1. The general rule is the same steps written with letters.
2. Always check the total and the ratio.

Question 5. A builder keeps the chapter's conversion table on his wall: 1 m = 3.281 ft, 1 m² = 10.764 ft², 1 hectare = 10,000 m² and 1 hectare = 2.471 acres.
(i) Convert the 20 m frontage of a plot into feet.
(ii) Convert a floor area of 150 m² into square feet.
(iii) A farm measures 2 hectares. Express this area in square metres and in acres.
Answer:
(i) 20 m = 20 × 3.281 = 65.62 ft.
(ii) 150 m² = 150 × 10.764 = 1614.6 ft². The area factor is not the same as the length factor, because area is a product of two lengths.
(iii) 2 hectares = 2 × 10,000 = 20,000 m², and 2 hectares = 2 × 2.471 = 4.942 acres.
In every part only the unit changes; the quantity and its converted value are proportional.
In simple words: Multiply by the right factor: 65.62 ft, 1614.6 ft², 20,000 m² and 4.942 acres.

Teacher's Note:
1. Use the length factor for lengths and the area factor for areas.
2. 10.764 is roughly 3.281 × 3.281.

Question 6. A tap pours water at a steady rate and fills a 750 mL jug in 18 seconds.
(i) Find the time the tap takes to fill 1 L.
(ii) Find the time it takes to fill a bucket of capacity 12 L, giving the answer in minutes and seconds.
(iii) Find the volume of water the tap delivers in 5 minutes, in litres.
Answer:
(i) 1 L = 1000 mL. 750 : 18 :: 1000 : t gives 750t = 18 × 1000 = 18000, so t = 24 s.
(ii) 12 × 24 = 288 s = 4 minutes 48 seconds.
(iii) 5 minutes = 300 seconds. The tap gives 1 L every 24 s, so in 300 s it gives \( \frac{300}{24} \) = 12.5 L.
Check: the rate is 750 ÷ 18 = \( 41\frac{2}{3} \) mL per second, and \( 41\frac{2}{3} \) × 300 = 12500 mL = 12.5 L.
In simple words: 1 L takes 24 s, 12 L takes 4 min 48 s, and 5 minutes gives 12.5 L.

Teacher's Note:
1. Volume and time are in direct proportion for a steady tap.
2. Convert seconds to minutes only at the end.

Question 7. A painter makes green paint by mixing blue and yellow paints in the ratio 3 : 5.
(i) Find the volume of blue paint and of yellow paint he must take to make 64 mL of green paint.
(ii) He then stirs in 16 mL of extra blue paint by mistake. Find the new ratio of blue to yellow in its simplest form.
(iii) State whether the new mixture is a lighter or a darker green than before, and give a reason using the two ratios.
Answer:
(i) 3 : 5 gives 8 parts; one part = 64 ÷ 8 = 8 mL. Blue = 3 × 8 = 24 mL; yellow = 5 × 8 = 40 mL. Check: 24 + 40 = 64.
(ii) Blue becomes 24 + 16 = 40 mL; yellow stays 40 mL. New ratio = 40 : 40 = 1 : 1.
(iii) The new green is darker. Blue was \( \frac{3}{8} \) of the paint; now it is \( \frac{1}{2} \). Since \( \frac{1}{2} \) > \( \frac{3}{8} \), there is more blue for the same yellow, so the shade deepens (3 : 5 has become 5 : 5).
In simple words: Adding blue changes 3 : 5 to 1 : 1, so the green becomes darker.

Teacher's Note:
1. Compare the fraction of blue in the whole mixture.
2. Yellow is unchanged, so only the blue decides the shade.

