# NCERT Solutions for Class 6 Maths Chapter 1

## NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers is prepared by well-experienced Maths teachers for the sake of 6th-grade students. It explains every concept of all chapters with plenty of solid questions and with a clarified explanation. It helps the students to understand slowly and to get practice well to become perfect and again a good score in their examination. Download Vedantu NCERT Book Solutions to get a better understanding of all the exercises questions. You can also register Online for NCERT Solutions Class 6 Science tuition on Vedantu.com to score more marks in your examination.

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Exercise 1.1

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

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) Seventy-three lakh seventy-five thousand three hundred seven  $=\text{ }73,75,307\text{ }$

(b) Nine crore five lakh forty-one $=\text{ }9,05,00,041$

(c) Seven crore fifty-two lakh twenty-one thousand three hundred two $=\text{ }7,52,21,302\text{ }$

(d) Fifty-eight million four hundred twenty-three thousand two hundred two $=\text{ }58,423,202$

(e) Twenty-three lakh thirty thousand ten $=\text{ }23,30,010$

3. Insert commas suitable 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) $9,84,32,701=$ Nine crore eighty-four lakh thirty-two thousand seven hundred one

4. Insert commas suitable and write the names according to International system of numeration:

(a) 78921092

(b) 7452283

(c) 99985102

(d) 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

Exercise 1.2

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 $\mathbf{1094},\mathbf{1812},\mathbf{2050}\text{ }\mathbf{and}\text{ }\mathbf{2751}$. Find the total number of tickets sold on all the four days.

Ans:

Tickets sold of first day $=\text{ }1,094$

Tickets sold of second day $=\text{ }1,812$

Tickets sold of third day = 2,050

Tickets sold of fourth day $=~2,751$

Total tickets sold \begin{align} & =\text{ }1,094+1,812+2,050+2,751 \\ & =\text{ }7,707 \\ \end{align}

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

Ans:

Runs scored by Shekhar $=\text{ }6980$

Total number of runs he need \begin{align} & =\text{ }10,000-6980 \\ & =\text{ }3020 \\ \end{align}.

3. In an election, the successful candidate registered $\mathbf{5},\mathbf{77},\mathbf{500}$ votes and his nearest rival secured $\mathbf{3},\mathbf{48},\mathbf{700}$ votes. By what margin did the successful candidate win the election?

Ans: Successful candidate secured votes $=\text{ }5,77,500$

Rival candidate secured votes $=\text{ }3,48,700$

Margin of votes \begin{align} & =\text{ }5,77,500-3,48,700 \\ & =\text{ }2,28,800 \\ \end{align}

4. Kirti Bookstore sold books worth $\mathbf{2},\mathbf{85},\mathbf{891}$ in the first week of June and books worth $\mathbf{4},\mathbf{00},\mathbf{768}$ 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: Books sold in week one $=\text{ }2,85,891$

Books sold in week second $=\text{ }4,00,768$

Total sale $=\text{ }4,00,768+2,85,891=6,86,659$

Clearly, sale of second week is more than the first week sale

Difference \begin{align} & =\text{ }4,00,768-2,85,891 \\ & =\text{ }1,14,877 \\ \end{align}

5. Find the difference between the greatest and the least number that can be written using the digits $\mathbf{6},\mathbf{2},\mathbf{7},\mathbf{4},\mathbf{3}$ each only once.

Ans: Largest five digit number  $=76432$

Smallest five-digit number  $=\text{ }23467$

Therefore, the difference between then  \begin{align} & =\text{ }76432-23467 \\ & =\text{ }52965 \\ \end{align}

6. A machine, on an average, manufactures $\mathbf{2},\mathbf{825}$screws a day. How many screws did it produce in the month of January $\mathrm{2006}$?

Ans:

Screws produced in one day $=\text{ }2,825$

Total Screws produced in $31$ days \begin{align} & =\text{ }2,825\times 31\text{ } \\ & =\text{ }87,575 \\ \end{align}

Hence, the machine produced $87,575$ screws in the month of January.

7. A merchant had $\mathbf{78},\mathbf{592}$with her. She placed an order for purchasing $\mathbf{40}$ radio sets at $\mathbf{1},\mathbf{200}$ each. How much money will remain with her after the purchase?

Ans:

Total money merchant had $=\text{ }Rs.\text{ }78,592$

Cost of one radio $=\text{ }Rs.\text{ }1200$

Cost of $40$ radios $=\text{ }1200\times 40=\text{ }Rs.\text{ }48,000$

Money left\begin{align} & =\text{ }78,592-48,000 \\ & =\text{ }30,592 \\ \end{align}

8. A student multiplied $\mathbf{7236}\text{ }\mathbf{by}\text{ }\mathbf{65}$ instead of multiplying $\mathbf{by}\text{ }\mathbf{56}$. By how much was his answer greater than the correct answer?

