**CHAPTER 1**

## NCERT SOLUTIONS FOR CLASS 6 MATHS CHAPTER 1, KNOWING OUR NUMBERS

Solutions for Knowing our Numbers Chapter 1 class 6 Maths NCERT are given.

You should study the textbook lesson Knowing our numbers very well.

Practice the example Problems and solutions given in the textbook.

### SOLUTIONS OF KNOWING OUR NUMBERS FOR VI CLASS MATHS CBSE

You can see the solutions of some chapters for class 6 th class maths.

NCERT maths class 6 solutions

Knowing our Numbers

## KNOWING OUR NUMBERS chapter 1 solutions class 6 maths

6th class maths textbook

**Exercise 1.1**

#### Problem 1

**6th class maths textbook solutions chapter 1 problem 1 exercise 1.1 Knowing our numbers **

1. Fill in the blanks:

a. 1 lakh = …….. ten thousand.

Solution: Ten

b. 1 million = …….. hundred thousand.

Solution: Ten

c. 1 crore = ……. ten lakh.

Solution: Ten

d. 1 crore = ……. million.

Solution: Ten

e. 1 million = ……. lakh.

Solution: Ten

#### Problem 2

**6th class maths textbook solutions chapter 1 problem 2 exercise 1.1**

2. Place commas correctly and write the numerals:

a. Seventy three lakh seventy five thousand three hundred seven.

Solution: 73,75,307

b. Nine crore five lakh forty- one.

Solution: 9,05,00,041

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

Solution: 7,52,21,302

d. Fifty- eight million four hundred twenty- three thousand two hundred two.

Solution: 5,84,23,202

e. Twenty- three lakh thirty thousand ten.

Solution: 23,30,010

#### Problem 3

**6th class maths textbook solutions chapter 1 problem 3 exercise 1.1**

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

*a. 87595762*

Solution: 8,75,95,762

Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.

*b. 8546283*

Solution: 85,46,283

Eighty-five lakh forty-six thousand two hundred eighty-three.

*c. 99900046*

Solution: 9,99,00,046

Nine crore ninety nine lakh forty six.

*d. 98432701*

Solution: 9,84,32,701

Nine crore eighty four lakh thirty two thousand seven hundred one.

#### Problem 4

**class 6 maths textbook solutions chapter 1 problem 4 exercise 1.1 Knowing our numbers **

1. Fill in the blanks:

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

*a. 78921092*

Solution: 78,921,092

Seventy eight million nine hundred twenty one thousand ninety two.

b. 7452283

Solution: 7,452,283

Seven million four hundred fifty-two thousand two hundred eighty-three.

c. 99985102

Solution: 99,985,102

Ninety-nine million nine hundred eighty-five thousand one hundred two.

d. 48049831

Solution: 48,049,831

Forty-eight million forty-nine thousand eight hundred thirty-one.

**CHAPTER 1**

#### NCERT SOLUTIONS FOR CLASS Vi MATHS CHAPTER 1

##### CBSE SOLUTIONS FOR CLASS Vi MATHS CHAPTER 1

##### KNOWING OUR NUMBERS SOLUTIONS

**EXERCISE 1.2**

**Problem 1**

**class 6 maths textbook solutions chapter 1 problem 1 exercise 1.2 Knowing our numbers **

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 the four days.

**Solution**:

The number of tickets sold on the first day = 1,094

The number of tickets sold on the second day = 1,812

The number of tickets sold on the third day. = 2,050

The number of tickets sold on the final day. = 2,751

The total number of tickets sold on all the four days = 1,094 + 1,812 + 2,050 + 2751 = 7,707

Therefore 7,707 tickets were sold on all the four days.

**Problem 2**

**class 6 maths textbook solutions chapter 1 problem 2 exercise 1.2 Knowing our numbers **

**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.

**Solution**:

The number of runs Shekhar scored = 6980

The number of runs he wishes to complete = 10,000

The number of runs he required. = 10,000 – 6980. = 3020

Therefore he needs 3020 more runs.

**Problem 3**

**class 6 maths textbook solutions chapter 1 problem 3 exercise 1.2 Knowing our numbers **

3. In an election, the successful candidate registered 5,77,500. votes and his nearest rival second 3,48,700. votes. By what margin did the successful candidate win the election?

**Solution**:

The number of votes secured by the successful candidate = 5,77,500

The number of votes secured by nearest rival candidate = 3,48,700

Margin between them = 5,77,500 – 3,48,700 = 2,28,800

Therefore, the successful candidate won by a margin of 2,28,800 votes.

