NCERT Solutions Class 8 Maths Chapter 7 Proportional Reasoning 1

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 8 · M AT H S

NCERT Solutions

Chapter 7: Proportional
Reasoning- 1

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

159 – 177 12 67 English

Solutions, notes, sample papers & more at 51 pages

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 7: Proportional Reasoning-
1
Chapter 7 closes Ganita Prakash Part I. It opens with five resized copies of the same photograph of a tiger and
asks a question you can answer by eye but not yet by arithmetic — why do three of them look right and two
look stretched? The answer is the ratio: a shape survives resizing only when every measurement is multiplied
by the same factor. From that one idea the chapter builds proportion, cross multiplication, the ancient
Trairasika (Rule of Three), sharing a quantity in a given ratio, and unit conversion.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 159 – 177

SECTIONS QUESTIONS

12 67

MEDIUM

English

In-text Questions — Pages 159–160
7.1 Observing Similarity in Change

MATH TALK

Q1 Which images look similar and which ones look different?

Images A, C and D look similar. Images B and E look different — even though every one of
the five shows the same tiger.

A → 60 mm × 40 mm → 3 : 2

C → 30 mm × 20 mm → 3 : 2

D → 90 mm × 60 mm → 3 : 2

B → 40 mm × 20 mm → 2 : 1

E → 60 mm × 60 mm → 1 : 1

Page 1 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

D E
A
B C

90 × 60 60 × 40 60 × 60 40 × 20 30 × 20
3:2 3:2 1:1 2:1 3:2

Widths and heights in mm, drawn to scale. The blue frames all reduce to 3 : 2; the red ones do not.

Why it happens: Looking similar has nothing to do with being the same size. A, C
and D are of three different sizes — C is half of A in each direction, D is one and a
half times A — yet all three share one number, the width-to-height ratio 3 : 2. That
single number is what decides the shape of the frame, and therefore whether the
tiger inside is drawn faithfully or squeezed.

Q2 Do images B and E look like the other three images?

No. They are distorted. In B the tiger appears stretched and elongated; in E it appears
compressed and fatter.

B → 40 : 20 = 2 : 1 (wider than 3 : 2 for its height)
E → 60 : 60 = 1 : 1 (not wide enough for its height)

A frame that is 2 : 1 is relatively wider than a 3 : 2 frame, so the same tiger has to be pulled
sideways to fill it. A 1 : 1 frame is relatively taller, so the tiger has to be squashed sideways — it
looks short and fat.

Q3 Why?

Because in B and E the width and the height were not changed by the same factor.

Page 2 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

A → B : width 60 → 40, factor 40⁄60 = 2⁄3 ; height 40 → 20, factor 20⁄40 = 1⁄2

A → E : width 60 → 60, factor 1 ; height 40 → 60, factor 60⁄40 = 3⁄2

In each case the two factors are different, so one direction was stretched more than the other.

Why the obvious explanation is not enough: It is tempting to say “E looks odd
because it is a square, and the others are rectangles.” But B is a rectangle too, and it
still looks wrong. So being rectangular is not the test. The test is whether the two
factors of change agree.

Q4 What makes images A, C, and D appear similar, and B and E different?

A, C and D were obtained from one another by multiplying both the width and the height by the
same factor. B and E were not.

From A to Width factor Height factor Same?

C (30 × 20) 30⁄60 = 1⁄2 20⁄40 = 1⁄2 Yes — similar

D (90 × 60) 90⁄60 = 3⁄2 60⁄40 = 3⁄2 Yes — similar

B (40 × 20) 40⁄60 = 2⁄3 20⁄40 = 1⁄2 No — distorted

E (60 × 60) 60⁄60 = 1 60⁄40 = 3⁄2 No — distorted

Why it happens: Look at how B was actually made: 20 mm was taken off the width
and 20 mm off the height. The difference is the same, yet the picture distorts. The
reason is that 20 mm is one-third of a 60 mm width but half of a 40 mm height — the
same subtraction eats a different share of each side. Shape is about shares, so it is
preserved by multiplying, and destroyed by adding or subtracting. Changes to width
and height that use the same factor are called proportional.

Page 3 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

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Can you check by what factors the width and height of image D change as
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Q5

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compared to image A? Are the factors the same?

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g l a
a

Both factors are 3⁄2, so yes, they are the same.

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Width: 60 × f = 90 ⇒ f = 90⁄60 = 3⁄2
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Height: 40 × f = 60 ⇒ f = 60⁄40 =a3⁄2

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D is A enlarged one and a half times in every direction. That is why D, though the biggest of the

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five, still looks exactly like A.

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a s em it yourself: Go the other way — from D to A the factor is 2⁄3 in both directions
gl
Check
a (90 × 2⁄3 = 60 and 60 × 2⁄3 = 40). Reversing a proportional change gives another
proportional change.
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In-text Questions — Page 161

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7.2 Ratios · 7.3 Ratios in their Simplest Form

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By what factor should we multiply the ratio 60 : 40 (image A) to get 90 : 60 (image

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60 × 3⁄2 : 40 × 3⁄2

= 90 : 60
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only one term is multiplied the comparison itself changes. Multiplying both terms by
the same factor leaves the comparison untouched — exactly as multiplying the
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numerator and the denominator of a fraction by the same number gives an
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equivalent fraction.
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m ase
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a g l Page 4 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q2 What is the simplest form of the ratios of images B and E?

Divide each pair of terms by their HCF.

Image B: 40 : 20, HCF = 20 → 40⁄20 : 20⁄20 = 2 : 1

Image E: 60 : 60, HCF = 60 → 60⁄60 : 60⁄60 = 1 : 1

Neither of these is 3 : 2, so the ratios of B and E are not proportional to those of A, C and D —
which is the arithmetic behind what your eye already noticed.

Tip: Dividing by any common factor moves you towards the simplest form, but only
dividing by the HCF gets you there in one step. 40 : 20 divided by 4 is 10 : 5, which is
correct but not yet simplest.

In-text Questions — Pages 162–163
7.4 Problem Solving with Proportional Reasoning (Examples 1–5)

MATH TALK

Q1 Example 1: Are the ratios 3 : 4 and 72 : 96 proportional?

Yes. Both have the same simplest form.

3 : 4 is already simplest (HCF of 3 and 4 is 1)

72 : 96, HCF = 24 → 72⁄24 : 96⁄24 = 3 : 4

So 3 : 4 :: 72 : 96

Check it yourself: Cross multiply — 3 × 96 = 288 and 4 × 72 = 288. Equal products
confirm the proportion without simplifying anything.

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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q2 What is the HCF of 72 and 96?

24.

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

Common part = 2³ × 3 = 24

Dividing both terms of 72 : 96 by 24 gives 3 : 4.

Q3 Example 2: Kesang wanted to make lemonade for a celebration. She made 6 glasses
of lemonade in a vessel and added 10 spoons of sugar to the drink. Her father
expected more people to join the celebration. So he asked her to make 18 more
glasses of lemonade. To make the lemonade with the same sweetness, how many
spoons of sugar should she add?

30 spoons of sugar for the 18 extra glasses.

glasses : spoons must stay proportional

6 : 10 :: 18 : ?

First term: 6 → 18, factor = 18 ÷ 6 = 3

Second term must change by 3 as well: 10 × 3 = 30

6 : 10 :: 18 : 30

Why the factor must be the same: Sweetness is the amount of sugar per glass. In
the first vessel that is 10⁄6 spoons per glass. If the number of glasses is tripled but the
sugar is not, each glass gets a third as much sugar and the drink tastes weak.
Tripling both keeps 30⁄18 = 10⁄6 — the same sweetness.

Page 6 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Tip: Altogether Kesang will then have made 6 + 18 = 24 glasses using 10 + 30 = 40
spoons, and 24 : 40 also simplifies to 3 : 5. The whole batch is as sweet as the first
vessel.

Q4 How can we find the factor of change in the ratio?

Divide the new term by the old term of the same quantity.

factor = new first term ÷ old first term

= 18 ÷ 6 = 3

The factor need not be a whole number. If the first term had gone from 6 to 9, the factor would
be 9 ÷ 6 = 3⁄2, and the second term would have to be multiplied by 3⁄2 as well.

Q5 Example 3: Nitin and Hari were constructing a compound wall around their house.
Nitin was building the longer side, 60 ft in length, and Hari was building the shorter
side, 40 ft in length. Nitin used 3 bags of cement but Hari used only 2 bags of
cement. Nitin was worried that the wall Hari built would not be as strong as the
wall he built because she used less cement. Is Nitin correct in his thinking?

