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

Q1 Can you instantly find the greatest and the smallest numbers in each row?
2.   1473, 89423, 100, 5000, 310

3.   1834, 75284, 111, 2333, 450 

4.   2853, 7691, 9999, 12002, 124

Answer:

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

Q2 Find the greatest and the smallest numbers.
(a) 4536, 4892, 4370, 4452
(b) 15623, 15073, 15189, 15800
(c) 25286, 25245, 25270, 25210
(d) 6895, 23787, 24569, 24659

Answer:

(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

Q1 Use the given digits without repetition and make the greatest and smallest 4-digit numbers.
(a) 2, 8, 7, 4
(b) 9, 7, 4, 1
(c) 4, 7, 5, 0
(d) 1, 7, 6, 2
(e) 5, 4, 0, 3

Answer:

(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

Q2 Now make the greatest and the smallest 4-digit numbers by using any one digit twice.
(a) 3, 8, 7
(b) 9, 0, 5
(c) 0, 4, 9
(d) 8, 5, 1

Answer:

(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 _ _ _

Answer:

(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

Q4 Take two digits, say 2 and 3. Make 4-digit numbers using both the digits an equal number of times.
Which is the greatest number?
Which is the smallest number?
How many different numbers can you make in all?

Answer:

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

Q1 Arrange the following numbers in ascending order :
(a) 847, 9754, 8320, 571
(b) 9801, 25751, 36501, 38802

Answer:

(a) 847, 8320, 8320, 571

571 < 847 < 8320 < 8320

(b) 9801, 25751, 25751, 38802

9801 < 25751 < 25751 < 38802

Q2 Arrange the following numbers in descending order :
(a) 5000, 7500, 85400, 7861
(b) 1971, 45321, 88715

Answer:

(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

Q1 Read and expand the numbers wherever there are blanks.

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

5.     29485________________________________________
6.     29085________________________________________

Answer:

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

Q2 Read and expand the numbers wherever there are blanks.

4,57,928 _______________ _______________

4,07,928 _______________ _______________

4,00,829 _______________ _______________

4,00,029 _______________ _______________

Answer:

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

 

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Exercise: 1.1

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.

Answer:

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

Q2 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.

Answer:

(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

 

Q3 Insert commas suitably and write the names according to Indian System of Numeration :

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

Answer:

(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.

 

Solutions of NCERT for class 6 maths chapter 1 Knowing our numbers Topic: Large Numbers in Practise

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?

Answer:

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

 

Q2 A bus started its journey and reached different places with a speed of 60 km/hour.
 (i) Find the total distance covered by the bus from A to D.
(ii) Find the total distance covered by the bus from D to G.
(iii) Find the total distance covered by the bus, if it starts from A and returns back to A.
(iv) Can you find the difference of distances from C to D and D to E? 

                                                                                     
     

Answer:

(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

Q1 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.

Answer:

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

 

Q2 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?

Answer:

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

Q3 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?

Answer:

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.

Q4 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?

Answer:

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

 

Q5 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.

Answer:

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

 

Q6 A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Answer:

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

 

Q7 A merchant had rupees 78,592 with her. She placed an order for purchasing 40 radio sets at rupees 1200 each. How much money will remain with her after the purchase?

Answer:

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

 

Q8 A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)

Answer:

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

 

Q9 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? (Hint: convert data in cm.)

Answer:

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.

Q10 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?

Answer:

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.

Q11 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.

Answer:

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.

Q12 A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?

Answer:

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.

 

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

Q1 Estimate each of the following using general rule:
(a) 730 + 998      (b) 796 – 314       (c) 12,904 +2,888       (d) 28,292 – 21,496

Answer:

(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

 

Q2 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

Answer:

(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

 

Q3 Estimate the following products using the general rule:
(a) 578 × 161     (b) 5281 × 3491       (c) 1291 × 592      (d) 9250 × 29 

Answer:

(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

 

NCERT solutions for class 6 maths chapter 1 Knowing our numbers Topic: Using Brackets

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.

Answer:

(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

 

Q2 Write in Roman numerals.
1. 73
2. 92

Answer:

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!

 

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