Important Topics under NCERT Solutions for Class 6 Maths Chapter 1
Chapter 1 of the class 6 maths syllabus is on ‘Knowing Our Numbers’. This is a very important chapter in class 6 that develops a student’s number sense. This crucial chapter in the class 6 maths syllabus is divided into 5 major sections or topics. The following is a list of the important topics covered under NCERT Solutions for Class 6 Maths Chapter 1 – Knowing Our Numbers. We recommend that students carefully read through each one of these topics to get a clear understanding of the concepts introduced in the chapter and be able to utilize the provided solutions appropriately and efficiently.
- Introduction
- Comparing Numbers
- Large Numbers in Practice
- Using Brackets
- Roman Numerals
Study Studio’s expert teachers have meticulously curated these solutions for the betterment of clarity in internalizing the concepts included in this chapter and to ensure that students are able to score well in exams easily after going through and practicing these solutions.
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 Study Studio 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 Study Studio.com to score more marks in your examination.
The scholars use an easy and straightforward language which is understandable to all the levels of students. It is also available in PDF format to download at Study Studio for free.
Knowing Our Numbers Class 6 Maths Chapter 1
Given below are some of the important concepts that students can learn in Class 6 Chapter 1: Knowing Our Numbers
- Numbers are basically the arithmetic values.
- Numbers generally convey the magnitude of everything that is around us.
- Ascending Order: If numbers are arranged from the smallest to the greatest.
- Descending Order: If numbers are arranged from the greatest to the smallest number.
For example, if we consider a group of numbers: 31, 13, 95, 466, 9678 and 10802. They can be arranged in descending order as 10802, 9678, 466, 95, 31 and 13, and in ascending order as 13, 31, 95, 466, 9678 and 10802.
- BODMAS Rule: It describes the sequence of operations to be carried out while solving an expression. If an expression has brackets ((), ), we must first solve the bracket, then ‘order’ (powers and roots, etc.), then division, multiplication, addition, and subtraction from left to right. If you solve the problem in the wrong order, you’ll get the wrong answer.
Access NCERT Solutions for Class 6 Maths Chapter 1- Knowing Our Numbers
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 =
1010
ten thousand
(b) 1 million =
1010
hundred thousand
(c) 1 crore =
1010
ten lakh
(d) 1 crore =
1010
million
(e) 1 million =
1010
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
= 73,75,307 = 73,75,307
(b) Nine crore five lakh forty-one
= 9,05,00,041= 9,05,00,041
(c) Seven crore fifty-two lakh twenty-one thousand three hundred two
= 7,52,21,302 = 7,52,21,302
(d) Fifty-eight million four hundred twenty-three thousand two hundred two
= 58,423,202= 58,423,202
(e) Twenty-three lakh thirty thousand ten
= 23,30,010= 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=8,75,95,762=
Eight-crore seventy five lakh ninety-five thousand seven hundred sixty two
(b)
85,46,283=85,46,283=
Eighty-five lakh forty-six thousand two hundred eighty three
(c)
9,99,00,046=9,99,00,046=
Nine crore ninety-nine lakh forty six
(d)
9,84,32,701=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=78,921,092=
Seventy eight million nine hundred twenty one thousand ninety two
(b)
7,452,283=7,452,283=
Seven million four hundred fifty two thousand two hundred eighty three
(c)
99,985,102=99,985,102=
Ninety nine million nine hundred eighty five thousand one hundred two
(d)
48,049,831=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
1094,1812,2050 and 27511094,1812,2050 ��� 2751
. Find the total number of tickets sold on all the four days.
Ans:
Tickets sold of first day
= 1,094= 1,094
Tickets sold of second day
= 1,812= 1,812
Tickets sold of third day = 2,050
Tickets sold of fourth day
= 2,751= 2,751
Total tickets sold = 1,094+1,812+2,050+2,751= 7,707= 1,094+1,812+2,050+2,751= 7,707
2. Shekhar is a famous cricket player. He has so far scored
69806980
runs in test matches. He wishes to complete
10,00010,000
runs. How many more runs does he need?
