How do you check if two ratios are proportional?

Use cross-multiplication: for ratios a to b and c to d, check if a times d equals b times c. If the two cross-products are equal, the ratios are proportional.

What does a map's Representative Fraction mean?

A representative fraction like 1 to 60 lakh means 1 unit of distance on the map equals 60 lakh of the same units on the ground. Measure the map distance with a ruler, then multiply by this scale factor to get the real distance.

How do you find the angle for a slice of a pie chart?

Divide the category value by the total, then multiply by 360 degrees. All the angles in a pie chart must add up to exactly 360 degrees.

What is the difference between direct and inverse proportion?

In a direct proportion, two quantities change by the same factor, so their ratio stays constant. In an inverse proportion, when one quantity changes by a factor, the other changes by the reciprocal factor, so their product stays constant.

How do you divide a quantity in a given ratio with more than two terms?

Add up all the terms of the ratio to get the total number of parts. Each share of the quantity is the whole multiplied by that term's value divided by the sum of all terms.

Class 8 Maths NCERT Solutions Part 2 Chapter 3: Proportional Reasoning-2 (Ganita Prakash, Part 2)

Key Concepts & Formulae at a Glance

Two ratios \(a:b\) and \(c:d\) are proportional if \(a\times d=b\times c\), equivalently \(\dfrac{a}{c}=\dfrac{b}{d}\) (cross-multiplication).
A Representative Fraction (RF) on a map, like 1 : 60,00,000, means 1 unit on the map equals that many of the same units on the ground.
Ratios can have more than two terms, e.g. \(a:b:c:d\); two such ratios are proportional when \(\dfrac{a}{p}=\dfrac{b}{q}=\dfrac{c}{r}=\dfrac{d}{s}\).
To divide a quantity \(x\) in the ratio \(p:q:r:\ldots\), each part is \(x\times\dfrac{p}{(p+q+r+\ldots)}\), \(x\times\dfrac{q}{(p+q+r+\ldots)}\), and so on.
In a pie chart, each category's angle is proportional to its share of the whole: \(\text{angle}=\dfrac{\text{category value}}{\text{total}}\times360^\circ\).
Direct proportion: \(x\) and \(y\) change by the same factor, so \(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=k\) (constant ratio).
Inverse proportion: when one quantity changes by a factor \(n\), the other changes by \(\dfrac1n\); so \(x_1y_1=x_2y_2=k\) (constant product).

\[a\times d=b\times c \qquad \text{angle}=\frac{\text{part}}{\text{total}}\times360^\circ \qquad \underbrace{\frac{x_1}{y_1}=\frac{x_2}{y_2}}_{\text{direct}} \qquad \underbrace{x_1y_1=x_2y_2}_{\text{inverse}}\]

3.1 Proportionality — A Quick Recap

When two or more related quantities change by the same factor, we call that relationship a proportional relationship. For example, idli batter mixes rice and urad dal; one common proportion is 2 cups of rice to 1 cup of urad dal, written as the ratio 2 : 1.

MTMath Talk — Viswanath mixes 6 cups rice with 3 cups urad dal; Puneet mixes 4 cups rice with 2 cups urad dal. If cooked the same way, would their idlis taste the same?

Viswanath's mixture: 6 : 3. Puneet's mixture: 4 : 2. To check if these are proportional, cross-multiply:

\(6\times2=12\) and \(3\times4=12\) — the two cross-products are equal, so the ratios are proportional.

Yes, it's likely the idlis would taste the same (assuming all other ingredients are also kept proportional), since 6:3 and 4:2 are proportional ratios — both simplify to 2:1.

In general, two ratios \(a:b\) and \(c:d\) are proportional if \(a\times d=b\times c\), or equivalently \(\dfrac{a}{c}=\dfrac{b}{d}\).

