Mathematics Class 8 — Ganita…
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Similarity and Trairasika Exercises and Summary

🎓 Class 8 Mathematics CBSE Theory Ch 7 — Playing with Polynomials ⏱ ~35 min
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This MCQ module is based on: Similarity and Trairasika Exercises and Summary

This mathematics assessment will be based on: Similarity and Trairasika Exercises and Summary
Targeting Class 8 level in Algebra, with Basic difficulty.

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7.6 Unit Conversions

We have noticed earlier that solving problems with proportionality often requires us to convert units from one system to another. Here are a few important unit conversions for your reference.

Key Conversions
Length: 1 metre ≈ 3.281 feet
Area: 1 square metre ≈ 10.764 sq ft · 1 acre = 43,560 sq ft · 1 hectare = 10,000 sq m · 1 hectare ≈ 2.471 acres
Volume: 1 millilitre (mL) = 1 cubic centimetre (cc) · 1 litre = 1,000 mL = 1,000 cc
Temperature: \(\text{Fahrenheit} = \tfrac{9}{5}\text{Celsius} + 32\)  and  \(\text{Celsius} = \tfrac{5}{9}(\text{Fahrenheit} - 32)\). For example, 25°C = 77°F.
0°C 10°C 25°C 37°C 50°C 32°F 50°F 77°F 98.6°F 122°F Celsius ↔ Fahrenheit (same temperature, different units)
Fig 7.6 — Temperature is a classic proportional conversion
Activity 2: Counter Conversion
L3 Apply
Materials: A small recipe you like
Predict: If 1 cup = 240 mL, and your recipe asks for \(\tfrac34\) cup of milk, will the milk in mL be more or less than 240?
  1. Pick a recipe and list every ingredient in cups / tablespoons.
  2. Convert each quantity to millilitres using 1 cup = 240 mL and 1 tablespoon = 15 mL.
  3. Verify: the ratio of any two ingredients must stay the same in both units (cups vs mL).

Why ratios stay the same: Since we multiply each ingredient by the same factor (240 for cups → mL), the ratios between ingredients are unchanged. This is proportional reasoning applied to unit conversion.

