Figure it Out — Ratios, Rates and Unit Conversions — Proportional Reasoning

1. 11 markGanita Prakash Cl-8 Part 1, Figure it Out, page 176

   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.

   Hint. Divide both quantities by their highest common factor.

   Step 1 — Write the ratio as given. Orange : apple = 600 : 900

   Step 2 — Simplify. The HCF of 600 and 900 is 300. Dividing both parts: 600 ÷ 300 = 2 900 ÷ 300 = 3

   = 2 : 3

   Step 3 — Check. 2 : 3 scaled up by 300 gives back 600 : 900 ✓ Cross-multiplying: 600 × 3 = 1800 and 900 × 2 = 1800 ✓

   Both quantities are in the same unit (millilitres) before the ratio is formed, which is essential — a ratio between 600 mL and 0.9 L is meaningless until one is converted.

   What the simplest form tells you. 2 : 3 says that for every 2 parts of orange there are 3 of apple, so the drink is 2/5 orange and 3/5 apple. That is far easier to read off than 600 : 900, and it lets you scale the recipe to any size: 200 mL and 300 mL, or 1 L and 1.5 L, give exactly the same drink.

   ✦ 2 : 3, obtained by dividing both 600 and 900 by their HCF of 300.

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 176

   Last year we hired 3 buses for the school trip and 162 students and teachers went, with all buses full. This year we have 204 students and teachers. How many buses will we need? Will all the buses be full?

   Hint. Find the capacity of one bus first, then see how many buses 204 people need — remembering you cannot hire part of a bus.

   (Note: the book's printed working for this question contains a garbled fraction; the clean derivation is given below and reaches the same answer.)

   Step 1 — Find the capacity of one bus. All 3 buses were full with 162 people, so capacity = 162 ÷ 3 = 54 people per bus

   Step 2 — Find how many buses 204 people need. 204 ÷ 54 = 3.78

   Step 3 — Round in the right direction. You cannot hire 0.78 of a bus, and 3 buses would seat only 3 × 54 = 162 people — leaving 42 people behind. So the number must be rounded up: 4 buses

   This is the crucial step. Ordinary rounding would give 4 here anyway, but if the answer had been 3.1 buses the arithmetic rounding would say 3 while the real answer is still 4. Whenever the units are indivisible — buses, boxes, tickets — always round up.

   Step 4 — Will they all be full? Total seats in 4 buses = 4 × 54 = 216 People travelling = 204 Empty seats = 216 − 204 = 12

   So no, the buses will not all be full — three will be full and the fourth will carry 42 people with 12 seats to spare.

   ✦ 4 buses are needed. They will not all be full: 4 buses seat 216 but only 204 are travelling, leaving 12 empty seats.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 176

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

   Hint. Population alone will not do — you need people per unit of area.

   Step 1 — Recognise that raw population is the wrong comparison. Delhi has more people (30 million against 20 million), but it also has far more land. "Crowded" means people packed into each square kilometre, so the right measure is population density:

   density = population ÷ area

   This is a rate — a ratio between quantities in different units — and reducing to a per-unit figure is what makes the two cities comparable.

   Step 2 — Compute each density.

   Delhi: 30,000,000 ÷ 1,484 ≈ 20,216 people per sq km

   Mumbai: 20,000,000 ÷ 550 ≈ 36,364 people per sq km

   Step 3 — Compare. 36,364 > 20,216, so Mumbai is more crowded — by a factor of about 1.8.

   Step 4 — Say why, in words. Mumbai has two-thirds of Delhi's population living on barely a third of the land. Because area shrinks faster than population between the two cities, the people per square kilometre rises sharply.

   A quick mental check. Delhi's area is roughly 2.7 times Mumbai's, while its population is only 1.5 times as large. Since 1.5 < 2.7, Delhi must be the less dense of the two — no long division needed.

   ✦ Mumbai, at about 36,364 people per sq km against Delhi's 20,216 — Delhi has more people but spread over nearly three times the area.

4. 42 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 176

   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 had this ratio, how tall would your neck be?

   Hint. The ratio 4 : 6 covers the whole body — so how many parts is the total height?

   Step 1 — Find what fraction of the total the neck is. Neck : rest of body = 4 : 6, so the whole creature is 4 + 6 = 10 parts. The neck is therefore 4/10 of the total height, which simplifies to 2/5.

   Note the ratio is not neck : total. It compares the neck with the rest of the body, so the denominator must be the sum, not the 6.