3.2 Ratios in Maps

Bengaluru–Chennai (dashed purple) and Mangaluru–Chennai (dashed gold) lines shown for the exercise below. City positions are plotted proportionally to their real latitude/longitude, so ruler measurements on this figure scale consistently with real-world distances.

A Representative Fraction (RF) shows the ratio between a distance on the map and the corresponding actual distance on the ground. If the RF is 1 : 60,00,000, a distance of 1 cm on the map equals a geographical distance of 60,00,000 cm on the ground.

MTMath Talk — Convert 60,00,000 cm to kilometres. Verify it is 60 km.

\(60{,}00{,}000\text{ cm} \div 100 = 60{,}000\text{ m}\) (cm to m). Then \(60{,}000\text{ m}\div1000=60\text{ km}\) (m to km).

Yes — 60,00,000 cm = 60 km, confirming 1 cm on the map represents 60 km on the ground.

MTMath Talk — Using the map, find the geographical distance between Bengaluru and Chennai, and between Mangaluru and Chennai. [Hint: measure with a ruler, then use the map's RF.]

Measuring the Bengaluru–Chennai line on the map gives approximately 4.8 cm. Using the RF (1 cm = 60 km): \(4.8\times60\approx288\) km.

Measuring the Mangaluru–Chennai line gives approximately 9.8 cm. Using the same scale: \(9.8\times60\approx588\) km.

Your own ruler measurement on the printed textbook map may differ slightly from these figures depending on the exact print size and measuring precision — small variations of a few km are normal and expected. The real geographical (straight-line) distances are approximately 290 km (Bengaluru–Chennai) and 587 km (Mangaluru–Chennai), which is what a careful ruler measurement should approximate.

Bengaluru–Chennai ≈ 288–290 km. Mangaluru–Chennai ≈ 587–588 km (roughly double the Bengaluru–Chennai distance).

MTMath Talk — Try finding the distance between the same two cities using different maps with different scales. Do they give approximately the same geographical distance?

Yes — as long as each map is drawn accurately to its own stated scale, the ruler distance will differ from map to map (a map with a smaller RF number, like 1:10,00,000, will show cities farther apart on paper than one with 1:60,00,000), but multiplying by each map's own RF should bring you back to approximately the same real-world distance every time. Small differences can arise from measurement precision, map projection distortion, or rounding.

Yes, approximately — different scaled maps should agree on the real geographical distance once each map's own RF is correctly applied, though small measurement/projection differences are normal.

Map Making Activity: Sketch your classroom at a scale of 1 : 50 (so 1 cm on your sketch = 50 cm in the real room), marking the teacher's desk, blackboard, fans, lights, and other furniture to scale using appropriate symbols.

3.3 Ratios with More than 2 Terms

Viswanath's spice mix powder uses 8 spoons coriander seeds, 4 red chillies, 2 spoons toor dal, and 1 spoon fenugreek seeds — a ratio of 8 : 4 : 2 : 1 (four terms). Ratios can have many terms, as long as each quantity changes by the same factor to keep the proportional relationship.

MTMath Talk — Puneet has only 2 red chillies. How much of the other ingredients should he use to match Viswanath's spice mix?

Puneet's 2 chillies are half of Viswanath's 4, so every other ingredient must also be halved: coriander \(8\div2=4\) spoons, toor dal \(2\div2=1\) spoon, fenugreek \(1\div2=0.5\) spoon.

Puneet needs 4 spoons coriander, 2 red chillies, 1 spoon toor dal, and 0.5 spoon fenugreek — ratio 4 : 2 : 1 : 0.5, proportional to 8 : 4 : 2 : 1.

In general, when two multi-term ratios are proportional, \(a:b:c:d::p:q:r:s\), then \(\dfrac{a}{p}=\dfrac{b}{q}=\dfrac{c}{r}=\dfrac{d}{s}\).