No, Nitin is not correct. The two walls are equally strong.

Nitin: length : bags = 60 : 3, HCF = 3 → 20 : 1

Hari: length : bags = 40 : 2, HCF = 2 → 20 : 1

60 : 3 :: 40 : 2

Why it happens: Strength depends on how much cement goes into each foot of
wall, not on the total number of bags. Both are building at 1 bag for every 20 feet.
Hari used fewer bags only because she had less wall to build. Comparing the totals
alone (3 against 2) compares two things of different sizes; the ratio compares them
fairly.

Page 7 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q6 Example 4: In my school, there are 5 teachers and 170 students. The ratio of
teachers to students in my school is 5 : 170. Count the number of teachers and
students in your school. What is the ratio of teachers to students in your school?
Write it below. ______ : ______ Is the teacher-to-student ratio in your school
proportional to the one in my school?

This one needs your own school's figures, but here is exactly how to settle it.

Given school: 5 : 170, HCF = 5 → 1 : 34

(one teacher for every 34 students)

Count the teachers and the students in your school, write the ratio, and reduce it. Then
compare:

If your simplest form is also 1 : 34, the two ratios are proportional.
If it is 1 : 25, each teacher in your school handles fewer students.
If it is 1 : 40, each teacher handles more.

Check it yourself: Suppose your school has 12 teachers and 408 students. Then 12 :
408 has HCF 12 and reduces to 1 : 34 — proportional to 5 : 170, even though both
numbers are much larger. Cross multiplication says the same thing: 5 × 408 = 2040 =
170 × 12.

Q7 Example 5: Measure the width and height (to the nearest cm) of the blackboard in
your classroom. What is the ratio of width to height of the blackboard? ______ : ______
Can you draw a rectangle in your notebook whose width and height are
proportional to the ratio of the blackboard?

Measure with a metre scale and reduce the ratio you get. A very common classroom size is 240
cm by 120 cm.

Width : height = 240 : 120, HCF = 120 → 2 : 1

To draw a proportional rectangle in your notebook, pick any convenient factor and multiply both
terms by it:

Page 8 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
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Factor Width Height Ratio

m a e
2s: 1
× 4 cm
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8 cm 4 cm
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× 5 cm
a 10 cm 5 cm 2:1
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× 6.5 cm 13 cm 6.5 cm 2:1

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Tip: If your blackboard measures, say, 210 cm by 90 cm, the ratio is 210 : 90 = 7 : 3,
g proportional to it.
and a 14 cm by 6 cm rectangleais

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Compare the rectangle you have drawn to those drawn by your classmates. Do they
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all look the same?
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They will be of different sizes but the same shape — every one of them is a scaled copy of the
blackboard.
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Why it happens: Each of you was free to choose the factor, so the sizes differ. But
nobody was free to change the ratio, so the shape does not differ. Place any two of
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the drawings one on top of the other with a corner matched, and the diagonals will
m l as
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lie along the same line — the sure sign of two rectangles with equal ratios.
m a
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different slant. That is a quick way to spot the odd one out in the whole class without
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measuring anything.
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In-text Questions — Pages 164–165
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7.4 Problem Solving with Proportional Reasoning (Examples 6–7) · Filter Coffee!
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MATH TALK
a g
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Q1

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What is the ratio of Neelima’s age to her mother’s age? What would be the ratio of
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their ages when Neelima is 12 years old? Would it remain the same?
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ANSWER a
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At age 3 the ratio is 1 : 10; at age 12 it is 4 : 13. No, it does not remain the same.

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Page 9 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Neelima 3, mother 3 × 10 = 30 → 3 : 30 = 1 : 10

12 − 3 = 9 years later, mother is 30 + 9 = 39

12 : 39, HCF = 3 → 4 : 13

Why it happens: Nine years were added to both ages. Adding the same number to
both terms of a ratio does not keep the ratio — only multiplying both terms by the
same factor does. For a proportional ratio the mother would have had to be 120
when Neelima turned 12, which is impossible. Ages simply do not stay in proportion;
the gap of 27 years stays fixed instead.

Did you know? As both ages grow, the ratio keeps creeping towards 1 : 1. At 30 and
57 it is 10 : 19; at 60 and 87 it is 20 : 29. This is why an age gap feels enormous
between a child and a parent and hardly noticeable between two adults.

Q2 Example 7: Fill in the missing numbers for the following ratios that are proportional
to 14 : 21. ______ : 42 6 : ______ 2 : ______

28 : 42, 6 : 9, 2 : 3.

14 : 21 in simplest form is 2 : 3 (HCF = 7)

___ : 42 → 42 ÷ 21 = 2, so first term = 14 × 2 = 28

6 : ___ → 6 ÷ 14 = 3⁄7, so second term = 21 × 3⁄7 = 9

2 : ___ → 14 ÷ 7 = 2, so second term = 21 ÷ 7 = 3

Check it yourself: Each answer must reduce to 2 : 3 — and 28 : 42, 6 : 9 and 2 : 3 all
do. Cross multiplication is a second check: 14 × 42 = 588 = 21 × 28.

Page 10 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q3 What factor should we multiply 14 by to get 6? Can it be an integer? Or should it be
a fraction?

The factor is 3⁄7 — it must be a fraction, not an integer.

14y = 6

y = 6⁄14 = 3⁄7

Why it cannot be a whole number: 6 is smaller than 14, so the factor has to be less
than 1, and the only whole number less than 1 that is of any use here is 0. A factor
smaller than 1 shrinks a quantity; there is nothing unusual about it. Applying the
same 3⁄7 to the other term gives 21 × 3⁄7 = 9, so the ratio is 6 : 9.

Q4 Why is this coffee stronger?

Because a larger share of the cup is decoction.

Regular: 15 mL decoction + 35 mL milk = 50 mL cup → decoction is 15⁄50 = 30%

Stronger: 20 mL decoction + 30 mL milk = 50 mL cup → decoction is 20⁄50 = 40%

Both cups hold the same 50 mL, but the second carries 5 mL more decoction and 5 mL less milk.
In ratio language, 15 : 35 = 3 : 7 has become 20 : 30 = 2 : 3, and 2 : 3 has more decoction for
every part of milk than 3 : 7 does.

Tip: To compare two ratios quickly, make one term match. 3 : 7 is the same as 3 : 7,
and 2 : 3 is the same as 4.67 : 7 — clearly more decoction for the same milk.

Q5 Why is this coffee lighter?

Because a smaller share of the cup is decoction.

Page 11 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Lighter: 10 mL decoction + 40 mL milk = 50 mL cup → decoction is 10⁄50 = 20%

Ratio 10 : 40 = 1 : 4, against the regular 15 : 35 = 3 : 7

For every 4 parts of milk this cup carries only 1 part of decoction, whereas the regular cup
carries 1 part for every 2⅓ parts of milk. Less decoction per unit of milk means a lighter taste.

Why the ratio and not the amount: Manjunath could pour 100 mL of decoction
and 400 mL of milk into a jug. That is much more decoction than a regular cup
contains, yet the coffee is still light — because strength is decided by the ratio 1 : 4,
not by the raw quantity.

Q6 The following table shows the different ratios in which Manjunath mixes coffee
decoction with milk. Write in the last column if the coffee is stronger or lighter than
the regular coffee.

Reduce each row and compare it with the regular mix, 15 : 35 = 3 : 7.

Coffee decoction (mL) Milk (mL) Simplest form Decoction in the cup Regular / Strong / Light

300 600 1:2 300⁄900 = 33.3% Strong

150 500 3 : 10 150⁄650 = 23.1% Light

200 400 1:2 200⁄600 = 33.3% Strong

24 56 3:7 24⁄80 = 30% Regular

100 300 1:3 100⁄400 = 25% Light

Why comparing simplest forms works: Regular coffee is 3 parts decoction to 7
parts milk, that is 30% decoction. A mix is stronger exactly when its decoction share
is above 30%. Row 4 reduces to 3 : 7 itself, so 24 mL and 56 mL is just a smaller cup
of ordinary filter coffee — the quantities are different but the drink is identical.

Tip: Rows 1 and 3 have different quantities (300 : 600 and 200 : 400) but the same
simplest form 1 : 2, so they must taste exactly alike. That is the whole point of
reducing a ratio — it strips away the size and leaves the recipe.

Page 12 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Figure it Out — Pages 165–167
7.4 Problem Solving with Proportional Reasoning

MATH TALK

Q1 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

True: (i), (iv) and (vi).

