NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers is a helpful resource for students starting their Class 6 Maths journey. This chapter is part of the CBSE Syllabus Class 6 Maths and introduces important number concepts that form the base for future mathematics. In Class 6 Maths Chapter 1, students learn about large numbers, place value, comparing numbers, using commas in big numbers, expanding numbers, and rounding off numbers to the nearest tens, hundreds, and thousands.

The Class 6 Maths Chapter 1 Solution is designed to make these topics easy to understand. With clear explanations and many practice questions, NCERT Solutions for Class 6 Maths Chapter 1 helps students practice regularly and build strong problem-solving skills. These Class 6 Maths NCERT Solutions Chapter 1 are prepared by expert teachers and follow the latest CBSE syllabus to ensure students are well-prepared for exams.

For easy access, students can download the Class 6 Maths Chapter 1 PDF from Infinity Learn’s website. Along with Class 6 Maths Chapter 1 Solutions, students can also find NCERT Solutions Class 6 Science and many other study materials to boost their learning.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF Download

Get comprehensive understanding of basic number concepts with the NCERT Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers. PDF of NCERT solutions for class 6 Maths chapter 1 resource is crafted by expert Maths teachers to help you grasp the fundamentals of large numbers, place values, rounding off numbers, and more. Perfect for self-paced learning and thorough practice, these solutions are designed to boost your confidence and performance in exams.

NCERT Solutions for Class 6 Maths Chapter 1 Questions and Answers

1.1 Introduction

This section acquaints students with the notion of counting and representation of large numbers. It provides historical insight into how numerical systems have developed and underscores the importance of numbers in various contexts.

1.2 Comparing Numbers

The chapter progresses to comparing numbers, where students engage with exercises designed to recognize the greatest and smallest numbers in a set. They also dive into the concept of place values and how these determine the size of a number.

• 1.2.1 How Many Numbers Can You Make? This part invites students to explore permutations of numbers to understand which combinations produce the largest or smallest possible numbers.
• 1.2.2 Shifting of Numbers: Students learn the effects of changing the order of digits within a number and how this can affect its value.
• 1.2.3 Introducing 10,000: The concept of five-digit numbers is introduced here, laying down the foundation for understanding the base-10 system.
• 1.2.4 Revisiting Place Value: This reinforces the place value system for numbers, extending to five-digit numbers, which is essential for comprehending the number system.
• 1.2.5 Introducing 1,00,000: Six-digit numbers are discussed, with emphasis on the numerical term ‘lakh’, a critical milestone in the Indian number system.
• 1.2.6 Larger Numbers: The journey of numbers continues as the chapter expands on the idea of creating and understanding numbers beyond lakhs.
• 1.2.7 An Aid in Reading and Writing Large Numbers: This subsection provides students with tools and methods to read and write large numbers effectively, using place values and commas.

1.3 Large Numbers in Practice: Practical applications of large numbers are showcased here, from estimating the number of grains in a sack to the population of a city.

• 1.3.1 Estimation: Teaches students how to estimate large numbers and make informed guesses.
• 1.3.2-1.3.7 Estimating to the Nearest Tens, Hundreds, and Thousands by Rounding Off: Students learn rounding off numbers to the nearest ten, hundred, and thousand, which is a useful skill in everyday calculations.

1.4 Using Brackets: This concept introduces the hierarchy in operations, teaching students the correct order to solve mathematical expressions.

1.5 Roman Numerals: The chapter wraps up with an introduction to Roman numerals, adding historical context to the number system and providing a different perspective on numerical representation.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers All Exercise

NCERT solutions for class 6 Maths chapter 1 Knowing Our Numbers cover the 3 exercises which are exercise 1.1, exercise 1.2, and exercise 1.3.

NCERT Solutions for Class 6 Chapter 1 Knowing Our Numbers Exercise 1.1

Here is the NCERT solutions for class 6 Maths exercise 1.1 textbook questions on Page No. 12

1. Fill in the blanks:

(a) 1 lakh = ………….. ten thousand.

(b) 1 million = ………… hundred thousand.

(c) 1 crore = ………… ten lakhs.

(d) 1 crore = ………… million.

(e) 1 million = ………… lakhs.

Sol.

(a) 1 lakh =ten ten thousand.

(b) 1 million ten hundred thousand.

(c) 1 crore ten ten lakh

(d) 1 crore ten million

(e) 1 million ten lakh

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

Sol.