   Step 2 — Check the rule on the crane itself. Neck = (4/10) × 155 = 62 cm Rest of body = (6/10) × 155 = 93 cm Check: 62 + 93 = 155 ✓ and 62 : 93 = 2 : 3 = 4 : 6 ✓

   Step 3 — Apply it to your own height. Neck length = (4/10) × your height = 0.4 × your height

   Worked examples: • If you are 150 cm tall, your neck would be 0.4 × 150 = 60 cm • If you are 165 cm tall, it would be 0.4 × 165 = 66 cm

   Step 4 — Notice how unreasonable that is. A real human neck is about 10-12 cm, roughly 1/14 of height, not 2/5. So a person built to a crane's proportions would have a neck well over half a metre long. That contrast is the point of the question: proportions are what make a shape recognisable, and changing them changes the creature entirely.

   ✦ Neck = 4/10 (that is, 0.4) of your height — about 60 cm for a 150 cm person, since the ratio 4 : 6 makes the neck 4 parts out of 10.

5. 53 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177 (from the Lilavati)

   An ancient problem from the Līlāvatī: "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?"

   Hint. Find how much saffron one niska buys, then scale up. Keep the mixed number and the fraction as fractions.

   Step 1 — Write the given quantities as fractions. 2½ palas = 5/2 palas Cost = 3/7 niskas

   Step 2 — Find how much saffron one niska buys. saffron per niska = (5/2) ÷ (3/7) = (5/2) × (7/3) = 35/6 palas per niska

   Dividing by a fraction means multiplying by its reciprocal, which is why the 7 moves up and the 3 down.

   Step 3 — Scale up to 9 niskas. Saffron for 9 niskas = 9 × 35/6 = 315/6 = 52.5 palas

   Step 4 — Check by proportion. Set up the proportion directly: (5/2) : (3/7) :: x : 9 Cross-multiplying: (3/7)x = (5/2) × 9 = 45/2 x = (45/2) × (7/3) = 315/6 = 52.5 ✓

   A rough check on the size. 9 niskas is 21 times the cost of 3/7 niskas, since 9 ÷ (3/7) = 21. So the saffron should be 21 times 2½ palas, and 21 × 2.5 = 52.5 ✓

   About the source. The Līlāvatī was written by Bhāskara II around 1150 CE and poses its arithmetic in verse, addressed to the reader. Problems of exactly this type — a rate given, a new total asked for — have been the standard test of proportional reasoning for nearly nine centuries.

   ✦ 52.5 palas of saffron can be bought for 9 niskas.

6. 63 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177

   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?

   Hint. Both ages grow by the same amount each year — so add the same unknown to both.

   Step 1 — Set up the unknown correctly. Let x be the number of years from now. Both children age at the same rate, so after x years: Harmain's age = 1 + x Brother's age = 5 + x

   The key insight is that the same x is added to both. It is the gap of 4 years that never changes, while the ratio keeps changing as they grow.

   Step 2 — Write the condition. We want (1 + x) : (5 + x) = 1 : 2, that is

   (1 + x)/(5 + x) = 1/2

   Step 3 — Solve by cross-multiplying. 2(1 + x) = 1(5 + x) 2 + 2x = 5 + x 2x − x = 5 − 2 x = 3

   Step 4 — Answer the question asked. After 3 years, Harmain's age = 1 + 3 = 4 years (and her brother will be 8).

   Check: 4 : 8 = 1 : 2 ✓ and the age gap is still 8 − 4 = 4 years ✓

   Read the question carefully — it asks for Harmain's age, not the number of years that pass. Answering "3" is the standard error here.

   Why the ratio keeps changing. Right now the ratio is 1 : 5; in 3 years it is 1 : 2; in 15 years it will be 16 : 20 = 4 : 5. As both ages grow, the fixed 4-year gap matters less and less, so the ratio creeps towards 1 : 1 without ever reaching it.

   ✦ Harmain will be 4 years old (in 3 years' time), when her brother is 8.

7. 72 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177

   The masses of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water has a mass of 1 kg, what is the mass of 1 litre of gold?

   Hint. The volumes are equal, so the ratio applies directly to the masses.

   Step 1 — Note that the volumes match. The ratio 37 : 2 holds for equal volumes, and we are comparing 1 litre with 1 litre. So the ratio applies directly, with no scaling needed.

   Step 2 — Set up the proportion. mass of gold : mass of water = 37 : 2 x : 1 = 37 : 2

   Step 3 — Solve. x/1 = 37/2 x = 18.5 kg

   Step 4 — Check against reality. This says gold is 18.5 times as dense as water — and the true density of gold is about 19.3 g/cm³ against water's 1 g/cm³, so the figure is right to within a few per cent ✓

   What this means physically. A one-litre carton of milk weighs about a kilogram. The same carton filled with gold would weigh 18.5 kg — as much as a well-loaded school bag. This is why even small gold bars are surprisingly heavy, and why density, not size, is what makes gold hard to carry in quantity.