Statement Cross multiply (ad and bc) Simplest forms True?

(i) 4 : 7 :: 12 : 21 4 × 21 = 84, 7 × 12 = 84 4 : 7 and 4 : 7 True

(ii) 8 : 3 :: 24 : 6 8 × 6 = 48, 3 × 24 = 72 8 : 3 and 4 : 1 False

(iii) 7 : 12 :: 12 : 7 7 × 7 = 49, 12 × 12 = 144 7 : 12 and 12 : 7 False

(iv) 21 : 6 :: 35 : 10 21 × 10 = 210, 6 × 35 = 210 7 : 2 and 7 : 2 True

(v) 12 : 18 :: 28 : 12 12 × 12 = 144, 18 × 28 = 504 2 : 3 and 7 : 3 False

(vi) 24 : 8 :: 9 : 3 24 × 3 = 72, 8 × 9 = 72 3 : 1 and 3 : 1 True

Why (iii) is a trap: Reversing the terms of a ratio does not give a proportional ratio.
7 : 12 and 12 : 7 are equal only if 7 × 7 = 12 × 12, which would need 7 = 12. In general
a : b :: b : a only when a = b.

Tip: In (ii) the first terms were tripled (8 → 24) but the second terms were doubled (3
→ 6). Two different factors — so the statement fails, and you can see that without
multiplying anything out.

Q2 Give 3 ratios that are proportional to 4 : 9. ______ : ______ ______ : ______ ______ : ______

8 : 18, 12 : 27, 16 : 36 — and there are infinitely many more.

Page 13 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
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4 : 9 × 2 → 8 : 18
m l as
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4 : 9 × 3 → 12 : 27
m a g
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4 : 9 × 4 → 16 : 36
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Any factor works, whole or fractional: × 10 gives 40 : 90, × 1⁄2 gives 2 : 4.5, × 25 gives 100 : 225.

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Why 4 : 9 itself never changes: 4 and 9 have HCF 1, so 4 : 9 is already in its simplest
a
form. Every ratio proportional to it must reduce back to 4 : 9 — which is exactly why

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the simplest form works as a fingerprint. Check any answer by cross multiplication: 4
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× 18 = 72 = 9 × 8.

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a glQ3 Fill in the missing numbers for these ratios that are proportional to 18 : 24. 3 : ______
12 : ______ 20 : ______ 27 : ______
m a s
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a s

l Every answer must reduce to 3 : 4, so the second term is
First reduce: 18 : 24 = 3 : 4 (HCF =g6).
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always 4⁄3 of the first.

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3 : ___ → 3 × 4⁄3 = 4 → 3 : 4
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27 : ___ → 27 × 4⁄3 = 36 → 27 : 36

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Why one answer is not a whole number: Going from 3 to 20 needs the factor 20⁄3,
which is not a whole number, so the second term 4 × 20⁄3 = 80⁄3 cannot be one either.

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Nothing is wrong — a ratio's terms are allowed to be fractions. If you would rather
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avoid the fraction, multiply both terms by 3 and write the same
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m Check it yourself: 20 × 4 = 80 and 3 × 80⁄3 = 80. Equal cross products, so 3 : 4 :: 20 : 80⁄3
l a se
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co m
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m ase
.co


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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q4 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.

A and D are similar to each other, and C and E are similar to each other. B is not similar to
any of them.
Measure each rectangle's two sides with your scale — three of them are printed at a slant, so
measure along the sides, not across the page. Rounded to the nearest millimetre:

Rectangle Shorter side Longer side Longer : shorter Simplest form

A 6 mm 18 mm 3.0 3:1

B 12 mm 18 mm 1.5 3:2

C 20 mm 51 mm 2.55 5:2

D 13 mm 39 mm 3.0 3:1

E 7 mm 17 mm 2.43 5:2

A C
B D
E

6 × 18 mm 18 × 12 mm 51 × 20 mm 39 × 13 mm 17 × 7 mm
1:3 3:2 5:2 3:1 5:2

The five rectangles redrawn upright, to scale, with the sides measured to the nearest millimetre. Long
side : short side is printed under each.

Why turning a rectangle does not matter: C, D and E are printed tilted, but
rotating a figure moves it without changing any length. The pair of side lengths —
and therefore the ratio — is exactly what it was before the turn. So compare side
ratios and ignore the tilt entirely. A is 3 : 1 and so is D; C is about 5 : 2 and so is E; B at
3 : 2 has no partner in this set.

Page 15 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Tip: Your own measurements may be a millimetre off here and there, so read 2.43
and 2.55 as “both about 2.5”. When two ratios agree to within your measuring error,
treat the rectangles as similar and, if you can, check by laying one diagonal against
the other.

Q5 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?

The printed rectangle measures about 35 mm × 21 mm, so its width : height is 5 : 3. Multiply
both terms by the same factor to get as many copies as you like.

Factor Width Height Ratio

× 0.4 cm (smaller) 2 cm 1.2 cm 5:3

the printed one 3.5 cm 2.1 cm 5:3

× 1.6 cm (bigger) 8 cm 4.8 cm 5:3

No, your classmates' rectangles will not all be the same size — and no, they are not wrong.

Why different answers are all correct: The question fixes the shape but leaves the
size free. Each of you chose your own factor, so the drawings differ in size; but
nobody changed the ratio, so they do not differ in shape. A question with infinitely
many correct answers is not a defective question — it is a question about a family of
figures rather than a single figure.

Check it yourself: Place any two of the drawings so that their bottom-left corners
meet and their sides lie along each other. The two top-right corners and the shared
corner will lie on one straight line. If a classmate's rectangle misses that line, its ratio
is not 5 : 3.

Page 16 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q6 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.

(a) 9 : 6 = 3 : 2 (b) 16 : 12 = 4 : 3
Since the pattern repeats for ever, you do not count the whole wall. Find one repeating block,
count inside it, and that ratio holds for the entire wall.
Wall (a). The block is 5 columns wide and 3 rows tall — 15 bricks. The coloured bricks form a
triangle pointing down:

Top row: 3 coloured, 2 grey

Middle row: 2 coloured, 3 grey

Bottom row: 1 coloured, 4 grey

Coloured = 3 + 2 + 1 = 6, grey = 15 − 6 = 9

grey : coloured = 9 : 6 = 3 : 2

One repeating block of wall (a): 5 columns × 3 rows = 15 bricks, of which 3 + 2 + 1 = 6 are coloured and
9 are grey.

Wall (b). Here the coloured bricks outline diamonds, and the pattern repeats every 4 columns
across 7 rows — 28 bricks in a block:

Page 17 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Rows from the top: 1, 2, 2, 2, 2, 2, 1 coloured

Coloured = 1 + 2 + 2 + 2 + 2 + 2 + 1 = 12, grey = 28 − 12 = 16

grey : coloured = 16 : 12 = 4 : 3

Wall (b): the coloured bricks outline diamonds that repeat every 4 columns (dashed line). One block is
4 columns × 7 rows = 28 bricks — 12 coloured, 16 grey.

Why one block settles the whole wall: A wall 20 blocks long has 20 × 9 grey and 20
× 6 coloured bricks in pattern (a). Both counts are multiplied by 20, which is the same
factor for both terms — so the ratio is unchanged. This is the reason a repeating
pattern can be described by a single ratio however long the wall runs.

Tip: Choosing where the block starts does not matter, as long as the block is one full
period wide. Start counting one brick further along and you will still find 9 grey to 6
coloured in (a).

