Ncert Solutions for Class 6 Maths Chapter 1 Pdf

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Chapter 1 Knowing Our Numbers NCERT Solutions for Class 6 Maths is available here which will be helpful in covering the entire syllabus and solving the difficult problems given in exercise. You can also Download PDF of Class 6 Maths Chapter 1 Knowing Our Numbers NCERT Solutions which will prove useful guide in making a student confident.

These NCERT Solutions for Class 6 will develop you understanding of the chapter and help in gaining good marks in the examinations. Chapter 1 NCERT Solutions will help you in completing your homework on time. These NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily.

Page No: 12

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

Answer

a) 100
b) 10
c) 10
d) 10
e) 100

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.

Answer

a) 73,75,307
b) 9,05,000,41
c) 7,52,21,302
d) 58,423,202
e) 23,30,010

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

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

Answer

(a) 8,75,95,762: Eight crore seventy five lakh ninety five 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 International System of Numeration:

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

Answer

(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.
(e) 48,049,831: Forty eight million forty nine thousand eight hundred thirty one.

Page No: 16

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 the four days.

Answer

Number of tickets sold first day = 1094
Number of tickets sold second day = 1812
Number of tickets sold third day = 2050
Number of tickets sold fourth day = 2751
Total number of tickets sold on all the four days  =  1094 + 1812 + 2050 + 2751 = 7707

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?

Answer

Shekhar scored = 6980 runs
Shekhar wants to score = 10,000 runs
He need to score 10,000 – 6980 = 3020

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?

Answer

Successful candidate registered 5,77,500 votes
Score secured by his rival = 3,48,700 votes

5,77,500  -  3,48,700 = 22880
Successful candidate need 22880 margin to win the election.

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?

Answer

Books sold in the first week = 2,85,891
Books sold in the second week = 4,00,768
The sale of two weeks together = 2,85,891 + 4,00,768 = 686659
Second week of the month books sale 4,00,768  was greater than first week 2,85,891.

4,00,768 – 2,85,891 = 114877
Book sale was greater than by 114877

Page No: 17

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.

Answer

Greatest number = 76432
Smallest number = 23467
The difference between greatest and smallest number = 76432 – 23467 = 52965

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

Answer

Number of screw produced in one days = 2,825
As we know in the month of January there is 31 days.
Number of screw produced in 31 days = 2,825 × 31 = 87575.
So, the number of screw produced in January  2006 = 87575.

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?

Answer

Merchant had 78,592 with her
Cost of one radio = 1200
Cost of 40 radio = 12000 × 40 = 48000
Money spent by merchant by = 48000
Money left = 78,592 – 48000 = 30592

Rs. 30592 will left after purchasing.

8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
(Hint: Do you need to do both the multiplications?)

Answer

Differences between 65 × 56 = 9
Correct answer is greater than by 7236 × 9 = 65124

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?
(Hint: convert data in cm.)

Answer

Cloth available = 40 m
As we know 1 m = 100 cm
40 m = 40 × 100 = 4000
2 m 15 cm = 215 cm
Required cloth for one t-shirt = 215 cm
Number of cloths stitched = 4000 ÷ 215

Therefore, 18 shirts can be made.
Remaining cloths is 1 m 130 cm.

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?

Answer

Weight of each boxes is = 4 kg 500 g

1 kg = 1000 g

4 kg 500 g = 4500 g

So, 800 kg = 800 × 1000 = 800000 g

Number of boxes required for 800000 ÷ 4500

So, the number of boxes required for medicine = 177.

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.

Answer

Distance between the school and the house = 1 km 875 m.

As we know 1 km = 1000 m

1 km 875 m = 1875 m.

There are two ways

Distance covered by her in each day was 1875 × 2 = 3750 m

And the distance covered in six days = 3750 × 6 = 22500

∴ Distance covered in six days = 22500 m = 22 km 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?

Answer

Vessel can store = 4 litres and 500 ml

As we know, 1 l = 1000 ml

We have to find number of classes that can store 25 ml of curd.