**Problem 4**

**CBSE class 6 maths solutions chapter 1 problem 4 exercise 1.2 Knowing our numbers **

4. Kirti bookstore sold books worth Rs. 2,85,891 in the first week of June and books worth Rs. 4,00,786 in the 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?

**Solution**:

The worth of books sold in the first week = Rs. 2,85,891

The worth of books sold in the second week = Rs. 4,00,786

The worth of total books sold in two weeks = Rs.2,85,891 + Rs. 4,00,768 =Rs.6,86,659

Therefore, sale of second week is greater than that of first week.

The worth of more books sold in the second week = Rs. 4,00,768 – Rs. 2,85,891 = Rs. 1,14,877

Therefore, the worth of Rs. 1,14,877 books were sold in the second week.

**Problem 5**

**CBSE class 6 maths solutions chapter 1 problem 5 exercise 1.2 Knowing our numbers **

5. Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once*.*

**Solution**:

Using the digits 6, 2, 7, 4, 3

The greatest five digit number = 76,432

The smallest five digit number = 23,467

The difference between the greatest and smallest number = 76,432 -23,467 = 52,965

Therefore, the difference is 52,965

**Problem 6**

**CBSE class 6 maths solutions chapter 1 problem 6 exercise 1.2 Knowing our numbers **

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

**Solution**:

The number of screws manufactured in a day = 2,825

The number of days in the month of January = 31

The number of screws manufactured in 31 days = 2,825 × 31 = 87,575

Therefore, the machine produced 87,575 screws in the month of January 2oo6.

**Problem 7**

**CBSE class 6 maths solutions chapter 1 problem 7 exercise 1.2 Knowing our numbers **

7. A merchant had Rs.78,592 with her. She placed an order for purchasing 40 radio sets at Rs.1,200 each. How much money will remain with her after the purchase ?

**Solution**:

The cost of one radio set = Rs. 1,200

The cost of 40 radio sets = Rs. 1,200 × 40 = Rs. 48,000

The merchant contained the money = Rs.78,592

She spent the money for purchasing radio sets = Rs. 48,000

She contains the money after the purchase = Rs. 78,592 – Rs. 48,000 = Rs. 30,592

Therefore, Rs. 30,592 will remain with her after the purchase.

**Problem 8**

**Std 6 maths textbook solutions chapter 1 problem 8 exercise 1.2 Knowing our numbers **

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

**Solution**:

The multiplication done by the student = 7236 × 65 = 470,340

The multiplication required by the student = 7236 × 56 = 4,05,216

The difference between the multiplications = 4,70,340 – 4,05,516 = 65,124

Therefore, 65,124 is greater than the correct answer.

**Problem 9**

**Std 6 maths textbook solutions chapter 1 problem 9 exercise 1.2 Knowing our numbers **

9. To stitch a shirt, 2m 15 cm cloth is needed. Out of 40m cloth, how many shirts can be stitched and how much cloth will remain ?

**Solution**:

The cloth needed to stitch one shirt = 2 m 15 cm = 215 cm

Length of the cloth = 40 m = 4000 cm

The number of shirts can be stitched = 4000 ÷ 215

[ 4000 /215 = 18 and the remainder130 ]

Therefore, 18 shirts can be stitched and 130 cm or 1m 30 cm will remain.

**Problem 10**

**Std 6 maths textbook solutions chapter 1 problem 10 exercise 1.2 Knowing our numbers **

10. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg ?

**Solution**:

The weight of one box = 4 kg 500g = 4500 g

Maximum weight can be loaded in a van = 800 kg = 800000 g

The number of boxes can be loaded = 800000 g ÷ 4500 g

[ 8,00,000/4,500 = 177 and the remainder 3,500 ]

Therefore, 177 boxes can be loaded in the van.

**Problem 11**

**Std 6 maths textbook solutions chapter 1 problem 11 exercise 1.2 Knowing our numbers **

11. The distance between the school and the house of 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.

**Solution**:

The distance between the school and the house of a student = 1 km 875 m = 1875 m

The distance between the house and school of the student = 1875 m

The total distance covered in one day = 1875 + 1875 = 3750 m

The distance covered in six days = 3750 m × 6 = 22500 m= 22 km 500 m

( 22500 ÷ 1000 = 22.5 )

Therefore, 22 km 500m distance covered by her in six days.

**Std 6 maths textbook solutions chapter 1 problem 12 exercise 1.2 Knowing our numbers **

**Problem 12**

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

**Solution**:

The quantity of curd in a vessel = 4 litres 500 ml= 4500 ml

The capacity of one glass = 25 ml

The number of glasses can be filled = 4500 ÷ 25 = 180

Therefore, 180 glasses can be filled by curd.