   The ratio 37 : 2 is just density expressed as whole numbers. Dividing gives 18.5, the relative density (or specific gravity) of gold — the number of times heavier it is than the same volume of water.

   ✦ 18.5 kg, since equal volumes give mass in the ratio 37 : 2 and the water weighs 1 kg.

8. 84 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177

   It is good farming practice to apply 10 tonnes of cow manure per acre of land. A farmer plans to grow tomatoes in a plot of 200 ft by 500 ft. How much manure should he buy?

   Hint. The rate is per acre but the plot is in square feet — convert before comparing. 1 acre = 43,560 sq ft.

   Step 1 — Find the plot area. Area = 200 × 500 = 100,000 sq ft

   Step 2 — Note the unit mismatch. The rate is given per acre but the plot is in square feet, so one must be converted before any proportion can be set up: 1 acre = 43,560 sq ft

   Step 3 — Convert the manure figure. 10 tonnes = 10 × 1,000 = 10,000 kg per 43,560 sq ft

   Step 4 — Set up and solve the proportion. 10,000 kg : 43,560 sq ft :: x kg : 100,000 sq ft

   x = (10,000 × 100,000) ÷ 43,560 = 1,000,000,000 ÷ 43,560 ≈ 22,956.8 kg

   In tonnes: 22,956.8 ÷ 1,000 ≈ 22.96 tonnes

   Step 5 — Check the size of the answer. The plot is 100,000 ÷ 43,560 ≈ 2.3 acres, and at 10 tonnes per acre that is about 23 tonnes ✓ — which is the same calculation done in a friendlier order, and worth using as the main method.

   The step that decides this question is the unit conversion. Working the proportion without converting acres to square feet would give an answer wrong by a factor of more than forty thousand.

   Practical note: manure is sold by the tonne, so the farmer would order 23 tonnes.

   ✦ About 22,957 kg, or roughly 23 tonnes — the plot is about 2.3 acres, since 100,000 sq ft ÷ 43,560 sq ft per acre ≈ 2.3.

9. 92 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177

   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 10-litre bucket?

   Hint. Convert the bucket's capacity to millilitres so both volumes are in the same unit.

   Step 1 — Put both volumes in the same unit. 1 litre = 1,000 mL, so 10 litres = 10,000 mL

   Step 2 — Find how many mugfuls fill the bucket. 10,000 ÷ 500 = 20 mugfuls

   Step 3 — Scale the time. The tap flows at a steady rate, so time is proportional to volume: Time = 20 × 15 = 300 seconds

   Step 4 — Convert to a friendlier unit. 300 ÷ 60 = 5 minutes

   Check by unit rate: the tap delivers 500 mL in 15 s, which is 500/15 ≈ 33.3 mL per second. Then 10,000 ÷ 33.3 ≈ 300 s ✓

   Why the answer is exact rather than approximate. The bucket's capacity is a whole number of mugfuls, so no rounding enters anywhere. Had the bucket held 9 litres, the answer would be 18 mugfuls and 270 seconds — still exact, since 9,000 ÷ 500 = 18.

   The assumption being made is that the tap's flow rate stays constant. In practice water pressure drops as a tank empties, so a real bucket might take a little longer — but a constant rate is exactly what "proportional" means, and it is what the question intends.

   ✦ 300 seconds, or 5 minutes — the 10-litre bucket holds 20 mugfuls, each taking 15 seconds.

10. 102 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 177

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

    Hint. Convert the acre to square feet first: 1 acre = 43,560 sq ft.

    Step 1 — Convert to a common unit. 1 acre = 43,560 sq ft, so ₹15,00,000 buys 43,560 sq ft.

    Step 2 — Set up the proportion. ₹15,00,000 : 43,560 sq ft :: ₹x : 2,400 sq ft

    Step 3 — Solve. x = (15,00,000 × 2,400) ÷ 43,560 = 3,600,000,000 ÷ 43,560 ≈ ₹82,645

    Step 4 — Check by unit rate. Cost per square foot = 15,00,000 ÷ 43,560 ≈ ₹34.44 per sq ft Then 2,400 × 34.44 ≈ ₹82,656 — agreeing with the answer above to within rounding ✓

    Working with the unit rate first is usually safer, because it gives a figure you can sanity-check against what land actually costs, and it makes any further question (the cost of 1,000 sq ft, say) immediate.