Page 18 of 51

as e
Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
m.
Let us draw some human figures. Measure your friend’s body — the lengths of their
se
Q7

o m l
head, torso, arms, and legs. Write the ratios as mentioned below— head : torso
a
g draw a figure with
______ .: c______ torso : arms ______ : ______ torso : legs ______ : ______ Now,
m a
l a se torso, arms, and legs with equivalent ratios as above. Does the drawing look
head,
a g more realistic if the ratios are proportional? Why? Why not?

co m
m . ag

l a se
Measure with a tape and record the lengths, then reduce each pair. Here is one real set of
ag150 cm tall:
measurements from a student about

co m
m.
Part Length Ratio asked for Simplest form

m as e
head
.co
22 cm head : torso = 22 : 55
a g l2:5

se m
g l a
torso 55 cm torso : arms = 55 : 60 11 : 12

a arms 60 cm torso : legs = 55 : 80 11 : 16

m a s
legs 80 cm
em
—
.co —
agl
a s
gl
To draw the figure, choose oneaconvenient factor and apply it to every measurement. Using 1⁄5 of
the real lengths: head 4.4 cm, torso 11 cm, arms 12 cm, legs 16 cm.

co m
.
Yes, the drawing looks far more realistic when the ratios are kept.

se m
o m l a
g sizes of its parts, not
m .c it happens: We recognise a human figure by the relative
Why
a
aseby its absolute height — which is why a 5 cm sketch and a life-size poster can both
agl look convincingly human. Enlarge just the head and keep everything else, and the
two factors disagree; the figure reads as a cartoon. Keep one factor throughout and
se m
c m distance scaled by that factor, so nothing
every distance in the drawing is theoreal
g l a
. a
looks out of place.
a s em
agl
m
Did you know? Artists often measure a figure in “head lengths”. An adult is roughly
. co
m
7 to 8 heads tall, while a small child is about 5 — and that difference in ratio, not the

m as e
.co
difference in height, is what makes a child's drawing look like a child.
a g l
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g l a
a c
m .
In-text Questions — Pages 167–170
m a s e
em . co agl
g l as
a

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m .
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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Trairasika — The Rule of Three (Examples 8–10) · Activity 1

Q1 Example 8: For the mid-day meal in a school with 120 students, the cook usually
makes 15 kg of rice. On a rainy day, only 80 students came to school. How many
kilograms of rice should the cook make so that the food is not wasted?

10 kg of rice.

students : rice must stay proportional

120 : 15 :: 80 : ?

Factor of change = 80⁄120 = 2⁄3

Rice = 15 × 2⁄3 = 10 kg

Why it works: The cook is really working at a fixed rate — 15 kg for 120 students is
15⁄120 = 0.125 kg per student. Fewer students do not change that rate, so 80 × 0.125 =
10 kg. Multiplying the rice by the same 2⁄3 that the students were multiplied by is a
quicker way of saying the same thing.

Check it yourself: Cross multiply — 120 × 10 = 1200 and 15 × 80 = 1200. Equal, so
the proportion holds.

Q2 What is the factor of change in the first term?

2⁄3 — the number of students has fallen to two-thirds of what it usually is.

80⁄120 = 8⁄12 = 2⁄3

A factor smaller than 1 means the quantity has decreased. The second term must be multiplied
by the very same 2⁄3.

Page 20 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q3 Example 9: A car travels 90 km in 150 minutes. If it continues at the same speed,
what distance will it cover in 4 hours?

144 km.

4 hours = 4 × 60 = 240 minutes

150 : 90 :: 240 : x

Cross multiplying: 150 × x = 240 × 90

x = (240 × 90) ÷ 150

x = 21600 ÷ 150 = 144 km

Why time and distance may be paired this way: The speed is unchanged, so
distance grows exactly as time grows — double the time, double the distance.
Quantities that rise together in this way are the only ones the Rule of Three applies
to.

Check it yourself: The car's speed is 90 ÷ 150 = 0.6 km per minute. In 240 minutes
that is 240 × 0.6 = 144 km. Same answer by a different route.

Q4 Is this the right way to formulate the question?

No. The statement 150 : 90 :: 4 : ? mixes units — 150 is in minutes but 4 is in hours.

Wrong: 150 min : 90 km :: 4 h : ?

Right: 4 h = 240 min, so 150 : 90 :: 240 : ?

Page 21 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why the units must match: The factor of change is found by dividing the first
terms, and 4 ÷ 150 is a meaningless number when one is hours and the other
minutes. Once both are minutes, 240 ÷ 150 = 8⁄5 is a genuine factor, and multiplying
90 by 8⁄5 gives 144. Using the raw 4 would have given 90 × 4⁄150 = 2.4 km for four hours
of driving — an answer that fails the common-sense test instantly.

Q5 How can you find the distance covered in 240 minutes? Discuss with your
classmates and find the answer using different strategies.

All three routes below give 144 km. Use whichever you find clearest.

Unitary method. In 1 minute the car covers 90 ÷ 150 = 0.6 km. In 240 minutes it covers 240
× 0.6 = 144 km.
Scaling the ratio. 240 ÷ 150 = 8⁄5, so multiply the distance by 8⁄5 as well: 90 × 8⁄5 = 144 km.
Splitting the time. 150 min → 90 km, so 300 min → 180 km and 30 min → 18 km. Then 240
= 300 − 60, and 60 min → 36 km, giving 180 − 36 = 144 km.

Cross multiplication, for comparison:

150 : 90 :: 240 : x ⇒ 150x = 240 × 90 ⇒ x = 144 km

Why they must agree: Every one of these methods is applying the single fact that
the car covers 0.6 km each minute. Cross multiplication is not a fourth idea — it is
the same equation with the division postponed to the last step.

Q6 Example 10: A small farmer in Himachal Pradesh sells each 200 g packet of tea for
₹200. A large estate in Meghalaya sells each 1 kg packet of tea for ₹800. Are the
weight-to-price ratios in both places proportional?

No, they are not proportional. First put both weights in the same unit.

Himachal: 200 g : ₹200 → HCF 200 → 1 : 1

Meghalaya: 1 kg = 1000 g, so 1000 : 800 → HCF 200 → 5 : 4

1 : 1 is not 5 : 4, so the ratios are not proportional

Page 22 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why writing 1 : 800 would be wrong: The Himachal ratio was formed with the
weight in grams. If Meghalaya's weight is left in kilograms, the two ratios are
measuring different things and cannot be compared at all. Converting to a common
unit is not a formality — it is what makes the comparison mean anything.

Check it yourself: Cross multiply the corrected ratios — 200 × 800 = 160000 while
200 × 1000 = 200000. Unequal products confirm that the ratios are not proportional.

Q7 Which tea is more expensive? Why?

The Himachal tea is more expensive — ₹1,000 per kg against ₹800 per kg.

Meghalaya: 1 kg costs ₹800

Himachal: 200 g is 1⁄5 of a kg, so if 1 kg costs ₹x,

1⁄5 × x = 200

x = 200 × 5 = ₹1,000

Why the packet prices cannot be compared directly: ₹200 looks far less than
₹800, but the two packets are not the same size. Prices only become comparable
once they are quoted for the same weight. Bringing both to “rupees per kilogram”
does exactly that, and then the Himachal tea is ₹200 per kg dearer.

Did you know? The price per unit printed on a shop shelf — per kg, per litre, per 100
g — exists for precisely this reason. It converts every packet, whatever its size, to one
comparable ratio.

Page 23 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
m.
Activity 1: Take your favourite dish. Find out all the ingredients and their respective
se
Q8

o m l a
quantities needed to make the dish for your family. Suppose you are celebrating a
g of the ingredients
m .c and you want to invite 15 guests. Find out the quantities
festival a
l a se
required to cook the same dish for them.
a g
m

. co ag
m
Write the family recipe first, then scale every ingredient by one factor. Here is a worked example

ase
l
for upma cooked for a family of 5, with 15 guests invited so that 20 people must be fed.
a g
m
Factor of change = 20⁄5 = 4
co
e m.
m l as
m
Ingredient
.co For 5 people ×4
a g For 20 people

l a se
a g Rava (semolina) 250 g 250 × 4 1000 g = 1 kg

m a s
Water 600 mL 600 × 4 2400 mL = 2.4 L

em
.co agl
s
Onion 1 medium 1×4 4 medium

l a
Oil ag
2 tablespoons 2×4 8 tablespoons

co m
.
Salt 1 teaspoon 1×4 4 teaspoons

se m
o m l a
m
Why.c every ingredient needs the same factor: The tasteaofg a dish is fixed by the
aseratios between its ingredients — rava to water, salt to rava. Scale them all by 4 and
agl every one of those ratios survives, so the dish tastes the same, only there is more of

se m
a
it. Quadruple the rava but merely double the water and the ratio 250 : 600 becomes
1000 : 1200, and the upma turns out
. com
dry.
a g l
a s em
aglyou must feed 6 + 15 = 21, the factor is 21⁄6 = 3.5 — a
Tip: If your family is 6 and
fraction, and perfectly usable. 250 g of rava becomes 875 g.
co m
m .
m ase
.co a g l
a s em it Out — Pages 170–171
gl
Figure
a c
m .
m a s e
em . co agl
g l as
a

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m .
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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Trairasika — The Rule of Three

Q1 The Earth travels approximately 940 million kilometres around the Sun in a year.
How many kilometres will it travel in a week?

About 1,80,76,923 km, that is roughly 1.81 crore kilometres in one week.