Ans:

Wrong answer \begin{align} & =\text{ }7236\times 65\text{ } \\ & =\text{ }470340 \\ \end{align}

Correct answer \begin{align} & =\text{ }7236\times 56\text{ } \\ & =\text{ }405216 \\ \end{align}

Difference in answers \begin{align} & =\text{ }470340-405216\text{ } \\ & =\text{ }65,124 \\ \end{align}

9. To stitch a shirt $\mathbf{2}\text{ }\mathbf{m}\text{ }\mathbf{15}\text{ }\mathbf{cm}$ cloth is needed. Out of $\mathbf{40}\text{ }\mathbf{m}$ cloth, how many shirts can be stitched and how much cloth will remain?

Ans:

Cloth required for stitch one shirt

\begin{align} & =\text{ }2\text{ }m\text{ }15\text{ }cm \\ & =\text{ }2\times 100\text{ }cm+15\text{ }cm \\ & =215cm \\ \end{align}

Total length of cloth \begin{align} & =\text{ }40\text{ }m\text{ } \\ & =\text{ }40\times 100\text{ }cm \\ & =4000\text{ }cm \\ \end{align}

Number of shirts that can be stitched = $\frac{4000}{215}$ 215\overset{18}{\overline{\left){\begin{align} & 4000 \\ & \frac{-215}{\begin{align} & 1850 \\ & \frac{-1720}{130} \\ \end{align}} \\ \end{align}}\right.}}

Hence, $18$ shirts can be stitched and $130\text{ }cm\text{ }\left( 1\text{ }m\text{ }30\text{ }cm \right)$ cloth will be left.

10. Medicine is packed in boxes, each weighing $\mathbf{4}\text{ }\mathbf{kg}\text{ }\mathbf{500}\text{ }\mathbf{g}$ . How many such boxes can be loaded in a can which cannot carry beyond $\mathbf{800}\text{ }\mathbf{kg}$?

Ans:

Weight of a box \begin{align} & =\text{ }4\text{ }kg\text{ }500\text{ }g\text{ } \\ & =\text{ }4500\text{ }g \\ \end{align}

Number of boxes $=\text{ }\frac{800000}{4500}$

4500\overset{177}{\overline{\left){\begin{align} & 800000 \\ & \frac{-4500}{\begin{align} & 35000 \\ & \frac{-31500}{\begin{align} & 35000 \\ & \frac{-31500}{3500} \\ \end{align}} \\ \end{align}} \\ \end{align}}\right.}}

Hence, $177$ boxes can be loaded in the Van.

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

Ans:

Distance between school and her house $=\text{ }1875\text{ }m$

Total distance covered \begin{align} & =\text{ }2\times 1875\text{ } \\ & =\text{ }3750\text{ }m \\ \end{align}

Distance covered in $6\text{ }days$ \begin{align} & =\text{ }6\times 3750\text{ } \\ & =\text{ }22500\text{ }m \\ \end{align}

Thus, she covers $22\text{ }km\text{ }500\text{ }m$ distance in $6\text{ }days$

12. A vessel has $\mathbf{4}\text{ }\mathbf{liters}\text{ }\mathbf{and}\text{ }\mathbf{500}\text{ }\mathbf{ml}$ of curd. In how many glasses each of $\mathbf{25}\text{ }\mathbf{ml}$ capacity, can it be filled?

Ans:

Capacity of vessel\begin{align} & =\text{ }4\text{ }liters\text{ }500\text{ }ml\text{ } \\ & =\text{ }4500\text{ }ml \\ \end{align}

Capacity of a glass $=\text{ }25\text{ }ml$

Number of glasses can be filled $=\text{ }\frac{4500}{25}$ 25\overset{180}{\overline{\left){\begin{align} & 4500 \\ & \frac{-25}{\begin{align} & 200 \\ & \frac{-200}{0} \\ \end{align}} \\ \end{align}}\right.}}

Therefore, $180$ glasses are required.

