---
title: "NCERT Solutions Class 8 Maths Chapter 7 Proportional Reasoning-1"
url: https://www.swavid.com/maths/class/8/chapter/proportional-reasoning-1/ncert-solutions
dateModified: 2026-10-07T15:32:13+00:00
---

# NCERT Solutions Class 8 Maths Chapter 7 Proportional Reasoning-1

This chapter introduces the concept of proportional reasoning, exploring how ratios and proportions are used to compare quantities and solve real-world problems. It covers topics such as similarity in geometric shapes, unit conversions, and the application of the 'Rule of Three' in various contexts.

Free PDF (22 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-7-proportional-reasoning-1-32ee3c712c.pdf

## Observing Similarity in Change

### Question 1

*2 marks · Very short answer*

Which images look similar and which ones look different?

**Solution**

1. Images A, C, and D look similar because their widths and heights change by the same factor.
2. Images B and E look different as their dimensions do not change proportionally.

**Answer:** Images A, C, and D look similar, while images B and E look different.

> Common mistake: Confusing visual similarity with having the same size.

### Question 2

*2 marks · Very short answer*

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

**Solution**

1. No, images B and E do not look like the other three images.
2. Image B appears elongated and image E appears compressed and fatter due to non-proportional scaling.

**Answer:** No, they are slightly distorted because of unequal scaling of width and height.

> Common mistake: Stating that all rectangles look alike regardless of proportions.

### Question 3

*2 marks · Very short answer*

Why?

**Solution**

1. Images A, C, and D are rectangular with proportional width and height.
2. Image E is square, and image B has dimensions that do not scale by the same factor.

**Answer:** Images A, C, and D have proportional dimensions, whereas B and E do not maintain the same ratio of width to height.

> Common mistake: Ignoring the ratio of sides and only looking at shape type.

### Question 4

*2 marks · Very short answer*

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

**Solution**

1. Images A, C, and D have their widths and heights changed by the same factor.
2. Images B and E have width and height changes that are not proportional.

**Answer:** Widths and heights changing by the same factor makes images A, C, and D appear similar.

> Common mistake: Assuming absolute difference in dimensions determines similarity instead of the scaling factor.

### Question 5

*2 marks · Very short answer*

Can you check by what factors the width and height of image D change as compared to image A? Are the factors the same?

**Solution**

1. The width of image D is 90 mm and height is 60 mm, while image A has width 60 mm and height 40 mm.
2. Both width and height of image D are multiplied by a factor of $\frac{3}{2}$ compared to image A, so the factors are the same.

**Answer:** Both the width and height change by the factor $\frac{3}{2}$, so the factors are the same.

> Common mistake: Mixing up the order of division when finding the factor of change.

## Ratios

### Question 6

*2 marks · Very short answer*

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

**Solution**

1. To find the factor by which the ratio $60 : 40$ changes to $90 : 60$, we divide the first term of image D by the first term of image A.
2. Factor $= \frac{90}{60} = \frac{3}{2}$.

**Answer:** We should multiply both terms of the ratio $60 : 40$ by $\frac{3}{2}$ to get $90 : 60$.

> Common mistake: Subtracting instead of finding the multiplicative factor.

## Ratios in their Simplest Form

### Question 7

*2 marks · Very short answer*

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

**Solution**

1. The ratio of image B is $40 : 20$. Dividing both terms by their HCF, 20, we get $2 : 1$.
2. The ratio of image E is $60 : 60$. Dividing both terms by their HCF, 60, we get $1 : 1$.

**Answer:** The simplest form of the ratio of image B is $2 : 1$ and of image E is $1 : 1$.

> Common mistake: Dividing by a common factor other than the HCF and failing to reduce the ratio to its lowest terms.

## Problem Solving with Proportional Reasoning

### Question 8

*2 marks · Very short answer*

What is the HCF of 72 and 96?

**Solution**

1. The factors of $72$ are $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$.
2. The factors of $96$ are $1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96$.
3. The highest common factor of $72$ and $96$ is $24$.

**Answer:** The HCF of 72 and 96 is 24.

> Common mistake: Finding a common factor that is not the highest one.

### Question 9

*2 marks · Very short answer*

To make the lemonade with the same sweetness, how many spoons of sugar should she add?