1 year ≈ 52 weeks

940 million km = 940 × 10⁶ = 94,00,00,000 km

52 : 940000000 :: 1 : x

x = 940000000 ÷ 52

x = 1,80,76,923 km (approximately)

Why this is a proportion: The Earth's orbital speed is very nearly steady, so distance
and time rise together. Weeks and kilometres can therefore be paired as 52 :
940000000, and dividing by 52 gives the distance for one week.

Check it yourself: 52 × 1,80,76,923 = 93,99,99,996 km — back to 940 million, apart
from the rounding. Using 365 days instead of 52 weeks gives 940000000 × 7 ÷ 365 ≈
1,80,27,397 km; the small difference appears because a year is 52 weeks and one
extra day.

Q2 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.

15,660 bricks. First add up every foot of wall in the plan, then scale.

Page 25 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

15 ft

12 ft 9 × 12 15 × 12

9 ft
6×9 9 ft

6 ft
inner wall = 12 ft

The plan: two rooms 9 ft and 15 ft wide share a 12 ft inner wall (red), and a 6 ft × 9 ft room hangs
below. Every segment of the outline is a wall.

Top block (9 ft and 15 ft rooms, both 12 ft deep):

top edge 9 + 15 = 24 ft

bottom edge 9 + 15 = 24 ft

left side 12 ft, right side 12 ft

inner dividing wall 12 ft
Lower room (6 ft × 9 ft):

left 9 ft, right 9 ft, bottom 6 ft

Total = 24 + 24 + 12 + 12 + 12 + 9 + 9 + 6 = 108 ft

Page 26 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

10 : 1450 :: 108 : x

10 × x = 108 × 1450

x = 156600 ÷ 10 = 15,660 bricks

Why the shared edge is counted once: The top edge of the small lower room is the
same piece of masonry as part of the bottom edge of the big block — the mason
lays those bricks once. Counting each rectangle's perimeter separately would charge
for that wall twice and give 132 ft instead of 108 ft. Always trace the outline of the
actual walls, then add the inner ones.

Tip: 108 ft is 10.8 lots of 10 ft, so the answer must be 10.8 × 1450. Doing it that way
— 1450 + 4350 + 5800 + … — is easy to slip on; the cross-multiplication form 108 ×
1450 ÷ 10 keeps the arithmetic in whole numbers.

In-text Questions — Pages 171–172
Trairasika — The Rule of Three · Activity 2

MATH TALK

Q1 Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle
at a speed of 50 km/h. If he drives at 75 km/h, how long will it take him to reach
Kanpur? Can we form this problem as a proportion — 50 : 2 :: 75 : __ Would it take
Puneeth’s father more time or less time to reach Kanpur? Think about it.

1 hour 20 minutes, and no — this problem cannot be written as 50 : 2 :: 75 : __ .

Distance = speed × time = 50 × 2 = 100 km

At 75 km/h: time = 100 ÷ 75 = 4⁄3 hours

4⁄3 h = 1 h + 1⁄3 × 60 min = 1 hour 20 minutes

Page 27 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why the Rule of Three fails here: The Rule of Three needs two quantities that rise
together. Here the distance is fixed, so a faster ride is a shorter ride — as speed goes
up, time comes down. Writing 50 : 2 :: 75 : x would give x = 3 hours, saying that riding
faster takes longer. What stays constant is not the ratio speed : time but the product
speed × time, which is the 100 km of road between Lucknow and Kanpur.

Tip: Ask yourself before setting up any proportion: if the first quantity increases,
should the second increase too? Students and rice — yes. Speed and time over a
fixed distance — no. That single question tells you whether the Rule of Three applies.

Q2 Activity 2: Go to the market and collect the prices of different sizes of shampoo
containers of the same shampoo and create a table like the one given below. See if
the volume of shampoo is proportional to the price. (Sachet 6 mL ₹2; Small Bottle
180 mL ₹154; Medium Bottle 340 mL ₹276; Large Bottle 1000 mL ₹540)

Collect real prices from a shop, then work out the price of one millilitre for each size — that
single number makes all four containers comparable. For the sample table in the book:

Container Volume Price Price ÷ volume Cost of 1 mL

Sachet 6 mL ₹2 2÷6 ₹0.33

Small Bottle 180 mL ₹154 154 ÷ 180 ₹0.86

Medium Bottle 340 mL ₹276 276 ÷ 340 ₹0.81

Large Bottle 1000 mL ₹540 540 ÷ 1000 ₹0.54

The volume is not proportional to the price. If it were, every row of the last column would
show the same figure; instead it runs 0.33, 0.86, 0.81, 0.54.

Why the unit rate is the right test: Two ratios are proportional exactly when they
reduce to the same simplest form, and volume : price reduces to “1 mL for so many
rupees”. Four different unit rates mean four different ratios, so no single proportion
covers the whole table. Notice too that the rate does not even fall steadily with size
here — among the three bottles it does, but the tiny sachet is the cheapest per
millilitre of all. In the market you will usually find the opposite; make your own table
and see.

Page 28 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
m.
The ratio of the volume of a sachet to a small bottle is 6 : 180. The ratio of their
e
Q3

m l as
.co
prices is 2 : 154. Are these ratios proportional?

a g
sem
g l a
a

No.

com
e m . ag
as
Volumes: 6 : 180, HCF = 6 → 1 : 30

a
Prices: 2 : 154, HCF = 2 → 1 : 77 g l

m
1 : 30 is not 1 : 77
co
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m l as
.co g
Cross multiplication says the same thing: 6 × 154 = 924 while 180 × 2 = 360.

em a
a s
agl What the two numbers mean: The small bottle holds 30 times as much shampoo
as the sachet, but it costs 77 times as much. For the price to have been proportional

m a s
.co agl
the bottle would have to cost 30 × ₹2 = ₹60. At ₹154 it is far dearer, millilitre for
millilitre.
se m
g l a
a
Q4
. c om
Why do you think that the ratio of the prices is not proportional to the ratio of the
volumes? Discuss the pros and cons of different size bottles form
s e the company and

. cfor gla you recommend to the
omcustomers. For reducing ecological footprint, whatawould
a sem company and to the customer? Does the same occur for other products?
agl
se m
com g l a
. a
The price of a container is not only the price of what is inside it.
m
ase
agl
Fixed costs per pack. The bottle, the cap, the label, the printing, the filling and the transport
cost roughly the same whether the pack holds 180 mL or 1000 mL. Those costs are spread
over more shampoo in a big bottle, so its rate per millilitre falls.

co m
.
Bulk discounts. A company sells a large bottle at a lower rate to persuade you to buy more

em
m as
at one time.

.co a g l
m can still buy shampoo. Pricing here follows what a customer can pay, not only what the
Affordability. A sachet can be priced low so that someone who cannot spend ₹540 today

l a se
ag shampoo costs.
.c
s e m
m a
e m . co agl
g l as
a

com
m .
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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Small packs Large packs

For the Low price today, easy to carry, no waste if used Usually cheaper per mL, fewer trips to the
customer rarely shop

For the Reaches more buyers, but far more packaging and Less packaging and handling per litre,
company handling per litre sold larger sale at one go

For a smaller ecological footprint: a customer should buy the largest size they will actually
finish, and refill it where refill packs are sold; a company should offer refill pouches, take back
and reuse bottles, and keep the per-millilitre price of the large size genuinely lower so that the
low-waste choice is also the cheap one.

Why plastic is the real cost here: Getting 1 litre of shampoo through sachets
means roughly 167 sachets — 167 caps, seals and pieces of film, almost none of
which is recycled, against a single bottle. The money ratio and the waste ratio are
two different ratios, and they do not point the same way.

Try This: Build the same table for rice or atta at 500 g, 1 kg and 10 kg. Staples are
often priced very nearly in proportion to weight, so their unit rates come out almost
equal — quite unlike shampoo. Compare and discuss why.

In-text Questions — Pages 172–174
7.5 Sharing, but Not Equally! · Activity 3

Q1 Activity 3: Form a pair. Collect 12 countable objects or counters (it can be coins,
seeds, or pebbles). Now, share them between the two of you in different ways. If
you divide them equally, what is the ratio of the number of counters with each of
you?

6 : 6, which in its simplest form is 1 : 1.

12 ÷ 2 = 6 counters each

6 : 6, HCF = 6 → 1 : 1

Page 30 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why 1 : 1 is what “equally” means: The simplest form throws away the total and
keeps only the comparison. Sharing 12 counters equally, or 50, or 400, always gives
1 : 1 — one counter for you for every one for your partner. That is exactly the
information the word “equally” carries.

Q2 If your partner gets 5 counters, how many objects will you get? What is the ratio of
the counters?

