NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers
NCERT Solutions for Class 6 Maths offer comprehensive explanations for the questions found within the NCERT textbooks endorsed by the Central Board of Secondary Education (CBSE). Orchids the international school provides these NCERT Class 6 Maths Solutions on a chapter-by-chapter basis, aiming to assist students in resolving any uncertainties and acquiring a profound comprehension of the subject matter. These resources, including NCERT Solutions, are conveniently accessible in PDF format, allowing students to download them for offline learning. Students can employ these NCERT Solutions for Class 6 to practise various question types featured in the textbooks, which are expected to appear in their final examinations. These solutions have been meticulously crafted by our experts, adhering to a well-structured format to present the most effective problem-solving methods and ensure a thorough grasp of the underlying concepts. It is strongly recommended that students engage in practising the problems from the CBSE Class 6 Maths textbook as this endeavour will not only prepare them for their exams but also establish a robust foundation for advanced coursework.
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The NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers
Students can access the NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Introduction
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.
(a) The numeral of seventy-three lakh seventy-five thousand three hundred seven is 73,75,307
(b) The numeral of nine crore five lakh forty-one is 9,05,00,041
(c) The numeral of seven crore fifty-two lakh twenty-one thousand three hundred two is 7,52,21,302
(d) The numeral of fifty-eight million four hundred twenty-three thousand two hundred two is 5,84,23,202
(e) The numeral of twenty-three lakh thirty thousand ten is 23,30,010
Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
(a) 87595762 – Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two
(b) 8546283 – Eighty-five lakh forty-six thousand two hundred eighty-three
(c) 99900046 – Nine crore ninety-nine lakh forty-six
(d) 98432701 – Nine crore eighty-four lakh thirty-two thousand seven hundred one
Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831
(a) 78921092 – Seventy-eight million nine hundred twenty-one thousand ninety-two
(b) 7452283 – Seven million four hundred fifty-two thousand two hundred eighty-three
(c) 99985102 – Ninety-nine million nine hundred eighty-five thousand one hundred two
(d) 48049831 – Forty-eight million forty-nine thousand eight hundred thirty-one
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.
(a) 1 lakh = 10 ten thousand
= 1,00,000
(b) 1 million = 10 hundred thousand
= 10,00,000
(c) 1 crore = 10 ten lakh
= 1,00,00,000
(d) 1 crore = 10 million
= 1,00,00,000
(e) 1 million = 10 lakh
= 1,000,000
Comparing Numbers
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.
Number of tickets sold on the 1st day = 1094
Number of tickets sold on the 2nd day = 1812
Number of tickets sold on the 3rd day = 2050
Number of tickets sold on the 4th day = 2751
Hence, number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets
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?
Shekhar scored = 6980 runs
He wants to complete = 10000 runs
Runs he needs to score = 10000 – 6980 = 3020
Hence, he needs to score 3020 more runs
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?
No. of votes secured by the successful candidate = 577500
No. of votes secured by his rival = 348700
Margin by which he won the election = 577500 – 348700 = 228800 votes
∴ Successful candidate won the election by 228800 votes
Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 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?
Price of books sold in June first week = Rs 285891
Price of books sold in June second week = Rs 400768
No. of books sold in both weeks together = Rs 285891 + Rs 400768 = Rs 686659
The sale of books is the highest in the second week
Difference in the sale in both weeks = Rs 400768 – Rs 285891 = Rs 114877
∴ Sale in second week was greater by Rs 114877 than in the first week.
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.
Digits given are 6, 2, 7, 4, 3
Greatest 5-digit number = 76432
Least 5-digit number = 23467
Difference between the two numbers = 76432 – 23467 = 52965
∴ The difference between the two numbers is 52965
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Number of screws manufactured in a day = 2825
Since January month has 31 days
Hence, number of screws manufactured in January = 31 × 2825 = 87575
Hence, machine produce 87575 screws in the month of January 2006
A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Total money the merchant had = Rs 78592
Number of radio sets she placed an order for purchasing = 40 radio sets
Cost of each radio set = Rs 1200
So, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000
Money left with the merchant = Rs 78592 – Rs 48000 = Rs 30592
Hence, money left with the merchant after purchasing radio sets is Rs 30592
A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Difference between 65 and 56 i.e (65 – 56) = 9
The difference between the correct and incorrect answer = 7236 × 9 = 65124
Hence, by 65124, the answer was greater than the correct answer
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?
Given
Total length of the cloth = 40 m
= 40 × 100 cm = 4000 cm
Cloth required to stitch one shirt = 2 m 15 cm
= 2 × 100 + 15 cm = 215 cm
Number of shirts that can be stitched out of 4000 cm = 4000 / 215 = 18 shirts
Hence, 18 shirts can be stitched out of 40 m and 1m 30 cm of cloth is left out
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?