Example 1 — Purple paint. Mixing Red : Blue : White in ratio 2 : 3 : 5. Yasmin has 10 L of white paint (5 parts). Since 5 parts = 10 L, 1 part = 2 L. Red \(=2\times2=4\) L, Blue \(=3\times2=6\) L. Total volume \(=4+6+10=20\) L.

Example 2 — Concrete mix. Cement : Sand : Gravel = 1 : 1.5 : 3. With 3 bags of cement, multiply every term by 3: Sand \(=1.5\times3=4.5\), Gravel \(=3\times3=9\). Total \(=3+4.5+9=16.5\) bags of concrete.

3.4 Dividing a Whole in a Given Ratio

Recall dividing a whole in a 2-term ratio, e.g. 12 in the ratio 2 : 1 — add the terms (\(2+1=3\)), divide the whole by this sum (\(12\div3=4\)), then multiply each term by this quotient: \(2\times4=8\) and \(1\times4=4\). So 12 divided in ratio 2:1 is 8:4.

This extends naturally to multi-term ratios: when dividing \(x\) in the ratio \(a:b:c:\ldots\), the terms are \(x\times\dfrac{a}{(a+b+c+\ldots)}\), \(x\times\dfrac{b}{(a+b+c+\ldots)}\), \(x\times\dfrac{c}{(a+b+c+\ldots)}\), and so on.

Example 3 — Concrete, 110 units. Ratio 1 : 1.5 : 3 sums to 5.5 "parts." Since \(110\div5.5=20\), multiply every term by 20: Cement \(=1\times20=20\), Sand \(=1.5\times20=30\), Gravel \(=3\times20=60\) units.

Example 4 — Purple paint, 50 mL. Ratio 2 : 3 : 5 (sum 10). Red \(=50\times\frac{2}{10}=10\) mL, Blue \(=50\times\frac{3}{10}=15\) mL, White \(=50\times\frac{5}{10}=25\) mL.

MTMath Talk — Example 5: Construct a triangle with angles in the ratio 1 : 3 : 5.

The angle sum of a triangle is always 180°, and the ratio sums to \(1+3+5=9\):

\(\angle A=180^\circ\times\frac19=20^\circ\), \(\angle B=180^\circ\times\frac39=60^\circ\), \(\angle C=180^\circ\times\frac59=100^\circ\).

∠A = 20°, ∠B = 60°, ∠C = 100° (checking: 20+60+100 = 180° ✓).

Figure it Out — Set 1

Five questions from page 60 of the textbook.

1A cricket coach schedules practice in the ratio warm-up/cool-down : batting : bowling : fielding :: 3 : 4 : 3 : 5. If each session is 150 minutes, how much time is spent on each activity?

Sum of parts \(=3+4+3+5=15\). Each part \(=150\div15=10\) minutes.

Warm-up/cool-down \(=3\times10=30\) min. Batting \(=4\times10=40\) min. Bowling \(=3\times10=30\) min. Fielding \(=5\times10=50\) min.

Warm-up/cool-down: 30 min. Batting: 40 min. Bowling: 30 min. Fielding: 50 min. (Total: 150 min ✓)

2A school library has books in ratio Odiya : Hindi : English :: 3 : 2 : 1. If there are 288 Odiya books, how many Hindi and English books are there?

Odiya corresponds to 3 parts \(=288\), so 1 part \(=288\div3=96\).

Hindi \(=2\times96=192\) books. English \(=1\times96=96\) books.

Hindi: 192 books. English: 96 books.

3I have 100 coins in the ratio ₹10 : ₹5 : ₹2 : ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in total?

Sum of parts \(=4+3+2+1=10\). Each part \(=100\div10=10\) coins.

₹10 coins: \(4\times10=40\) coins → ₹400. ₹5 coins: \(3\times10=30\) coins → ₹150. ₹2 coins: \(2\times10=20\) coins → ₹40. ₹1 coins: \(1\times10=10\) coins → ₹10.

Total \(=400+150+40+10=\)₹600.

