Class 8 Maths Ganita Prakash Chapter 7 Solutions (NCERT 2026–27) – Proportional Reasoning

These Class 8 Maths Ganita Prakash Chapter 7 solutions cover Proportional Reasoning from the new NCF-2023 textbook (Part I, Reprint 2026–27). In the book this chapter is titled “Proportional Reasoning-1” because it is the Part I half of the topic; Part II continues it as “Proportional Reasoning-2.” The worked answers below include original rectangle comparisons, the missing house plan, ratio bars and explanations of the assumptions behind each calculation.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 7 Exercises: 4 “Figure it Out” sets Session: 2026–27
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Chapter 7 Overview

Chapter 7 of Ganita Prakash, Proportional Reasoning-1, starts from a simple everyday observation — why some resized images look natural while others look stretched — and builds it into the powerful idea of proportion. You learn what a ratio a : b means, how to reduce a ratio to its simplest form using the HCF, and how to test whether two ratios are proportional (a : b :: c : d) using cross multiplication (ad = bc). The chapter then applies this to real problems through the ancient Indian Trāirāśika or Rule of Three, to sharing a quantity in a given ratio, and to unit conversions. The Class 8 Maths Ganita Prakash Chapter 7 solutions below solve every exercise so each method is crystal clear.

Key Concepts & Definitions

Ratio (a : b): a comparison of two quantities of the same kind — for every ‘a’ units of the first quantity there are ‘b’ units of the second. The numbers a and b are the terms of the ratio.

Simplest form: a ratio with its terms divided by their HCF (e.g. 60 : 40 = 3 : 2).

Proportion: two ratios are proportional (written a : b :: c : d) when they are equal in their simplest forms, i.e. their terms change by the same factor.

Cross multiplication: a : b :: c : d is true exactly when ad = bc. This lets us find an unknown fourth term.

Rule of Three (Trāirāśika): given three of the four terms in a proportion, the fourth is found by Āryabhaṭa’s rule — multiply the phala by the ichchhā and divide by the pramāṇa.

Sharing in a ratio: a quantity x divided in the ratio m : n gives parts of size m × x/(m+n) and n × x/(m+n).

Direct proportion: as one quantity increases, the other increases by the same factor (more students ⇒ proportionally more rice). Not every situation is proportional — e.g. higher speed gives less travel time.

Important Formulas (Chapter 7)

Simplest form: a : b = (a÷HCF) : (b÷HCF)

Proportion test: a : b :: c : d  ⇔  a × d = b × c (cross multiplication)

Missing term (Rule of Three): if a : b :: c : d, then d = (b × c) ÷ a

Sharing x in the ratio m : n: first part = m × x/(m+n), second part = n × x/(m+n)

Useful conversions: 1 m ≈ 3.281 ft • 1 acre = 43,560 sq ft • 1 hectare = 10,000 m2 ≈ 2.471 acres • 1 L = 1000 mL = 1000 cc • 1 tonne = 1000 kg

Five rectangles use the textbook widths and heights A 60 by 40, B 40 by 20, C 30 by 20, D 90 by 60 and E 60 by 60. A, C and D share ratio three to two.
Pages 159–161: A to C uses factor 1/2; A to D uses factor 3/2. Subtracting the same length from both sides does not preserve the ratio.

Figure it Out (Pages 165–167)

1. Circle the following statements of proportion that are true. (i) 4 : 7 :: 12 : 21    (ii) 8 : 3 :: 24 : 6 (iii) 7 : 12 :: 12 : 7    (iv) 21 : 6 :: 35 : 10 (v) 12 : 18 :: 28 : 12    (vi) 24 : 8 :: 9 : 3

SOLUTION Use cross multiplication: a : b :: c : d is true when a×d = b×c. (i) 4×21 = 84 and 7×12 = 84 → equal, TRUE. (ii) 8×6 = 48 and 3×24 = 72 → not equal, false. (iii) 7×7 = 49 and 12×12 = 144 → not equal, false. (iv) 21×10 = 210 and 6×35 = 210 → equal, TRUE. (v) 12×12 = 144 and 18×28 = 504 → not equal, false. (vi) 24×3 = 72 and 8×9 = 72 → equal, TRUE. ∴ The true statements are (i), (iv) and (vi).