Weight of one box = 4 kg 500 g = 4 × 1000 + 500
= 4500 g
Maximum weight carried by the van = 800 kg = 800 × 1000
= 800000 g
Hence, number of boxes that can be loaded in the van = 800000 / 4500 = 177 boxes
The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Distance covered between school and house = 1 km 875 m = 1000 + 875 = 1875 m
Since, the student walk both ways.
Hence, distance travelled by the student in one day = 2 × 1875 = 3750 m
Distance travelled by the student in 6 days = 3750 m × 6 = 22500 m = 22 km 500 m
∴ Total distance covered by the student in six days is 22 km and 500 m
A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Quantity of curd in the vessel = 4 l 500 ml = 4 × 1000 + 500 = 4500 ml
Capacity of 1 glass = 25 ml
∴ Number of glasses that can be filled with curd = 4500 / 25 = 180 glasses
Hence, 180 glasses can be filled with curd.
Large Numbers in Practice
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) 108734 – 47599
(c) 8325 – 491
(d) 489348 – 48365
Make four more such examples.
(a) 439 + 334 + 4317
Rounding off to nearest hundreds
439 + 334 + 4317 = 400 + 300 + 4300
= 5000
Rounding off to nearest tens
439 + 334 + 4317 = 440 + 330 + 4320
= 5090
(b) 108734 – 47599
Rounding off to nearest hundreds
108734 – 47599 = 108700 – 47600
= 61100
Rounding off to nearest tens
108734 – 47599 = 108730 – 47600
= 61130
(c) 8325 – 491
Rounding off to nearest hundreds
8325 – 491 = 8300 – 500
= 7800
Rounding off to nearest tens
8325 – 491 = 8330 – 490
= 7840
(d) 489348 – 48365
Rounding off to nearest hundreds
489348 – 48365 = 489300 – 48400
= 440900
Rounding off to nearest tens
489348 – 48365 = 489350 – 48370
= 440980
Four more examples are as follows
(i) 4853 + 662
Rounding off to nearest hundreds
4853 + 662 = 4800 + 700
= 5500
Rounding off to nearest tens
4853 + 662 = 4850 + 660
= 5510
(ii) 775 – 390
Rounding off to nearest hundreds
775 – 390 = 800 – 400
= 400
Rounding off to nearest tens
775 – 390 = 780 – 400
= 380
(iii) 6375 – 2875
Rounding off to nearest hundreds
6375 – 2875 = 6400 – 2900
= 3500
Rounding off to nearest tens
6375 – 2875 = 6380 – 2880
= 3500
(iv) 8246 – 6312
Rounding off to nearest hundreds
8246 – 6312 = 8200 – 6300
= 1900
Rounding off to nearest tens
8246 – 6312 = 8240 – 6310
= 1930
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12904 + 2888
(d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
(a) 730 + 998
Round off to hundreds
730 rounds off to 700
998 rounds off to 1000
Hence, 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Round off to hundreds
796 rounds off to 800
314 rounds off to 300
Hence, 796 – 314 = 800 – 300 = 500
(c) 12904 + 2888
Round off to thousands
12904 rounds off to 13000
2888 rounds off to 3000
Hence, 12904 + 2888 = 13000 + 3000 = 16000
(d) 28292 – 21496
Round off to thousands
28292 round off to 28000
21496 round off to 21000
Hence, 28292 – 21496 = 28000 – 21000 = 7000
Ten more such examples are
(i) 330 + 280 = 300 + 300 = 600
(ii) 3937 + 5990 = 4000 + 6000 = 10000
(iii) 6392 – 3772 = 6000 – 4000 = 2000
(iv) 5440 – 2972 = 5000 – 3000 = 2000
(v) 2175 + 1206 = 2000 + 1000 = 3000
(vi) 1110 – 1292 = 1000 – 1000 = 0
(vii) 910 + 575 = 900 + 600 = 1500
(viii) 6400 – 4900 = 6000 – 5000 = 1000
(ix) 3731 + 1300 = 4000 + 1000 = 5000
(x) 6485 – 4319 = 6000 – 4000 = 2000
Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
(a) 578 × 161
Rounding off by general rule
578 and 161 rounded off to 600 and 200, respectively
600
× 200
____________
120000
_____________
(b) 5281 × 3491
Rounding off by general rule
5281 and 3491 rounded off to 5000 and 3500, respectively
5000
× 3500
_________
17500000
_________
(c) 1291 × 592
Rounding off by general rule
1291 and 592 rounded off to 1300 and 600, respectively
1300
× 600
_____________
780000
______________
(d) 9250 × 29
Rounding off by general rule
9250 and 29 rounded off to 9000 and 30, respectively
9000
× 30
_____________
270000
Frequently Asked Questions
The NCERT solution for Class 6 Chapter 1: Knowing Our Numbers is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 6 Chapter 1: Knowing Our Numbers is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 6 Maths Chapter 1 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 6 Maths Chapter 1: Knowing Our Numbers as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 6 Maths Chapter 1 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.