Exercise 1.3

1. Estimate each of the following using general rules:

$\begin{array}{*{35}{l}} \left( \mathrm{a} \right)\mathrm{ 730+998} \\ \left( \mathrm{b} \right)\mathrm{ 796--314} \\ \left( \mathrm{c} \right)\mathrm{ 12,904+2,888} \\ \left( \mathrm{d} \right)\mathrm{ 28,292--21,496} \\ \end{array}$

Ans:

(a) $730$ round off to $=700$

$998$ round off to $=1000$

Estimated sum $=\text{ }1700$

(b) $796$round off to $=800$

$314$ round off to $=300$

Estimated sum $=\text{ }500$

(c) $12904$ round off to $=13000$

$2888$ round off to $=3000$

Estimated sum $=\text{ }16000$

(d) $28292$ round off to $=28000$

$21496$ round off to $=21000$

Estimated difference $=\text{ }7000$

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

\begin{align} & \begin{array}{*{35}{l}} \left( \mathrm{a} \right)\mathrm{ 439+334+4317} \\ \left( \mathrm{b} \right)\mathrm{ 1,08,737--47,599} \\ \left( \mathrm{c} \right)\mathrm{ 8325--491} \\ \end{array} \\ & \left( \mathrm{d} \right)\mathrm{ 4,89,348--48,365} \\ \end{align}

Ans:

(a) Rough estimate by rounding off to nearest hundreds:

$439$ round off to $=400$

$334$ round off to $=300$

$4317$ round off to $=4300$

Estimated sum $=\text{ }5000$

Rough estimate by rounding off to nearest tens:

$439$ round off to $=440$

$334$ round off to $=330$

$4317$ round off to $=4320$

Estimated sum $=\text{ }5090$

(b) Rough estimate by rounding off to nearest hundreds:

$108734$ round off to $=108700$

$47599$ round off to $=47600$

Estimated difference $=\text{ }61100$

Rough estimate by rounding off to nearest tens:

$108734$ round off to $=108730$

$47599$ round off to $=47600$

Estimated difference $=\text{ }61130$

(c) Rough estimate by rounding off to nearest hundreds:

$8325$ round off to $=8300$

$491$ round off to $=500$

Estimated difference $=\text{ }7800$

Rough estimate by rounding off to nearest tens:

$8325$ round off to $=8330$

$491$ round off to $=490$

Estimated difference $=\text{ }7840$

(d) Rough estimate by rounding off to nearest hundreds:

$489348$ round off to $=489300$

$48365$ round off to $=48400$

Estimated difference $=\text{ }440900$

Rough estimate by rounding off to nearest tens:

$489348$ round off to $=489350$

$48365$ round off to $=48370$

Estimated difference $=\text{ 440980}$

3. Estimate the following products using general rule:

$\begin{array}{*{35}{l}} \left( \mathrm{a} \right)\mathrm{ 578 }\!\!\times\!\!\text{ 161} \\ \left( \mathrm{b} \right)\mathrm{ 5281 }\!\!\times\!\!\text{ 3491} \\ \left( \mathrm{c} \right)\mathrm{ 1291 }\!\!\times\!\!\text{ 592} \\ \left( \mathrm{d} \right)\mathrm{ 9250 }\!\!\times\!\!\text{ 29} \\ \end{array}$

Ans:

(a) $578\times 161$

$578$ rounds off to $=600$

$161$ rounds off to $=200$

Product \begin{align} & =\text{ }600\times 200\text{ } \\ & =\text{ }1,20,000 \\ \end{align}

(b) $5281\times 3491$

$5281$ rounds of to $=5,000$

$3491$ rounds off to $=3,500$

Product \begin{align} & =\text{ }5,000\times 3,500\text{ } \\ & =\text{ }1,75,00,000 \\ \end{align}

(c) $1291\times 592$

$1291$ rounds off to $=1300$

$592$ rounds off to $=600$

Product \begin{align} & =\text{ }1300\times 600\text{ } \\ & =\text{ }7,80,000 \\ \end{align}

(d) $9250\times 29$

$9250$ rounds off to $=9,000$

$29$ rounds off to $=30$

Product\begin{align} & =\text{ }9,000\times 30\text{ } \\ & =\text{ }2,70,000 \\ \end{align}

## NCERT Solutions Class 6 of Mathematics chapter-Wise PDF

The NCERT solutions Class 6 Maths Chapter 1 PDF is available on our official website ABC.com for free. It allows students to practice themselves at their feasible timings and also prevents them from sitting in front of the electronic gadget. It is useful to recall and revise during the time of examinations as well as no need to get worried about the internet connection. Also, the professors were available to clarify the doubts of students through live chat, or they can share their questions in the chat box.

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

The concept of numbers was developed to define the quantity of things, people, etc. Apart from simple numerical concepts, the Class 6 students will learn what kind of number exists and how they are identified. With the help of NCERT Solutions Class 6 Maths Chapter 1, the students will be able to understand these new concepts. A clear foundation will be constructed that will help them to learn advanced concepts in the next chapters and higher classes.