Total money = ₹600 (from 40+30+20+10 = 100 coins ✓).

4Math Talk — Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all such triangles be congruent? Why or why not?

Any triangle with sides in this ratio, at any scale (3k, 4k, 5k for a chosen unit length k), automatically has a right angle between the sides of length 3k and 4k, since \((3k)^2+(4k)^2=(5k)^2\) — this is the well-known 3-4-5 Baudhāyana/Pythagorean triple.

No, not all such triangles are congruent — they are all similar (same shape, same angles) but can be scaled to any size \(k\) (e.g. sides 3,4,5 cm, or 6,8,10 cm, or 1.5,2,2.5 cm are all valid "3:4:5" triangles, but none of these are congruent to each other since congruent triangles must have exactly equal — not just proportional — sides).

No — triangles with sides in a given ratio are all similar to each other (same angles, same shape), but only congruent if they're also the same actual size.

5Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?

For any three lengths to form a triangle, the triangle inequality must hold: the sum of any two sides must exceed the third side. For sides \(k, 3k, 5k\) (any scale \(k\)): the two shorter sides sum to \(k+3k=4k\), which is less than the third side \(5k\).

Even laid end-to-end in a straight line, k + 3k = 4k never reaches the length of the third side (5k) — so the two shorter sides can never tilt upward far enough to meet and close the triangle.

No — impossible for any scale, since 1+3=4 is less than 5, violating the triangle inequality (the two shorter sides can never "reach across" to meet, however the triangle is scaled).

3.5 A Slice of the Pie

Pie charts show proportions of a whole. Consider grades scored by 40 students:

Grade	A	B	C	D	E
Students	12	10	8	6	4

MTMath Talk — How do we mark the different slices? Can the ratio 12:10:8:6:4 be simplified first?

Each grade's angle should be proportional to its student count, out of the full 360° circle. Dividing 360 in the ratio 12:10:8:6:4 works directly, but it's easier to simplify first: the HCF of 12,10,8,6,4 is 2, giving the simplified ratio 6:5:4:3:2 (sum = 20).

Grade A \(=\frac{6}{20}\times360^\circ=6\times18^\circ=108^\circ\). Grade B \(=5\times18^\circ=90^\circ\). Grade C \(=4\times18^\circ=72^\circ\). Grade D \(=3\times18^\circ=54^\circ\). Grade E \(=2\times18^\circ=36^\circ\).

Grade A: 108°, Grade B: 90°, Grade C: 72°, Grade D: 54°, Grade E: 36° (sum = 360° ✓).

Constructing the pie chart: draw a circle with radius AB, then successively measure each angle anticlockwise from the previous radius: 108° for slice A (giving radius AC), then 90° more for slice B (radius AD), then 72° more for slice C (radius AE), then 54° for slice D and 36° for slice E to complete the circle.

All five radii drawn: A (108°, region B-C), B (90°, C-D), C (72°, D-E), D (54°, E-F), E (36°, F-back to B) — together completing the full 360°.

The finished pie chart, coloured and labelled by grade.

Figure it Out — Set 2

Three questions from pages 61–63 of the textbook.

1360 people voted for their favourite season: 90 liked summer, 120 liked rainy, and the rest liked winter. Draw a pie chart.

Winter \(=360-90-120=150\) people.

Since the total is exactly 360, each person's angle contribution is simply \(1^\circ\) — angles equal the vote counts directly: Summer \(=90^\circ\), Rainy \(=120^\circ\), Winter \(=150^\circ\).

Summer: 90°, Rainy: 120°, Winter: 150° (sum = 360° ✓).

2Draw a pie chart for favourite TV channel type: Entertainment 50%, Sports 25%, News 15%, Information 10%.

Converting each percentage to degrees (\(\%\times3.6\), since \(360\div100=3.6\)):

Entertainment \(=50\times3.6=180^\circ\). Sports \(=25\times3.6=90^\circ\). News \(=15\times3.6=54^\circ\). Information \(=10\times3.6=36^\circ\).