Question 8. Two villages are being compared for crowding. Village Amla has 4,800 people living in an area of 12 km². Village Beeja has 7,350 people living in an area of 21 km².
(i) Explain why comparing only the populations, or only the areas, is not enough.
(ii) Find the number of people per square kilometre in each village.
(iii) State which village is more crowded and write the ratio of the two crowding figures in its simplest form.
Answer:
(i) Beeja has more people, but it also has more land. Crowding depends on how many people share each unit of land, so population and area must be compared together, as a ratio.
(ii) Amla: 4800 ÷ 12 = 400 people per km². Beeja: 7350 ÷ 21 = 350 people per km².
(iii) Amla is more crowded (400 against 350 per km²).
Ratio = 400 : 350. HCF is 50, so the simplest form is 8 : 7: for every 8 people on a km² of Amla there are 7 on a km² of Beeja.
In simple words: Amla has 400 people per km² and Beeja 350, so Amla is more crowded; ratio 8 : 7.

Teacher's Note:
1. Population density = people ÷ area.
2. A bigger village is not always a more crowded one.

Question 9. Āryabhaṭa called problems of proportionality "Rule of Three" problems.
(i) Starting from a : b :: c : d and the fact that both terms change by the same factor f, show by algebra that ad = bc.
(ii) Use the cross-multiplication test to decide whether 36 : 48 and 51 : 68 are proportional.
(iii) Solve this Līlāvatī-style question: if \( 3\frac{1}{2} \) palas of saffron cost \( \frac{4}{5} \) niskas, what quantity of saffron can be bought for 8 niskas?
Answer:
(i) Both terms of a : b are multiplied by f to give c : d, so c = fa … (1) and d = fb … (2).
From (1), f = \( \frac{c}{a} \); from (2), f = \( \frac{d}{b} \). So \( \frac{c}{a} = \frac{d}{b} \).
Multiplying both sides by ab gives bc = ad, that is, ad = bc.
(ii) ad = 36 × 68 = 2448 and bc = 48 × 51 = 2448. Since ad = bc, the ratios are proportional (both reduce to 3 : 4).
(iii) \( 3\frac{1}{2} = \frac{7}{2} \) palas. Pramāṇa = \( \frac{4}{5} \) niskas, phala = \( \frac{7}{2} \) palas, ichchhā = 8 niskas. Let the answer be x palas.
\( \frac{4}{5} : \frac{7}{2} :: 8 : x \) gives \( \frac{4}{5}x = \frac{7}{2} \times 8 = 28 \), so \( x = 28 \times \frac{5}{4} = 35 \).
So 35 palas of saffron can be bought for 8 niskas. Check: \( \frac{7}{2} \div \frac{4}{5} = \frac{35}{8} \) palas per niska, and \( \frac{35}{8} \) × 8 = 35.
In simple words: Cross multiplication gives ad = bc, the two ratios are proportional, and 8 niskas buy 35 palas.

Teacher's Note:
1. Phala × ichchhā ÷ pramāṇa gives the answer directly.
2. Fractions follow the same Rule of Three as whole numbers.

Question 10. A brass rod of mass 4.2 kg is made of copper and zinc mixed in the ratio 5 : 2. Copper costs ₹820 per kg and zinc costs ₹290 per kg.
(i) Find the mass of copper and the mass of zinc in the rod.
(ii) Find the cost of the copper, the cost of the zinc, and the total cost of the metal in the rod.
(iii) Write the ratio of the cost of the copper to the cost of the zinc in its simplest form, and explain why this ratio is not the same as 5 : 2.
Answer:
(i) 5 : 2 gives 7 parts; one part = 4.2 ÷ 7 = 0.6 kg. Copper = 5 × 0.6 = 3 kg; zinc = 2 × 0.6 = 1.2 kg. Check: 3 + 1.2 = 4.2 and 3 : 1.2 = 30 : 12 = 5 : 2.
(ii) Copper: 3 × 820 = ₹2,460. Zinc: 1.2 × 290 = ₹348. Total = 2460 + 348 = ₹2,808.
(iii) Cost ratio = 2460 : 348. HCF is 12, so the simplest form is 205 : 29.
The masses are in 5 : 2, but each mass is multiplied by a different rate (820 and 290). Multiplying the terms by different factors changes the ratio; only equal rates would keep it 5 : 2.
In simple words: Copper 3 kg (₹2,460) and zinc 1.2 kg (₹348); the cost ratio 205 : 29 differs from 5 : 2 because the rates differ.

