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

NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers- Numbers are very important for real life as well as for mathematics. In the previous classes, you have learnt to work with numbers. In the first chapter, you will deal with the number systems and some basic operations of numbers. You have added, subtracted, divided and multiplied them. This chapter will move forward on interesting things like patterns in number sequences with a brief review and revision as well. CBSE NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers is also there to provide you assistance in solving the practice problems. In NCERT class 6 maths chapter 1 Knowing Our Numbers, you will learn Indian arrangement of numeration, the universal arrangement of numeration, learn large numbers up to 1 crore ( 10000000 ), estimation of large numbers and roman numerals. Along with all these, you will also learn how to write and read these large numbers, comparing numbers for example which number is greatest and which number is the smallest among the given numbers. CBSE NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers are covering the solutions for the problems from all the concepts. In this particular chapter, there is a total of 19 questions in 3 exercises. Solutions of NCERT for class 6 maths chapter 1 Knowing Our Numbers are covering all the problems' solutions in a very comprehensive manner. Apart from this chapter, you can check the NCERT solutions for other classes and subjects using the given link.

Questions from the following exercises are discussed here

## NCERT solutions for class 6 maths chapter 1 Knowing our numbers Topic: Comparing Numbers

3.   1834, 75284, 111, 2333, 450

4.   2853, 7691, 9999, 12002, 124

By Observing the number of digits and digit in the leftmost place can straight away tell us about the greatest and smallest number

(2)  1473, 89423, 100, 5000, 310

89423 is the greatest while 100 is the smallest

(3) 1834, 75284, 111, 2333, 450

75284 is the greatest 111 while is the smallest

(4)  2853, 7691, 9999, 12002, 124

12002 is the greatest while 124 is the smallest

(a) 4536, 4892, 4370, 4452

4892 is the greatest while 4370 is the smallest

(b) 15623, 15073, 15189, 15800

15800 is the greatest while 15073 is the smallest
(c) 25286, 25245, 25270, 25210

25286 is the greatest while 25245 is the smallest
(d) 6895, 23787, 24569, 24659

24659 is the greatest while 6895 is the smallest

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Topic: Number Formation

(a) 2, 8, 7, 4

8742 is the greatest while 2478 is the smallest

(b) 9, 7, 4, 1

9741 is the greatest 1479 while is the smallest

(c) 4, 7, 5, 0

4750 is the greatest while 4057 is the smallest

(d) 1, 7, 6, 2

7621  is the greatest while 1267 is the smallest

(e) 5, 4, 0, 3

5430 is the greatest while 3045 is the smallest

(a) 3, 8, 7

8873 is the greatest while 3387 is the smallest

(b) 9, 0, 5

9950 is the greatest while 5009 is the smallest

(c) 0, 4, 9

9940 is the greatest while 4009 is the smallest

(d) 8, 5, 1

8851 is the greatest while 1158 is the smallest

Q3 Make the greatest and the smallest 4-digit numbers using any four different digits with conditions as given.

(a) Digit 7 is always at one's place
Greatest - 9 8 6 7
Smallest - 1 0 2 7
(Note, the number cannot begin with the digit 0. Why?)

(b) Digit 4 is always at tens place
Greatest - _ _ 4 _
Smallest - _ _ 4 _

(c) Digit 9 is always at hundreds place
Greatest - _ 9 _ _
Smallest - _ 9 _ _
(d) Digit 1 is always at thousands place
Greatest- 1 _ _ _
Smallest - 1 _ _ _

(b) Digit 4 is always at tens place
Greatest - 9847
Smallest - 1042

(c) Digit 9 is always at hundreds place
Greatest - 8976
Smallest - 1902

(d) Digit 1 is always at thousands place
Greatest- 1987
Smallest - 1023

Greatest number - 3322

Smallest number - 2233

Different numbers are

2222, 2232, 2233, 2322, 2333, 2332, 2323, 2223, 3222, 3223, 3232, 3233, 3322, 3333, 3332, 3323

There are 16 in total

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Topic: Arrangement of Numbers

(a) 847, 8320, 8320, 571

571 < 847 < 8320 < 8320

(b) 9801, 25751, 25751, 38802

9801 < 25751 < 25751 < 38802

(a) 5000, 7500, 85400, 7861

85400 > 7861 > 5000 > 75000

(b) 1971, 45321, 88715

88715  > 45321 > 1971

Solutions of NCERT for class 6 maths chapter 1 Knowing our numbers Topic: Revisiting Place Value

1.   50000- ________________________________________
2.    41000- ________________________________________
3.    47300- ________________________________________
4.     57630- ________________________________________

5.     29485________________________________________
6.     29085________________________________________

1.   50000-Fifty thousand
2.    41000-forty one thousand
3.    47300- forty-seven thousand and three hundred
4.     57630- fifty-seven thousand six hundred

5.     29485- twenty-nine thousand and four hundred eighty-five
6.     29085- twenty-Nine Thousand and Eighty-Five

4,57,928 _______________ _______________

4,07,928 _______________ _______________

4,00,829 _______________ _______________

4,00,029 _______________ _______________

4,57,928   Four Lakhs Fifty Seven Thousand and Nine Hundred Twenty Eight

4,07,928    Four Lakhs Seven Thousand and Nine Hundred Twenty-Eight

4,00,829    Four Lakhs Eight Hundred Twenty-Nine

4,00,029     Four Lakhs and Twenty-Nine

Using them, make five numbers each with 6 digits.