2. Give 3 ratios that are proportional to 4 : 9.

SOLUTION Multiply both terms by the same number to keep the ratio proportional. 4×2 : 9×2 = 8 : 18;   4×3 : 9×3 = 12 : 27;   4×4 : 9×4 = 16 : 36. (Use the same positive factor for both terms in these quantity ratios; multiplying by zero does not give a valid ratio.)

3. Fill in the missing numbers for these ratios that are proportional to 18 : 24. 3 : ______    12 : ______    20 : ______    27 : ______

SOLUTION First reduce 18 : 24 to its simplest form: HCF(18, 24) = 6, so 18 : 24 = 3 : 4. Every proportional ratio has its second term = (4/3) × first term. (i) First term 3: second = 4/3 × 3 = 4 → 3 : 4. (ii) First term 12: second = 4/3 × 12 = 16 → 12 : 16. (iii) First term 20: second = 4/3 × 20 = 80/3 → 20 : 80/3. (iv) First term 27: second = 4/3 × 27 = 36 → 27 : 36.

4. Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios. (Rectangles A, B, C, D, E are shown in the textbook figure.)

SOLUTIONThe similar pairs are A and D, and C and E. Rectangle B has no matching partner among the other four.Compare each rectangle’s long side : short side, allowing rotation. From the printed figure, A and D are about 3 : 1, C and E about 5 : 2, and B about 3 : 2. Small ruler differences are expected; a rotated rectangle retains its shape.The labelled comparison drawing below turns the slanted rectangles to make corresponding sides easier to compare. The letters here belong to Question 4, not the introductory image examples.
Rectangles A and D have long-to-short side ratio about three to one; C and E about five to two. B has ratio about three to two and no match.
Page 166, Question 4: an original comparison drawing with C, D and E turned upright for easy measurement. Compare corresponding sides, not page-aligned widths.

5. Look at the following rectangle. Can you draw a smaller rectangle and a bigger rectangle with the same width to height ratio in your notebooks? Compare your rectangles with your classmates’ drawings. Are all of them the same? If they are different from yours, can you think why? Are they wrong?

SOLUTION Yes. Keep the width-to-height ratio the same and just multiply both measurements by a factor smaller than 1 (for a smaller rectangle) or greater than 1 (for a bigger one). Classmates’ rectangles may be different sizes but all share the same shape because the ratio is unchanged. They satisfy the proportion, so they are not wrong — they are all similar to the given rectangle.
Three example rectangles have dimensions five by three, ten by six and fifteen by nine, all in ratio five to three.
Question 5: illustrative dimensions only. Measure the supplied rectangle in your book, then apply a common scale factor to both dimensions.

6. The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.

SOLUTION Wall (a): in one repeating block, grey bricks = 2 + 3 + 4 = 9 and coloured bricks = 3 + 2 + 1 = 6. Ratio = 9 : 6 = 3 : 2 (dividing both by HCF 3). Wall (b): in one repeating block, grey bricks = 16 and coloured bricks = 12. Ratio = 16 : 12 = 4 : 3 (dividing both by HCF 4).
A rearranged counting block for wall a has rows with three, two and one coloured bricks among five bricks. Wall b has seven rows of four bricks with coloured counts one, two, two, two, two, two and one.
Page 166, Question 6: these are counting blocks with rows aligned for clarity, not drawings of the wall’s staggered joints. Grey-to-coloured ratios are 3:2 and 4:3.

7. Let us draw some human figures. Measure your friend’s body—the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below— head : torso, torso : arms, torso : legs. Now, draw a figure with head, torso, arms, and legs with equivalent ratios as above. Does the drawing look more realistic if the ratios are proportional? Why? Why not?