Teacher's Note:
1. A ratio stays the same only when both terms use the same factor.
2. Divide by 12 carefully: 2460 ÷ 12 = 205 and 348 ÷ 12 = 29.

Case Study Based Questions

Question 11. Kheer for the festival. Devi cooks kheer for her family every Pongal. Her usual recipe serves 8 people and needs 1 L of milk, 160 g of rice and 200 g of sugar. This year her cousins are visiting, so 20 people will eat the kheer. Devi wants the kheer to taste exactly the same as always, so she decides that every ingredient must be scaled by the same factor. Her younger brother says she should simply add 12 to each quantity because 12 more people are coming. Devi disagrees and uses proportional reasoning instead.
(i) Write the ratio of rice to sugar in the usual recipe in its simplest form.
(ii) Find the factor by which the number of people has changed.
(iii) Find the quantity of milk Devi needs for 20 people.
(iv) Find the quantity of rice she needs for 20 people, and say why her brother's "add 12" idea is wrong.
Answer:
(i) Rice : sugar = 160 : 200. HCF is 40, so the simplest form is 4 : 5.
(ii) Factor = \( \frac{20}{8} \) = \( \frac{5}{2} \) = 2.5.
(iii) Milk = 1 × 2.5 = 2.5 L (2500 mL).
(iv) Rice = 160 × 2.5 = 400 g. The brother is wrong because adding the same number to every quantity changes the ratios; only multiplying every quantity by the same factor keeps the taste the same.
In simple words: Scale every ingredient by 2.5: 2.5 L of milk and 400 g of rice.

Teacher's Note:
1. Recipes scale by multiplying, not by adding.
2. Adding 12 g of rice and 12 L of milk would make no sense.

Question 12. Printing a photograph. A photo studio has an original picture that measures 15 cm in width and 10 cm in height. A customer orders two enlargements. The first print comes out 24 cm wide and 16 cm high. The second print comes out 20 cm wide and 12 cm high. When the customer places the three pictures side by side, one of the enlargements looks stretched and does not match the original. The studio owner explains that a picture keeps its shape only when the width and the height are multiplied by the same factor.
(i) Write the width-to-height ratio of the original picture in its simplest form.
(ii) Write the width-to-height ratio of the first print in its simplest form.
(iii) Identify the print that looks stretched, giving its simplest-form ratio as the reason.
(iv) The studio now prints a third copy of width 36 cm that is similar to the original. Find its height.
Answer:
(i) 15 : 10. HCF is 5, so the simplest form is 3 : 2.
(ii) 24 : 16. HCF is 8, so the simplest form is 3 : 2, the same as the original.
(iii) The second print looks stretched. 20 : 12 = 5 : 3 (HCF 4), which is not 3 : 2.
(iv) 3 : 2 :: 36 : h, so 3h = 2 × 36 = 72 and h = 24 cm.
In simple words: Prints with ratio 3 : 2 match the original; a 36 cm wide copy must be 24 cm high.

Teacher's Note:
1. Compare simplest forms to spot the odd print.
2. Use cross multiplication to find a missing side.