Entertainment: 180°, Sports: 90°, News: 54°, Information: 36° (sum = 360° ✓).

3Try This — Prepare a pie chart of your own class's favourite subjects (Language, Arts Education, Vocational Education, Social Science, Physical Education, Maths, Science).

This is an open-ended data-collection activity — survey your own classmates (each choosing exactly one favourite subject), record the counts in the table, then use the same method as above: find the total number of students, compute each subject's angle as \(\dfrac{\text{count}}{\text{total}}\times360^\circ\), and construct the pie chart using a protractor.

Method: angle for each subject = (number of students who chose it ÷ total students) × 360°, then draw each slice with a protractor as shown in the worked example above.

3.6 Inverse Proportions

Recall the rule of three: when \(a:b::c:d\), then \(d=\dfrac{bc}{a}\). We call such proportions direct proportions, used to find a fourth quantity given the other three.

Example 1. If 5 workers move 4500 bricks in a day, how many workers move 18000 bricks in a day? Setting up \(4500:18000::5:x\): \(x=\dfrac{18000\times5}{4500}=20\) workers.

MTMath Talk — Example 2: Puneeth's father rides a motorcycle at 30 km/h for 3 hours from Lucknow to Kanpur. At 60 km/h by car, how long will it take? Can this be represented as 30:60::3:x? Will travel time increase or decrease as speed increases?

No — this is not a direct proportion of the form \(a:b::c:d\), because speed and time move in opposite directions: as speed increases, travel time decreases (not increases in the same ratio, as a direct proportion would require).

	Walk	Bicycle	Motorcycle	Car
Speed (km/h)	5	15	30	60
Time (hours)	18	6	3	1.5

Travel time decreases as speed increases — this is an inverse relationship, not a direct proportion.

MTMath Talk — Does the travel time decrease by the same factor that speed increases? Check for all modes of transport.

Bicycle is 3× faster than walking (\(15\div5=3\)); travel time is 3× shorter (\(18\div6=3\)) — matches. Motorcycle is 6× faster than walking (\(30\div5=6\)); time is 6× shorter (\(18\div3=6\)) — matches. Car is 12× faster than walking (\(60\div5=12\)); time is 12× shorter (\(18\div1.5=12\)) — matches.

In every case, speed×time = 90 (the fixed distance between Lucknow and Kanpur, in km).

Yes — in every case, speed and time change by exactly reciprocal factors, and their product (90 km) stays constant. This is an inverse proportion.

Two quantities \(x\) and \(y\) are inversely proportional if \(xy=k\) for some constant \(k\). If \(x_1,x_2\) are values of \(x\) with corresponding \(y_1,y_2\), then \(x_1y_1=x_2y_2=k\), which also means \(\dfrac{x_1}{x_2}=\dfrac{y_2}{y_1}\).

Example 3. 20 workers take 4 days to lay a road. How many days for 10 workers? Fewer workers means more days needed (inverse): \(x_1y_1=x_2y_2 \Rightarrow 20\times4=10\times y_2 \Rightarrow y_2=8\) days.

Example 4. 2 pumps fill a tank in 18 hours. With 2 more pumps (4 total), how long? \(2\times18=4\times x \Rightarrow x=9\) hours.

Example 5. Food for 80 students lasts 15 days. If 20 more students join (100 total), how long will it last? \(80\times15=100\times x \Rightarrow x=12\) days.

MTMath Talk — Example 6: Ram cuts a quantity of vegetables in 1 hour; Shyam takes 1.5 hours for the same quantity. Working together, how long will they take?

Treat the job as 1 unit of work. In 1 hour, Ram does \(1\) unit; Shyam does \(\frac{1}{1.5}=\frac23\) unit. Together in 1 hour: \(1+\frac23=\frac53\) units.

