# NCERT Solutions Class 6 Maths Chapter 1

## NCERT Solutions for Class 6 Maths Chapter 1 – Knowing Our Numbers

Chapter 1 of the NCERT Textbook “Knowing Our Numbers”begins with comparing numbers, shifting digits, revisiting place value, larger numbers, and so on.  The NCERT solutions of this chapter will help them to strengthen basic concepts and help them excel in the examination. Subject experts at Extramarks create the solutions for Class 6 Maths Chapter 1. This will help boost the student’s confidence in learning mathematical concepts at an early age.

As all the solutions are prepared based on the CBSE guidelines and as per the marking scheme provided by NCERT, these will help clear the concepts about use of commas, revisiting place value, large numbers, reading and writing large numbers etc. Additionally, students will get a fair idea of the exam pattern, get enough practice and need not look for any kind of help elsewhere. The NCERT Solutions for Class 6 Chapter 1 Maths can be accessed from Extramarks.

## NCERT Solutions for Class 6 Maths Chapter 1

The first chapter of NCERT’s Class 6 Maths book is “Knowing Our Numbers’. It is not so that students learn about numbers for the first time. They are already aware of various operations such as addition, multiplication, division and subtraction. In this particular chapter, students will be learning some advanced concepts such as ideas on larger numbers, operations using more significant numbers, comparisons, shifting of digits, commas, etc. Students will also be learning about the process of estimating numbers and rounding off. The beginning of the chapter includes topics on comparing numbers, commas, ascending and descending order, large numbers, reading and writing large numbers and  other concepts.

## Access NCERT Solutions for Class 6 Maths Chapter 1- Knowing Our Numbers

Subject experts and experienced teachers at Extramarks who draft the NCERT solutions  prepare NCERT solutions after conducting extensive research. Students can rely on these solutions to understand every topic with examples, gamified content and graphics to make it colourful and interesting. These solutions also help students in completing their assignments and adequately prepare for their exams. The solutions are available at the official website of Extramarks.

## NCERT Solutions Class 6 of Maths Chapter-Wise

If you can practise the NCERT books exercise regularly, it will be easier to understand the concepts. You will cover all the topics equally and have a better idea about every topic given in the book. The total number of questions in every exercise of Chapter 1 is given below::

 Chapter 1 – Knowing Our Numbers Exercises Exercise 1.1 4 Questions & Solutions Exercise 1.2 12 Questions & Solutions Exercise 1.3 3 Questions & Solutions

As the chapter includes more significant numbers, students need to learn the usage of commas and should have a clear idea about the different number systems: Indian and International. As soon as students understand the formation of the numbers, they can move on to comparisons and arrangement  of numbers. Usually, this is done by ascending and descending order, shifting digits to form new numbers. Students will be learning about reading and writing large numbers using tablets and other indicators. The chapter-wise solutions are  available on Extramarks.

## Knowing Our Numbers: NCERT Class 6 Maths Chapter 1 Solutions Summary

Numbers involve counting and measuring. With the help of numbers, it becomes easy to perform simple operations such as addition, subtraction, multiplication and division. In this chapter, students will learn how to compare numbers and expand the numbers. Also, make the largest and the smallest numbers using digits.

### Chapter 1 – Knowing Our Numbers

Chapter 1 “knowing our numbers” covers many important topics. These include ideas on larger numbers, operations using more significant numbers, comparisons, shifting of digits, commas, etc. The NCERT Solutions for Chapter 1 Maths help students in understanding all the concepts easily.

#### 1.1 Introduction

The first part is the basic introduction of the chapter “Knowing Our Numbers”. Here one can revise the old concepts of numbers, such as the Indian system, the International system and many more. Solved sums are given in the NCERT textbook and the exercise to ensure that students understand the concepts and the topics mentioned.

#### 1.2 Comparing Numbers

Once the students know the number system and its structure, then they learn about comparisons. The comparison of more significant numbers using the number systems and the arrangement of numbers in ascending and descending order.