(a) 73,75,307

(b) 9,05,41,000

(c) 7,52,21,302

(d) 58,423,202

(e) 23,30,010

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

(a) 87595762

(b) 8546283

(c) 99900046

(d) 98432701

Sol.

(a) 8,75,95,762

(b) 85,46,283

(c) 9,99,00,046

(d) 9,84,32,701

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

(a) 78921092

(b) 7452283

(c) 99985102

(d) 48049831

Sol.

(a) 78,921,092

(b) 7,452,283

(c) 99,985,102

(d) 48,049,831

NCERT Solutions for Class 6 Chapter 1 Knowing Our Numbers Exercise 1.2

Below is the simple and easy to understand NCERT solutions for class 6 Maths exercise 1.2 textbook questions on Page No. 16

1. 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 four days.

Sol.

Number of tickets sold on the first day – 1094

Number of tickets sold on the second day – 1812

Number of tickets sold on the third day – 2050

Number of tickets sold on the final day – 2751

Total number of tickets sold on all the four days 1094 +1812 +2050 + 2751 = 7707.

2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Sol.

Let

x be the additional runs needed.

Given that Shekhar has already scored 6980 runs and wishes to complete 10,000 runs, we can write the equation:

6980+ x=10000

Subtracting 6980 from both sides, we get:

x=10000−6980x=3020

So, Shekhar needs to score 3020 more runs to reach his goal of 10,000 runs.

3. 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?

Sol.

Margin of victory = Votes secured by the successful candidate – Votes secured by the rival

Margin of victory = 577,500 – 348,700

Margin of victory = 228,800 votes

Therefore, the successful candidate won the election by 228,800 votes.

4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 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?

Sol.

Total sale for both weeks = Sale in the first week + Sale in the second week

Total sale for both weeks = Rs 285,891 + Rs 400,768

Total sale for both weeks = Rs 686,659

The sale was greater in the second week.

Difference in the sale in both weeks = Sale in the second week – Sale in the first week

Difference in the sale = Rs 400,768 – Rs 285,891

Difference in the sale = Rs 114,877

Therefore, the sale in the second week was greater by Rs 114,877 than in the first week.

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

Sol.

Greatest 5-digit number = 76,432

Least 5-digit number = 23,467

Difference between the two numbers = 76,432 – 23,467

Difference between the two numbers = 52,965

Therefore, the difference between the greatest and the least 5-digit number is 52,965.

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

Sol.

Number of screws manufactured in January = Number of screws manufactured in a day x Number of days in January

Number of screws manufactured in January = 2,825 x 31

Number of screws manufactured in January = 87,575

Therefore, the machine produced 87,575 screws in January 2006.

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

Sol.

Total cost of purchasing radio sets = Number of radio sets x Cost per radio set

Total cost of purchasing radio sets = 40 x Rs 1,200

Total cost of purchasing radio sets = Rs 48,000

Money remaining with the merchant = Total money – Total cost of purchasing radio sets

Money remaining with the merchant = Rs 78,592 – Rs 48,000

Money remaining with the merchant = Rs 30,592

Therefore, money remaining with the merchant after purchasing radio sets is Rs 30,592.

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

Sol.

Difference between the multipliers = 65 – 56 = 9

Difference between the student’s answer and the correct answer = 7,236 x 9

Difference between the answers = 65,124

Therefore, the student’s answer was greater than the correct answer by 65,124.

9. 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?

Sol.

Total length of cloth in cm = Total length of cloth x 100

Total length of cloth in cm = 40 x 100 = 4000 cm

Length of cloth required for one shirt = 2 m 15 cm = 2 x 100 + 15 = 215 cm

Number of shirts that can be stitched = Total length of cloth / Length of cloth required for one shirt

Number of shirts = 4000 / 215 = 18 shirts

Remaining cloth = Total length of cloth – (Number of shirts x Length of cloth required for one shirt)

Remaining cloth = 4000 – (18 x 215)

Remaining cloth = 4000 – 3870

Remaining cloth = 130 cm

Therefore, 18 shirts can be stitched out of 40 m, and 1 m 30 cm of cloth is left.

10. 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?

Sol.

Weight of one box in g = (4 x 1000) + 500 = 4500 g

Maximum weight the van can carry in g = 800 x 1000 = 800,000 g

Number of boxes that can be loaded in the van = Maximum weight the van can carry / Weight of one box

Number of boxes = 800,000 / 4500 = 177 boxes

Therefore, 177 boxes can be loaded in the van.

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

Sol.