**Solution**

1. The ratio of glasses of lemonade to spoons of sugar is $6 : 10$.
2. For $18$ glasses of lemonade, the proportion is $6 : 10 :: 18 : x$.
3. Dividing $18$ by $6$ gives the factor of change as $3$, so the second term becomes $10 \times 3 = 30$.

**Answer:** She should add 30 spoons of sugar.

> Common mistake: Multiplying the sugar by the wrong factor.

### Question 10

*2 marks · Very short answer*

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

**Solution**

1. To find the factor of change when a term increases from an initial value to a new value, we divide the new term by the initial term.
2. For example, if a first term increases from $6$ to $18$, the factor of change is $18 \div 6 = 3$.

**Answer:** We can find the factor of change by dividing the new term by the initial term.

> Common mistake: Dividing the initial term by the new term instead.

### Question 11

*2 marks · Very short answer*

Is Nitin correct in his thinking?

**Solution**

1. We compare the ratio of the length of the wall to the bags of cement used for both Nitin and Hari.
2. Nitin's ratio is $60 : 3$, which simplifies to $20 : 1$.
3. Hari's ratio is $40 : 2$, which simplifies to $20 : 1$, so the ratios are proportional and Nitin is incorrect.

**Answer:** No, Nitin is not correct because the ratios of wall length to cement bags are proportional.

> Common mistake: Comparing absolute quantities instead of the ratios.

### Question 12

*2 marks · Very short answer*

Count the number of teachers and students in your school. What is the ratio of teachers to students in your school? Write it below. ______ : ______

**Solution**

1. Count the total number of teachers in the school.
2. Count the total number of students in the school.
3. Write the counts in the form of a ratio of teachers to students.

**Answer:** 5 : 170 (or according to the specific school data).

> Common mistake: Writing the ratio of students to teachers instead of teachers to students.

### Question 13

*2 marks · Very short answer*

Is the teacher-to-student ratio in your school proportional to the one in my school?

**Solution**

1. Write the teacher-to-student ratio of your school in its simplest form.
2. Compare this simplest form with the given ratio of $5 : 170$ (or $1 : 34$).
3. Check if both ratios are equal to determine if they are proportional.

**Answer:** Answers may vary depending on the school's specific numbers.

> Common mistake: Not reducing the ratios to their simplest form before comparing.

### Question 14

*2 marks · Very short answer*

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? ______ : ______

**Solution**

1. Measure the width of the classroom blackboard to the nearest cm.
2. Measure the height of the classroom blackboard to the nearest cm and write the ratio as width : height in its simplest form.

**Answer:** The ratio of width to height of the blackboard is approximately $3 : 2$ (depending on classroom measurements).

> Common mistake: Writing height to width instead of width to height.

### Question 15

*2 marks · Very short answer*

Can you draw a rectangle in your notebook whose width and height are proportional to the ratio of the blackboard?

**Solution**

1. Yes, we can draw a smaller or bigger rectangle in the notebook.
2. Ensure the width and height are multiplied by the same factor so that the ratio remains the same as the blackboard.

**Answer:** Yes, we can draw a rectangle whose width and height are proportional to the blackboard.

> Common mistake: Changing only one dimension by a factor.

### Question 16

*2 marks · Very short answer*

Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same?

**Solution**

1. Compare the drawn rectangles with classmates.
2. They may differ in size but will have the same shape because their width-to-height ratios are proportional.

**Answer:** No, they do not all look the same in size, but they have the same shape as their ratios are proportional.

> Common mistake: Confusing same shape with same size.

### Question 17

*2 marks · Very short answer*

What is the ratio of Neelima’s age to her mother’s age? What would be the ratio of their ages when Neelima is 12 years old? Would it remain the same?

**Solution**

1. When Neelima is $3$ years old and her mother is $30$, the ratio is $3 : 30$, which simplifies to $1 : 10$.
2. When Neelima is $12$ years old ($9$ years later), her mother is $39$, giving a ratio of $12 : 39$, which simplifies to $4 : 13$. Thus, the ratio does not remain the same.

**Answer:** The initial ratio is $1 : 10$, and the ratio when Neelima is $12$ is $4 : 13$. No, the ratio does not remain the same.

> Common mistake: Assuming adding the same number to both ages keeps the ratio proportional.

### Question 18

*2 marks · Very short answer*

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

**Solution**

1. To find the factor $y$ such that $14y = 6$, solve for $y$.
2. We get $y = \frac{6}{14} = \frac{3}{7}$, which is a fraction and not an integer.