Figure it Out — End-of-Chapter Exercises

Q1. Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21   (ii) 11 : 7 :: 55 : 35   (iii) 33 : 12 :: 7 : 4   (iv) 21 : 6 :: 35 : 10   (v) 12 : 18 :: 28 : 12   (vi) 24 : 8 :: 9 : 3
(i) 4:7 and 12:21 → both simplify to 4:7 ✓ True.
(ii) 11:7 (already simplest) and 55:35 → 11:7 ✓ True.
(iii) 33:12 → 11:4 and 7:4 — different. False.
(iv) 21:6 → 7:2 and 35:10 → 7:2 ✓ True.
(v) 12:18 → 2:3 and 28:12 → 7:3 — different. False.
(vi) 24:8 → 3:1 and 9:3 → 3:1 ✓ True.
Q2. Give 3 ratios that are proportional to 4 : 9.
Multiply both terms by 2, 3, 5: 8 : 18, 12 : 27, 20 : 45. (Many more are valid — any \(4k:9k\) for \(k\ne 0\).)
Q3. Fill in the missing numbers for these ratios that are proportional to 18 : 24.
3 : ___   12 : ___   20 : ___   27 : ___
Simplest form of 18 : 24 = 3 : 4.
3 : 4 (×1)   12 : 16 (×4)   20 : \(\tfrac{20}{3}\times 4=\) \(\tfrac{80}{3}\)   27 : 36 (×9).
Q4. Look at the rectangles A, B, C, D. Which rectangles are similar to each other? Verify by measuring width and height and comparing ratios.
A 40×90 B 50×50 C 160×60 D 120×45
A: 40:90 = 4:9. B: 50:50 = 1:1 (square). C: 160:60 = 8:3. D: 120:45 = 8:3. Rectangles C and D are similar (ratio 8:3). A and B are not similar to any other, and not to each other.
Q5. Look at the rectangle shown. Can you draw a smaller rectangle and a bigger rectangle with the same width-to-height ratio in your notebook? Are all of them the same? If they are different from each other, can you think why? Are they wrong?
Measure the given rectangle and reduce the width:height to its simplest form \(a:b\). Then draw \(\tfrac12 a \times \tfrac12 b\) (smaller) and \(2a \times 2b\) (bigger). All three rectangles look similar because they share the same simplest-form ratio. Any classmate drawings with the same ratio are not wrong — they are simply scaled versions.
Q6. The figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues the same pattern throughout. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.
(a) Count grey vs coloured bricks in a single repeating unit; e.g., if a unit has 5 grey and 3 coloured, the ratio is 5 : 3 in simplest form. (b) Repeat the count for pattern (b); e.g., 2 coloured per 6 grey gives 3 : 1. Exact counts depend on the printed figure — use the repeating unit as your sample.
Q7. Measure the lengths of head, torso, arms and legs of a friend. Write the ratios head : torso, torso : arms, torso : legs. Now draw a figure with the same ratios. Does the drawing look more realistic?
Typical adult proportions: head : torso ≈ 1 : 3, torso : arms ≈ 1 : 1, torso : legs ≈ 2 : 3. Drawing with these ratios produces a realistic human figure. Drawings that ignore these ratios (very short legs, very big head) look cartoonish or distorted.
Q8. The Earth travels approximately 940 million km around the Sun in a year. How many km does it travel in a week?
1 year ≈ 52 weeks. So distance per week = \(\tfrac{940}{52} \approx \mathbf{18.08}\) million km.
Q9. A mason is building a house in the shape shown. To build the outer walls and the inner wall (10 ft height) he needs bricks. The mason separates two rooms by a wall 10 ft high. To build the wall at speed 50 km/h, if he drives at 75 km/h, how long will it take him to reach Kanpur? (Puneeth's father went from Lucknow to Kanpur in 2 hours riding at 50 km/h.)
Distance = speed × time = 50 × 2 = 100 km. At 75 km/h, time = \(\tfrac{100}{75}=\tfrac{4}{3}\) h = 1 h 20 min. This is an inverse proportion: higher speed → lower time.
Q10. A company sells shampoo in four container sizes. Sachet = 6 mL (₹2), Small = 180 mL (₹154), Medium = 400 mL (₹276), Large = 1000 mL (₹540). Is the volume proportional to the price?
Compute price per mL: Sachet ₹0.333, Small ₹0.856, Medium ₹0.690, Large ₹0.540. The ratios differ — volume is not proportional to price. Bigger bottles give better value per mL, but the relationship is not a strict proportion.
Q11. Divide ₹4,500 into two parts in the ratio 2 : 3.
Total groups = 5. Each group = \(4500 \div 5 = 900\). Parts: \(2 \times 900 = ₹1800\) and \(3 \times 900 = ₹2700\).
Q12. In a science lab, acid and water are mixed in the ratio 1 : 5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain?
Total groups = 6. Acid = \(\tfrac{1}{6}\times 240 = 40\) mL. Water = \(\tfrac{5}{6}\times 240 = 200\) mL.
Q13. Blue and yellow paints are mixed in the ratio 1 : 5 to produce green paint. To produce 40 mL of green paint, how much of these two colours are needed? To make the paint a lighter shade of green, I added 20 mL of yellow to the mixture. What is the new ratio of blue and yellow in the paint?
Original 40 mL: blue = \(\tfrac16 \times 40 \approx 6.67\) mL; yellow = \(\tfrac56 \times 40 \approx 33.33\) mL. After adding 20 mL yellow: blue ≈ 6.67 mL, yellow ≈ 53.33 mL. New ratio ≈ \(6.67 : 53.33 = 1 : 8\). (If original amounts are 8 : 32 using integer blocks, new ratio = 8 : 52 = 2 : 13.)
Q14. To make soft idlis, you need to mix rice and urad dal in the ratio 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need?
Total groups = 3. Each group = 2 cups. Rice = \(2 \times 2 = 4\) cups. Urad dal = \(1 \times 2 = 2\) cups.
Q15. I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio 3 : 5. I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
If the original bucket had 8 parts (3 red + 5 yellow) and the added bucket also had 8 parts (all yellow), the total is 3 red : 13 yellow. New ratio = 3 : 13.
Q16. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in simplest form.
600 : 900 → divide by HCF 300 → 2 : 3.
Q17. Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers for the trip, and all the buses were full. This year we have 200 students. How many buses will we need? Will all the buses be full?
Each bus holds \(162 \div 3 = 54\) people. For 200: \(200 \div 54 \approx 3.7\). So we need 4 buses. Three will be fully packed (54+54+54 = 162) and the 4th will have 38 people (not full).
Q18. Delhi is 1,484 sq km and Mumbai is 550 sq km. The population of Delhi is approximately 30 million and Mumbai is 20 million. Which city is more crowded?
Population density (per sq km): Delhi \(\tfrac{30{,}000{,}000}{1484} \approx 20{,}215\); Mumbai \(\tfrac{20{,}000{,}000}{550} \approx 36{,}364\). Mumbai is more crowded — about 1.8× as densely populated as Delhi.
Q19. A crane 4 ft tall casts a shadow 4 ft 6 in long. If its body is in the ratio 4 : 6, how tall is the crane's neck and the rest of the body when the body also casts a shadow proportionally?
If the full crane is 4 ft and split in 4 : 6 ratio (neck : body), then neck = \(\tfrac{4}{10}\times 4 = 1.6\) ft, body = \(\tfrac{6}{10}\times 4 = 2.4\) ft. (Original NCERT version includes a shadow comparison; use similar proportional reasoning.)
Q20. Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named pala and niska. If 2 palas of saffron cost \(\tfrac{3}{7}\) niska, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?
Set up: \(\tfrac{3}{7} : 2 :: 9 : x\). Cross multiply: \(\tfrac{3}{7} \cdot x = 2 \times 9 = 18\). So \(x = 18 \times \tfrac{7}{3} = 42\) palas. 42 palas of saffron.
Q21. Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain's age when his brother's age is 11?
Brother goes 5 → 11, that is 6 years later. Harmain also ages 6 years: \(1 + 6 = \mathbf{7}\) years. (Note: ratio of ages is NOT preserved because adding a constant to both changes the ratio.)
Q22. The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
\(37:2 :: x:1\). So \(x = \tfrac{37}{2}=18.5\) kg. 1 litre of gold ≈ 18.5 kg.
Q23. It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy? (1 acre = 43,560 sq ft)
Plot area = 200 × 500 = 100,000 sq ft. In acres: \(\tfrac{100000}{43560}\approx 2.296\) acres. Manure = \(10 \times 2.296 \approx \mathbf{22.96}\) tonnes.
Q24. A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 mL. How much time does the same tap take to fill a bucket of water if the bucket has a 10 litre capacity?
10 L = 10,000 mL. Time ratio: \(500:15 :: 10{,}000:x\). Cross multiply: \(500x = 15 \times 10{,}000 = 150{,}000\); \(x = 300\) s = 5 minutes.
Q25. One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land? (1 acre = 43,560 sq ft)
\(43560 : 1500000 :: 2400 : x\). \(x = \tfrac{1500000 \times 2400}{43560} \approx \mathbf{₹82{,}645}\).
Q26. A tractor can plough the same area of a field 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an acre of land. How much time would it take if the farmer used a pair of oxen to plough the field? How much time would it take him if he decides to use a tractor instead?
Oxen: 20 acres × 6 h = 120 hours. Tractor: 4× faster → \(\tfrac{120}{4}=\mathbf{30}\) hours.
Q27. The ₹10 coin is an alloy of copper and nickel mixed in a 3 : 1 ratio by mass. The mass of the coin is 7.74 grams. If the cost of copper is ₹906 per kg and of nickel is ₹1,341 per kg, what is the cost of these metals in a ₹10 coin?
Copper mass = \(\tfrac{3}{4}\times 7.74 = 5.805\) g = 0.005805 kg. Copper cost = \(906 \times 0.005805 \approx \mathbf{₹5.26}\). Nickel mass = \(\tfrac{1}{4}\times 7.74 = 1.935\) g. Nickel cost = \(1341 \times 0.001935 \approx \mathbf{₹2.60}\). Total ≈ ₹7.86 — less than the ₹10 face value.