Total distance covered in one trip = Distance to school + Distance to house

Total distance covered in one trip = 1 km 875 m + 1 km 875 m = 3 km 750 m

Total distance covered in six days = Total distance covered in one day x Number of trips in six days

Total distance covered = 3 km 750 m x 6

Total distance covered = 22 km 500 m

Therefore, the total distance covered by the student in six days is 22 km 500 m.

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

Sol.

Total quantity of curd in ml = (4 x 1000) + 500 = 4500 ml

Capacity of one glass = 25 ml

Number of glasses that can be filled = Total quantity of curd / Capacity of one glass

Number of glasses = 4500 / 25 = 180 glasses

Therefore, 180 glasses can be filled with curd.

NCERT Solutions for Class 6 Chapter 1 Knowing Our Numbers Exercise 1.3

NCERT solutions for class 6 Maths exercise 1.3 focuses on practicing the concept of comparing and ordering numbers. This exercise helps students understand the relative values of different numbers and enhances their ability to arrange numbers in ascending or descending order.

1. Estimate each of the following using the general rule:

(a) 730 + 998

(b) 796 – 314

(c) 12904 + 2888

(d) 28292 – 21496

Sol.

(a) 730 + 998

Rounded to nearest hundreds:

730 is about 700

998 is about 1000

Estimated sum is about 700 + 1000 = 1700

(b) 796 – 314

Rounded to nearest hundreds:

796 is about 800

314 is about 300

Estimated difference is about 800 – 300 = 500

(c) 12904 + 2888

Rounded to nearest thousands:

12904 is about 13000

2888 is about 3000

Estimated sum is about 13000 + 3000 = 16000

(d) 28292 – 21496

Rounded to nearest thousands:

28292 is about 28000

21496 is about 21000

Estimated difference is about 28000 – 21000 = 7000

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

(a) 439 + 334 + 4317

(b) 108734 – 47599

(c) 8325 – 491

(d) 489348 – 48365

Sol.

(a) 439 + 334 + 4317

Rounded to nearest hundreds:

439 is about 400

334 is about 300

4317 is about 4300

Estimated sum is about 400 + 300 + 4300 = 5000

Rounded to nearest tens:

439 is about 440

334 is about 330

4317 is about 4320

Estimated sum is about 440 + 330 + 4320 = 5090

(b) 108734 – 47599

Rounded to nearest hundreds:

108734 is about 108700

47599 is about 47600

Estimated difference is about 108700 – 47600 = 61100

Rounded to nearest tens:

108734 is about 108730

47599 is about 47600

Estimated difference is about 108730 – 47600 = 61130

(c) 8325 – 491

Rounded to nearest hundreds:

8325 is about 8300

491 is about 500

Estimated difference is about 8300 – 500 = 7800

Rounded to nearest tens:

8325 is about 8330

491 is about 490

Estimated difference is about 8330 – 490 = 7840

(d) 489348 – 48365

Rounded to nearest hundreds:

489348 is about 489300

48365 is about 48400

Estimated difference is about 489300 – 48400 = 440900

Rounded to nearest tens:

489348 is about 489350

48365 is about 48370

Estimated difference is about 489350 – 48370 = 440980

3. Estimate the following products using the general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Sol.

(a) 578 × 161

Rounded to nearest hundreds:

578 is about 600

161 is about 200

Estimated product is about 600 × 200 = 120000

(b) 5281 × 3491

Rounded to nearest thousands:

5281 is about 5000

3491 is about 3500

Estimated product is about 5000 × 3500 = 17500000

(c) 1291 × 592

Rounded to nearest hundreds:

1291 is about 1300

592 is about 600

Estimated product is about 1300 × 600 = 780000

(d) 9250 × 29

Rounded to nearest hundreds:

9250 is about 9000

29 is about 30

Estimated product is about 9000 × 30 = 270000

NCERT Class 6 Maths Chapter 1 Solutions Summary

NCERT Class 6 Maths Chapter 1 – Knowing Our Numbers teaches many basic but important ideas about numbers. In this chapter, students will learn how to compare numbers, arrange them in ascending and descending order, understand place value, work with larger numbers, and estimate values. The chapter also explains how to use commas in large numbers, how to shift digits, estimate sums and products, and follow rules like BODMAS and the use of brackets in calculations.

The main goal of this chapter is to make sure students understand the basics of numbers well. This will help them solve problems more easily and do better in exams. The NCERT Solutions for Chapter 1 give clear explanations, simple examples, and practice exercises. These solutions are made by expert teachers who follow the latest CBSE syllabus and NCERT guidelines, so students get full and correct preparation.