**CBSE SOLUTIONS FOR CLASS 6 MATHS CHAPTER 1**

**NCERT SOLUTIONS FOR CLASS 6TH MATHS**

**CBSE SOLUTIONS FOR CLASS 6 MATHS**

**SOLUTIONS FOR KNOWING OUR NUMBERS CLASS 6 MATHS NCERT**

**EXERCISE 1.3 (Eliminated for 2023 – 2024)**

**Problem 1**

1. Estimate each of the following using general rule:

a. 730 + 998

**Solution**:

730 + 998

Round off to hundreds,

730 rounds off to 700

998 rounds off to 1000

Estimated sum = 700 + 1000 = 1700

b. 796 – 314

**Solution**:

796 – 314

Round off to nearest hundreds,

796 rounds off to 800

314 rounds off to 300

Estimated difference = 800 – 300 = 500

c. 12,904 + 2,888

**Solution**:

12,904 + 2,888

Round off to nearest thousands,

12,904 rounds off to 13,000

2,888 rounds off to 3,000

Estimated sum = 13,000 + 3,000 = 16,000

**d. 28,292 + 21,496**

**Solution** :

28,292 + 21,496

Round off to nearest thousands,

28,292 rounds off to 28,000

21,496 rounds off to 21,000

Estimated difference = 28,000 – 21,000 = 7,000

**Problem 2**

*2*. Give a rough estimate ( by rounding off to nearest hundreds ) and also a closure estimate ( by rounding off to nearest tens* ) :*

a. 439 + 334 + 4,317

**Solution**:

439 + 334 + 4,317

Round off to nearest hundreds

439 rounds off to 400

334 rounds off to 300

4,317 is rounds off to 4,300

Estimated sum = 400 + 300 + 4,300 = 5000

439 + 334 + 4,317

Round off to tens

439 rounds off to 440

334 rounds off to 330

4,317 rounds off to 4,320

Estimated sum = 440 + 330 + 4,320 = 5090

b. 1,08,734 – 47,599

**Solution** :

1,08,734 – 47,599

Round off to hundreds

1,08,734 rounds off to 1,08,700

47,599 rounds off to 47,600

Estimated difference = ,108,700 – 47,600 = 61,100

1,08,734 – 47,599

Round off to tens

1,08,734 rounds off to 1,08,730

47,599 rounds off to 47,600.

Estimated difference = 1,08,730 – 47,600 = 61,130

c. 8325 – 491

**Solution**:

8325 – 491

Round off to hundreds.

8,325 rounds off 8,300.

491 rounds off to 500.

Estimated difference = 8300 – 500 = 7800

8325 – 491

Round off to tens.

8325 rounds off to 8330.

491 rounds off to 490.

Estimated difference = 8,330 – 490 = 7,840

d. 4,89,348 – 48,365

**Solution:**

4,89,348 – 48,365

Round off to hundreds.

4,89,348 rounds off to 4,89,300

48,365 rounds off to 48,400.

Estimated difference = 4,89,300 – 48,400 = 4,40,900

4,89,348 – 48,365

Round off to tens

4,89,348 rounds off to 4,89,350

48,365 rounds off to 48,370.

Estimated difference = 4,89,350 – 48,370 = 40,980

**Problem 3**

3. Estimate the following products using general rule:

**a. 578 × 161**

**Solution** :

578 × 161

Round off to nearest hundreds

578 rounds off to 600

161 rounds off to 200

Estimated product = 600 × 200 = 1,20,000

b. 5281 × 3491

**Solution** :

5281 × 3491

Round off to nearest hundreds

5281 rounds off to 5000

3491 rounds off to 3,000

Estimated product = 5000 × 3,000 = 1,50,00,000

c. 1291 × 592

**Solution :**

1201 × 592

Round off to nearest hundreds

1291 rounds off to 1300

592 rounds off to 600.

Estimated product = 1300 × 600 = 7,80,000

d. 9750 × 29

**Solution:**

9,750 rounds off to nearest hundreds.

9,750 rounds off to 9,000.

29 rounds off to tens.

29 rounds off to 30.

Estimated product = 9,000 × 30 = 2,70,000

**Note:**** **Observe the solutions of knowing our numbers and try them in your own methods.

**Chapter 4 Basic Geometrical Ideas**

**Integers**

Ncert solutions for class 6th chapter 6

**Fractions**

**Chapter 14 Practical Geometry class 6**

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