Question 13. Petrol and distance. Harish fills his car's tank and notes the odometer reading. On a long highway drive at a steady speed, the car covers 240 km and uses exactly 16 L of petrol. Harish records this in a small notebook so that he can plan his next trips. He assumes that on the same highway, at the same speed, the distance covered and the petrol used stay proportional. His next journey is 360 km long, and later in the month he wants to know how far he can go on a full tank of 25 L.
(i) Find the distance the car covers on 1 L of petrol.
(ii) Find the petrol needed for the 360 km journey.
(iii) Find the distance the car can cover on 25 L of petrol.
(iv) Write the ratio of distance covered to petrol used in its simplest form.
Answer:
(i) 240 ÷ 16 = 15 km per litre.
(ii) 360 ÷ 15 = 24 L. Check: 16 × 360 = 5760 = 240 × 24.
(iii) 15 × 25 = 375 km.
(iv) 240 : 16. HCF is 16, so the simplest form is 15 : 1, that is, 15 km for every litre.
In simple words: The car does 15 km per litre, so 360 km needs 24 L and 25 L gives 375 km.

Teacher's Note:
1. The unit rate (km per litre) answers every part.
2. A simplest form of 15 : 1 is the unit rate itself.

Question 14. Partners in a juice stall. Nandini and Farid open a fresh juice stall outside their college. Nandini puts in ₹18,000 to buy the juicer and the cart, while Farid puts in ₹12,000 for the first stock of fruit and the licence. They agree at the start that any profit will be shared in the same ratio as the money each of them invested, and not equally. At the end of the first season the stall has made a profit of ₹7,500. The two friends now sit down to work out each person's share using the rule for dividing a quantity in a given ratio.
(i) Write the ratio of Nandini's investment to Farid's investment in its simplest form.
(ii) Find the number of equal parts the profit must be divided into.
(iii) Find the value of one part.
(iv) Find the share of the profit that each friend receives.
Answer:
(i) 18000 : 12000. Dividing by the HCF 6,000 gives 3 : 2.
(ii) Number of parts = 3 + 2 = 5.
(iii) One part = 7500 ÷ 5 = ₹1,500.
(iv) Nandini = 3 × 1500 = ₹4,500; Farid = 2 × 1500 = ₹3,000. Check: 4500 + 3000 = 7500 and 4500 : 3000 = 3 : 2.
In simple words: The profit is split 3 : 2, so Nandini gets ₹4,500 and Farid ₹3,000.

Teacher's Note:
1. Profit follows the investment ratio.
2. Check both the total and the ratio at the end.

Question 15. Mixing wall paint. A painter prepares a pale blue shade for a classroom by mixing white paint and blue paint in the ratio 7 : 3. For the classroom he needs 25 L of the mixed paint in all. He measures the two paints carefully into a drum and stirs well. Half-way through the job the teacher asks for a slightly deeper shade, so the painter stirs in another 5 L of blue paint and no extra white paint. He then writes the new ratio of white to blue on the lid of the drum so that the same shade can be made again next year.
(i) Find the total number of equal parts in the ratio 7 : 3 and the volume of one part.
(ii) Find the volume of white paint in the first mixture.
(iii) Find the volume of blue paint in the first mixture.
(iv) Find the ratio of white paint to blue paint after the extra 5 L of blue is added, in its simplest form.
Answer:
(i) Total parts = 7 + 3 = 10; one part = 25 ÷ 10 = 2.5 L.
(ii) White = 7 × 2.5 = 17.5 L.
(iii) Blue = 3 × 2.5 = 7.5 L. Check: 17.5 + 7.5 = 25 L.
(iv) Blue becomes 7.5 + 5 = 12.5 L; white stays 17.5 L. Ratio = 17.5 : 12.5 = 175 : 125 = 7 : 5 (dividing by 25).
In simple words: The first mix has 17.5 L white and 7.5 L blue; adding 5 L blue makes it 7 : 5.

Teacher's Note:
1. Multiply by 10 to clear decimals before simplifying.
2. More blue for the same white gives a deeper shade.

