NCERT solutions for class 6th maths Chapter 1, Knowing our Numbers Solutions

Knowing our Numbers solutions

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.

You must 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

Exercise 1.1

Exercise 1.2

Exercise 1.3

KNOWING OUR NUMBERS chapter 1 solutions class 6 maths NCERT

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

Exercise 7.1