#### 1.3 Large Numbers in Practise

Large numbers in practice involve practising through real-life examples.

#### 1.4 Using Brackets

The use of brackets and the Indian and international numeration system are covered in this section.

#### 1.5 Roman Numerals

The Roman numerals involve the Hindu Arabic system of numerals. This is a particular system of writing numerals in the form of the Roman number system. There are specific rules which are listed in this section. Students are advised to keep these rules in mind while writing or reading the roman numerals.

Q.1 Fill in the blanks:

(a) 1 lakh = _______ ten thousand.

(b) 1 million = _______ hundred thousand.

(c) 1 crore = _______ ten lakh.

(d) 1 crore = _______ million.

(e) 1 million = _______ lakh.

Ans-

(a) 1 lakh = __10___ ten thousand.

(b) 1 million = __10__ hundred thousand.

(c) 1 crore = __10__ ten lakh.

(d) 1 crore = _10__ million.

(e) 1 million = _10__ lakh.

Q.2 Place commas correctly and write the numerals:

(a) Seventy three lakh seventy five thousand three hundred seven.

(b) Nine crore five lakh forty one.

(c) Seven crore fifty two lakh twenty one thousand three hundred two.

(d) Fifty eight million four hundred twenty three thousand two hundred two.

(e) Twenty three lakh thirty thousand ten.

Ans-

(a) 73,75,307

(b) 9,05,00,041

(c) 7,52,21,302

(d) 58,423,202

(e) 23,30,010

Q.3 Insert commas suitably and write the names according to Indian System of Numeration:

(a) 87595762 (b) 8546283

(c) 99900046 (d) 98432701

Ans-

(a) 8,75,95,762

Eight crore seventy five lakh ninety five thousand seven hundred sixty two

(b) 85,46,283

Eighty five lakh forty six thousand two hundred eighty three

(c) 9,99,00,046

Nine crore ninety nine lakh forty six

(d) 9,84,32,701

Q.4 Nine crore eighty four lakh thirty two thousand seven hundred one

Insert commas suitably and write the names according to International System of Numeration:

(a) 78921092 (b) 7452283

(c) 99985102 (c) 48049831

Ans-

(a) 78,921,092

Seventy eight million nine hundred twenty one thousand ninety two

(b) 7,452,283

Seven million four hundred fifty two thousand two hundred eighty three

(c) 99,985,102

Ninety nine million nine hundred eighty five thousand one hundred two

(d) 48, 049,831

Forty eight million forty nine thousand eight hundred thirty one

Q.5 A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Ans-

$Tickets sold on 1 st day=1094 Tickets sold on 2 nd day=1812 Tickets sold on 3 rd day=2050 Tickets sold on 4 th day=2751 Total tickets sold=1094+1812+2050+2751 Now, 1094 1812 2050 + 2751 ¯ 7707 ∴Total tickets sold in four days at book exhibition=7,707 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@112E@$

Q.6 Shekhar is a famous cricket player. He has so far scored 6,980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Ans-

Runs scored = 6,980

Runs desired = 10,000

Therefore,

Runs needed = 10,000 – 6,980 = 3,020

Q.7 In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

Ans-

$\begin{array}{l}\text{Votes secured by successful candidate}=\text{5},\text{77},\text{5}00\\ \text{Votes secured by nearest rival}=\text{3},\text{48},\text{7}00\\ \text{Margin}=\text{5},\text{77},\text{5}00-\text{3},\text{48},\text{7}00\\ \text{Now,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5},\text{77},\text{5}00\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{-\text{3},\text{48},\text{7}00}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,28,800\\ \text{Hence successful candidate won the election by}\\ \text{2,28,800 votes}\text{.}\end{array}$

Q.8 Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.

Ans-

Let us form the greatest and the least number using the digits 6, 2, 7, 4, 3 only once.

So, we have

Greatest number = 76432

Least number = 23467

Difference = 76432 – 23467 = 52965

Therefore, the required difference is 52,965.