Question 16. The moving walkway. An airport has a long moving walkway that carries passengers at a steady speed. A notice near the entrance says the walkway moves 48 m in one minute. Passengers who stand still on it are carried forward at exactly this rate, so the distance travelled and the time spent on the walkway stay proportional. A tourist standing on the walkway wants to work out how long she will be on it, and an engineer checking the notice wants the same distance written in feet, using 1 m = 3.281 ft.
(i) Find the distance the walkway moves in 1 second.
(ii) Find the distance a passenger standing on it travels in 5 minutes.
(iii) Express 48 m in feet.
(iv) Find the time, in minutes and seconds, taken to travel the full 300 m length of the walkway.
Answer:
(i) 48 ÷ 60 = 0.8 m per second.
(ii) 48 × 5 = 240 m.
(iii) 48 × 3.281 = 157.488 ft.
(iv) 48 : 1 :: 300 : t gives 48t = 300, so t = 6.25 minutes = 6 minutes 15 seconds. Check: 48 × 6.25 = 300 m.
In simple words: The walkway moves 0.8 m each second, so 300 m takes 6 minutes 15 seconds.

Teacher's Note:
1. Change 0.25 minute to seconds: 0.25 × 60 = 15 s.
2. Distance and time are in direct proportion at a steady speed.

Question 17. Manure for a tomato plot. A farmer in Nashik is preparing a rectangular plot for tomatoes. The plot measures 180 ft by 484 ft. Good farming practice in his district is to spread 10 tonnes of well-rotted cow manure on every acre of land before planting. The farmer buys the manure from a nearby dairy co-operative at ₹2,500 per tonne, delivered to the field. Before he places the order he must change the area of his plot from square feet into acres, using 1 acre = 43,560 square feet.
(i) Find the area of the plot in square feet.
(ii) Express this area in acres.
(iii) Find the quantity of manure, in tonnes, the farmer should buy.
(iv) Find the total amount he must pay for the manure.
Answer:
(i) Area = 180 × 484 = 87,120 ft².
(ii) 87120 ÷ 43560 = 2 acres exactly.
(iii) Manure = 10 × 2 = 20 tonnes.
(iv) Cost = 20 × 2500 = ₹50,000.
In simple words: The plot is 2 acres, so it needs 20 tonnes of manure costing ₹50,000.

Teacher's Note:
1. Convert the area to acres before using the per-acre rate.
2. Each step uses the answer of the step before.

Question 18. Which tea is dearer? A small grower in the Nilgiris sells tea in 250 g packets at ₹180 a packet. A large estate in Assam sells the same kind of tea in 1 kg packets at ₹700 a packet. A shopkeeper who buys from both wants to know which tea is really more expensive. He knows that weight-to-price ratios can only be compared when the weights are written in the same unit, so he changes the Assam packet's mass into grams before he compares anything.
(i) Write the weight-to-price ratio for the Nilgiris tea, in grams and rupees, in its simplest form.
(ii) Write the weight-to-price ratio for the Assam tea, in grams and rupees, in its simplest form.
(iii) Find the cost of 1 kg of the Nilgiris tea.
(iv) State which tea is more expensive and by how much per kilogram.
Answer:
(i) 250 : 180. HCF is 10, so the simplest form is 25 : 18.
(ii) 1 kg = 1000 g, so 1000 : 700. HCF is 100, giving 10 : 7.
(iii) 250 g is \( \frac{1}{4} \) of a kilogram, so 1 kg costs 4 × 180 = ₹720.
(iv) 25 : 18 and 10 : 7 are different, so the prices are not proportional. Per kilogram, Nilgiris costs ₹720 and Assam ₹700, so the Nilgiris tea is dearer by ₹20 per kg.
In simple words: Nilgiris tea costs ₹720 a kg against ₹700 for Assam tea, so it is ₹20 dearer.

Teacher's Note:
1. Bring both to the same weight before comparing prices.
2. Per-kilogram price is the fairest comparison.