SOLUTIONMeasure the selected head, torso, arm and leg lengths consistently, using the same unit. Record head : torso, torso : arm, and torso : leg, then simplify each ratio.Illustrative measurements only: head 20 cm, torso 50 cm, arm 60 cm and leg 80 cm give ratios 2 : 5, 5 : 6 and 5 : 8. At a scale of 1/10, draw lengths 2 cm, 5 cm, 6 cm and 8 cm. Replace these example numbers with your friend’s measurements.Using one scale factor preserves that person’s relative proportions. Real people have different proportions; there is no single correct set of measurements for everyone.

Figure it Out (Pages 170–171)

1. The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?

SOLUTION Use the textbook approximation of 52 weeks per year and an average distance per week. Also, 940 million km = 940,000,000 km. Distance per week = 940,000,000 ÷ 52 ≈ 18,076,923 km (about 1.81 crore km) in one week.

2. A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness. (The plan shows segments of 12 ft, 15 ft, 9 ft, 9 ft and 6 ft, with the inner dividing wall.)

SOLUTIONThe upper rectangle is 9 + 15 = 24 ft wide and 12 ft high. Its lower projection is 6 ft wide and 9 ft deep. The remaining lower-right outer segment is 15 − 6 = 9 ft.Outer walls: 24 + 12 + 9 + 9 + 6 + 9 + 9 + 12 = 90 ft.Internal walls: the vertical 12-ft divider and horizontal 6-ft segment add 18 ft. Count each drawn wall once, giving 90 + 18 = 108 ft.Bricks per foot = 1450 ÷ 10 = 145. Required bricks ≈ 108 × 145 = 15,660 bricks, using the question’s equal-height and equal-thickness assumptions.
A house plan has a top rectangle 24 feet wide and 12 feet high and a lower projection 6 feet wide and 9 feet deep. Two orange inner wall segments measure 12 feet and 6 feet.
Pages 170–171: the exterior totals 90 ft; the two internal segments add 18 ft. At 145 bricks per foot, the estimate is 15,660 bricks.

Figure it Out (Page 175)

1. Divide ₹4,500 into two parts in the ratio 2 : 3.

SOLUTION Total number of groups = 2 + 3 = 5. Size of one group = 4500 ÷ 5 = ₹900. First part = 2 × 900 = ₹1,800;   second part = 3 × 900 = ₹2,700. (Check: 1800 + 2700 = 4500.)
A five-part bar has two blue blocks and three yellow blocks. Dividing 4500 rupees by five gives 900 rupees per block and shares 1800 and 2700.
Page 175, Question 1: the two shares add back to ₹4500.

2. In a science lab, acid and water are mixed in the ratio of 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?

SOLUTION Total groups = 1 + 5 = 6. Size of one group = 240 ÷ 6 = 40 mL. Acid = 1 × 40 = 40 mL;   water = 5 × 40 = 200 mL. (Check: 40 + 200 = 240 mL.)

3. Blue and yellow paints are mixed in the ratio of 3 : 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?

SOLUTION Total groups = 3 + 5 = 8. Size of one group = 40 ÷ 8 = 5 mL. Blue = 3 × 5 = 15 mL;   yellow = 5 × 5 = 25 mL. After adding 20 mL of yellow: yellow = 25 + 20 = 45 mL, blue is still 15 mL. New ratio blue : yellow = 15 : 45 = 1 : 3 (dividing both by HCF 15).
The original paint bar has three blue and five yellow equal parts. Adding twenty millilitres of yellow adds four equal parts, producing three blue and nine yellow.
Page 175, Question 3: the total rises from 40 mL to 60 mL, while blue remains 15 mL.

4. To make soft idlis, you need to mix rice and urad dal in the ratio of 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?