(a) Put commas for easy reading.

(b) Arrange them in ascending and descending order.

Given, digits : $4, 5, 6, 0, 7, 8$

Five 6-digit numbers are: $456078, 45708, 548760, 750486, 876540$

## Q1 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.

Let us take this as a physical world problem:

You went to a shop and you bought a toy of Rs. 100. You checked your pocket and found many notes but of only Rs.10

How many Rs.10 notes will you give to shopkeeper:

Well, it’s simple.

You will pay 10 notes.

How?

100/10 =10

The similar easy approach can be done in all the blanks,

Hint: Always remember to keep your eyes on a number of zeros

(a)1 lakh= 100000

1thousand= 1000

Ten thousand= 10*1000 = 10000

Therefore,

1 lakh/ 10 thousand =100000/10000 =10

So, 1lakh= 10 ten thousand

(b) 1 million= 1000000

1thousand= 1000

Hundred thousand= 100*1000 = 100000

Therefore,

1 million/ hundred thousand =1000000/100000 =10

So, 1milion= 10 hundred thousand

(c) 1 crore= 10000000 (7 zeros)

1lakh= 100000 (5 zeros)

Ten lakh= 10*1000 = 1000000( 6 zeros)

Therefore,

1 crore/ Ten lakh=10000000/1000000 =10

So, 1 crore= 10 Ten lakh

(d)1 crore= 10000000 (7 zeros)

1million= 100000 (6 zeros)

Therefore,

1 crore/ 1million=10000000/1000000 =10

So, 1 crore= 10 million

(e) 1 million= 1000000

1lakh= 100000

Therefore,

1million/ 10 lakh = 1000000/100000 =10

So, 1 million= 10 lakhs

See, all answers came out to be 10 but not everytime

(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.

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

$73,75,307$

(b) Nine crore five lakh forty one.

$9,05,00,041$

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

$7,52,21,302$

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

$58,423,302$

(e) Twenty three lakh thirty thousand ten.

$23,30,010$

(a) 87595762        (b) 8546283          (c) 99900046       (d) 98432701

(a) $8,75,95,762$ : Eight crores seventy five lakhs ninety five thousand seven hundred sixty two

(b) $85,46,283$ : Eighty five lakhs forty six thousand two hundred eighty three.

(c) $9,99,00,046$ : Nine crores ninety nine lakhs and forty six.

(d) $9,84,32,701$ : Nine crores eighty four lakhs thirty two thousand seven hundred and one.

## Q1 A box contains 2,00,000 medicine tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams and in kilograms?

Number of medicine tablets = $2,00,000$

Weight of each medicine tablet = $20\ mg = \frac{20}{1000}\ g$

Therefore, total weight = $2,00,000\times\frac{20}{1000}\ g$

$= 4,000\ g$

$= 4\ kg$

(i) Total distance from A to D = AB + BC + CD

$= (4170 + 3410 + 2160)\ km = 9740\ km$

(ii) Total distance between D to G = DE + EF + FG

$= (8140 + 4830 + 2550)\ km = 15520\ km$

(iii) Required distance = AD + DG + GA

$= (4170 + 3410 + 2160 + 8140 + 4830 + 2550 + 1290)\ km = 26550 km$

(iv) The difference of distances from C to D and D to E $= (8140 - 2160)\ km = 5980\ km$

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Exercise: 1.2

Given,

Tickets sold on the first day of exhibition = 1094

Tickets sold on the second day = 1812

Tickets sold on the third day  = 2050

Tickets sold on the last day = 2751

Total number of tickets sold =

$\\ = 1094 + 1812 + 2050 + 2751 \\ = 7707$

Total number of tickets sold = $7,707$

The number of runs scored by Shekhar so far = 6980

The number of runs Shekhar wishes to score = 10,000

$\therefore$ The number of runs required = $10,000 - 6,980$

$= 3,020\ runs$

Therefore, Shekhar requires $3,020\ runs$

Given,

Number of votes registered by the successful candidate = 5, 77,500 votes

Number of votes registered by the rival candidate = 3, 48,700 votes

The margin $= 5,77,500 - 3,48,700$

$= 2,28,800\ votes$

Therefore, the successful candidate won the election by a margin of $2, 28,800$ votes.