Ans:
Runs scored by Shekhar
= 6980= 6980
Total number of runs he need = 10,000−6980= 3020= 10,000−6980= 3020.
3. In an election, the successful candidate registered
5,77,5005,77,500
votes and his nearest rival secured
3,48,7003,48,700votes. By what margin did the successful candidate win the election?
Ans: Successful candidate secured votes
= 5,77,500= 5,77,500
Rival candidate secured votes
= 3,48,700= 3,48,700
Margin of votes = 5,77,500−3,48,700= 2,28,800= 5,77,500−3,48,700= 2,28,800
4. Kirti Bookstore sold books worth
2,85,8912,85,891
in the first week of June and books worth
4,00,7684,00,768in 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
= 2,85,891= 2,85,891
Books sold in week second
= 4,00,768= 4,00,768
Total sale
= 4,00,768+2,85,891=6,86,659= 4,00,768+2,85,891=6,86,659
Clearly, sale of second week is more than the first week sale
Difference = 4,00,768−2,85,891= 1,14,877= 4,00,768−2,85,891= 1,14,877
5. Find the difference between the greatest and the least number that can be written using the digits
6,2,7,4,36,2,7,4,3
each only once.
Ans: Largest five digit number
=76432=76432
Smallest five-digit number
= 23467= 23467
Therefore, the difference between then = 76432−23467= 52965= 76432−23467= 52965
6. A machine, on an average, manufactures
2,8252,825
screws a day. How many screws did it produce in the month of January
20062006
?
Ans:
Screws produced in one day
= 2,825= 2,825
Total Screws produced in
3131
days
= 2,825×31 = 87,575= 2,825×31 = 87,575
Hence, the machine produced
87,57587,575
screws in the month of January.
7. A merchant had
78,59278,592
with her. She placed an order for purchasing
4040
radio sets at
1,2001,200
each. How much money will remain with her after the purchase?
Ans:
Total money merchant had
= Rs. 78,592= ��. 78,592
Cost of one radio
= Rs. 1200= ��. 1200
Cost of
4040
radios
= 1200×40= Rs. 48,000= 1200×40= ��. 48,000
Money left= 78,592−48,000= 30,592= 78,592−48,000= 30,592
8. A student multiplied
7236 by 657236 �� 65
instead of multiplying
by 56�� 56
. By how much was his answer greater than the correct answer?
Ans:
Wrong answer = 7236×65 = 470340= 7236×65 = 470340
Correct answer = 7236×56 = 405216= 7236×56 = 405216
Difference in answers = 470340−405216 = 65,124= 470340−405216 = 65,124
9. To stitch a shirt
2 m 15 cm2 � 15 ��
cloth is needed. Out of
40 m40 �
cloth, how many shirts can be stitched and how much cloth will remain?
Ans:
Cloth required for stitch one shirt = 2 m 15 cm= 2×100 cm+15 cm=215cm= 2 � 15 ��= 2×100 ��+15 ��=215��
Total length of cloth = 40 m = 40×100 cm=4000 cm= 40 � = 40×100 ��=4000 ��
Number of shirts that can be stitched = 40002154000215
215⎞⎠⎟⎟⎟⎟⎟4000−2151850−1720130¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯18215)4000−2151850−1720130¯18
Hence,
1818
shirts can be stitched and
130 cm (1 m 30 cm)130 �� (1 � 30 ��)
cloth will be left.
10. Medicine is packed in boxes, each weighing
4 kg 500 g4 �� 500 �
. How many such boxes can be loaded in a can which cannot carry beyond
800 kg800 ��?
Ans:
Weight of a box = 4 kg 500 g = 4500 g= 4 �� 500 � = 4500 �
Number of boxes
= 8000004500= 80000045004500⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟800000−450035000−3150035000−315003500¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯1774500)800000−450035000−3150035000−315003500¯177
Hence,
177177
boxes can be loaded in the Van.
11. The distance between the school and the house of a student’s house is
1 km 875 m1 �� 875 �
. Every day she walks both ways. Find the total distance covered by her in six days.