Q.9 A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Ans-

Screws manufactured in one day = 2,825

Days in January = 31

Screws produced in 31 days = 2825 × 31 = 87575

Therefore, screws produced in the month of January = 87,575

Q.10 A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?

Ans-

(7236 × 65) – (7236 × 56) = 7236 × (65 – 56)

= 7236 × 9

= 65,124

Q.11 To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?

Ans-

Cloth required for one shirt = 2 m 15 cm

= 200 cm + 15 cm

= 215 cm

Total cloth = 40 m = 40 × 100 = 4000 cm

Number of shirts that can be stitched out of 4000 cm =

$4000÷215= 18 130 215 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeaabaWaaqaafeaakeaacaqG0aGaaGimaiaaicdacaaIWaGaey49aGRaaeOmaiaabgdacaqG1aGaeyypa0JaaiiOaiaaigdacaaI4aWaaSaaaeaacaaIXaGaaG4maiaaicdaaeaacaaIYaGaaGymaiaaiwdaaaaaaa@486D@$

Therefore, 18 shirts can be stitched and 130 cm cloth will remain.

Q.12 Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?

Ans-

$We know that 1 kg=1000 g Therefore, 4 kg 500 g=(4×1000) g+500 g =4000+500=4500 g and 800 kg=800×1000 g=8,00,000 g Number of boxes that can be loaded in the van =800000÷4500 =177 3500 4500 Hence, 177 boxes at maximum can be loaded in the van. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3961@$

Q.13 The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.

Ans-

Distance between school and house = 1 km 875 m

Now, 1 km = 1000 m

So, 1 km 875 m = (1000 + 875) m = 1875 m

Distance covered in 1 day = 1875 × 2 = 3750 m

Distance covered in 6 days = 3750 × 6 = 22,500

Therefore, distance covered in 6 days = 22,500 m

= 22 km 500 m

Q.14 A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?

Ans-

We know that

1 litre = 1000 ml

Therefore,

Capacity of vessel = 4 litres and 500 ml

= (4000 + 500) ml

= 4500 ml

Capacity of a glass = 25 ml

Number of glasses that can be filled = 4500/25 = 180

Therefore, 180 glasses can be filled.

Estimate each of the following using general rule:

(a) 730 + 998 (b) 796 – 314 (c) 12,904 +2,888 (d) 28,292 – 21,496

Q.15 Make ten more such examples of addition, subtraction and estimation of their outcomes.

Ans-

(a) 730 + 998

We round off the given numbers to hundreds.

Number 730 rounds off to 700 and 998 rounds off to 1000.

Therefore,

Estimated sum = 700 + 1000 = 1700

(b) 796 – 314

We round off the given numbers to hundreds.

Number 796 rounds off to 800 and 314 rounds off to 300.

Therefore,

Estimated difference = 800 – 300 = 500

(c) 12904 + 2888

We round off the given numbers to thousands.

Number 12904 rounds off to 13000 and 2888 rounds off to 3000.

Therefore,

Estimated sum = 13000 + 3000 = 16000

(d) 28,292 − 21,496

We round off the given numbers to thousands.

Number 28,2962 rounds off to 28000 and 21,496 rounds off to 21000.

Therefore,

Estimated difference = 28000 − 21000 = 7000

Q.16 Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :

(a) 439 + 334 + 4,317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365

Make four more such examples.

Ans-

(a) 439 + 334 + 4317

We round off the given numbers to hundreds.

Number 439 rounds off to 400, 334 rounds off to 300 and 4317 rounds off to 4300.

Therefore,

Estimated sum = 400 + 300 + 4300 = 5000

Now, we round off the given numbers to tens.

Number 439 rounds off to 440, 334 rounds off to 330 and 4317 rounds off to 4320.

Therefore,

Estimated sum = 440 + 330 + 4320= 5090

(b) 1,08,734 − 47,599

We round off the given numbers to hundreds.

