NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Edited By Ramraj Saini | Updated on Apr 01, 2024 10:13 AM IST

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers are discussed here in detailed. These NCERT solutions are created as per the latest syllabus and pattern of the CBSE considering the need of the students. Numbers are very important for real life as well as for Mathematics. In NCERT textbook for Class 6 chapter 1, students can 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. CBSE NCERT Solutions for Class 6 Maths are covering the solutions for the problems from all the concepts. In NCERT Solutions for Class 6 Math chapter 1, there is a total of 19 questions in 3 exercises.

This Story also Contains

NCERT Solutions for Class 6 Maths Chapter 1 - Important Formula
NCERT Solutions for Class 6 Maths Chapter 1 - Important Topics
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers (Intext Questions and Exercise)
NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.1
NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.2
NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.3
NCERT Solutions for Class 6 Maths Chapter 1 - Topics
NCERT Solutions for Class 6 Maths - Chapter-Wise
Key features of 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

NCERT 6th Class solutions are covering all the answers or solutions in a very comprehensive manner. CBSE NCERT solutions for Class 6 Maths chapter 1 also provides assistance in solving the practice problems. Along with NCERT Class 6 Maths solutions, students can check the NCERT solutions for other subjects prepared based on NCERT syllabus.

NCERT Solutions for Class 6 Maths Chapter 1 - Important Formula

Natural numbers: 1, 2, 3, 4, and so forth.
Sum of first n natural numbers = n(n+1)/2
A number is formed by combining digits, which can only be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Place Value of a digit in a number = Face Value × Position Value
When comparing two numbers, the one with more digits is always greater. If two numbers have the same digit, the comparison starts from the leftmost digit.
To create the smallest number, begin with 1 in the leftmost position and add zeroes. For instance, the smallest four-digit number is 1000.
To create the largest number, start with 9 in the leftmost position. For example, the largest four-digit number is 9999.
There are specific quantity weights for conversions, such as:

1 kilometre (km) = 1000 metres (m)
1 metre (m) = 100 centimetres (cm)
1 centimetre (cm) = 10 millimetres (mm)
1 kilogram (kg) = 1000 grams (gm)
1 litre (l) = 1000 millilitres (ml)

Roman numerals are another system of representation, where each symbol corresponds to a specific value:

I – 1
II – 2
V – 5
X – 10
L – 50
C – 100
D – 500
K – 1000

In Roman numerals, you can perform subtraction or addition by placing the desired quantity to the left or right. For example, 21 can be written as XXI, and 49 can be written as ILIX.

NCERT Solutions for Class 6 Maths Chapter 1 - Important Topics

Place Value: Place value refers to the value of a digit based on its position in a number. It helps understand the significance of each digit and determines the overall value of a number.

Comparing Numbers: Comparing numbers involves determining the relationship between two or more numbers, whether they are equal, greater than, or less than each other. It helps in understanding the order and magnitude of numbers.

Estimation: Estimation is the process of making approximate calculations or predictions based on available information. It helps in getting a quick idea of the magnitude or value of a number or quantity without performing precise calculations.

Rounding Off: Rounding off involves approximating a number to a specified place value. It simplifies numbers and makes them easier to work with by reducing the number of digits while retaining a reasonable degree of accuracy.

Brackets: Brackets are symbols used in mathematical expressions to indicate grouping or prioritizing operations. They help in clarifying the order of operations and ensure correct calculation.

Roman Numerals: Roman numerals are a numeral system used in ancient Rome. They use specific symbols to represent different values, such as I, V, X, L, C, D, and M. Understanding Roman numerals helps in interpreting historical dates, numbering lists, and understanding some traditional numeral systems.

Free download NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers PDF for CBSE Exam.

Download PDF

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers (Intext Questions and Exercise)

Knowing Our Numbers Class 6 NCERT 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

Knowing Our Numbers Class 6 NCERT 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

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

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 - 98 47
Smallest - 10 42

(c) Digit 9 is always at hundreds place
Greatest - 8 976
Smallest - 1 902

(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

Solutions For Knowing Our Numbers Class 6 NCERT 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 For Knowing Our Numbers Class 6 NCERT 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

Q3 You have the following digits 4, 5, 6, 0, 7 and 8.

Using them, make five numbers each with 6 digits.

(a) Put commas for easy reading.

(b) Arrange them in ascending and descending order.

Answer:

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

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

NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.1

Q1 Fill in the blanks:

(a) 1 lakh = _______ ten thousand.

(b) 1 million = _______ hundred thousand.

(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

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.

(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.

$9,05,00,041$

$7,52,21,302$

(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.

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

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers: 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?
1643027706759 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 - 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 $65124$

Ques 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? (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.