You get 7 counters, and the ratio of your partner's counters to yours is 5 : 7.

12 − 5 = 7

partner : you = 5 : 7, and 5 and 7 have HCF 1, so it is already simplest

Tip: Order matters. Your counters to your partner's is 7 : 5, a different ratio from 5 :
7. Always say which quantity is named first.

Q3 Now, if you want to share the counters between the two of you in the ratio of 3 : 1,
how many counters would each of you get?

Your partner gets 9, you get 3.
Deal the counters out 3 and 1 at a time and see where it lands:

Round Partner takes You take Counters left

start — — 12

1 3 1 8

2 3 1 4

3 3 1 0

total 9 3 —

Page 31 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Or in one step: 3 + 1 = 4 groups, each group is 12 ÷ 4 = 3 counters

Partner: 3 × 3 = 9 You: 1 × 3 = 3

Check: 9 : 3 = 3 : 1 ✓ and 9 + 3 = 12 ✓

Why the whole becomes 4, not 3 or 1: The ratio 3 : 1 describes the counters as four
equal groups — three for one person and one for the other. The number 12 is 3
times 4, so the ratio 3 : 1 is scaled up by 3 to become 9 : 3. Dividing by the sum of the
terms is what turns a comparison into actual quantities.

Q4 Now, if you want to share 42 counters between the two of you in the ratio of 4 : 3,
how will you do it?

Your partner gets 24, you get 18.

Total groups = 4 + 3 = 7

Size of each group = 42 ÷ 7 = 6

Partner: 4 × 6 = 24 You: 3 × 6 = 18

Check: 24 + 18 = 42 ✓ and 24 : 18 = 4 : 3 ✓

Why dealing them out one round at a time is a poor plan: Handing over 4 and 3
repeatedly works, but it takes six rounds and you only discover at the end whether
the counters come out even. Splitting into 7 equal groups gets there in one division,
and it also tells you at once when a share is impossible — 40 counters in the ratio 4 :
3 would need groups of 40⁄7, which is not a whole counter.

Tip: In general, to divide x in the ratio m : n, the parts are m × x⁄(m+n) and n × x⁄
(m+n). Here 4 × 42⁄7 = 24 and 3 × 42⁄7 = 18.

Page 32 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Q5 What is the size of each group?

6 counters.

Number of groups = 4 + 3 = 7

Size of each group = 42 ÷ 7 = 6

Every one of the seven groups holds 6 counters; four of them go to your partner and three to
you.

Figure it Out — Page 175
7.5 Sharing, but Not Equally!

Q1 Divide ₹4,500 into two parts in the ratio 2 : 3.

₹1,800 and ₹2,700.

Total groups = 2 + 3 = 5

Each group = 4500 ÷ 5 = ₹900

First part = 2 × 900 = ₹1,800

Second part = 3 × 900 = ₹2,700

Check: 1800 + 2700 = 4500 ✓ and 1800 : 2700 = 2 : 3 ✓

Tip: Think of the parts as fractions of the whole — 2⁄5 of ₹4,500 and 3⁄5 of ₹4,500. The
two fractions add to 1, which is another way of saying nothing is left over.

Page 33 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
m.
In a science lab, acid and water are mixed in the ratio of 1 : 5 to make a solution. In
e
Q2

com
a bottle that has 240 mL of the solution, how much acid and water does the solution
g l as
. a
em
contain?

a s
a gl

m
40 mL of acid and 200 mL of water.
. co ag
e m
g l as
a
Total groups = 1 + 5 = 6

Each group = 240 ÷ 6 = 40 mL

co m
m.
Acid = 1 × 40 = 40 mL

m as e
.co
Water = 5 × 40 = 200 mL
a g l
a s em 40 + 200 = 240 ✓ and 40 : 200 = 1 : 5 ✓
gl
Check:
a
om a s
c agl
Why 240 is not split into 1 and 5: The ratio 1 : 5 does not mean 1 mL and 5 mL; it
.
mshares, one of acid and five of water. The acid is
means the bottle is made of 6 equal
s e
a glaa slip worth guarding against, and in a laboratory a
1⁄6 of the solution, not 1⁄5 of it —
dangerous one.

co m
m .
o m l a se
g
Q3 .cBlue and yellow paints are mixed in the ratio of 3 : 5 toaproduce
m
green paint. To
e produce 40 mL of green paint, how much of these two colours are needed? To make
aglas 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?
se m
com g l a
m . a
ase

agl
15 mL of blue and 25 mL of yellow. After adding the yellow, the new ratio is 1 : 3.

co m
Total groups = 3 + 5 = 8
m .
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Each group = 40 ÷ 8 = 5 mL
a g l
se m Blue = 3 × 5 = 15 mL Yellow = 5 × 5 = 25 mL
g l a
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m a s e
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Add 20 mL of yellow: yellow = 25 + 20 = 45 mL, blue is still 15 mL

em
l as
blue : yellow = 15 : 45, HCF = 15 → 1 : 3
a g

com
m .
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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why the ratio moves so much: Only one term changed. The blue is untouched, so
the comparison shifts from 3 : 5 to 1 : 3 — for every part of blue there are now three
parts of yellow instead of 1⅔. The total is also up, from 40 mL to 60 mL. Adding to
one term of a ratio always changes the ratio; that is precisely how the painter
lightens the shade.

Q4 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?

4 cups of rice and 2 cups of urad dal.

Total groups = 2 + 1 = 3

Each group = 6 ÷ 3 = 2 cups

Rice = 2 × 2 = 4 cups

Urad dal = 1 × 2 = 2 cups

Check: 4 + 2 = 6 ✓ and 4 : 2 = 2 : 1 ✓

Tip: If you needed 9 cups instead, each group would be 3 cups, giving 6 cups of rice
and 3 of dal. The ratio 2 : 1 fixes the recipe; the total decides the size of the batch.

Q5 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?

3 : 13.

Page 35 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

In one bucket of orange paint:

red = 3⁄8 bucket, yellow = 5⁄8 bucket

Add 1 whole bucket of yellow:
yellow = 5⁄8 + 1 = 13⁄8 bucket, red is unchanged at 3⁄8

red : yellow = 3⁄8 : 13⁄8

Multiply both terms by 8 → 3 : 13

Why no volume in litres is needed: The bucket itself is the unit. Whatever a bucket
holds, the orange paint is 3⁄8 red and 5⁄8 yellow, and the added bucket is 1 whole
bucket of yellow. Multiplying both terms of 3⁄8 : 13⁄8 by 8 clears the fractions without
disturbing the ratio.

Check it yourself: Take the bucket to be 8 litres. Then the orange paint is 3 L red
and 5 L yellow; adding 8 L of yellow makes 3 L red to 13 L yellow — the same 3 : 13,
as it must be.

Figure it Out — Pages 176–177
7.6 Unit Conversions (chapter-end set)

TRY THIS

Q1 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.

2 : 3.

orange : apple = 600 : 900

HCF of 600 and 900 = 300

600⁄300 : 900⁄300 = 2 : 3

Page 36 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Tip: Both quantities were already in millilitres, so no conversion was needed. Had
the apple juice been given as 0.9 litre, the first job would have been to write it as 900
mL.

Q2 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?

4 buses are needed, and they will not all be full — 12 seats will be empty.

Seats in one bus = 162 ÷ 3 = 54

Buses needed = 204 ÷ 54 = 3.78, so 4 buses

Seats in 4 buses = 4 × 54 = 216

Empty seats = 216 − 204 = 12

Why the answer is not 3.78: The proportion 3 : 162 :: x : 204 gives x = 3.78, and up
to that point the arithmetic is fine. But buses come whole. Rounding 3.78 down to 3
would leave 204 − 162 = 42 people standing on the road, so the answer must be
rounded up. Whenever a proportion produces a fraction of an object that cannot be
cut, ask which way the rounding has to go.

Q3 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?

Mumbai is more crowded — about 36,364 people in every square kilometre, against about
20,216 in Delhi.

Delhi: 30 million ÷ 1484 = 3,00,00,000 ÷ 1484 ≈ 20,216 people per sq km

Mumbai: 20 million ÷ 550 = 2,00,00,000 ÷ 550 ≈ 36,364 people per sq km

Page 37 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Why the bigger population is not the more crowded city: Delhi has 10 million
more people, but it also has almost three times the land to spread them over.
Crowding is people per unit of area — a ratio, not a count. Once both cities are
reduced to “people in one square kilometre”, they can be compared fairly, and
Mumbai packs nearly twice as many into the same space.