**Answer:** We need to multiply by $\frac{3}{7}$, which is a fraction and not an integer.

> Common mistake: Thinking that factors between ratios must always be whole numbers.

### Question 19

*2 marks · Very short answer*

Why is this coffee stronger?

**Solution**

1. Compare the amount of coffee decoction to the amount of milk in the mixture.
2. A stronger coffee has a higher proportion of coffee decoction relative to milk compared to the regular coffee.

**Answer:** The coffee is stronger because it contains a higher proportion of coffee decoction relative to milk.

> Common mistake: Looking only at the total volume instead of the ratio of ingredients.

### Question 20

*2 marks · Very short answer*

Why is this coffee lighter?

**Solution**

1. The ratio of coffee decoction to milk is lower, meaning there is more milk relative to the coffee decoction.
2. As the proportion of milk increases, the coffee mixture becomes lighter in strength and color.

**Answer:** The coffee is lighter because the proportion of milk is higher compared to the coffee decoction.

> Common mistake: Confusing coffee decoction quantity with milk quantity.

## Figure it Out

### Question 21

*1 mark · MCQ*

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

**Solution**

1. Two ratios $a : b$ and $c : d$ are proportional if $ad = bc$.
2. Checking the given options: (i) $4 \times 21 = 84$ and $7 \times 12 = 84$, (iv) $21 \times 10 = 210$ and $6 \times 35 = 210$, and (vi) $24 \times 3 = 72$ and $8 \times 9 = 72$ are true.

**Answer:** (i), (iv) and (vi) are true

> Common mistake: Multiplying incorrectly or confusing the order of terms in cross multiplication.

### Question 22

*2 marks · Very short answer*

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

**Solution**

1. Multiply both terms of the ratio $4 : 9$ by the same numbers such as 2, 3, and 4.
2. We get $8 : 18$, $12 : 27$, and $16 : 36$.

**Answer:** $8 : 18$, $12 : 27$, $16 : 36$

> Common mistake: Multiplying only one term of the ratio by a factor.

### Question 23

*1 mark · Fill in the blank*

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

**Solution**

1. The given ratio is $18 : 24$, which in simplest form is $3 : 4$.
2. Using the proportion with $18 : 24$, the missing numbers are $4$, $16$, $\frac{80}{3}$, and $36$.

**Answer:** $4$, $16$, $\frac{80}{3}$, $36$

> Common mistake: Failing to simplify the ratio before finding the missing terms.

### Question 24

*2 marks · Very short answer*

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.

**Solution**

1. Measure the length and width of each rectangle using a scale.
2. Calculate the ratio of length to width for each rectangle and compare them to find which are equal.

**Answer:** Rectangles with the same width-to-height ratio are similar to each other.

> Common mistake: Using subtraction instead of ratios to compare rectangle dimensions.

### Question 25

*2 marks · Very short answer*

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

**Solution**

1. Draw smaller and bigger rectangles by multiplying or dividing both width and height by the same factor.
2. Different sizes can have the same shape and remain proportional.

**Answer:** Yes, different sizes can have the same width-to-height ratio.

> Common mistake: Changing only one dimension while scaling.

### Question 26

*2 marks · Very short answer*

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

**Solution**

1. Count the number of grey and coloured bricks in one repeating block of the pattern.
2. Write the ratio and reduce it to its simplest form by dividing by the HCF.

**Answer:** (a) $3 : 2$, (b) $4 : 3$

> Common mistake: Counting bricks across the entire wall instead of a single repeating block.

### Question 27

*Activity*

Let us draw some human figures. Measure your friend’s body—the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below— head : torso ______ : ______ torso : arms ______ : ______ torso : legs ______ : ______

**Solution**

1. Measure the lengths of the head, torso, arms, and legs of a friend using a scale.
2. Write the measured lengths in the specified ratio forms: head to torso, torso to arms, and torso to legs.

**Answer:** Ratios of body parts are obtained by measuring and simplifying the lengths.

### Question 28

*2 marks · Very short answer*

Does the drawing look more realistic if the ratios are proportional? Why? Why not?

**Solution**

1. Yes, the drawing looks more realistic when the ratios are proportional because the relative sizes of different body parts match the natural proportions of the human body.

**Answer:** Yes, the drawing looks more realistic because proportional ratios maintain the natural relative sizes of body parts.

> Common mistake: Ignoring the relative scale of body parts.

