# NCERT Solutions For Class 6 Maths Chapter 1: Knowing Our Numbers

Contents

Chapter 1 NCERT Solutions for Class 6 Math Knowing Our Numbers assists students who want to achieve a high academic score on exams. Knowledge glow experts created these solutions to increase students’ confidence by supporting them in grasping the concepts taught in this chapter. NCERT Solutions for Class 6 include ways for quickly and efficiently solving problems in the book. These materials are based on the Class 6 NCERT syllabus and take into account the types of questions found in the NCERT textbook.

Using NCERT Solutions will help pupils comprehend the main topics quickly. Furthermore, all of the NCERT solutions adhere to the most recent CBSE norms and marking schemes. To get started, download the answers to this chapter in PDF format from the URL provided below.

NCERT Solutions for Class 6 Chapter 1: Understanding Our Numbers

## Exercise 1.1 PAGE NO: 12

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.

Solutions:

(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

2. Correctly place commas and write the numerals:

(a) Seventy-three lakh seventy-five thousand three hundred seventy-seventy-seventy-seventy-s

(b) 9 crore 5 lakh 441 rupees.

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

(d) 58 million four hundred twenty-three thousand two hundred two dollars.

(e) 23 lakh 30 thousand ten rupees.

Solutions:

(a) The numeral of seventy-three lakh seventy-five thousand three hundred seven is 73,75,307

(b) The numeral of nine crores five lakh forty-one is 9,05,00,041

(c) The numeral of seven crores 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 appropriate commas and write the names using the Indian System of Numeration:

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

Solutions:

(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

4. Use appropriate commas and write the names in accordance with the International System of Numeration:

(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831

Solutions:

(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

## Exercise 1.2 PAGE NO: 16

1. A four-day book display was conducted in a school. On the first, second, third, and final days, 1094, 1812, 2050, and 2751 tickets were sold at the counter. Determine the total number of tickets sold throughout all four days.

Solutions:

Number of tickets sold on 1st day = 1094

Number of tickets sold on 2nd day = 1812

Number of tickets sold on 3rd day = 2050

Number of tickets sold on 4th day = 2751

As a result, the total number of tickets sold throughout the four days is  = 1094 + 1812 + 2050 + 2751 = 7707 tickets

2. Shekhar is a well-known cricketer. In test matches, he has scored 6980 runs. He wants to run 10,000 miles. How many more runs will he require?

Solutions:

Shekhar scored = 6980 runs

He want to complete = 10000 runs

Runs need to score more = 10000 – 6980 = 3020

Hence, he need 3020 more runs to score

3. In an election, the winning candidate received 5,77,500 votes, while his nearest challenger received 3,48,700 votes. What was the margin of victory for the successful candidate?

Solutions:

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

∴ The successful candidate won the election by 228800 votes

4. In the first week of June, Kirti bookstore sold books worth Rs 2,85,891 and books worth Rs 4,00,768 in the second week of the month. How much did the sale for the two weeks total cost? Which week’s sales were higher, and by how much?

Solutions:

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 the second week was greater by Rs 114877 than in the first week.

5. Find the difference between the greatest and smallest 5-digit number that can be written just once using the numbers 6, 2, 7, 4, 3.

Solutions:

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

6. On average, a machine produces 2,825 screws every day. How many screws did it manufacture in January 2006?

Solutions:

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

7. A trader had Rs 78,592 on her person. She made an order for 40 radio sets, each costing Rs 1200. How much money will she have left over after the purchase?

Solutions:

Total money the merchant had = Rs 78592

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

8. A pupil multiplied 7236 by 65 rather than by 56. How much did his answer outnumber the correct answer?

Solutions:

Difference between 65 and 56 i.e (65 – 56) = 9

The difference between the correct and incorrect answer = 7236 × 9 = 65124

9. 2 m 15 cm material is required to stitch a shirt. How many shirts can be sewed from 40 metres of fabric, and how much fabric remains?

Solutions:

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

10. Medicine is packaged in boxes measuring 4 kg 500g apiece. How many of these boxes can be carried into a van that can only carry 800 kg?

Solutions:

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

11. The school is 1 kilometer 875 meters away from a student’s home. She walks in both directions every day. Determine her total distance traveled in six days.

Solutions:

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

12. A vessel holds 4 litres of curd and 500 ml of water. How many glasses, each with a capacity of 25 ml, can it hold?

Solutions:

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.

## Exercise 1.3 Page NO: 23

1. Estimate each of the following using general rule:

(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496

Make ten more addition, subtraction, and estimation of outcome examples.

Solutions:

(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

1. 330 + 280 = 300 + 300 = 600
2. 3937 + 5990 = 4000 + 6000 = 10000
3. 6392 – 3772 = 6000 – 4000 = 2000
4. 5440 – 2972 = 5000 – 3000 = 2000
5. 2175 + 1206 = 2000 + 1000 = 3000
6. 1110 – 1292 = 1000 – 1000 = 0
7. 910 + 575 = 900 + 600 = 1500
8. 6400 – 4900 = 6000 – 5000 = 1000
9. 3731 + 1300 = 4000 + 1000 = 5000
10. 6485 – 4319 = 6000 – 4000 = 2000

2. Give a rough estimate (rounding off to the closest hundreds) as well as a more precise estimate (rounding off to the nearest tens):

(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365

Make four more such examples.

Solutions:

(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

3. Using the general rule, estimate the following products:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Make four additional instances like this.

Solutions:

(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=70000

______________

Disclaimer:

Topics Removed – 1.3.1 Estimation, 1.3.2 Rounding off estimates to the nearest tens, 1.3.3 Rounding off estimates to the nearest hundreds, 1.3.4 Rounding off estimates to the nearest thousands, 1.3.5 Predicting the outcomes of number circumstances 1.3.6 To calculate the sum or difference, 1.3.7 In order to estimate products, 1.4 When Using Brackets, 1.5 Roman Numerals, 1.4.1 Expanding brackets

## Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 1

### What topics are addressed in Chapter 1 of the NCERT Solutions for Class 6 Maths?

These topics are covered in Chapter 1 of NCERT Solutions for Class 6 Maths are:

1. Introduction to numbers

2. Comparing numbers

3. Ascending order and Descending order

4. How many numbers can be formed using a certain number of digits?

5. Shifting digits

6. Place value

7. Larger Numbers and Estimates

8. Estimating sum or difference

9. Estimating products of numbers

10. BODMAS

11. Using Brackets

### How many problems are there in each NCERT Solutions for Class 6 Maths Chapter 1 exercise?

Regular repetition of the exercise-based issues allows students to better understand the concepts covered. The number of questions in each Chapter 1 practise is as follows:

Exercise 1.1 – 4 questions

Exercise 1.2 – 12 questions

Exercise 1.3 – 3 questions

### What is the message of Chapter 1 of NCERT Solutions for Class 6 Maths?

A number is a mathematical value that is used to count and measure various objects. We can all add, subtract, divide, and multiply with the help of numbers. Here we will learn how to compare numbers, extend numbers, and discover the largest and smallest numbers. Roman numerals and the Hindu-Arabic numeral system are the two main forms of number system used for writing numbers in numerous areas. We can all see roman numerals in clocks, page numbers, school timetables on syllabus pages, and so on.