Did you know? This ratio is called population density. It is why a small island city
can feel far more crowded than a large state with many more people in it.

Q4 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?

Your neck would be 4⁄10 = 2⁄5 of your height. Measure yourself and multiply.

Total groups = 4 + 6 = 10

Neck = 4⁄10 × your height = 0.4 × your height

For the crane: 4⁄10 × 155 = 62 cm of neck

If you are 150 cm tall: 4⁄10 × 150 = 60 cm
If you are 160 cm tall: 4⁄10 × 160 = 64 cm

Why the answer is worth pausing over: A 60 cm neck on a 150 cm person is longer
than the whole of your head and chest together. The ratio 4 : 6 is perfectly ordinary
for a crane, whose long neck is built for fishing in shallow water; imposed on a
human body it is absurd. Ratios always belong to something — copying one from
one creature to another has to be done knowing what it will look like.

Page 38 of 51

as e
Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
m.
Let us try an ancient problem from Lilavati. At that time weights were measured in
se
Q5

o m l a
a unit named palas and niskas was a unit of money. “If 2½ palas of saffron costs 3⁄7

m .cO expert businessman! tell me quickly what quantity aofgsaffron can be
niskas,

l a se for 9 niskas?”
bought
a g
m

. co ag
m
52.5 palas of saffron.

ase
a g l
2½ palas = 5⁄2 palas, cost = 3⁄7 niskas

co m
m.
3⁄7 : 5⁄2 :: 9 : x

m as e
o l
Cross multiplying: 3⁄7 × x = 5⁄2 × 9
. c
3⁄7 ×mx = 45⁄2 a g
ase
agl x = 45⁄2 × 7⁄3 = 315⁄6 = 52.5 palas
m a s
e
. c o
m Compare the money first — 9 ÷ 3⁄7 = 9 × 7⁄3 = 21, agl
s
Why the answer is so much bigger:
a times as much as before. The saffron must
agl
so the buyer is spending twenty-one
therefore be twenty-one times as much: 5⁄2 × 21 = 105⁄2 = 52.5 palas. Fractions in the
data change nothing about the method; they only mean the factor of change is itself
co m
m .
e
a fraction.
m l as
.co a g
m you know? Bhāskarāchārya wrote the Līlāvatī in 1150 CE, and problems in it are
a s eDid
a gl addressed to the reader as though in conversation — “O expert businessman!”. Its
trairāśika problems are exactly the Rule of Three of this chapter.
se m
com g l a
m . a
ase
Q6
agl
Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s

m
age when the ratio of her age to her brother’s age is 1 : 2?

. co
se m
o m l a
g3 years.
.c

m a
se
Harmain will be 4 years old (and her brother 8), that is, after
a
agl c
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m a s e
em . co agl
g l as
a

co m
m .
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.co


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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Let the number of years from now be x.

Harmain: 1 + x, brother: 5 + x

(1 + x) : (5 + x) = 1 : 2
Cross multiplying: 2(1 + x) = 1(5 + x)

2 + 2x = 5 + x

x=3

Harmain's age = 1 + 3 = 4 years, brother = 5 + 3 = 8 ✓ and 4 : 8 = 1 : 2 ✓

Why the ratio of ages keeps changing: The age gap of 4 years never changes, but
each year adds 1 to both terms — and adding the same number to both terms of a
ratio does change it. Today it is 1 : 5, in three years 1 : 2, and later 10 : 14 = 5 : 7,
always drifting towards 1 : 1. The moment when it is exactly 1 : 2 is when the
brother's age is twice Harmain's, and since the gap is 4, that must be 4 and 8.

Q7 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?

18.5 kg.

gold : water = 37 : 2 for equal volumes

37 : 2 :: x : 1

2 × x = 37 × 1

x = 37⁄2 = 18.5 kg

Why “equal volumes” is essential: The ratio 37 : 2 compares masses only when the
two samples occupy the same space. It says nothing about a kilogram of gold
against a kilogram of water — those have equal mass by definition. What it tells you
is that gold is 18.5 times as dense as water, so one litre of gold is as heavy as
eighteen and a half litres of water.

Page 40 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Did you know? A one-litre bottle full of gold would weigh 18.5 kg — heavier than a
full school bag. This is the property Archimedes is said to have used to test whether
a crown was pure gold.

Q8 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).

About 22.96 tonnes, that is roughly 22,957 kg of cow manure.

Area of plot = 200 × 500 = 1,00,000 sq ft

1 acre = 43,560 sq ft

Plot in acres = 100000 ÷ 43560 = 2.2957 acres

Manure = 10 × 2.2957 = 22.957 tonnes ≈ 22.96 tonnes

In kilograms: 22.957 × 1000 ≈ 22,957 kg

Why the conversion cannot be skipped: The recommended dose is quoted per
acre, but the plot is measured in feet. Pairing 10 tonnes with 1,00,000 square feet
directly would compare tonnes-per-acre with tonnes-per-square-foot — two
different rates. Converting the plot to acres puts both quantities on the same scale
before the Rule of Three is applied.

Tip: Manure is sold by the tractor-load, so a farmer would round up and buy 23
tonnes. As with the buses, the mathematics gives the exact figure and the situation
decides how to round it.

Q9 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?

300 seconds, that is 5 minutes.

Page 41 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

10 litres = 10 × 1000 = 10,000 mL

Number of mugfuls = 10000 ÷ 500 = 20

Time = 20 × 15 = 300 seconds = 5 minutes

By the Rule of Three:

500 : 15 :: 10000 : t

500 × t = 10000 × 15

t = 150000 ÷ 500 = 300 seconds

Why volume and time are proportional here: The tap runs at a steady rate — 500
÷ 15 = 33⅓ mL per second. Nothing about the tap changes when a bigger vessel is
put under it, so twenty times the water needs exactly twenty times the time.

Q10 One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same
land?

About ₹82,645.

1 acre = 43,560 sq ft costs ₹15,00,000

43560 : 1500000 :: 2400 : x

43560 × x = 2400 × 15,00,000

x = 36,00,00,00,000 ÷ 43560

x = ₹82,644.63 ≈ ₹82,645

Why the answer is so small a share: 2,400 square feet is only 2400⁄43560 ≈ 5.5% of an
acre, so it should cost about 5.5% of ₹15,00,000 — near ₹82,500, which matches.
Doing that rough check first tells you at once whether the long division has gone
astray.

Page 42 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Tip: The rate per square foot is 1500000 ÷ 43560 ≈ ₹34.44. Multiplying 2400 × 34.44
gives ₹82,656 — the small difference is only the rounding of the rate, which is why it
is safer to divide at the very end.

Q11 A tractor can plough the same area of a field 4 times faster than a pair of oxen. A
farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an
acre of land. How much time would it take if the farmer used a pair of oxen to
plough the field? How much time would it take him if he decides to use a tractor
instead?

Oxen: 120 hours. Tractor: 30 hours.

Oxen: 6 hours for 1 acre

1 : 6 :: 20 : t ⇒ t = 20 × 6 = 120 hours

The tractor is 4 times faster, so it needs one-fourth of the time:

Per acre: 6 ÷ 4 = 1.5 hours

For 20 acres: 20 × 1.5 = 30 hours, or simply 120 ÷ 4 = 30 hours

Two different relationships in one question: Area and time rise together — twice
the field, twice the hours — so that part is a straight Rule of Three. Speed and time
do the opposite: four times the speed means one-quarter of the time. Handling both
correctly in the same problem is the point of the question, and confusing them
would give the tractor 480 hours instead of 30.

Did you know? 120 hours is fifteen working days of eight hours; 30 hours is under
four. This is the arithmetic behind mechanisation changing the rhythm of the
farming year.