Given,

Worth of books sold in the first week = $Rs \ 2,85,891$

Worth of books sold in the second week = $Rs\ 4, 00,768$

Total sale in the two weeks = $Rs. \ (2, 85,891 + 4, 00,768)$

$= Rs. \ 6, 86,659$

Clearly, sales in the second week is greater than the first year by

$Rs.\ (4, 00,768 - 2, 85,891) = Rs.\ 1, 14,877$

Given, digits: 6, 2, 7, 4 and 3

Since, the digits have to be used only once, arrange them in ascending and descending order to get the minimum and maximum number.

Greatest number = $76,432$

Smallest number = $23,467$

$\therefore$ The difference between the greatest and the least number
$\\ = 76,432 - 23,467 \\ = 52,965$

Given,
Screws produced by machine in one day = $2, 825$

We know, there are 31 days in January

$\therefore$ Screws produced in 31 days = $2, 825 \times 31$

$= 87,575\ screws$

Therefore, screws produced in January 2006 = $87,575$

Total money merchant has = $Rs.\ 78, 592$

Cost of one radio set = $Rs.\ 1200$

$\therefore$ Cost of 40 radio sets = $Rs.\ (1200 \times 40 )= Rs.\ 48,000$

$\therefore$ Money left with the merchant = $Rs.\ (78,592 - 48,000) = Rs. \ 30,592$

Therefore, money left with her after purchase = $Rs. \ 30,592$

Given,

The student multiplied $7236\ by\ 65$ instead of $56$.

Wrong answer = $7236 \times 65$

Correct answer = $7236 \times 56$

$\therefore$ The difference in the answers = $7236 \times 65 - 7236 \times 56$

$= 7236 \times (65 - 56) = 7236 \times 9$

$= 65124$

Hence, his answer was greater than the correct answer by $65124$

We know, $(1\ m = 100\ cm)$

Given,

Length of cloth required to stitch a shirt = $2\ m\ 15\ cm = 215\ cm$

Also, $40\ m = (40\times100)\ cm = 4000\ cm$

Now, the cloth required for one shirt = $215\ cm$

Number of shirts that can be stitched out = $4000\div215$

$4000 = (18\times215) + 130$

Therefore, 18 shirts can be made from the given cloth.

And, $130\ cm = 1\ m\ 30\ cm$ cloth will remain unused.

We know, $(1\ kg = 1000\ g)$

Weight of each medicine box= $4\ kg\ 500\ g = 4500\ g$

Weight limit of the van = $800\ kg = (800\times1000)\ g = 8,00,000\ g$

$\therefore$ The number of boxes that can be loaded in the van = $800000\div4500$

$800000 = (4500\times177) + 3500$

Hence, 177 boxes can be loaded in the van.

We know, $(1\ km = 1000\ m)$

Given,
Distance between the school and her house = $1\ km\ 875\ m = 1875\ m$

The distance she covers each day = $[(1875\times2)\times6]\ m$

$(1875\times12)\ m = 22500\ m$

$= 22000\ m\ 500\ m = 22\ km\ 500\ m$

Therefore, she will cover  $22\ km\ 500\ m$ in six days.

We know, $(1\ l = 1000\ ml)$

Given, Capacity of vessel = $4\ l\ 500\ ml = 4500\ ml$

The capacity of a single glass = $25\ ml$

$\therefore$ Number of glasses required to fill the vessel = $4500\div25$

$4500 = (180\times25)+0$

Hence, 180 glasses are needed to fill the vessel completely.

## Q1 Round off the given numbers to the nearest tens, hundreds and thousands.

Given Number               Approximate to Nearest       Rounded Form
75847                                     Tens                            ________________
75847                                   Hundreds                      ________________
75847                                   Thousands                    ________________
75847                                   Ten thousands               ________________

 Given Number Approximate to Nearest Rounded Form 75847 Tens $75850$ 75847 Hundreds $75800$ 75847 Thousands $76000$ 75847 Ten thousands $80000$

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Exercise: 1.3

(a) $730 + 998$
By rounding off to hundreds,

730 rounds off to 700,

998 rounds off to 1000.

Required sum = $700+1000$

$= 1700$

(b) $796-314$

By rounding off to hundreds,

796 rounds off to 800,

314 rounds off to 300.

Required difference = $800-300$

$= 500$

(c) $12904+2822$

By rounding off to thousands,

2904 rounds off to 13000,

2822 rounds off to 3000.