Competency-Based Questions

Scenario: A village water-tanker fills a storage tank of 12,000 litres in 8 hours using a pump flowing at a steady rate. A new, more powerful pump can fill the same tank in 5 hours. Water is distributed to 240 households in proportion to family size: families of 3 members get 1 share, 5 members get 2 shares, 7 members get 3 shares. Total shares across the village are 1,200.
Q1. How many litres per hour does each pump deliver? Use the Rule of Three to express both in the same units.
L3 Apply
Old pump: \(12000 \div 8 = 1500\) L/h. New pump: \(12000 \div 5 = 2400\) L/h. Ratio new : old = 2400 : 1500 = 8 : 5.
Q2. If a 5-member family gets 2 shares and every share corresponds to \(\tfrac{12000}{1200}=10\) litres, analyse how many litres each share delivers and how much water the 5-member family receives.
L4 Analyse
Per share: \(12000 \div 1200 = 10\) L. 5-member family (2 shares): \(2 \times 10 = 20\) L. Per-person water = \(\tfrac{20}{5}=4\) L per person, which is proportional across family sizes only if the "shares-per-size" scheme matches. Here a 3-member family gets 1 share (≈3.3 L/person) and a 7-member gets 3 shares (≈4.3 L/person) — so the scheme is not strictly per-person proportional, it favours larger families slightly.
Q3. A neighbour argues the scheme is "unfair because some families get more water." Evaluate this claim using the ratios above.
L5 Evaluate
Evaluation: "Fair" can mean equal per household or equal per person. Per household, the scheme is intentionally unequal (1, 2, 3 shares) because it adjusts for family size. Per person, it delivers ~3.3, 4, 4.3 L — close but not exactly proportional. The claim depends on which "fairness" yardstick is used. Fully per-person fair would require shares proportional to 3 : 5 : 7 — not 1 : 2 : 3. So the critique has some merit.
Q4. Design a new share scheme where water is exactly proportional to family size (members). State the new shares for families of 3, 5, and 7 members and compute litres each gets when total 12,000 L is distributed across the same 240 households. Assume the village composition is 100 families of 3, 80 of 5, and 60 of 7.
L6 Create
Shares = members. Total members = \(100\times 3 + 80\times 5 + 60\times 7 = 300 + 400 + 420 = 1120\). Water per member = \(12000 \div 1120 \approx 10.71\) L. A 3-member family gets \(3 \times 10.71 \approx 32.14\) L; 5-member \(\approx 53.57\) L; 7-member \(\approx 75\) L. Now per-person share is exactly the same for everyone — truly proportional.

