**RBSE Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers** is designed to provide students with a strong foundation in understanding numbers, their properties, and their uses. This chapter deals with large numbers, how to represent them, and perform basic operations on them. These solutions will help students build a strong number sense and are crucial for solving advanced problems in higher classes.

In this chapter, students will learn about the Indian and International number systems, rounding off numbers, and using the estimation technique for faster calculations. The RBSE solutions for this chapter offer step-by-step explanations for each problem, making it easier for students to grasp the concepts.

## Table of Contents

### Topics Covered in Chapter 1: Knowing Our Numbers

- Understanding large numbers
- Comparing and ordering numbers
- Estimating numbers
- Indian and International number systems
- Rounding off numbers
- Arithmetic operations on large numbers

Below are the detailed solutions for **RBSE Class 6 Maths Chapter 1: Knowing Our Numbers**.

## Class 6 Chapter 1 Exercise 1.1

**1. Fill in the blanks:**

(a) 1 lakh = ………….. ten thousand.

(b) 1 million = ………… hundred thousand.

(c) 1 crore = ………… ten lakhs.

(d) 1 crore = ………… million.

(e) 1 million = ………… lakhs.

**Solutions:**

(a) 1 lakh = 10 ten thousand

= 1,00,000

(b) 1 million = 10 hundred thousand

= 10,00,000

(c) 1 crore = 10 ten lakhs

= 1,00,00,000

(d) 1 crore = 10 million

= 1,00,00,000

(e) 1 million = 10 lakhs

= 1,000,000

**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

**Solutions:**

**(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

**3. Insert commas suitably and write the names according to the Indian System of Numeration:**

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

**Solutions:**

(a) 8,75,95,762 – Eight crore seventy five lakh ninety five thousand seven hundred sixty two

(b) 85,46,283 – Eighty five lakh forty six thousand two hundred eighty three

(c) 9,99,00,046 – Nine crore ninety nine lakh forty six

(d) 9,84,32,701 – Nine crore eighty four lakh thirty two thousand seven hundred one

**4. Insert commas suitably and write the names according to the International System of Numeration:**

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

**Solutions:**

(a) 78,921,092 – Seventy eight million nine hundred twenty one thousand ninety two

(b) 7,452,283 – Seven million four hundred fifty-two thousand two hundred eighty three

(c) 99,985,102 – Ninety-nine million nine hundred eighty five thousand one hundred two

(d) 48,049,831 – Forty-eight million forty-nine thousand eight hundred thirty-one

## Class 6 Chapter 1 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 2751. Find the total number of tickets sold on 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

Hence, the total number of tickets sold on all four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets

**2. 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?**

**Solutions:**

Shekhar scored = 6980 runs

He wants to complete = 10000 runs

Runs needed to score more = 10000 – 6980 = 3020

Hence, he needs 3020 more runs to score

**3. 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?**

**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. 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?**

**Solutions:**

Price of books sold in the first week of June = Rs 285891

Price of books sold in the second week of June = 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 the least 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 each only once.**

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

**Solutions:**

Number of screws manufactured in a day = 2825

Since January month has 31 days,

The number of screws manufactured in January = 31 × 2825 = 87575

Hence, the machine produced 87575 screws in the month of January 2006.

**7. 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?**

**Solutions:**

Total money the merchant had = Rs 78592

The 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.

**8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than 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

Hence, by 65124, the answer was greater than the correct answer.

**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?**

**Solutions:**

Given

The 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 1 m 30 cm of cloth is left.

**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?**

**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, the number of boxes that can be loaded in the van = 800000/4500 = 177 boxes

**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.**

**Solutions:**

Distance covered between the school and her house = 1 km 875 m = 1000 + 875 = 1875 m

Since the student walks both ways,

The 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

∴ The total distance covered by the student in six days is 22 km and 500 m.

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

**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.

## Class 6 Chapter 1 Exercise 1.3

**1. Estimate each of the following using the 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.**

**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

(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

**2. Give a rough estimate (by rounding off to the nearest hundreds) and also a closer estimate (by 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 the nearest hundreds

439 + 334 + 4317 = 400 + 300 + 4300

= 5000

Rounding off to the nearest tens

439 + 334 + 4317 = 440 + 330 + 4320

= 5090

(b) 108734 – 47599

Rounding off to the nearest hundreds

108734 – 47599 = 108700 – 47600

= 61100

Rounding off to the nearest tens

108734 – 47599 = 108730 – 47600

= 61130

(c) 8325 – 491

Rounding off to the nearest hundreds

8325 – 491 = 8300 – 500

= 7800

Rounding off to the nearest tens

8325 – 491 = 8330 – 490

= 7840

(d) 489348 – 48365

Rounding off to the nearest hundreds

489348 – 48365 = 489300 – 48400

= 440900

Rounding off to the nearest tens

489348 – 48365 = 489350 – 48370

= 440980

Four more examples are as follows:

(i) 4853 + 662

Rounding off to the nearest hundreds

4853 + 662 = 4800 + 700

= 5500

Rounding off to the nearest tens

4853 + 662 = 4850 + 660

= 5510

(ii) 775 – 390

Rounding off to the nearest hundreds

775 – 390 = 800 – 400

= 400

Rounding off to the nearest tens

775 – 390 = 780 – 400

= 380

(iii) 6375 – 2875

Rounding off to the nearest hundreds

6375 – 2875 = 6400 – 2900

= 3500

Rounding off to the nearest tens

6375 – 2875 = 6380 – 2880

= 3500

(iv) 8246 – 6312

Rounding off to the nearest hundreds

8246 – 6312 = 8200 – 6300

= 1900

Rounding off to the nearest tens

8246 – 6312 = 8240 – 6310

= 1930

**3. Estimate the following products using the general rule:**

**(a) 578 × 161**

**(b) 5281 × 3491**

**(c) 1291 × 592**

**(d) 9250 × 29**

**Make four more such examples.**

**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

_____________

270000

______________

# RBSE Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers (2023-24 Edition)

**RBSE Solutions for Class 6 Maths Chapter 1: Knowing Our Numbers** is designed to provide students with a strong foundation in understanding numbers, their properties, and their uses. This chapter deals with large numbers, how to represent them, and perform basic operations on them. These solutions will help students build a strong number sense and are crucial for solving advanced problems in higher classes.

In this chapter, students will learn about the Indian and International number systems, rounding off numbers, and using the estimation technique for faster calculations. The RBSE solutions for this chapter offer step-by-step explanations for each problem, making it easier for students to grasp the concepts.

### Topics Covered in Chapter 1: Knowing Our Numbers

- Understanding large numbers
- Comparing and ordering numbers
- Estimating numbers
- Indian and International number systems
- Rounding off numbers
- Arithmetic operations on large numbers

Below are the detailed solutions for **RBSE Class 6 Maths Chapter 1: Knowing Our Numbers**.

- RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 13: Symmetry | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 11: Algebra | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 10: Mensuration | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 9: Data Handling | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 8: Decimals | Updated for 2024-25
- Rbse Solutions to Class 6 Chapter 7: Fractions
- Rbse Solutions for Class 6 Chapter 6: Integers
- Rbse Solutions for Class 6 Chapter 5: Understanding Elementary Shapes
- Rbse Solutions for Class 6 Chapter 4: Basic Geometrical Ideas
- Rbse Solutions For Class 6 Maths Chapter 3 Playing with Numbers
- Rbse Solutions for Class 6 Chapter 2: Whole Numbers
- Rbse Solutions for Class 6 Chapter 1: Knowing Our Numbers

### Key Takeaways from Chapter 1:

**Understanding the Indian and International Number Systems**: Students learn the difference between the two number systems, including how numbers are read and written in both formats.**Rounding Off and Estimation**: Estimation helps students make quick and approximate calculations, which are useful in real-life situations.**Basic Arithmetic with Large Numbers**: Students gain proficiency in adding, subtracting, multiplying, and estimating large numbers.

## Benefits of RBSE Solutions for Class 6 Maths Chapter 1:

**Concept Clarity**: The step-by-step solutions provide clear explanations, helping students understand the fundamentals of numbers.**Strengthens Basics**: This chapter lays a strong foundation for future mathematical concepts and advanced calculations.**Practice-Oriented**: Regular practice with these solutions helps in better preparation for exams.

## FAQs (Frequently Asked Questions)

**Where can I download RBSE Solutions for Class 6 Maths Chapter 1?**

You can download the solutions for RBSE Class 6 Maths Chapter 1 in PDF format from this page for free.

**How can these solutions help me in exams?**

These solutions are designed to explain complex problems in a simplified way, making it easier for students to prepare and perform well in exams.**What topics are covered in Chapter 1: Knowing Our Numbers?**

This chapter covers topics like understanding large numbers, comparing and ordering numbers, rounding off, estimation, and basic arithmetic with large numbers.**Is it important to practice the exercises in Chapter 1?**

Yes, practicing the exercises helps in building a strong number sense and is crucial for understanding more complex topics in higher classes.**How can I improve my performance in Maths using these solutions?**

Regular practice using these solutions will help you understand the concepts better, improve problem-solving skills, and boost your confidence in Maths.