Required sum = $13000+3000$

$= 16000$

(d) $28,296 - 21, 496$

By rounding off to nearest thousands,

28,296 rounds off to 28000,

21,496 rounds off to 21,000

Required difference = $28,000 - 21,000$

$= 7,000$

(a) $439 + 334 + 4,317$

Rounding off to nearest hundreds,

439 rounds off to 400

334 rounds off to 300

4317 rounds off to 4300,

Required sum = $400+300+4300$

$= 5000$

Again, by rounding off to nearest tens,

439 rounds off to 440

334 rounds off to 330

4317 rounds off to 4320,

Required sum = $440+330+4320$

$= 5090$

(b) $1,08,734 - 47,599$

Rounding off to nearest hundreds,

$1, 08,734$ rounds off to $1, 08,700$

$47,599$ rounds off to $47,600$

Required difference = $1, 08,700 - 47,600$

$= 61100$

Also, rounding off to nearest tens,

$1, 08,734$  rounds off to $1, 08,730$

$47,599$  rounds off to $47,600$

Required difference = $1, 08,730 - 47,600$

$= 61130$

(c) $8325 - 491$

Rounding off to nearest hundreds,

8325 rounds off to 8300

491 rounds off to 500

Required difference = $8300 - 500$

$= 7800$

Also, rounding off to nearest tens,

8325 rounds off to 8330

491 rounds off to 490

Required difference = $8330 - 490$

$= 7840$

(d) $4,89,348 - 48,365$

Rounding off to nearest hundreds,

$4,89,348$ rounds off to $4,89,300$

$48,365$ rounds off to $48,400$

Required difference = $489300 - 48400$

$= 440900$

Also, rounding off to nearest tens,

$4,89,348$ rounds off to $4,89,350$

$48,365$ rounds off to $48,370$

Required difference = $489350 - 48370$

$= 440980$

(a) $578\times161$

Rounding off by general rule,

578 rounds off to 600

161 rounds off to 200,

Required product =  $600\times200$

$= 120000$

(b) $5281\times3491$

Rounding off by the general rule,

5281 rounds off to 5000

3491 rounds off to 3500

Required product = $5000\times3500$

$= 1,75,00,000$

(c) $1291\times592$

Rounding off by general rule,

1291 rounds off to 1300

592 rounds off to 600

Required product = $1300\times600$

$= 780000$

(d)  $9250\times29$

Rounding off by general rule,

9250 rounds off to 10000

29 rounds off to 30Required product = $10000 \times 30$

$= 300000$

## Q1 Write the expressions for each of the following using brackets. (a) Four multiplied by the sum of nine and two. (b) Divide the difference of eighteen and six by four. (c) Forty five divided by three times the sum of three and two.

(a) Four multiplied by the sum of nine and two.

$= 4\times(9+2) = 4\times11 = 44$
(b) Divide the difference of eighteen and six by four.

$= \frac{18-6}{4} = \frac{12}{4} = 3$
(c) Forty-five divided by three times the sum of three and two.

$= \frac{45}{3(3+2)}$

$= \frac{45}{15} = 3$

In Roman numerals:
$\\ 1.\ 73 - LXXIII \\ 2.\ 92 - XCII$

## NCERT solutions for class 6 mathematics chapter-wise

 Chapters No. Chapters Name Chapter - 1 NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers Chapter - 2 Solutions of NCERT for class 6 maths chapter 2 Whole Numbers Chapter - 3 CBSE NCERT solutions for class 6 maths chapter 3 Playing with Numbers Chapter - 4 NCERT solutions for class 6 maths chapter 4 Basic Geometrical Ideas Chapter - 5 Solutions of NCERT for class 6 maths chapter 5 Understanding Elementary Shapes Chapter - 6 CBSE NCERT solutions for class 6 maths chapter 6 Integers Chapter - 7 NCERT solutions for class 6 maths chapter 7 Fractions Chapter - 8 Solutions of NCERT for class 6 maths chapter 8 Decimals Chapter - 9 CBSE NCERT solutions for class 6 maths chapter 9 Data Handling Chapter -10 NCERT solutions for class 6 maths chapter 10 Mensuration Chapter -11 Solutions of NCERT for class 6 maths chapter 11 Algebra Chapter -12 CBSE NCERT solutions for class 6 maths chapter 12 Ratio and Proportion Chapter -13 NCERT solutions for class 6 maths chapter 13 Symmetry Chapter -14 Solutions of NCERT for class 6 maths chapter 14 Practical Geometry

## NCERT solutions for class 6 subject wise

 NCERT Solutions for class 6 maths Solutions of NCERT for class 6 science

## How to use NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers?

• Before coming to these please ensure that you should have knowledge of addition, subtraction, multiplication, and division.
• Go through the concepts given in the textbook.
• Learn the application of the concepts by going through some examples given.
• Once you are done with all the above-written processes, you can practice the questions given in the practice exercises.
• During the practice, you can take the help of NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers to enhance the preparation.

Keep working hard and happy learning!