SOLUTION Total groups = 2 + 1 = 3. Size of one group = 6 ÷ 3 = 2 cups. Rice = 2 × 2 = 4 cups;   urad dal = 1 × 2 = 2 cups. (Check: 4 + 2 = 6 cups.)

5. I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 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?

SOLUTION Take the orange bucket as 1 whole. Red = 3/8 of a bucket and yellow = 5/8 of a bucket. Adding one full bucket of yellow: yellow = 5/8 + 1 = 13/8; red stays 3/8. New ratio red : yellow = (3/8) : (13/8) = 3 : 13.
One bucket is split into three red and five yellow parts. Adding another equal-sized bucket of yellow gives three red and thirteen yellow parts.
Page 175, Question 5: assume both buckets have equal capacity and are full.

Figure it Out (Pages 176–177)

1. 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 its simplest form.

SOLUTION Ratio = 600 : 900. HCF(600, 900) = 300. 600÷300 : 900÷300 = 2 : 3.

2. Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers who went on that trip and all the buses were full. This year we have 204 students. How many buses will we need? Will all the buses be full?

SOLUTIONEach bus holds 162 ÷ 3 = 54 people, assuming the same bus size this year.Treating 204 as the total travelling group, 204 ÷ 54 = 34/9 = 3.777… buses. Round up to 4 buses.Four buses provide 216 seats, so 12 seats remain empty. One possible distribution is 54 + 54 + 54 + 42.The question says “204 students”, while the worked textbook answer treats 204 as the group total. If accompanying teachers are additional, include them before dividing by 54; the empty-seat count then changes.
Four buses carry 204 people: three buses hold 54 each and the fourth holds 42, leaving twelve empty seats.
Page 176, Question 2: use the same 54-person bus capacity. If 204 counts only students, include accompanying teachers before calculating actual transport needs.

3. The area of Delhi is 1,484 sq. km and the area of Mumbai is 550 sq. km. The population of Delhi is approximately 30 million and that of Mumbai is 20 million people. Which city is more crowded? Why do you say so?

SOLUTION Using the populations and areas supplied in this textbook question, compare population density (people per sq. km). These are exercise data, not current population estimates. Delhi: 30,000,000 ÷ 1,484 ≈ 20,216 people/sq km. Mumbai: 20,000,000 ÷ 550 ≈ 36,364 people/sq km. Mumbai has more people in each square kilometre, so Mumbai is more crowded.

4. A crane of height 155 cm has its neck and the rest of its body in the ratio 4 : 6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be?

SOLUTION Neck : rest of body = 4 : 6, so neck is 4 out of every (4 + 6) = 10 parts of the total height. That is, neck = (4/10) × height. Neck height = (4/10) × your height. For example, if your height is 150 cm, your neck would be (4/10) × 150 = 60 cm. (Use your own measured height.) (For the crane itself this gives neck = (4/10) × 155 = 62 cm.)

5. Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If 2½ palas of saffron costs 3/7 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”

SOLUTION Set up the proportion (palas) : (niskas) → 2.5 : (3/7) :: x : 9. By cross multiplication: (3/7) × x = 2.5 × 9 = 22.5. x = 22.5 × (7/3) = 22.5 × 7 ÷ 3 = 157.5 ÷ 3 = 52.5 palas. So 52.5 palas of saffron can be bought for 9 niskas.

6. Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2?

SOLUTION Let this happen after x years. Then Harmain is (1 + x) and her brother is (5 + x). (1 + x) : (5 + x) = 1 : 2 ⇒ 2(1 + x) = 1(5 + x) ⇒ 2 + 2x = 5 + x ⇒ x = 3. So after 3 years their ages are 4 and 8 (ratio 4 : 8 = 1 : 2). Harmain will be 4 years old.
Harmain has four equal year-blocks and her brother eight, showing a four-year age gap and age ratio one to two after three years.
Page 177, Question 6: both people gain the same number of years. Do not multiply their present ages by a common factor.

7. 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?