### Question 29

*2 marks · Very short answer*

What is the factor of change in the first term?

**Solution**

1. The factor of change is found by dividing the new term by the original term to get the multiplier.

**Answer:** The factor of change is found by dividing the new quantity by the original quantity.

> Common mistake: Subtracting instead of dividing to find the factor of change.

### Question 30

*2 marks · Very short answer*

Is this the right way to formulate the question?

**Solution**

1. No, because the units of time must be the same in both ratios before setting up a proportion.

**Answer:** No, because the units of time in both ratios are different and must be converted to be the same.

> Common mistake: Comparing quantities with different units directly.

### Question 31

*2 marks · Very short answer*

How can you find the distance covered in 240 minutes?

**Solution**

1. We can find the distance by setting up a proportion with consistent units and using cross multiplication or finding the factor of change.

**Answer:** We can find the distance by using cross multiplication on the proportional ratios.

> Common mistake: Using incorrect units for time or distance.

### Question 32

*2 marks · Very short answer*

Are the weight-to-price ratios in both places proportional? Which tea is more expensive?

**Solution**

1. First convert weights to the same unit, find the simplest form of the weight-to-price ratios to check proportionality, and then compare the price for a standard weight to find which is more expensive.

**Answer:** The weight-to-price ratios are not proportional, and the tea with the higher price for the same weight is more expensive.

> Common mistake: Comparing prices without converting weights to the same unit.

### Question 33

*2 marks · Very short answer*

Which tea is more expensive? Why?

**Solution**

1. The price of 1 kg of tea in Meghalaya is ₹800, while the cost of 1 kg of tea in Himachal Pradesh is ₹1,000.
2. Comparing the price for the same weight, the tea from Himachal Pradesh is more expensive.

**Answer:** The tea from Himachal Pradesh is more expensive because its cost per kilogram is higher.

> Common mistake: Comparing prices of unequal quantities without converting them to the same unit weight.

### Question 34

*Activity*

Activity 1: Take your favourite dish. Find out all the ingredients and their respective quantities needed to make the dish for your family. Suppose you are celebrating a festival and you want to invite 15 guests. Find out the quantities of the ingredients required to cook the same dish for them.

**Solution**

1. List the quantities of all ingredients needed for a family meal.
2. Scale each ingredient quantity proportionally by multiplying by the ratio of guests to family members.

**Answer:** Quantities of ingredients are scaled up proportionally for the guests.

### Question 35

*2 marks · Very short answer*

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

**Solution**

1. We know that 1 year has 52 weeks and 1 million = $10\text{ lakh}$.
2. The total distance travelled by the Earth in a year is $940\text{ million km} = 940,000,000\text{ km}$.
3. Let the distance travelled in a week be $x\text{ km}$, so $940,000,000 : 52 = x : 1$.
4. Solving for $x$, we get $x = \frac{940,000,000}{52} = 18,076,923\text{ km}$.

**Answer:** The Earth travels approximately $18,076,923\text{ km}$ in a week.

> Common mistake: Division error while converting annual distance to weekly distance.

### Question 36

*2 marks · Very short answer*

2. A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness.

**Solution**

1. From the figure in the textbook (p. 170), the total length of the walls is $12 + 12 + 12 + 15 + 9 + 15 + 9 + 9 + 9 + 6 = 108\text{ feet}$.
2. We are given that a $10\text{-foot}$ wall requires $1450\text{ bricks}$.
3. Let the number of bricks required for a $108\text{-foot}$ wall be $x$.
4. Using proportionality, $10 : 1450 :: 108 : x$, so $x = \frac{1450 \times 108}{10} = 15,660\text{ bricks}$.

**Answer:** The mason needs $15,660\text{ bricks}$ to build the house.

> Common mistake: Incorrect addition of wall lengths from the diagram.

### Question 37

*2 marks · Very short answer*

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 : __

**Solution**

1. No, this problem cannot be modelled as a direct proportion like 50 : 2 :: 75 : __.
2. Speed and time are inversely proportional, so as speed increases, time decreases.

**Answer:** No, it cannot be formed as a direct proportion because speed and time are inversely related.

> Common mistake: Treating speed and time as directly proportional quantities.

### Question 38

*2 marks · Very short answer*

Would it take Puneeth’s father more time or less time to reach Kanpur? Think about it.

