Key Concepts & Formulae at a Glance
- Ratio: \(a:b\) means for every \(a\) units of the first quantity, there are \(b\) units of the second. \(a\) and \(b\) are the terms of the ratio.
- Simplest form: divide both terms by their HCF.
- Proportion: two ratios \(a:b\) and \(c:d\) are proportional (written \(a:b::c:d\)) if their simplest forms match, or equivalently if \(ad=bc\) (cross multiplication).
- Trairasika (Rule of Three): given \(a:b::c:d\) with \(d\) unknown, \(d=\dfrac{bc}{a}\).
- Dividing a quantity in a ratio: to split \(x\) in the ratio \(m:n\), the two parts are \(m\times\dfrac{x}{m+n}\) and \(n\times\dfrac{x}{m+n}\).
- Caution 1: adding or subtracting the same number from both terms of a ratio changes it, and the new ratio is generally not proportional to the original.
- Caution 2: not every pair of changing quantities is directly proportional — e.g. speed and travel time move in opposite directions (inversely related), so such problems cannot be modelled with the Rule of Three the way direct-proportion problems can.
Common Unit Conversions
| Quantity | Conversion |
|---|---|
| Length | 1 m = 3.281 ft |
| Area | 1 sq. m = 10.764 sq. ft; 1 acre = 43,560 sq. ft; 1 hectare = 10,000 sq. m = 2.471 acres |
| Volume | 1 mL = 1 cc; 1 L = 1,000 mL = 1,000 cc |
| Temperature | °F = (9/5)×°C + 32; °C = (5/9)×(°F − 32) |
7.1 Observing Similarity in Change
Five resized images of the same tiger photo (A–E) were compared. Images A, C, and D look similar to each other, while B looks stretched and E looks squashed, even though all five are technically rectangles (or a square, for E).
| Image | Width (mm) | Height (mm) |
|---|---|---|
| A | 60 | 40 |
| B | 40 | 20 |
| C | 30 | 20 |
| D | 90 | 60 |
| E | 60 | 60 |
Comparing A and C: both the width and height of C are exactly half of A's — both dimensions changed by the same factor (multiplication), so the images look similar. Comparing A and B: the width and height of B are each 20 mm less than A's — the same difference (subtraction), but not the same factor, since the height of B is half of A's height while the width of B is not half of A's width. That's why B looks distorted: its width and height did not change proportionally.
MTMath Talk — Check by what factors the width and height of image D change compared to image A. Are the factors the same?
Width factor: \(\dfrac{90}{60}=1.5\). Height factor: \(\dfrac{60}{40}=1.5\).
Both dimensions of D changed by the exact same factor (1.5, or \(\tfrac{3}{2}\)) compared to A, so images A and D are proportional — which is why D looks similar to A, just larger.
7.2 & 7.3 Ratios and Their Simplest Form
The ratio of width to height of image A is \(60:40\); the numbers 60 and 40 are the terms of the ratio. For image C it's \(30:20\), and for D it's \(90:60\). In a ratio \(a:b\), for every '\(a\)' units of the first quantity there are '\(b\)' units of the second.
MTMath Talk — By what factor should we multiply the ratio 60:40 (image A) to get 90:60 (image D)?
Dividing corresponding terms: \(\dfrac{90}{60}=\dfrac{3}{2}\) and \(\dfrac{60}{40}=\dfrac{3}{2}\) — both terms scale by the same factor, \(\tfrac{3}{2}\) (i.e. 1.5).
A more systematic check is to reduce each ratio to its simplest form by dividing both terms by their HCF, and see if the simplest forms match.
| Image | Ratio | HCF | Simplest form |
|---|---|---|---|
| A | 60:40 | 20 | 3:2 |
| D | 90:60 | 30 | 3:2 |
| B | 40:20 | 20 | 2:1 |
| E | 60:60 | 60 | 1:1 |
A and D reduce to the same simplest form (3:2), so they are proportional. B (2:1) and E (1:1) don't match 3:2, so they are not proportional to A, C, or D — confirming why they look different.
7.4 Problem Solving with Proportional Reasoning
Example 1 — Are 3:4 and 72:96 proportional? 3:4 is already in simplest form. The HCF of 72 and 96 is 24; dividing both terms by 24 gives 3:4 as well. Since both simplest forms match, the ratios are proportional.