SOLUTION Using the mass ratio supplied in the exercise for equal volumes, gold : water = 37 : 2. Let 1 L of gold weigh x kg, while 1 L of water = 1 kg. 37 : 2 :: x : 1 ⇒ 2x = 37 × 1 ⇒ x = 37 ÷ 2 = 18.5 kg.

8. 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? (Please refer to the section on Unit Conversions earlier in this chapter).

SOLUTION 10 tonnes = 10,000 kg, and 1 acre = 43,560 sq ft. So manure : area = 10,000 : 43,560 (kg per sq ft). Area of plot = 200 × 500 = 100,000 sq ft. Manure x: x : 100,000 :: 10,000 : 43,560 ⇒ x = (100,000 × 10,000) ÷ 43,560 ≈ 22,956.8 kg (about 22.96 tonnes).

9. 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?

SOLUTION 10 litres = 10,000 mL, and one mug = 500 mL, so the bucket = 10,000 ÷ 500 = 20 mugs. Time = 20 × 15 = 300 seconds = 5 minutes.
At constant flow, a tap fills half a litre in fifteen seconds, one litre in thirty seconds, five litres in one hundred fifty seconds and ten litres in three hundred seconds.
Page 177, Question 9: convert litres and millilitres to the same unit before comparing volumes.

10. One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?

SOLUTION 1 acre = 43,560 sq ft costs ₹15,00,000. Cost is proportional to area: 43,560 : 15,00,000 :: 2,400 : x. x = (15,00,000 × 2,400) ÷ 43,560 = 360,00,00,000 ÷ 43,560 ≈ ₹82,645 (approx.).

11. 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?

SOLUTION Oxen take 6 hours per acre, so for 20 acres: 20 × 6 = 120 hours with a pair of oxen. The tractor is 4 times faster, so it needs one-fourth of the time: per acre = 6 ÷ 4 = 1.5 hours. For 20 acres with the tractor: 20 × 1.5 = 30 hours (equivalently 120 ÷ 4 = 30 hours).

12. The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel are mixed in a 3 : 1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the cost of copper is ₹906 per kg and the cost of nickel is ₹1,341 per kg, what is the cost of these metals in a ₹10 coin?

SOLUTION Split 7.74 g in the ratio 3 : 1. Total groups = 4, so one group = 7.74 ÷ 4 = 1.935 g. Copper = 3 × 1.935 = 5.805 g;   nickel = 1 × 1.935 = 1.935 g. Cost of copper = ₹906 per kg = ₹906 per 1000 g, so for 5.805 g: (906 ÷ 1000) × 5.805 ≈ ₹5.26. Cost of nickel = ₹1,341 per 1000 g, so for 1.935 g: (1341 ÷ 1000) × 1.935 ≈ ₹2.59. Total cost of metals in one coin ≈ 5.26 + 2.59 = ₹7.85 (approx.).

Math Talk and in-text activities

Pages 160–165: scaling images and comparing filter coffee.

SOLUTIONImage A to image D: 90/60 = 60/40 = 3/2, so both dimensions scale by the same factor. A, C and D all have width : height = 3 : 2; B has 2 : 1 and E has 1 : 1.Regular coffee has decoction : milk = 15 : 35 = 3 : 7, or 15/50 = 30% decoction. Strong coffee has 20/50 = 40%; light coffee has 10/50 = 20%. Compare the fraction of coffee in the whole drink, not just the absolute amount.
Three fifty-millilitre drinks have coffee-to-milk quantities fifteen to thirty-five, twenty to thirty and ten to forty. The coffee fractions are thirty, forty and twenty percent.
Pages 164–165: a greater fraction of coffee decoction gives a stronger drink.
Coffee (mL)Milk (mL)Simplest ratioCompared with regular
3006001 : 2Stronger
1505003 : 10Lighter
2004001 : 2Stronger
24563 : 7Regular
1003001 : 3Lighter

Page 163: school and blackboard measurements.