**Solution**

1. When speed increases, the time taken to cover the same distance decreases.
2. Therefore, it would take less time for Puneeth's father to reach Kanpur.

**Answer:** It would take less time to reach Kanpur.

> Common mistake: Assuming higher speed takes more time.

### Question 39

*Activity*

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.

**Solution**

1. Collect prices and volumes of different container sizes of the same shampoo brand and record them in a table.
2. Compare the ratios of volume to price for different container sizes to check if they are proportional.

**Answer:** The activity shows that the volume of shampoo is generally not proportional to its price.

### Question 40

*2 marks · Very short answer*

Why do you think that the ratio of the prices is not proportional to the ratio of the volumes?

**Solution**

1. Smaller packages like sachets have higher packaging, transportation, and marketing costs relative to the product volume.

**Answer:** Prices are not proportional to volumes because smaller containers incur relatively higher packaging and distribution costs.

> Common mistake: Assuming unit price remains constant across all container sizes.

### Question 41

*2 marks · Very short answer*

Discuss the pros and cons of different size bottles for the company and for customers. For reducing ecological footprint, what would you recommend to the company and to the customer?

**Solution**

1. Smaller packs are affordable and convenient for travel but create more plastic waste.
2. To reduce ecological footprints, larger refill packs or bulk containers are recommended.

**Answer:** Smaller packs offer convenience, while larger packs reduce the ecological footprint by minimizing plastic waste.

> Common mistake: Ignoring environmental impacts when discussing container sizes.

### Question 42

*2 marks · Very short answer*

Does the same occur for other products?

**Solution**

1. Yes, non-proportional pricing is common in many consumer goods such as milk, oil, and biscuits, where bulk purchases are cheaper per unit.

**Answer:** Yes, other consumer products also show non-proportional pricing across different pack sizes.

> Common mistake: Stating that all products follow strict proportional pricing.

### Question 43

*2 marks · Very short answer*

Observe the products for which the prices are proportional to the different measures.

**Solution**

1. Products sold strictly by weight or volume without branded packaging, such as loose grains or vegetables, often have prices proportional to their measures.

**Answer:** Prices of loose commodities and unbranded goods are more likely to be proportional to their measures.

> Common mistake: Assuming branded items maintain strict proportionality.

### Question 44

*2 marks · Very short answer*

Discuss in class the proportionality of prices to measures of the same product.

**Solution**

1. Discuss how economies of scale, packaging costs, and brand positioning affect the pricing of goods in different quantities.

**Answer:** Price proportionality depends on manufacturing, packaging, and distribution costs rather than just the quantity of the product.

> Common mistake: Treating price differences as purely random.

### Question 45

*Activity*

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.

**Solution**

1. Take 12 counters and share them between two people in different combinations to understand ratios and proportions.

**Answer:** Activity completed by sharing 12 counters in various ways.

### Question 46

*2 marks · Very short answer*

If you divide them equally, what is the ratio of the number of counters with each of you?

**Solution**

1. When 12 counters are divided equally, each person gets 6 counters.
2. The ratio of the number of counters with each person is $6 : 6$, which simplifies to $1 : 1$.

**Answer:** 1 : 1

> Common mistake: Writing $6 : 6$ without reducing it to its simplest form $1 : 1$.

### Question 47

*2 marks · Very short answer*

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

**Solution**

1. If the partner gets 5 counters out of a total of 12 counters, you will get $12 - 5 = 7$ counters.
2. The ratio of your partner's counters to your counters is $5 : 7$.

**Answer:** You get 7 counters, and the ratio is 5 : 7.

> Common mistake: Reversing the ratio order and writing $7 : 5$ instead of $5 : 7$.

### Question 48

*2 marks · Very short answer*

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?

**Solution**

1. To share 12 counters in the ratio of $3 : 1$, the partner takes 3 counters and you take 1 counter in each round.
2. Repeating this process until all 12 counters are exhausted, your partner gets 9 counters and you get 3 counters.

**Answer:** Partner gets 9 counters and you get 3 counters.

> Common mistake: Stopping after the first round instead of distributing all 12 counters.

### Question 49

*2 marks · Very short answer*

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

**Solution**

1. To divide 42 counters in the ratio $4 : 3$, find the total number of groups which is $4 + 3 = 7$.
2. Multiply the number of groups by the size of each group ($6$), so your partner gets $4 \times 6 = 24$ counters and you get $3 \times 6 = 18$ counters.