Page 43 of 51

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Class 8 Maths Chapter 7 Proportional Reasoning-1
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co m
m.
The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel
se
Q12

o m l a
are mixed in a 3 : 1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the
g kg, what is the cost
c copper is ₹906 per kg and the cost of nickel is ₹1,341 aper
cost .of
m
e these metals in a ₹10 coin?
las
of
ag
m

. co ag
m
Copper ≈ ₹5.26, nickel ≈ ₹2.59, together about ₹7.85.

as e
a g l
Total groups = 3 + 1 = 4, each group = 7.74 ÷ 4 = 1.935 g

co m
m.
Copper = 3 × 1.935 = 5.805 g

m as e
.co l
Nickel = 1 × 1.935 = 1.935 g
a g
a s em
a gl Copper: ₹906 per 1000 g → 5.805 × 906 ÷ 1000 = 5259.33 ÷ 1000 = ₹5.26
m
Nickel: ₹1341 per 1000 g → 1.935 × 1341 ÷ 1000 = 2594.835 ÷ 1000 = ₹2.59
a s
m .co agl
se
Total metal cost ≈ ₹7.85

g l a
a
m
Why a ₹10 coin is not made of ₹10 of metal: The metal in the coin is worth about
₹7.85, comfortably under its face value. That gap matters: if the metal were worth
. co
se m
o m l a
more than ₹10, it would pay to melt coins down rather than spend them, and the
g the mass with exactly
m .c would vanish from circulation. Mints choose the alloyaand
asethis ratio in mind.
coin

agl
se m
com
Check it yourself: Copper is 3⁄4 of the mass but only 5.26⁄7.85 ≈ 67% of the cost,
g l a
m . a
e
as ratios, and they need not agree.
because nickel is the dearer metal — ₹1,341 against ₹906 per kilogram. Mass ratio

a g l
and cost ratio are two different

co m
m .
m
Puzzle Time — Binairo
as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Binairo (Takuzu) — end of Part I

TRY THIS

Q1 Solve the following Binairo puzzles: (first grid — a vertical line in row 2 column 3;
horizontal lines in row 2 columns 5 and 6; a horizontal line in row 3 column 2;
vertical lines in row 4 columns 3 and 5; horizontal lines in row 5 columns 1 and 5; a
vertical line in row 6 column 3)

Here is the completed grid. Shaded cells are the ones printed in the book.

Puzzle 1 solved. The shaded cells are the ones the book gives; every row and every column has three
vertical and three horizontal lines, no three alike are adjacent, and all six rows and all six columns are
different.

Page 45 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

The chain of reasoning, using V for a vertical line and H for a horizontal line:

Row 2 already has H at columns 5 and 6, so column 4 cannot be H — three in a row is not
allowed. Column 4 is V. Now columns 3 and 4 are both V, so column 2 cannot be V; column 2
is H and column 1 is V. Row 2 is V H V V H H.
Column 3 now has V in rows 2, 4 and 6 — its full quota of three. Rows 1, 3 and 5 of column 3
are H.
Row 5 then has H at columns 1, 3 and 5 — three already — so columns 2, 4 and 6 are V: H V
H V H V.
Column 1 has V in rows 2 and 3, so rows 1 and 4 must be H; that leaves row 6 as V to make
three of each.
Carrying on the same way — count to three, never allow three alike side by side, and keep
every row and every column different — fills the rest with no choice left anywhere.

Why the solution is unique: Every step above was forced, never guessed. Rule 1
(three of each in a line) and rule 2 (no three alike adjacent) together mean that as
soon as two identical symbols sit side by side, the cells at both ends are decided.
Rule 3 then removes the last ambiguity between two rows that would otherwise be
interchangeable.

Q2 Solve the following Binairo puzzles: (second grid — horizontal lines in row 1 column
1, row 2 columns 1 and 6, row 5 column 6; vertical lines in row 3 column 4, row 4
columns 2 and 3, row 6 columns 2 and 5)

The completed grid:

Page 46 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Puzzle 2 solved. The shaded cells are the ones the book gives; every row and every column has three
vertical and three horizontal lines, no three alike are adjacent, and all six rows and all six columns are
different.

Row 1: H H V V H V

Row 2: H V V H V H

Row 3: V H H V H V

Row 4: H V V H H V

Row 5: V H H V V H

Row 6: V V H H V H

Page 47 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Column 1 begins with H in rows 1 and 2, so row 3 must be V.
Row 4 has V at columns 2 and 3, so columns 1 and 4 cannot be V — both are H.
Row 6 has V at columns 2 and 5; filling in the rest so that no three alike touch, and so that
row 6 differs from every other row, leaves only V V H H V H.
Each row and each column ends with exactly three vertical and three horizontal lines, and all
six rows and all six columns are different from one another.

Check it yourself: Read down each column — H H V H V V, H V H V H V, V V H V H H,
V H V H V H, H V H H V V, V H V V H H. Six different columns, three of each symbol in
every one, and nowhere three alike in a row.

Q3 Solve the following Binairo puzzles: (third grid — a horizontal line in row 1 column 5,
row 2 column 6, row 3 columns 1, 3 and 6; vertical lines in row 5 columns 3 and 4,
and row 6 column 1)

The completed grid:

Page 48 of 51

as e
Class 8 Maths Chapter 7 Proportional Reasoning-1
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g

co m
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
ag

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
.
Puzzle 3 solved. The shaded cells are the ones the book gives; every row and every column has three
m a
ase
vertical and three horizontal lines, no three alike are adjacent, and all six rows and all six columns are

agl
different.

co m
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Row 1: H V H V H V
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Row 2: V H V H V H
m Row 3: H V H V V H a g
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m
Row 4: V H V H H V

m a s e
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Row 5: H H V V H V
e m
Row 6: V V H H V H
g l as
a

co m
m .
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Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Row 3 is given H at columns 1, 3 and 6 — that is all three of its horizontal lines, so columns 2,
4 and 5 must be V.
Row 5 has V at columns 3 and 4, so column 2 and column 5 cannot be V; that starts the row
as H H V V H, and the count of three forces column 6 to be V.
Working down each column and keeping the three-of-each count gives the rest with no
choices left.

Why counting beats guessing: In a 6 by 6 Binairo every line holds exactly three of
each symbol. The instant three of one kind are placed in a line, every remaining cell
in it is settled. Looking for the line that is closest to its quota — as row 3 was here —
is the fastest way in, and it is the same habit that makes a proportion easy: find the
term you can pin down first.

Chapter at a glance
Two figures look alike when their measurements change by the same factor
(multiplication). Changing both by the same amount (subtraction) distorts them — that is
exactly why image B, made by taking 20 mm off both sides of A, looks elongated.
A ratio a : b says that for every a units of the first quantity there are b units of the second.
Dividing both terms by their HCF puts the ratio in its simplest form, which is the fingerprint
used to compare ratios.
Two ratios are in proportion, written a : b :: c : d, when their simplest forms agree.
Algebraically this is the same as ad = bc — the cross multiplication rule, which the chapter
derives rather than states.
Trairasika, the Rule of Three: given three of the four terms, the fourth is d = bc⁄a. Āryabhaṭa
put it as ichchhāphala = (phala × ichchhā) ÷ pramāṇa.
To divide a quantity x in the ratio m : n, cut x into m + n equal groups; the two parts are m ×
x⁄(m+n) and n × x⁄(m+n).
Not every problem with four numbers is a proportion. When speed rises the time for a fixed
journey falls, so 50 : 2 :: 75 : __ is the wrong model. And terms can only be compared after
the units are made to match — 4 hours must become 240 minutes, 1 kg must become 1000
g.

Page 50 of 51

Class 8 Maths Chapter 7 Proportional Reasoning-1 AglaSem · NCERT Solutions

Quick revision

IDEA IN SYMBOLS FROM THIS CHAPTER WATCH OUT FOR

Ratio a:b Image A, width : height = Order matters — 60 : 40 is not the
60 : 40 same as 40 : 60

Terms of a ratio a and b 60 and 40 are the terms Both terms must be in the same
unit before you compare

Simplest form divide both terms by 60 : 40 → 3 : 2 (HCF = 20) Any common factor helps, but only
their HCF the HCF finishes the job

Proportion a : b :: c : d 60 : 40 :: 30 : 20 :: 90 : 60 Same multiplying factor, not the
same difference

Cross ad = bc 150 : 90 :: 240 : x ⇒ 150x = Each ratio must compare the same
multiplication 240 × 90 two quantities, in the same order

Rule of Three d = bc⁄a 120 : 15 :: 80 : 10 (kg of Only for quantities that rise and
(Trairasika) rice) fall together

Sharing x in the m × x⁄(m+n) and n × ₹4,000 in 3 : 1 → ₹3,000 The whole is m + n groups — not
ratio m : n x⁄(m+n) and ₹1,000 m groups, not n

Similar figures every length × the Images A, C, D are all 3 : 2 Adding or subtracting the same
same factor length changes the shape

Inverse situation speed × time = 50 km/h for 2 h ⇒ 75 km/h More speed means less time — the
distance (fixed) for 1 h 20 min Rule of Three fails here

Unit conversion match the units first 1 acre = 43,560 sq ft; 1 L = 4 hours = 240 minutes before
1000 mL; 1 m = 3.281 ft pairing it with 150 minutes

Page 51 of 51