SOLUTIONThe given teacher-to-student ratio is 5 : 170 = 1 : 34. For your school, count teachers T and students S; the ratio matches if S = 34T. For example, 10 : 340 matches, while 10 : 300 does not.Measure your board’s width W and height H in the same unit. Draw dimensions kW and kH for a suitable positive scale factor k. Different sizes can all be correct if their corresponding side ratios agree.

Pages 169–171: rates, recipes and the motorcycle journey.

SOLUTIONCar: 4 hours = 240 minutes. At constant speed, distance = 90 × 240/150 = 144 km.Tea: ₹200 for 200 g means ₹1000 per kg; the other tea costs ₹800 per kg. On the given prices, the Himachal tea is more expensive by ₹200 per kg.Recipe activity: if a recipe serves f people and uses q units of an ingredient, 15 guests require 15q/f units. If the family also eats, use (f + 15)q/f. Keep serving sizes the same and use your own recipe measurements.Motorcycle: distance = 50 × 2 = 100 km. At 75 km/h, time = 100/75 = 4/3 hours = 1 hour 20 minutes. The proportion 50 : 2 :: 75 : t would incorrectly predict 3 hours. Instead, speed × time is constant.
For a fixed one-hundred-kilometre journey, fifty kilometres per hour takes two hours while seventy-five kilometres per hour takes one hour twenty minutes. Time bars show the second time is two thirds of the first.
Page 171: time = 100 ÷ 75 hours = 4/3 hours. This is inverse proportion.

Page 171: are the sample shampoo prices proportional to volume?

SOLUTIONCompare the price per mL using the textbook’s sample prices. The rates below differ, so price and volume are not proportional. These are example data, not current shop prices.
PackVolumePricePrice per mL
Sachet6 mL₹2₹0.333…
Small bottle180 mL₹154₹0.8556
Medium bottle340 mL₹276₹0.8118
Large bottle1000 mL₹540₹0.54

Pages 172–175: sharing and adding cement.

SOLUTIONTo share 42 in ratio 4 : 3, make 7 equal groups: 42 ÷ 7 = 6 per group. The shares are 24 and 18.A 40-kg sand-and-cement mixture in ratio 3 : 1 contains 30 kg sand and 10 kg cement. Sand stays 30 kg. For target ratio 5 : 2, cement must be 30 × 2/5 = 12 kg. Therefore add 2 kg cement.

Page 176: why Celsius and Fahrenheit are not directly proportional.

SOLUTIONF = (9/5)C + 32 and C = (5/9)(F − 32). Thus 25°C = 45 + 32 = 77°F. The added 32 means these readings are not directly proportional: 0°C corresponds to 32°F, not 0°F.

Common Mistakes to Avoid

Watch out for these

  • Thinking that equal differences keep things proportional — only equal factors (multiplication) do (image A vs B in the chapter).
  • Comparing two quantities in different units — convert first (e.g. 1 kg = 1000 g, 4 hours = 240 minutes) before forming the ratio.
  • Mixing up the order of the terms in cross multiplication for a : b :: c : d — the rule is ad = bc.
  • When sharing x in the ratio m : n, dividing by m or n instead of by (m + n) to get the group size.
  • Assuming every “more/less” situation is direct proportion — higher speed gives less time for a fixed distance. Use inverse proportion; the direct-proportion arrangement shown in this chapter does not apply.
  • Forgetting to reduce the final ratio to its simplest form.

Practice MCQs & Assertion–Reason

1. The simplest form of the ratio 60 : 40 is:

(a) 6 : 4    (b) 2 : 3    (c) 3 : 2    (d) 5 : 3

2. Two ratios a : b and c : d are proportional if:

(a) a + d = b + c    (b) ad = bc    (c) ac = bd    (d) a − b = c − d

3. Which of these is proportional to 4 : 9?

(a) 8 : 27    (b) 9 : 4    (c) 12 : 27    (d) 13 : 18

4. If ₹4,500 is divided in the ratio 2 : 3, the smaller part is:

(a) ₹900    (b) ₹1,800    (c) ₹2,700    (d) ₹2,250

5. In the proportion 6 : 10 :: 18 : x, the value of x is:

(a) 24    (b) 28    (c) 30    (d) 36

6. To divide a quantity x in the ratio m : n, the first part is:

(a) x/(m+n)    (b) m × x/(m+n)    (c) m × x/n    (d) (m+n)/x

7. The ancient Indian name for Rule of Three problems is:

(a) Lilavati    (b) Trāirāśika    (c) Bijaganita    (d) Aryabhatiya

8. 1 acre is equal to:

(a) 10,000 sq ft    (b) 43,560 sq ft    (c) 2.471 sq ft    (d) 1,000 sq ft

9. Acid and water are mixed in the ratio 1 : 5. In 240 mL of solution, the amount of acid is:

(a) 40 mL    (b) 48 mL    (c) 200 mL    (d) 60 mL

10. As the speed of a vehicle increases, the time taken to cover a fixed distance:

(a) increases proportionally    (b) decreases    (c) stays the same    (d) becomes zero

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The ratios 4 : 7 and 12 : 21 are proportional.

Reason: Two ratios a : b and c : d are proportional when ad = bc.

A-R 2. Assertion: Adding the same number to both terms of a ratio always keeps it proportional to the original.

Reason: A ratio stays proportional only when both terms are multiplied (or divided) by the same factor.

A-R 3. Assertion: When 12 counters are shared in the ratio 3 : 1, the two shares are 9 and 3.

Reason: The size of each group is the total divided by (m + n).

A-R 4. Assertion: Speed and time for a fixed distance can be modelled as 50 : 2 :: 75 : t (speed in km/h, time in hours).

Reason: Time decreases as speed increases, so speed and time are not in direct proportion.

A-R 5. Assertion: The ratio 60 : 40 in its simplest form is 3 : 2.

Reason: A ratio is reduced to simplest form by dividing both terms by their HCF.

Answer key: 1-(A), 2-(D), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • A ratio a : b compares two quantities; a and b are its terms.
  • Reduce a ratio to simplest form by dividing both terms by their HCF.
  • Two ratios are proportional (a : b :: c : d) when they are equal in simplest form, i.e. ad = bc.
  • Rule of Three: given three terms, the fourth = (b × c) ÷ a.
  • Sharing x in the ratio m : n → parts are m × x/(m+n) and n × x/(m+n); group size = x ÷ (m + n).
  • Equal multiplication keeps a ratio proportional; equal addition/subtraction usually does not.
  • Always convert to the same units before forming or comparing ratios.
  • Not every situation is direct proportion — e.g. speed vs time for a fixed distance is inverse.

How to score full marks in this chapter

State the test you are using (simplest form or cross multiplication) before you calculate, and always check the answer by substituting it back into the proportion. Convert units first, show the group size when sharing in a ratio, and reduce every final ratio to its simplest form. For “Rule of Three” word problems, ask yourself whether the quantities really grow together — if more of one means less of the other, do not use direct proportion.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 7 about?

Chapter 7, titled “Proportional Reasoning-1” in the textbook, teaches ratios, simplest form, proportion and cross multiplication, the ancient Rule of Three (Trāirāśika), sharing a quantity in a given ratio, and useful unit conversions. Part II of the book continues the topic as “Proportional Reasoning-2.”

How do you check if two ratios are proportional?

Reduce both ratios to simplest form — if they are equal, they are proportional. The quick test is cross multiplication: a : b :: c : d is true exactly when a × d = b × c.

How many exercises are there in Ganita Prakash Chapter 7?

There are four “Figure it Out” sets (on pages 165, 170, 175 and 176) along with several worked Examples, Math Talk and Try This boxes. All the Figure it Out questions are solved on this page.

Are these Class 8 Maths Ganita Prakash Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for 2026–27.

Textbook reference: NCERT Ganita Prakash, Chapter 7 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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