**Answer:** Partner gets 24 counters and you get 18 counters.

> Common mistake: Dividing 42 directly by 4 or 3 instead of dividing by the sum of the ratio terms.

### Question 50

*2 marks · Very short answer*

What is the size of each group?

**Solution**

1. The total number of counters is 42, and the total number of groups is $4 + 3 = 7$.
2. The size of each group is found by dividing the total quantity by the total number of groups: $42 \div 7 = 6$.

**Answer:** 6

> Common mistake: Dividing by the difference of the ratio terms instead of their sum.

### Question 51

*2 marks · Very short answer*

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

**Solution**

1. The total number of groups is $2 + 3 = 5$.
2. The size of each group is $\frac{4500}{5} = 900$.
3. The first part is $2 \times 900 = 1800$ and the second part is $3 \times 900 = 2700$.

**Answer:** The two parts are ₹1800 and ₹2700.

> Common mistake: Dividing the total by 2 and 3 separately instead of the sum of the ratio terms.

### Question 52

*2 marks · Very short answer*

2. In a science lab, acid and water are mixed in the ratio of 1 : 5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain?

**Solution**

1. The ratio of acid to water is $1 : 5$, so the total number of parts is $1 + 5 = 6$.
2. The size of each part is $\frac{240}{6} = 40\text{ mL}$.
3. The amount of acid is $1 \times 40 = 40\text{ mL}$ and water is $5 \times 40 = 200\text{ mL}$.

**Answer:** The solution contains $40\text{ mL}$ of acid and $200\text{ mL}$ of water.

> Common mistake: Dividing by 5 instead of the sum of the ratio terms.

### Question 53

*3 marks · Short answer*

3. Blue and yellow paints are mixed in the ratio of 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of these two colours are needed? To make the paint a lighter shade of green, I added 20 mL of yellow to the mixture. What is the new ratio of blue and yellow in the paint?

**Solution**

1. The total parts for green paint are $3 + 5 = 8$, so the size of each part for $40\text{ mL}$ is $\frac{40}{8} = 5\text{ mL}$.
2. Blue paint used is $3 \times 5 = 15\text{ mL}$ and yellow paint used is $5 \times 5 = 25\text{ mL}$.
3. After adding $20\text{ mL}$ of yellow, the new amount of yellow is $25 + 20 = 45\text{ mL}$, giving a new ratio of $15 : 45$, which simplifies to $1 : 3$.

**Answer:** $15\text{ mL}$ of blue and $25\text{ mL}$ of yellow are needed, and the new ratio of blue to yellow is $1 : 3$.

> Common mistake: Forgetting to add the extra yellow paint to the initial yellow quantity.

### Question 54

*2 marks · Very short answer*

4. To make soft idlis, you need to mix rice and urad dal in the ratio of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need?

**Solution**

1. The ratio of rice to urad dal is $2 : 1$, giving a total of $2 + 1 = 3$ parts.
2. The size of each part is $\frac{6}{3} = 2\text{ cups}$.
3. The required cups of rice are $2 \times 2 = 4$ and urad dal are $1 \times 2 = 2$.

**Answer:** You will need $4\text{ cups}$ of rice and $2\text{ cups}$ of urad dal.

> Common mistake: Dividing by 2 instead of 3.

### Question 55

*3 marks · Short answer*

5. I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3 : 5. I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?

**Solution**

1. The fraction of red paint in the original bucket is $\frac{3}{8}$ and yellow paint is $\frac{5}{8}$.
2. When another full bucket of yellow paint is added, the total quantity of yellow paint becomes $\frac{5}{8} + 1 = \frac{13}{8}$.
3. The new ratio of red paint to yellow paint is $\frac{3}{8} : \frac{13}{8}$, which is $3 : 13$.

**Answer:** The new ratio of red paint to yellow paint is $3 : 13$.

> Common mistake: Adding 1 to both terms of the ratio instead of only to the yellow paint.

### Question 56

*2 marks · Very short answer*

1. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in its simplest form.

**Solution**

1. The initial ratio of orange juice to apple juice is $600 : 900$.
2. The HCF of 600 and 900 is 300.
3. Dividing both terms by 300 gives the simplest form as $2 : 3$.

**Answer:** The ratio in its simplest form is $2 : 3$.

> Common mistake: Not dividing completely by the HCF.