The NCERT Solutions for Class 6 Maths Ch 1 will focus on the exercises related to the chapter. The chapter will discuss thoroughly the concepts used to define numbers, the mathematical operations we perform with numbers, and how these concepts are used to solve mathematical problems. Hence, the use of NCERT Solutions for Class 6th Maths Chapter 1 will be a smart step to complete the chapter efficiently.

## Chapter 1 - Knowing Our Numbers

### 1.1 Introduction

As students entered into the sixth standard, the NCERT solutions try to recall all their knowledge on numbers at the beginning of the chapter. As the students already learned different calculations like Asian, subtraction, multiplication, and division with the numbers, here, they may learn other ways to calculate or count the big numbers to express huge quantities.

### 1.2 Comparing Numbers

The NCERT solutions of Class 6 Maths Chapter 1 Knowing Our Numbers like to teach the students of grade 6 about comparing the larger numbers in different ways. Students can learn how to change the value of a number by shifting the digits, changing their places. Also, the subject experts explained the introduction of big numbers like 10,000 and 1,00,000. Not only numbers of students also taught how to find out the place value as well as keeping commas for the big numbers.

It ultimately deals with the reading and writing strategies of 5 - digit and 6 - digit numbers. So that the students can calculate more values while finding different problems as well as they are useful in their daily routine also.

• 1.2.1 How many numbers can you make?

• 1.2.2 Shifting of numbers.

• 1.2.3 Introducing 10,000.

• 1.2.4 Revisiting place value.

• 1.2.5 Introducing 1,00,000.

• 1.2.6 Larger numbers.

• 1.2.7 An aid in reading and writing large numbers.

### 1.3 Large Numbers in Practice

This part of Chapter 1 is essential for the students of Class 6 to improve their thinking skills. Here the NCERT Solutions of Class 6 Maths Chapter 1 has introduced the concept of estimation. That means students can learn how to estimate future things and how they can plan to reach that estimation because it can be faced in plenty of situations for them.

The NCERT Class 6 Maths Chapter 1 PDF Gail several solid examples and different scenarios to make the student understand and to make them perfect in the estimating and to forecast the future. They gave instances like, if the birthday party has been planned to celebrate in our house, the first thing we need to discuss that event is estimating the presence of members. Because it helps to plan for the food, return gifts, the quantity of cake, budget, etc. Also, the students are getting awareness of rounding those estimations into your perfect figure with various place values starting from tens to thousands. Students will practice addition, subtraction, multiplication with the large numbers as well as the estimating figures.

• 1.3.1 Estimation

• 1.3.2 Estimating to the nearest tens by rounding off.

• 1.3.3 Estimating to the nearest hundreds by rounding off.

• 1.3.4 Estimating to the nearest thousands by rounding off.

• 1.3.5 Estimating outcomes of number situations.

• 1.3.6 To estimate sum or difference.

• 1.3.7 To estimate products.

### 1.4 Using Brackets

Introducing the usage of brackets is essential for 6th-grade students. The usage of brackets will occur mostly and two or more different things with one common point.so the students need to place the different things in a bracket, and the common thing will be outside the bracket. Students also learn the distribution of common things, which is outside the bracket to the different parts available within the brackets.

### 1.5 Roman Numerals

As the students knew the Hindu Arabic numeral system till primary classes, here they are introduced to the Roman numeric system, which is widely spread in the society. So, it is mandatory to teach them with several examples and perfect explanations. It can be easily done by NCERT Solutions Class 6 Maths Chapter 1.

We cover all exercises in the chapter given below:-

1. What Makes NCERT Solutions the Best Choice?

Ans. Of course, the NCERT solutions are a perfect choice for the students with their distinct features. They are concerned more about the level of the students, and the subject experts will prepare the material in such a way which is easy to understand and good to practice.

The availability of PDF for free download is another positive feature that makes it more weight. Session of doubts clarification was fabulous for the students to get the clarification as well as confidence while solving problems.

2. 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 needed to stitch a single shirt  = 2m 15cm

= 200 cms + 15 cms

= 215 cm

Total cloth = 40 m

= 4000 cms

Number of shirts can be stitched with 4000 cm = 4000/ 215 cms

= 215*18 = 3870

= 18 shirts.

Remaining cloth = 4000 - 3870 = 130 cms.

So 18 shirts can be stitched with 40 m of cloth with 130 cm cloth will remain. SHARE TWEET SHARE SUBSCRIBE