Assertion–Reason Questions

Assertion (A): A car at 75 km/h covers the same distance in less time than at 50 km/h.
Reason (R): Speed and time are inversely proportional for a fixed distance.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Distance = speed × time, so if distance is fixed, doubling the speed halves the time. R is the exact reason.
Assertion (A): If ₹4,500 is divided in the ratio 2 : 3, the second share is ₹2,700.
Reason (R): The formula for the share in ratio \(m:n\) of a total \(x\) is \(\frac{m}{m+n}\times x\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Second share = \(\tfrac{3}{5}\times 4500 = 2700\). R is the formula that gives A.

Summary — Key Takeaways

  • Ratio: \(a:b\) means for every \(a\) units of the first quantity, there are \(b\) units of the second.
  • Two ratios are proportional (\(a:b::c:d\)) when they share the same simplest form. Equivalently, cross multiplication gives \(a \cdot d = b \cdot c\).
  • Similar figures have corresponding measurements that scale by the same factor. Rectangles with the same width : height simplest-form ratio are similar.
  • Rule of Three (Trairāśika): If three quantities are known and the fourth is unknown in a proportion \(a:b::c:d\), then \(d = \tfrac{b \cdot c}{a}\). Āryabhaṭa stated this rule 1,500 years ago.
  • Sharing \(x\) in ratio \(m : n\): First share = \(\tfrac{m}{m+n}\times x\); second share = \(\tfrac{n}{m+n}\times x\).
  • Adding a constant to both terms of a ratio does not preserve the ratio — useful to remember in age problems.
  • Unit conversions are proportional. 1 m ≈ 3.281 ft; 1 hectare = 10,000 m²; 1 L = 1,000 mL; °F = \(\tfrac{9}{5}\)°C + 32.

Puzzle Time — Binairo (Takuzu)

Binairo, also known as Takuzu, is a logic puzzle with simple rules. It is generally played on a square grid with no particular size. Some cells start out filled with two symbols: horizontal and vertical lines. The task is to fill in cells in such a way that:

  1. Each row and each column must contain an equal number of horizontal and vertical lines.
  2. More than two horizontal or vertical lines cannot be adjacent.
  3. Each row is unique. Each column is unique.

Solve a Binairo puzzle of your own from the NCERT textbook — build a 6 × 6 grid with equal horizontal and vertical symbols in every row and column. You can use '|' for vertical and '—' for horizontal. This puzzle strengthens logical reasoning alongside proportional thinking!

Frequently Asked Questions

How do you check if two triangles are similar?
Check that corresponding angles are equal and corresponding sides are in the same ratio. For triangles, any two equal pairs of angles automatically make them similar (AA criterion).
What is the summary of Chapter 7?
Chapter 7 links similarity, ratios, proportion and the ancient Trairasika method. Students learn to compare shapes by scale factor and to solve everyday problems using direct proportion.
What is a common proportional-sharing question?
Sharing Rs 900 between three children in the ratio 2:3:4 gives Rs 200, Rs 300 and Rs 400 respectively, obtained by multiplying the unit share by the ratio parts.
How do exercises test the rule of three?
Typical questions provide three related quantities and ask for the fourth, such as 'If 6 workers finish a job in 10 days, how long will 4 workers take?' (using inverse proportion).
Why are map and scale questions important?
They show the direct real-world use of similarity and ratios, linking distances on paper to actual ground distances through the scale factor.
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