So \(\frac53\) units take 1 hour — is 1 unit of work directly or inversely proportional to the time taken? It's directly proportional (more work needs proportionally more time at a fixed combined rate). So \(\frac53:1::1:x\), giving \(\frac53\times x=1\times1 \Rightarrow x=\dfrac{1}{5/3}=\dfrac35\) hours.

Working together, Ram and Shyam finish in 3/5 hour (= 36 minutes).

Figure it Out — Set 3

Twelve questions from pages 67–68 of the textbook.

1Which pairs of quantities are in inverse proportion? (i) Taps filling a tank vs. time to fill it. (ii) Painters hired vs. days to paint a wall. (iii) Distance a car can travel vs. petrol in the tank. (iv) Cyclist's speed vs. time for a fixed route. (v) Length of cloth bought vs. price paid at a fixed rate/metre. (vi) Pages in a book vs. time to read at a fixed reading speed.

(i) Inverse — more taps fill the tank faster (less time).

(ii) Inverse — more painters finish the wall in fewer days.

(iii) Direct — more petrol allows travelling a proportionally greater distance (these increase together, not inversely).

(iv) Inverse — a faster cyclist takes less time for the same fixed route.

(v) Direct — more cloth costs proportionally more at a fixed rate (these increase together).

(vi) Direct — more pages take proportionally more time to read at a fixed reading speed.

Inverse proportion: (i), (ii), (iv). Direct proportion (not inverse): (iii), (v), (vi).

2If 24 pencils cost ₹120, how much will 20 such pencils cost?

This is a direct proportion (more pencils cost proportionally more, not less): \(\dfrac{120}{24}\times20=5\times20=\)₹100.

20 pencils cost ₹100.

3Math Talk — A water tank supplies 20 families for 6 days. If 10 more families move in, how long will the water last? What assumptions are needed?

More families using the same fixed water supply means it runs out sooner — an inverse proportion. With 30 families total: \(20\times6=30\times x \Rightarrow x=\dfrac{120}{30}=4\) days.

Assumptions: every family uses water at the same average daily rate, no family uses noticeably more or less, and there's no additional water added to the tank during this period.

The water will last only 4 days (assuming all families use water at the same average rate).

4Fill in the average sleep hours (per day) for each animal shown, choosing from: 15, 2.5, 20, 8, 3.5, 13, 10.5, 18.

This question asks you to match values to specific pictured animals, which depends on exactly which 8 animals are shown in your printed copy. Based on commonly cited average sleep-hour data for animals frequently used in this kind of chart, here is a defensible matching — but do check it against your own textbook's icons.

Elephant: 3.5 hours (elephants are famous for sleeping very little). Human: 8 hours (typical adult average). Dog: 10.5 hours. Cat: 13 hours. Animals known as extreme sleepers such as bats and opossums: 18–20 hours each. A lighter-sleeping reptile or small mammal shown might take the remaining value of 15 hours, and any very-low-sleep herbivore (like a horse) would take 2.5 hours.

Best-effort matching: Elephant ≈ 3.5 hrs, Human ≈ 8 hrs, Dog ≈ 10.5 hrs, Cat ≈ 13 hrs, heavy sleepers (bat/opossum) ≈ 18–20 hrs each — please confirm exact icon-to-animal matching against your own textbook's images.

5A pie chart shows modes of transport to school: Walk 90°, Bus 120°, Two-wheeler 60°, Cycle 60°, and Car (unlabelled). (i) Most common mode? (ii) Fraction travelling by car? (iii) If 18 children travel by car, how many took part, and how many use taxis? (iv) Which two modes have equal numbers?

Since all five sectors must sum to 360°, the unlabelled Car sector \(=360-(90+120+60+60)=30^\circ\).

(i) The largest sector is Bus (120°) — the most common mode.

(ii) Car's fraction \(=\dfrac{30}{360}=\dfrac{1}{12}\).