    A rough mental check. 2,400 sq ft is about 2,400/43,560 ≈ 1/18 of an acre, and 15,00,000 ÷ 18 ≈ 83,000 ✓

    Why plots are quoted in square feet. An acre is far larger than a typical house plot — a 2,400 sq ft plot is only about a twentieth of an acre, yet is a common size for a residential site.

    ✦ About ₹82,645, since land costs ₹15,00,000 ÷ 43,560 ≈ ₹34.44 per square foot.

11. 113 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 178

    A tractor can plough the same area 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field, and a pair of oxen takes 6 hours to plough an acre. How long would the oxen take for the whole field? How long would the tractor take?

    Hint. 'Four times faster' means the time taken is one quarter — speed and time are inversely proportional.

    Step 1 — Interpret "4 times faster". If the tractor works 4 times as fast, it needs only one quarter of the time for the same job. Speed and time are inversely proportional: doubling the speed halves the time.

    This is the step that separates this question from the rest of the set. Everywhere else the quantities rise together; here one rises as the other falls, so you divide rather than multiply.

    Step 2 — Find each rate per acre. Oxen: 6 hours per acre (given) Tractor: 6 ÷ 4 = 1.5 hours per acre

    Step 3 — Scale to the whole 20-acre field. Oxen: 20 × 6 = 120 hours Tractor: 20 × 1.5 = 30 hours

    Step 4 — Check the relationship. 120 ÷ 30 = 4 ✓ — the tractor is indeed four times quicker over the whole field, as it was over one acre.

    What this means in practice. At 8 working hours a day, the oxen would need 15 days while the tractor needs under 4 days. In farming that difference decides whether a crop is sown before the rains arrive.

    The error to avoid: multiplying the tractor's time by 4 instead of dividing. Always ask which way the quantity moves — faster must mean less time.

    ✦ The oxen would take 120 hours and the tractor 30 hours, since being 4 times faster means taking one quarter of the time — 1.5 hours per acre instead of 6.

12. 124 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 178

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

    Hint. Split the mass in the given ratio first, then convert grams to kilograms before applying the per-kg prices.

    Step 1 — Split the coin's mass in the ratio 3 : 1. Total parts = 3 + 1 = 4 One part = 7.74 ÷ 4 = 1.935 g

    Copper = 3 parts = 3 × 1.935 = 5.805 g Nickel = 1 part = 1.935 g

    Check: 5.805 + 1.935 = 7.74 g ✓

    Step 2 — Cost the copper. Copper costs ₹906 per kg, that is per 1,000 g. Cost = 906 × (5.805 ÷ 1000) = 906 × 0.005805 ≈ ₹5.26

    Step 3 — Cost the nickel. Nickel costs ₹1,341 per kg. Cost = 1341 × (1.935 ÷ 1000) = 1341 × 0.001935 ≈ ₹2.59

    Step 4 — Find the total metal cost. 5.26 + 2.59 = ₹7.85

    The step that catches people out is the unit conversion. The prices are per kilogram but the masses are in grams, so every mass must be divided by 1,000 before multiplying. Skipping that gives an answer a thousand times too large.

    What the answer tells you. The metal in a ₹10 coin is worth about ₹7.85 — less than the coin's face value of ₹10, but not by much. Coins are deliberately minted so that the metal is worth less than the face value; if the metal were worth more, people would melt the coins down rather than spend them.

    Note on rounding. Both costs were rounded to the nearest paisa. Carrying full precision gives ₹5.259 and ₹2.595, totalling ₹7.854 — so ₹7.85 is right.

    ✦ Copper (5.805 g) costs about ₹5.26 and nickel (1.935 g) about ₹2.59, giving a total metal value of roughly ₹7.85 — just under the coin's face value.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp107.pdf). Four 'Figure it Out' blocks sit on pages 165, 170, 175 and 176-178. Every numeric answer here was independently recomputed before comparison with the book's printed answer key — including the acre-to-square-feet conversions (1 acre = 43,560 sq ft), the Lilavati saffron problem, the population densities of Delhi and Mumbai, and the cupro-nickel coin costings. THREE NOTES ARE FLAGGED IN PLACE: (1) question 6 on page 165 depends on counting bricks in a printed pattern, so its counts are cited from the key rather than derived; (2) questions 4, 5 and 7 on page 165 are measurement and drawing activities with no single numeric answer — the method is given instead; (3) the printed key's working for question 2 on page 176 contains a garbled fraction ('x = 34/9'), so the solution is re-derived cleanly here while reaching the same final answer of 4 buses.. Questions are referenced from the NCERT textbook for identification.