Number 1,08,734 rounds off to 1,08,700 and 47,599 rounds off to 47,600.

Therefore,

Estimated difference = 1,08,700 – 47,600 = 61,100

Now, we round off the given numbers to tens.

Number 1,08,734 rounds off to 1,08,730 and 47,599 rounds off to 47,600.

Therefore,

Estimated difference = 1,08,730 – 47,600 = 61,130

(c) 8325 − 491

We round off the given numbers to hundreds.

Number 8325 rounds off to 8300 and number 491 rounds off to 500.

Therefore,

Estimated difference = 8300 – 500 = 7,800

Now, we round off the given numbers to tens.

Number 8325 rounds off to 8330 and number 491 rounds off to 490.

Therefore,

Estimated difference = 8330 – 490 = 7,840

(d) 4,89,348 − 48,365

We round off the given numbers to hundreds.

Number 4,89,348 rounds off to 4,89,300 and number 48,365 rounds off to 48,400.

Therefore,

Estimated difference = 4,89,300 – 48,400 = 4,40,900

Now, we round off the given numbers to tens.

Number 4,89,348 rounds off to 4,89,350 and number 48,365 rounds off to 48,370.

Therefore,

Estimated difference = 4,89,350 – 48,370 = 4,40,980

Q.17 Estimate the following products using general rule:

(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29

Make four more such examples.

Ans-

(a) 578 × 161

We round off the given numbers to hundreds.

Number 578 rounds off to 600 and number 161 rounds off to 200.

Therefore,

Estimated product = 600 × 200 = 1,20,000

(b) 5281 × 3491

We round off 5281 to nearest thousands and so 5281 rounds off to 5000.

We round off 3491 to nearest hundreds and so 3491 rounds off to 3500.

Therefore,

Estimated product = 5000 × 3500 = 1,75,00,000

(c) 1291 × 592

We round off the given numbers to hundreds.

Number 1291 rounds off to 1300 and number 592 rounds off to 600.

Therefore,

Estimated product = 1300 × 600 = 7,80,000

(d) 9250 × 29

We round off 9250 to 9000 and 29 to 30.

Therefore,

Estimated product = 9,000 × 30 = 2,70,000

Q.18 Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Ans-

Worth of Books sold in 1st week = ₹2,85,891

Worth of books sold in 2nd week = ₹4,00,768

Total sale = Sale in 1st week + Sale in 2nd week

= ₹2,85,891 + ₹4,00,768 = ₹6,86,659

Thus, the sale for the two weeks together was ₹6,86,659.

Also, 4,00,768 – 2,85,891 = 1,14,877

Thus, the sale in 2nd week was greater than the sale in 1st week by ₹1,14,877.

Q.19 A merchant had ₹78,592 with her. She placed an order for purchasing 40 radio sets at ₹1200 each. How much money will remain with her after the purchase?

Ans-

Cost of one radio set = ₹1200

So, cost of 40 radio sets = ₹1200 × 40 = ₹48000

Merchant had ₹78,592 with her before purchase.

So, after purchase, she will have ₹(78,592 – 48,000),i.e. ₹30,592.

Therefore, ₹30,592 will remain with her after the purchase.

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### 1. Where can I get the NCERT Class 6 Maths syllabus in Hindi?

You will get the Class 6 Maths syllabus from Extramarks’ official website.

### 2. How many chapters are there in the NCERT Maths Class 6 syllabus?

There are 14 chapters in the NCERT  Maths Class 6 syllabus.

### 3. What are the different topics covered in Chapter 1 of Class 6 Maths?

The different topics covered in NCERT solutions for Class 6 Maths are:

1. Introduction to numbers
2. Comparing numbers
3. Ascending order and Descending order
4. How many numbers can be formed using a certain number of digits?
5. Shifting digits
6. Place value
7. Larger Numbers and Estimates
8. Estimating sum or difference
9. Estimating products of numbers
10. BODMAS
11. Using Brackets