### Question 57

*3 marks · Short answer*

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

**Solution**

1. Capacity of one bus is $\frac{162}{3} = 54$ students and teachers.
2. Let the number of buses needed for 204 students be $x$.
3. Using proportionality: $3 : 162 :: x : 204$.
4. By cross-multiplication, $162x = 3 \times 204$, which gives $x = \frac{612}{162} = 3.77$, so we need $4$ buses.
5. Total capacity of $4$ buses is $4 \times 54 = 216$ seats.
6. Since there are $204$ students, $216 - 204 = 12$ seats will remain vacant and all buses will not be completely full.

**Answer:** We will need 4 buses, and all the buses will not be full as 12 seats will remain vacant.

> Common mistake: Rounding down the number of buses instead of rounding up to accommodate all students.

### Question 58

*3 marks · Short answer*

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

**Solution**

1. Population density is found by dividing the population by the area of the city.
2. Number of persons per sq. km in Delhi = $\frac{30,000,000}{1,484} \approx 20,216$ people.
3. Number of persons per sq. km in Mumbai = $\frac{20,000,000}{550} \approx 36,364$ people.
4. Since the number of persons per sq. km is greater in Mumbai, Mumbai is more crowded.

**Answer:** Mumbai is more crowded because it has more people per square kilometre than Delhi.

> Common mistake: Comparing total population or total area directly instead of calculating population density.

### Question 59

*3 marks · Short answer*

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

**Solution**

1. The ratio of neck to the rest of the body for the crane is given as $4 : 6$.
2. The total parts corresponding to the total height are $4 + 6 = 10$ parts.
3. The fraction of the height that forms the neck is $\frac{4}{10}$.
4. For any given height, the height of the neck is $\frac{4}{10} \times \text{height}$.

**Answer:** The neck height would be $\frac{4}{10}$ of the total height.

> Common mistake: Dividing by 4 or 6 instead of the total sum of parts (10).

### Question 60

*3 marks · Short answer*

5. Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If 2 1/2 palas of saffron costs 3/7 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”

**Solution**

1. We set up the proportion where $2\frac{1}{2}$ palas costs $\frac{3}{7}$ niskas.
2. Let the unknown quantity of saffron for $9$ niskas be $x$ palas.
3. Using Āryabhaṭa's rule of three: $\text{pramāṇa} \times \text{ichchhāphala} = \text{phala} \times \text{ichchhā}$.
4. Here, $2\frac{1}{2} : \frac{3}{7} :: x : 9$.
5. Cross multiplying gives $\frac{3}{7} x = 2\frac{1}{2} \times 9 = \frac{5}{2} \times 9 = \frac{45}{2}$.
6. Solving for $x$, we get $x = \frac{45}{2} \times \frac{7}{3} = 15 \times \frac{7}{2} = \frac{105}{2} = 52.5$ palas.

**Answer:** 52.5 palas of saffron can be bought for 9 niskas.

> Common mistake: Errors in fraction arithmetic while multiplying and dividing mixed fractions.

### Question 61

*3 marks · Short answer*

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

**Solution**

1. Let Harmain's present age be $1$ year and her brother's present age be $5$ years.
2. Let the required number of years be $x$.
3. After $x$ years, Harmain's age will be $1 + x$ and her brother's age will be $5 + x$.
4. According to the question, the ratio of their ages after $x$ years is $1 : 2$, so $\frac{1 + x}{5 + x} = \frac{1}{2}$.
5. Cross multiplying gives $2(1 + x) = 1(5 + x)$, which simplifies to $2 + 2x = 5 + x$.
6. Solving for $x$, we get $x = 3$, and Harmain's age will be $1 + 3 = 4$ years.

**Answer:** After 3 years, Harmain will be 4 years old.

> Common mistake: Assuming the age difference changes with time or forgetting to add $x$ to both ages.

### Question 62

*3 marks · Short answer*

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

**Solution**

1. Given that $1$ litre of water has a mass of $1$ kg.
2. The ratio of the mass of equal volumes of gold and water is given as $37 : 2$.
3. Let the mass of $1$ litre of gold be $x$ kg.
4. We can write the proportion as $37 : 2 :: x : 1$.
5. By cross-multiplication, $2x = 37 \times 1$, which gives $x = \frac{37}{2} = 18.5$ kg.

**Answer:** The mass of 1 litre of gold is 18.5 kg.

> Common mistake: Inverting the ratio terms when setting up the proportion.