Number of glasses = 4500 ÷25

∴ 180 glasses are required for 25 ml of curd.

Page No: 23

Exercise 1.3

1. Estimate each of the following using general rule:

(a) 730 + 998

(b) 796 – 314

(c) 12,904 +2,888

(d) 28,292 – 21,496

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

Answer

a) 730 + 998

• Rounding off to nearest hundred. 730 round off to 700 and 998 round off to 1000.

700 + 1000 = 1700

b) 796 – 314

800 – 300 = 500

c) 12,904 + 2,888

13000 + 3000 = 16000

d) 28,292 – 21,496

28000 – 21000 = 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 + 4,317

(b) 1,08,734 – 47,599

(c) 8325 – 491

(d) 4,89,348 – 48,365

Make four more such examples.

Answer

a) 439 + 334 + 4,317

• Rounding of hundred = 400, 300, 4000

400 + 300 + 4000 = 5000

• Rounding of ten = 440, 330, 4,320

440 +330 + 4,320 = 5090.

b) 1,08,734 – 47,599

• Rounding of hundred = 1,08,700 and 47,600

1,08,700 - 47,600 = 61100

• Rounding of ten = 1,08,730 and  47,600

1,08,730  - 47,600

c) 8325 – 491

• Rounding of hundred 8300 and 500

8300 - 500 = 7800

• Rounding of ten 8330 and  490

83330 – 490 = 7840

d) 4,89,348 – 48,365

• Rounding of hundred = 4,89,300 – 48400 = 440900

• Rounding of ten = 489350 – 48370 = 440980

3. Estimate the following products using general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Answer

a) Rounding off to nearest hundred  ,578  round off to 600 and 161 round off to 200.

600 × 200 = 120000

b) 5000 × 3000 = 15000000

c) 1000 × 600 = 600000

d) 9000  × 30 = 270000

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Chapter 1 Knowing Our Numbers NCERT Solutions will develop your problem solving skills and make aware of the difficulty of questions. In this chapter, we will be comparing numbers and see the place value chart.

• Ascending order means arrangement from the smallest to the greatest.

• Descending order means arrangement from the greatest to the smallest.

• Some observation:

(i) Greatest single digit number + 1 = smallest 2-digit number.

(ii) Greatest 2-digit number + 1 = smallest 3-digit number.

(iii) Greatest 3-digit number + 1 = smallest 4-digit number.

Below you will find exercisewise Chapter 1 NCERT Solutions that will be updated according to the latest NCERT Class 6 Maths textbook and syllabus. By solving the questions from Class 6 NCERT textbook, a student will gain confidence that is going to help them in exams.

• Exercise 1.1 Chapter 1 Class 6 Maths NCERT Solutions
• Exercise 1.2 Chapter 1 Class 6 Maths NCERT Solutions
• Exercise 1.3 Chapter 1 Class 6 Maths NCERT Solutions

Studyrankers experts has prepared these Class 6 NCERT Solutions which are useful in the preparation of examinations and make you aware of concepts of the chapter. It will help you in analyzing the problems and answering it with precision and the right concepts.

NCERT Solutions for Class 6 Maths Chapters:

FAQ on Chapter 1 Knowing Our Numbers

What are the benefits of NCERT Solutions for Chapter 1 Knowing Our Numbers Class 6 NCERT Solutions?

NCERT Solutions for Class 6 are helpful in adopting a strategy that helps students operate and learn at maximum efficiency. It is very challenging to score good marks in tests that is why we have prepared NCERT Solutions.

How many exercises in Chapter 1 Knowing Our Numbers?

There are total 3 exercises in Chapter 1 Class 6 Maths which will help in understanding the key concepts of the chapter properly. You can figure out the latest marking scheme and prepare your answers as per the demand.

How to Compare Numbers?

When two numbers have different number of digits, the number having more number of digits is greater than the number having less number of digits. When two numbers have same number of digits, the number having greater left most digit is greater. If their left most digit is same then the number having greater next left most digit is greater and so on.

Ncert Solutions for Class 6 Maths Chapter 1 Pdf

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