Ans:
Distance between school and her house
= 1875 m= 1875 �
Total distance covered = 2×1875 = 3750 m= 2×1875 = 3750 �
Distance covered in 6 days6 ����
= 6×3750 = 22500 m= 6×3750 = 22500 �
Thus, she covers
22 km 500 m22 �� 500 �
distance in
6 days6 ����
12. A vessel has
4 liters and 500 ml4 ������ ��� 500 ��
of curd. In how many glasses each of
25 ml25 ��
capacity, can it be filled?
Ans:
Capacity of vessel= 4 liters 500 ml = 4500 ml= 4 ������ 500 �� = 4500 ��
Capacity of a glass
= 25 ml= 25 ��
Number of glasses can be filled = 450025= 450025
25⎞⎠⎟⎟⎟⎟⎟4500−25200−2000¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯18025)4500−25200−2000¯180
Therefore,
180180
glasses are required.
Exercise 1.3
1. Estimate each of the following using general rules:(a)730+998(b)796−−314(c)12,904+2,888(d)28,292−−21,496(a)730+998(b)796−−314(c)12,904+2,888(d)28,292−−21,496
Ans:
(a)
730730
round off to
=700=700
998998
round off to
=1000=1000
Estimated sum
= 1700= 1700
(b)
796796
round off to
=800=800
314314
round off to
=300=300
Estimated sum
= 500= 500
(c)
1290412904
round off to
=13000=13000
28882888
round off to
=3000=3000
Estimated sum
= 16000= 16000
(d)
2829228292
round off to
=28000=28000
2149621496
round off to
=21000=21000
Estimated difference
= 7000= 7000
2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):(a)439+334+4317(b)1,08,737−−47,599(c)8325−−491(d)4,89,348−−48,365(a)439+334+4317(b)1,08,737−−47,599(c)8325−−491(d)4,89,348−−48,365
Ans:
(a) Rough estimate by rounding off to nearest hundreds:
439439
round off to
=400=400
334334
round off to
=300=300
43174317
round off to
=4300=4300
Estimated sum
= 5000= 5000
Rough estimate by rounding off to nearest tens:
439439
round off to
=440=440
334334
round off to
=330=330
43174317
round off to
=4320=4320
Estimated sum
= 5090= 5090
(b) Rough estimate by rounding off to nearest hundreds:
108734108734
round off to
=108700=108700
4759947599
round off to
=47600=47600
Estimated difference
= 61100= 61100
Rough estimate by rounding off to nearest tens:
108734108734
round off to
=108730=108730
4759947599
round off to
=47600=47600
Estimated difference
= 61130= 61130
(c) Rough estimate by rounding off to nearest hundreds:
83258325
round off to
=8300=8300
491491
round off to
=500=500
Estimated difference
= 7800= 7800
Rough estimate by rounding off to nearest tens:
83258325
round off to
=8330=8330
491491
round off to
=490=490
Estimated difference
= 7840= 7840
(d) Rough estimate by rounding off to nearest hundreds:
489348489348
round off to
=489300=489300
4836548365
round off to
=48400=48400
Estimated difference
= 440900= 440900
Rough estimate by rounding off to nearest tens:
489348489348
round off to
=489350=489350
4836548365
round off to
=48370=48370
Estimated difference
= 440980= 440980
3. Estimate the following products using general rule: (a)578× 161(b)5281× 3491(c)1291× 592(d)9250× 29(a)578× 161(b)5281× 3491(c)1291× 592(d)9250× 29
Ans:
(a)
578×161578×161
578578
rounds off to
=600=600
161161
rounds off to
=200=200
Product = 600×200 = 1,20,000= 600×200 = 1,20,000
(b)
5281×34915281×3491
52815281
rounds of to
=5,000=5,000
34913491
rounds off to
=3,500=3,500
Product = 5,000×3,500 = 1,75,00,000= 5,000×3,500 = 1,75,00,000
(c)
1291×5921291×592
12911291
rounds off to
=1300=1300
592592
rounds off to
=600=600
Product = 1300×600 = 7,80,000= 1300×600 = 7,80,000
(d)
9250×299250×29
92509250
rounds off to
=9,000=9,000
2929
rounds off to
=30=30
Product= 9,000×30 = 2,70,000= 9,000×30 = 2,70,000
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.