(iii) If 18 children \(=30^\circ\), then \(1^\circ=\dfrac{18}{30}=0.6\) children, so total \(=360\times0.6=216\) children took part. Since "taxi" isn't one of the five listed categories, 0 children in this survey use taxis.

(iv) Cycle and Two-wheeler both have 60° — equal numbers of children use these two modes.

(i) Bus. (ii) 1/12. (iii) 216 children total; 0 use taxis. (iv) Cycle and Two-wheeler.

6Three workers paint a fence in 4 days. If one more worker joins, how many days will it take? What assumptions are needed?

Inverse proportion (more workers, fewer days): \(3\times4=4\times x \Rightarrow x=3\) days.

Assumptions: every worker paints at the same steady rate, and the workers don't get in each other's way or slow each other down by working together.

It will take 3 days with 4 workers.

7A pump fills 2 tanks in 6 hours. How long to fill 5 such tanks?

This is direct proportion (more tanks need proportionally more time, at a fixed pump rate): \(\dfrac{6}{2}\times5=3\times5=15\) hours.

15 hours for 5 tanks.

8Chairs arranged in 25 rows of 12 chairs each are rearranged with 20 chairs per row. How many rows now?

Total chairs stay fixed, so rows and chairs-per-row are inversely proportional: \(25\times12=x\times20 \Rightarrow x=\dfrac{300}{20}=15\) rows.

15 rows.

9A school has 8 periods of 45 minutes each. How long is each period with 9 periods, keeping total school hours the same?

Total school time \(=8\times45=360\) minutes, fixed. More periods means each must be shorter (inverse): \(8\times45=9\times x \Rightarrow x=\dfrac{360}{9}=40\) minutes.

Each period would be 40 minutes long.

10A small pump fills a tank in 3 hours; a large pump fills it in 2 hours. Together, how long?

Small pump's rate: \(\frac13\) tank/hour. Large pump's rate: \(\frac12\) tank/hour. Combined rate: \(\frac13+\frac12=\frac{2}{6}+\frac{3}{6}=\frac56\) tank/hour.

Time to fill 1 tank together \(=\dfrac{1}{5/6}=\dfrac65=1.2\) hours.

Together, the tank fills in 6/5 hours = 1.2 hours (72 minutes).

11A factory needs 42 machines to make a batch of toys in 63 days. How many machines for the same batch in 54 days?

Inverse proportion (fewer days needs more machines): \(42\times63=x\times54 \Rightarrow x=\dfrac{2646}{54}=49\) machines.

49 machines are needed.

12A car takes 2 hours at 60 km/h. How long will it take at 80 km/h?

Distance is fixed \(=60\times2=120\) km. Speed and time are inversely proportional: \(60\times2=80\times x \Rightarrow x=\dfrac{120}{80}=1.5\) hours.

1.5 hours at 80 km/h.

Frequently Asked Questions

Use cross-multiplication: for ratios a:b and c:d, check if a×d equals b×c. If the two cross-products are equal, the ratios are proportional.

An RF like 1 : 60,00,000 means 1 unit of distance on the map equals 60,00,000 of the same units on the ground. Measure the map distance with a ruler, then multiply by this scale factor to get the real geographical distance.

Divide the category's value by the total, then multiply by 360°: angle = (category value ÷ total) × 360°. All the angles in a pie chart must add up to exactly 360°.

In a direct proportion, two quantities change by the same factor, so their ratio stays constant (x/y = k). In an inverse proportion, when one quantity changes by a factor n, the other changes by 1/n, so their product stays constant (xy = k).

Add up all the terms of the ratio to get the total number of parts. Each share of the quantity is then the whole multiplied by that term's value divided by the sum of all terms.

Continue with Part 2, Chapter 4

Move on to the next Part 2 chapter next.

Part 2, Chapter 4

Part 2, Chapter 3Proportional Reasoning-2