Example 2 — Kesang's lemonade. Kesang mixed 6 glasses of lemonade with 10 spoons of sugar (ratio 6:10). To make 18 more glasses with the same sweetness, we need \(6:10::18:?\). The first term's factor of change is \(18\div6=3\); multiplying the second term by the same factor: \(10\times3=30\). So \(6:10::18:30\) — she should use 30 spoons of sugar.
Example 3 — Nitin and Hari's wall. Nitin built a 60 ft wall with 3 bags of cement (ratio 60:3, simplest form 20:1); Hari built a 40 ft wall with 2 bags (ratio 40:2, simplest form 20:1). Since both simplest forms match, the two ratios are proportional — the walls are equally strong, and Nitin needn't worry.
Examples 4 & 5 (personal data) ask you to count the teacher-to-student ratio in your own school, and measure your own classroom blackboard's width-to-height ratio, then compare these to given references or draw a proportional rectangle. These depend on your own school/classroom measurements, so there's no single fixed numeric answer — the method is simply to write your counted/measured values as a ratio, reduce to simplest form, and check whether it matches the reference ratio (or draw a same-ratio rectangle at a different size to test proportionality).
MTMath Talk — Compare the rectangle you drew (proportional to your blackboard) with your classmates' rectangles. Do they all look the same?
Not necessarily the same absolute size — different students may draw their rectangles at different scales. But if every rectangle keeps the same width-to-height ratio as the blackboard, all the rectangles will look similar (same shape), even though their sizes differ, exactly like images A, C, and D from Section 7.1.
Example 6 — Neelima's age. When Neelima was 3, her mother was 10 times as old (30), giving a ratio of \(3:30\), simplest form \(1:10\). Nine years later, Neelima is 12 and her mother is 39, giving \(12:39\), simplest form \(4:13\). Since \(1:10\ne4:13\), the ratio changes over time — adding the same number (9) to both ages does not preserve the ratio, unlike multiplying both terms by the same factor.
Example 7 — Fill in ratios proportional to 14:21. (i) \(\_\_:42\): since \(42=2\times21\), the missing term is \(2\times14=28\), giving \(28:42\). (ii) \(6:\_\_\): the factor is \(6\div14=\tfrac{3}{7}\) (a fraction, not an integer!), so the second term is \(21\times\tfrac{3}{7}=9\), giving \(6:9\). (iii) \(2:\_\_\): dividing 14 by its HCF with 21 (which is 7) gives 2, so dividing 21 by 7 too gives 3, giving \(2:3\).
Filter Coffee! Manjunath's regular filter coffee mixes 15 mL decoction with 35 mL milk (ratio \(15:35\), simplest form \(3:7\)). For 'stronger' coffee he uses \(20:30\) (simplest form \(2:3\)); for 'lighter' coffee, \(10:40\) (simplest form \(1:4\)).
MTMath Talk — Why is the 20:30 coffee stronger, and the 10:40 coffee lighter, than the regular 15:35?
Comparing the decoction-to-milk ratio value in each: regular is \(\tfrac{15}{35}\approx0.43\), the stronger mix is \(\tfrac{20}{30}\approx0.67\) — a higher proportion of decoction relative to milk, making it stronger. The lighter mix is \(\tfrac{10}{40}=0.25\) — a lower proportion of decoction, making it weaker/lighter.
Classifying five more mixtures against the regular ratio (\(\approx0.43\)):
| Decoction (mL) | Milk (mL) | Simplest form | Decoction/Milk | Classification |
|---|---|---|---|---|
| 300 | 600 | 1:2 | 0.5 | Stronger |
| 150 | 500 | 3:10 | 0.3 | Lighter |
| 200 | 400 | 1:2 | 0.5 | Stronger |
| 24 | 56 | 3:7 | ≈0.43 | Regular (exact match) |
| 100 | 300 | 1:3 | ≈0.33 | Lighter |
Figure it Out — Set 1
Seven questions from pages 165–166 of the textbook.
1Circle the true statements of proportion: (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
Checking each by cross multiplication (\(ad\) vs. \(bc\)):
(i) \(4\times21=84\), \(7\times12=84\) → equal → True
(ii) \(8\times6=48\), \(3\times24=72\) → unequal → False
(iii) \(7\times7=49\), \(12\times12=144\) → unequal → False
(iv) \(21\times10=210\), \(6\times35=210\) → equal → True
(v) \(12\times12=144\), \(18\times28=504\) → unequal → False
(vi) \(24\times3=72\), \(8\times9=72\) → equal → True
2Give 3 ratios that are proportional to 4:9.