Question 19. A batch of sugarcane juice. A juice stall mixes sugarcane concentrate and water in the ratio 2 : 9 to make its standard drink. One busy morning the helper is asked to prepare a 44 L batch of the standard drink. Later in the day the helper is careless: he pours 10 L of concentrate into a drum that already holds 36 L of water. The owner tastes it at once and says the drink is too strong. Rather than throw the drum away, she decides to add only water until the mixture is back to the standard ratio.
(i) Find the volume of concentrate and the volume of water in the 44 L standard batch.
(ii) Write the ratio of concentrate to water in the careless mixture in its simplest form, and show by comparison with 2 : 9 that the drink is too strong.
(iii) Find the volume of water the drum should contain for the ratio to be 2 : 9, and hence the extra water that must be added.
(iv) Find the total volume of drink in the drum after the correction.
Answer:
(i) 2 : 9 gives 11 parts; one part = 44 ÷ 11 = 4 L. Concentrate = 2 × 4 = 8 L; water = 9 × 4 = 36 L. Check: 8 + 36 = 44 L.
(ii) Careless mixture = 10 : 36 = 5 : 18 (HCF 2). The standard ratio is 2 : 9 = 4 : 18. Since 5 > 4, there is more concentrate for the same water, so the drink is too strong.
(iii) Keep 10 L of concentrate: 2 : 9 :: 10 : w gives 2w = 90, so w = 45 L. Extra water = 45 − 36 = 9 L.
(iv) Total = 10 + 45 = 55 L. Check: 10 : 45 = 2 : 9.
In simple words: Adding 9 L of water brings the drink back to 2 : 9, giving 55 L in all.

Teacher's Note:
1. Make the second terms equal to compare strengths.
2. Only water is added, so the concentrate stays fixed at 10 L.

Question 20. Reading the weather in two scales. A school weather station records the temperature every day in degrees Celsius, but the old chart on the wall is printed in degrees Fahrenheit. The students use the chapter's conversion rules, \( \text{Fahrenheit} = \frac{9}{5} \times \text{Celsius} + 32 \) and \( \text{Celsius} = \frac{5}{9} \times (\text{Fahrenheit} - 32) \), to move between the two scales. On the hottest day of May the station recorded 40°C. A news report from a hill station in another country gave a reading of 5°F. On one spring day the school's own temperature rose from 22°C in the morning to 31°C in the afternoon.
(i) Express 40°C in degrees Fahrenheit.
(ii) Express 5°F in degrees Celsius.
(iii) Find the rise in temperature of that spring day in degrees Fahrenheit.
(iv) A student claims that the Fahrenheit reading is always proportional to the Celsius reading. Decide whether the claim is correct, giving a reason supported by two of your answers above.
Answer:
(i) F = \( \frac{9}{5} \) × 40 + 32 = 72 + 32 = 104, so 40°C = 104°F.
(ii) C = \( \frac{5}{9} \) × (5 − 32) = \( \frac{5}{9} \) × (−27) = −15, so 5°F = −15°C.
(iii) Morning: \( \frac{9}{5} \) × 22 + 32 = 39.6 + 32 = 71.6°F. Afternoon: \( \frac{9}{5} \) × 31 + 32 = 55.8 + 32 = 87.8°F.
Rise = 87.8 − 71.6 = 16.2°F. The same comes from \( \frac{9}{5} \) × 9 = 16.2, because the 32 cancels in a difference.
(iv) The claim is wrong. From (i), 104 : 40 = 13 : 5. From (iii), 71.6 : 22 = 716 : 220 = 179 : 55, which is not 13 : 5.
The "+ 32" in the rule adds a fixed number, and adding changes a ratio, so Fahrenheit is not proportional to Celsius even though the two are linked by a fixed rule.
In simple words: 40°C = 104°F, 5°F = −15°C, the rise is 16.2°F, and °F is not proportional to °C because of the + 32.

Teacher's Note:
1. A negative Celsius reading is fine for very cold places.
2. Proportional means a fixed multiplying factor with nothing added.