Ques 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?

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.

Ques 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.

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.

Ques 12- 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 - Estimating to the Nearest Thousand Rounding Off

Ques- 1 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 ________________

Answer:

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 - Exercise 1.3

Ques 1- 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) :
$\\(a) 439 + 334 + 4,317 \\\\(b) 1,08,734 - 47,599 \\\\ (c) 8325- 491 \\\\ (d) 4,89,348 ï¿½ 48,365$

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 = $489350 - 48370$

$= 440980$

Ques 3- 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 = $5000\times3500$

$= 1,75,00,000$

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

Ques 1- 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.

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

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

Ques 2- Write in Roman numerals.
1. 73
2. 92

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

NCERT Solutions for Class 9 - Subject Wise

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 6 Maths Chapter 1 - Topics

1.1 Introduction

1.2 Comparing Numbers

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

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

1.5 Roman Numerals

NCERT Solutions for Class 6 Maths - Chapter-Wise

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

Key features of NCERT Solutions for Class 6 Maths chapter 1

Expertly Crafted Solutions: The NCERT Solutions for Chapter 1 are carefully developed to provide clear and accurate explanations to all the problems and concepts discussed in the chapter.

Conceptual Clarity: The NCERT maths book class 6 chapter 1 solutions aim to enhance the conceptual understanding of students by breaking down complex topics into simpler and more understandable explanations.

Comprehensive Coverage: The maths class 6 chapter 1 cover all the important topics and subtopics of Chapter 1, ensuring that students have a complete understanding of the chapter.

Exam-Focused Approach: The ncert class 6 maths chapter 1 are designed with an exam-focused approach, providing students with the necessary tools and techniques to tackle questions effectively and score well in exams.

Also Check -

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. What are the features of NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers provide a comprehensive and student-friendly approach to learning mathematics, helping students build a strong foundation in the subject. There are 3 exercise including exercise 1 which has 4 questions, exercise 2 that has 12 questions and exercise 3 that has 3 questions. Students can practice these exercise to command the concepts.

2. How to download Class 6 Maths NCERT solutions for Chapter 1?

Knowing Our Numbers is an important chapter from NCERT syllabus of Class 6. Students can download Knowing Our Numbers solutions to use it offline. they can study knowing our numbers class 6 pdf after download. On clicking the download button the complete page will be downloaded and can be used for offline preparation.

Also Read

NCERT Solutions for Class 6 Science Chapter 3 Fibre to Fabric

NCERT Solutions for Class 6 Science Chapter 2 Components of Food

NCERT Solutions for Class 6 Science Chapter 1 Food Where Does It Come From

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

Articles

IGNOU BEd Answer Key 2025: Download OMR Sheet Here

Dec 13, 2024

TOSS Admission 2024-25: Apply for SSC & Inter TS Open School Admission @telanganaopenschool.org

Dec 13, 2024

IGNOU Registration 2025 (Statred), January Session, Last Date, Fess, Link, Apply Online

Dec 12, 2024

10 Ways To Think Straight, Without Bias

Dec 12, 2024

5 Apps To Help You Avoid Distractions, Become More Productive

Dec 12, 2024

Class 9 NSTSE 5-Year Analysis: Know Which Topics Get More Weightage And Why

Dec 12, 2024

How to Make APAAR ID via School, DigiLocker : Step-by-Step Guide

Dec 11, 2024

DBSE Admit Card 2025 for Class 10th and 12th

Dec 11, 2024

Upcoming School Exams

National Institute of Open Schooling 10th examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Mock Test

National Institute of Open Schooling 12th Examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Admit Card

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Eligibility Criteria

Application Process

Exam Pattern

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:02 December,2024 - 16 December,2024

Exam Pattern

Admit Card

Result

West Bengal Board 12th Examination

Exam Date:02 December,2024 - 20 December,2024

Result

Exam Pattern

Admit Card

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 1 - Important Formula

NCERT Solutions for Class 6 Maths Chapter 1 - Important Topics

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers (Intext Questions and Exercise)

NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.1

NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.2

NCERT Solutions for Class 6 Maths Chapter 1 - Exercise 1.3

NCERT Solutions for Class 9 - Subject Wise

NCERT Solutions for Class 6 Maths Chapter 1 - Topics

NCERT Solutions for Class 6 Maths - Chapter-Wise

Key features of NCERT Solutions for Class 6 Maths chapter 1

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 10th examination

National Institute of Open Schooling 12th Examination

National Means Cum-Merit Scholarship

Himachal Pradesh Board of Secondary Education 10th Examination

West Bengal Board 12th Examination

Popular Questions

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

JEE Main Important Mathematics Formulas

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

Schools By Medium

By Ownership

By City

By State

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app