Multiplying both terms by 2, 3, and 4 respectively: \(8:18\), \(12:27\), \(16:36\).
3Fill in the missing numbers for these ratios that are proportional to 18:24.
18:24 simplifies to 3:4. Scaling each given first term to match:
\(3:\_\_\): factor \(=3\div18=\tfrac16\), so second term \(=24\times\tfrac16=4\) → 3:4
\(12:\_\_\): factor \(=12\div18=\tfrac23\), so second term \(=24\times\tfrac23=16\) → 12:16
\(20:\_\_\): factor \(=20\div18=\tfrac{10}{9}\), so second term \(=24\times\tfrac{10}{9}=\tfrac{80}{3}\) → 20: 80/3 (not a whole number)
\(27:\_\_\): factor \(=27\div18=\tfrac32\), so second term \(=24\times\tfrac32=36\) → 27:36
4Which of the rectangles (A–E, shown at various sizes and rotations) are similar to each other?
Rectangles are similar exactly when the ratio of their width to height (in simplest form) is the same. Measure each rectangle's two sides with a ruler, write each as a ratio, reduce to simplest form, and group together the rectangles whose simplest forms match — those are the similar ones.
5Draw a smaller and a bigger rectangle with the same width-to-height ratio as a given rectangle. Compare with classmates — are they all the same? If different, are they wrong?
They won't all be the same absolute size, but they don't need to be — as long as everyone's rectangle keeps the same ratio of width to height as the original, every rectangle drawn is correct (proportional), regardless of its size.
6Find the ratio of grey bricks to coloured bricks (in simplest form) for wall patterns (a) and (b).
(a) In one repeating block of the pattern: grey bricks \(=2+3+4=9\); coloured bricks \(=3+2+1=6\). Ratio \(=9:6=3:2\).
(b) In one repeating block: grey bricks \(=16\); coloured bricks \(=12\). Ratio \(=16:12=4:3\).
7Measure a friend's head, torso, arms, and legs; find head:torso, torso:arms, and torso:legs. Draw a new figure with equivalent ratios.
This depends on the actual measurements taken, so there's no single fixed answer — the method is to measure each body part's length, write the three ratios, reduce them to simplest form, and then draw a new figure whose corresponding parts are in exactly the same simplest-form ratios (just scaled up or down).
MTMath Talk — Does the drawing look more realistic if the ratios are proportional? Why or why not?
Yes. Real human bodies have fairly consistent head:torso:limb proportions, so a figure that keeps those same ratios — just scaled to a different overall size — looks like a natural, correctly-proportioned person. A figure where the ratios are changed (e.g. torso stretched but arms not) looks distorted, exactly like the elongated tiger in image B from Section 7.1.
Trairasika — The Rule of Three
Example 8 — Mid-day meal rice. A school normally makes 15 kg of rice for 120 students. On a day with only 80 students, the ratio of students to rice must stay proportional: \(120:15::80:?\). The factor of change in the first term is \(\dfrac{80}{120}=\dfrac{2}{3}\); applying the same factor to rice: \(15\times\dfrac23=10\). The cook should make 10 kg of rice.
In general, for \(a:b::c:d\), term \(c\) must be a multiple of \(a\) by some factor \(f\), and \(d\) must be the same multiple of \(b\): \(c=fa\) and \(d=fb\). So \(f=\dfrac{c}{a}=\dfrac{d}{b}\), which rearranges (multiplying both sides by \(ab\)) to give cross multiplication:
Ancient Indian mathematicians, including Āryabhaṭa (499 CE), called this a Trairasika (Rule of Three) problem. Three quantities were named: pramāṇa (measure, \(a\)), phala (fruit, \(b\)), and ichchhā (requisition, \(c\)); the unknown ichchhāphala (yield, \(d\)) was found by "multiply the phala by the ichchhā and divide by the pramāṇa" — exactly \(d=\dfrac{bc}{a}\).
Example 9 — Car travel distance. A car travels 90 km in 150 minutes; how far in 4 hours? Since 4 hours \(=240\) minutes (matching units is essential — modelling it as \(150:90::4:?\) directly would be wrong, since 4 is in hours but 150 is in minutes), the correct proportion is \(150:90::240:x\). By cross multiplication: \(150x=240\times90 \Rightarrow x=\dfrac{240\times90}{150}=144\). The car covers 144 km in 4 hours.