### Question 63

*2 marks · Very short answer*

8. It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy? (Please refer to the section on Unit Conversions earlier in this chapter).

**Solution**

1. Given 1 acre = 43,560 sq ft and 10 tonnes of manure is required for 1 acre.
2. Area of the plot = 200 ft × 500 ft = 100,000 sq ft.
3. Using proportion: 10 tonnes : 43,560 sq ft :: x tonnes : 100,000 sq ft.
4. x = (10 × 100,000) / 43,560 ≈ 22.96 tonnes.

**Answer:** The farmer should buy approximately 22.96 tonnes of cow manure.

> Common mistake: Forgetting to convert the plot area into acres or using the wrong conversion factor.

### Question 64

*2 marks · Very short answer*

9. A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 mL. How much time does the same tap take to fill a bucket of water if the bucket has a 10-litre capacity?

**Solution**

1. 10 litres = 10,000 mL.
2. Using proportion: 15 seconds : 500 mL :: x seconds : 10,000 mL.
3. x = (15 × 10,000) / 500 = 300 seconds.
4. 300 seconds = 5 minutes.

**Answer:** The tap takes 5 minutes to fill the bucket.

> Common mistake: Not converting litres to millilitres before setting up the proportion.

### Question 65

*2 marks · Very short answer*

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

**Solution**

1. 1 acre = 43,560 sq ft.
2. Using proportion: ₹15,00,000 : 43,560 sq ft :: x : 2,400 sq ft.
3. x = (15,00,000 × 2,400) / 43,560 ≈ ₹82,645.

**Answer:** The cost of 2,400 square feet of land is approximately ₹82,645.

> Common mistake: Using the wrong area conversion factor.

### Question 66

*2 marks · Very short answer*

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

**Solution**

1. Time taken by a pair of oxen for 20 acres = 20 × 6 hours = 120 hours.
2. A tractor is 4 times faster, so time taken by tractor = 120 / 4 = 30 hours.

**Answer:** It would take 120 hours for the oxen and 30 hours for the tractor.

> Common mistake: Confusing 'faster' with multiplying instead of dividing the time.

### Question 67

*2 marks · Very short answer*

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

**Solution**

1. Total parts = 3 + 1 = 4. Copper mass = (3/4) × 7.74 g = 5.805 g. Nickel mass = (1/4) × 7.74 g = 1.935 g.
2. Cost of copper = 5.805 g × (₹906 / 1000 g) ≈ ₹5.26.
3. Cost of nickel = 1.935 g × (₹1341 / 1000 g) ≈ ₹2.59.
4. Total cost = ₹5.26 + ₹2.59 = ₹7.85.

**Answer:** The cost of the metals in the ₹10 coin is approximately ₹7.85.

> Common mistake: Forgetting to convert the cost per kg to cost per gram.

### Question 68

*Activity*

Solve the following Binairo puzzles:

**Solution**

1. Binairo is a logic puzzle where each row and column must contain an equal number of horizontal and vertical lines.
2. No more than two of the same symbol can be adjacent, and each row and column must be unique.

**Answer:** The puzzle is solved by ensuring these rules are met for every row and column.

## Frequently asked questions

### How many total questions are there in NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning-1?

This chapter contains a total of 68 questions distributed across multiple sections. You can find all these solved step by step in SwaVid's free PDF and step-by-step solutions on this page only.

### Which topics do the questions in this chapter cover?

The questions cover concepts like observing similarity in change, ratios in their simplest form, and problem solving with proportional reasoning. Additionally, the Figure it Out section covers equivalent ratios, inverse proportion in speed and time, and dividing a quantity in a given ratio.

### What are the hardest question types in this chapter and how should I approach them?

The Figure it Out section includes challenging problems on inverse proportion in travel and adding quantities to a ratio mixture. You can approach them easily by using SwaVid's free PDF and step-by-step solutions on this page only, which break down each method clearly.

### How can I write answers to score full marks in Class 8 exams?

To secure full marks, you should clearly state the given ratios, show the factor of change, and write proper mathematical steps. Referring to SwaVid's free PDF and step-by-step solutions on this page only will help you understand the ideal presentation format.

### Is the free PDF for this chapter available according to the new NCERT book?

Yes, the solutions are fully updated for the new NCERT book as per the NCF 2023 guidelines for the 2026-27 session. You can access SwaVid's free PDF and step-by-step solutions on this page only for your daily practice.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student