Example 10 — Which tea is more expensive? A Himachal farmer sells 200 g of tea for ₹200 (ratio 200:200, simplest form 1:1). A Meghalaya estate sells 1 kg (1000 g) for ₹800 (ratio 1000:800, simplest form 5:4). Since 1:1 ≠ 5:4, the weight-to-price ratios are not proportional. To compare prices fairly, convert both to price-per-kg: for Himachal, let the price of 1 kg be \(x\) rupees; since 200 g is \(\tfrac15\) of a kg, \(\tfrac15 x=200 \Rightarrow x=1000\). So Himachal tea costs ₹1,000/kg versus Meghalaya's ₹800/kg — the Himachal tea is more expensive.
Activity 1 asks you to scale up a favourite recipe's ingredient quantities proportionally to serve 15 guests instead of your usual family size — an open, personal exercise: multiply every ingredient amount by the same factor (guests needed ÷ usual servings).
Figure it Out — Set 2
Two questions from page 170 of the textbook.
1Earth travels approximately 940 million km around the Sun in a year. How many km will it travel in a week?
Using 52 weeks in a year: \(940{,}000{,}000:52::x:1\), so \(x=\dfrac{940{,}000{,}000}{52}\approx18{,}076{,}923\).
2A mason needs 1,450 bricks for a 10-ft wall (outer walls + inner dividing wall, all same height/thickness, dimensions as in the diagram: 12, 15, 9, 9, 9, 6 ft segments). How many bricks for the whole house?
Total wall length \(=12+12+12+15+9+15+9+9+9+6=108\) ft.
Using \(10:1450::108:x\): \(x=\dfrac{1450\times108}{10}=15{,}660\).
Not Everything Is Directly Proportional
MTMath Talk — Puneeth's father rides 50 km/h and takes 2 hours to reach Kanpur. If he rides at 75 km/h instead, can we model the new time as \(50:2::75:?\)
No — this cannot be solved using the Rule of Three. Riding faster decreases travel time, not increases it — speed and time move in opposite directions for a fixed distance, so they are inversely related, not directly proportional. The Rule of Three only applies when both quantities increase (or decrease) together by the same factor; here one goes up while the other must go down.
Activity 2 — Shampoo pricing. Comparing volume and price for different container sizes of the same shampoo (sachet 6 mL for ₹2; small bottle 180 mL for ₹154; medium 340 mL for ₹276; large 1000 mL for ₹540):
| Container | Volume (mL) | Price | Price per mL |
|---|---|---|---|
| Sachet | 6 | ₹2 | ≈₹0.33 |
| Small Bottle | 180 | ₹154 | ≈₹0.86 |
| Medium Bottle | 340 | ₹276 | ≈₹0.81 |
| Large Bottle | 1000 | ₹540 | ₹0.54 |
MTMath Talk — Why isn't the ratio of prices proportional to the ratio of volumes? Discuss the pros and cons of different bottle sizes for the company, customers, and the environment.
The price per mL is not constant across sizes — the tiny sachet actually costs less per mL than the small or medium bottles (likely a promotional or trial-size price), while the large bottle offers the best per-mL value of the regular sizes. Since price-per-unit isn't fixed, the volume:price ratios (e.g. sachet's \(6:2=3:1\) vs. large's \(1000:540\approx1.85:1\)) don't match — they are not proportional.
Larger containers usually reduce packaging waste per mL of product (better for the environment) and give customers a lower per-unit cost, but require more money upfront, which can be a barrier for some customers; smaller sachets let customers try a product or manage tight budgets, at a higher cost per mL and more packaging waste overall. A reasonable recommendation is for companies to price larger sizes proportionally cheaper per unit (to reward bulk buying and reduce waste), while still offering low-cost trial sizes for accessibility.
7.5 Sharing, but Not Equally!
Activity 3. Sharing 12 counters equally gives 6 each — ratio \(6:6\), simplest form \(1:1\). If a partner takes 5 counters (and you take the remaining 7), the ratio of partner's counters to yours is \(5:7\).
To share 12 counters in the ratio \(3:1\): give 3 to one side and 1 to the other, repeatedly, until all are used. Doing this 3 times uses all 12 counters, giving 9 and 3. In general, dividing 12 in the ratio \(3:1\) means splitting into \(3+1=4\) equal groups of size \(12\div4=3\); one side gets \(3\times3=9\), the other gets \(1\times3=3\).
Sharing 42 in the ratio \(4:3\): total groups \(=4+3=7\); group size \(=42\div7=6\); parts are \(4\times6=24\) and \(3\times6=18\).
Example 11 — Business profit sharing. Prashanti invested ₹75,000 and Bhuvan invested ₹25,000 (ratio 75000:25000, simplest form 3:1). Sharing a profit of ₹4,000 in this ratio: \(3+1=4\) groups, group size \(=4000\div4=1000\). Prashanti's share \(=3\times1000=\)₹3,000; Bhuvan's share \(=1\times1000=\)₹1,000.
Example 12 — Sand and cement mixture. A 40 kg mixture has sand:cement \(=3:1\), so sand \(=\dfrac{3}{4}\times40=30\) kg and cement \(=\dfrac14\times40=10\) kg. To change the ratio to \(5:2\) while keeping the sand fixed at 30 kg: \(5:2::30:?\); since the second term is \(\tfrac25\) of the first, new cement \(=\dfrac25\times30=12\) kg. Since 10 kg of cement is already present, 2 kg of cement must be added.
Figure it Out — Set 3
Five questions from page 175 of the textbook.
1Divide ₹4,500 into two parts in the ratio 2:3.
Group size \(=\dfrac{4500}{2+3}=900\). Parts: \(2\times900=1800\) and \(3\times900=2700\).
2Acid and water are mixed in the ratio 1:5. In 240 mL of solution, how much acid and water are there?
Group size \(=\dfrac{240}{1+5}=40\). Acid \(=1\times40=40\) mL; water \(=5\times40=200\) mL.
3Blue and yellow are mixed 3:5 to make 40 mL of green. How much of each is needed? If 20 mL more yellow is added, what's the new blue:yellow ratio?
Group size \(=\dfrac{40}{3+5}=5\). Blue \(=3\times5=15\) mL; yellow \(=5\times5=25\) mL.
Adding 20 mL more yellow: new yellow \(=25+20=45\) mL. New ratio blue:yellow \(=15:45=1:3\).
4Rice and urad dal are mixed 2:1 for soft idlis. For 6 cups of mixture, how much of each is needed?
Group size \(=\dfrac{6}{2+1}=2\). Rice \(=2\times2=4\) cups; urad dal \(=1\times2=2\) cups.
5A bucket of orange paint is made from red and yellow in the ratio 3:5. One more full bucket of yellow is added. What's the new red:yellow ratio?
In the original bucket, red \(=\dfrac38\) of a bucket and yellow \(=\dfrac58\) of a bucket. Adding one whole bucket of yellow: new yellow \(=\dfrac58+1=\dfrac{13}{8}\) of a bucket.
New ratio red:yellow \(=\dfrac38:\dfrac{13}{8}=3:13\).
7.6 Unit Conversions
Solving proportionality problems often needs quantities in matching units first. Key conversions: Length — 1 m = 3.281 ft. Area — 1 sq. m = 10.764 sq. ft; 1 acre = 43,560 sq. ft; 1 hectare = 10,000 sq. m = 2.471 acres. Volume — 1 mL = 1 cc; 1 L = 1,000 mL. Temperature — °F = (9/5)×°C + 32 and °C = (5/9)×(°F − 32); e.g. 25°C = 77°F.
Figure it Out — Set 4
Twelve questions from pages 176–177 of the textbook.
1Anagh mixes 600 mL orange juice with 900 mL apple juice. Ratio of orange to apple juice in simplest form?
HCF of 600 and 900 is 300. \(600:900 = 2:3\).
2Last year, 3 full buses carried 162 students and teachers. This year there are 204 students. How many buses are needed, and will they all be full?
Capacity per bus \(=162\div3=54\). Using \(3:162::x:204\): \(x=\dfrac{3\times204}{162}=\dfrac{34}{9}\approx3.78\), so 4 buses are needed (round up, since a fraction of a bus isn't possible).
Total capacity in 4 buses \(=4\times54=216\); with only 204 students, \(216-204=12\) seats will be empty.
3Delhi: area 1,484 sq. km, population ≈30 million. Mumbai: area 550 sq. km, population ≈20 million. Which city is more crowded?
Population density (people per sq. km): Delhi \(=\dfrac{30{,}000{,}000}{1484}\approx20{,}216\). Mumbai \(=\dfrac{20{,}000{,}000}{550}\approx36{,}364\).
4A 155 cm crane has neck:rest-of-body in the ratio 4:6. If a person's height followed this same ratio, how tall would their neck be?
Neck fraction of total height \(=\dfrac{4}{4+6}=\dfrac{4}{10}=0.4\), i.e. the neck would be 40% of a person's total height.
For example, for someone 150 cm tall, the neck would be \(0.4\times150=60\) cm — for your own height, multiply it by 0.4 to get the equivalent neck length under this ratio.
5Lilavati problem: If \(2\tfrac12\) palas of saffron costs \(\tfrac37\) niskas, what quantity of saffron can be bought for 9 niskas?
Setting up \(2.5:\tfrac37::x:9\), cross multiplying: \(2.5\times9=x\times\tfrac37\), so \(x=\dfrac{2.5\times9\times7}{3}=\dfrac{157.5}{3}=52.5\).
6Harmain is 1 year old; her brother is 5. In how many years will the ratio of her age to her brother's age be 1:2?
Let \(x\) be the number of years from now: \(\dfrac{1+x}{5+x}=\dfrac12\). Cross multiplying: \(2(1+x)=5+x \Rightarrow 2+2x=5+x \Rightarrow x=3\).
After 3 years, Harmain will be \(1+3=4\) and her brother \(5+3=8\) — ratio \(4:8=1:2\). ✓
7Equal volumes of gold and water have masses in ratio 37:2. If 1 litre of water is 1 kg, what is the mass of 1 litre of gold?
\(37:2::x:1\), so \(x=\dfrac{37\times1}{2}=18.5\).
810 tonnes of manure is needed per acre. How much manure for a 200 ft × 500 ft tomato plot?
Plot area \(=200\times500=100{,}000\) sq. ft. Since 1 acre \(=43{,}560\) sq. ft: \(43{,}560:10::100{,}000:x\), so \(x=\dfrac{10\times100{,}000}{43{,}560}\approx22.9568\) tonnes \(\approx22{,}956.8\) kg.
9A tap fills a 500 mL mug in 15 seconds. How long to fill a 10-litre bucket?
10 litres \(=10{,}000\) mL. \(500:15::10{,}000:x\), so \(x=\dfrac{15\times10{,}000}{500}=300\) seconds \(=5\) minutes.
10One 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, costing ₹15,00,000. \(43{,}560:1{,}500{,}000::2{,}400:x\), so \(x=\dfrac{1{,}500{,}000\times2{,}400}{43{,}560}\approx82{,}645\).
11A tractor ploughs 4× faster than a pair of oxen, which take 6 hours per acre. How long for oxen, and for a tractor, to plough a 20-acre field?
Oxen: \(20\times6=120\) hours. Tractor (4× faster, so \(\tfrac14\) the time per acre \(=1.5\) hours/acre): \(20\times1.5=30\) hours.
12Try This — A ₹10 coin (cupro-nickel, copper:nickel = 3:1, mass 7.74 g) uses copper at ₹906/kg and nickel at ₹1,341/kg. What is the cost of the metals in one coin?
Copper mass \(=\dfrac34\times7.74=5.805\) g; nickel mass \(=\dfrac14\times7.74=1.935\) g.
Cost of copper: \(5.805\text{ g}\times\dfrac{906}{1000}\)/g \(\approx\)₹5.26. Cost of nickel: \(1.935\text{ g}\times\dfrac{1341}{1000}\)/g \(\approx\)₹2.59.
Puzzle Time — Binairo
Binairo (Takuzu)
Binairo is a logic puzzle played on a square grid, where cells are filled with one of two symbols (here, horizontal and vertical lines). The rules:
- Each row and each column must contain an equal number of horizontal and vertical lines.
- More than two horizontal or vertical lines in a row cannot be adjacent to each other.
- Every row must be unique, and every column must be unique.
The textbook provides a worked example (a 6×6 grid with a full solution shown) followed by three unsolved puzzle grids to complete on your own. Since these are printed grids with specific starting symbols in specific cells, solving them means working directly from the printed page — apply the three rules above one cell at a time, starting from rows/columns that already have the most symbols filled in, since those constrain their neighbours the most.
Frequently Asked Questions
Continue with Part 2, Chapter 1
Move on to Fractions in Disguise next.