# CBSE Class 6 Maths Chapter 2 Important Questions with Solutions

Maths is a vital subject taught in school. We need Maths in our daily lives. From calculating household expenses to constructing bridges, Maths is applied everywhere.

Chapter 2 of Class 6 Maths under the CBSE syllabus is the Study of Whole Numbers. In simple words, whole numbers include 0 and natural numbers. These numbers do not include fractions or decimal numbers and are placed along the number line. In other words, the numbers we generally deal with in real life are mostly whole numbers. Students must practise questions from this chapter to ensure that they understand the material.

Extramarks is one of the leading educational platforms in India. We help teachers and students by providing a wide range of study materials. Our maths experts have created important chapter-wise questions for every subject. The Important Questions from Class 6 Maths Chapter 2 include all the vital questions from this chapter. The experts have collected the questions from various sources like the NCERT textbook exercises, different reference books, CBSE sample papers, and CBSE past year’s question papers.

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## Get Access to CBSE Class 6 Maths Important Questions with Solutions

Also, get access to CBSE Class 6 Maths Important Questions for other chapters too:

### CBSE Important Questions for Class 6 Maths

Sr No Chapter No Chapter Name
1 Chapter 1 Knowing Our Numbers
2 Chapter 2 Whole Numbers
3 Chapter 3 Playing with Numbers
4 Chapter 4 Basic Geometrical Ideas
5 Chapter 5 Understanding Elementary Shapes
6 Chapter 6 Integers
7 Chapter 7 Fractions
8 Chapter 8 Decimals
9 Chapter 9 Data Handling
10 Chapter 10 Mensuration
11 Chapter 11 Algebra
12 Chapter 12 Ratio and Proportion
13 Chapter 13 Symmetry
14 Chapter 14 Practical Geometry

## Whole Numbers Class 6 Questions and Answers With Solutions

Our experts believe in continuous practice. For this purpose, they have created this question series. The experts have collected vital questions from the textbook and several reference books. They have also taken help from CBSE’s past years’ question papers and CBSE sample papers. Apart from this, they have provided solutions which are checked by experts. Thus, students will find different questions and answers in the Important Questions Class 6 Maths Chapter 2. It will help the students to improve their grades in exams. The following are some of the important questions for a quick revision and self-assessment:

### Question 1. Write the 3 whole numbers which occur right before 10001.

The 3 whole numbers occurring right before 10001 are 10000, 9999 and 9998.

### Question 2. Which is the smallest whole number?

The smallest whole number is 0.

### Question 3. How many whole numbers come between 32 and 53?

The whole numbers that come between 32 and 53 are –

(33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51 and 52)

Hence, there are exactly 20 whole numbers  between 32 and 53.

### Question 4. Write the successor of the following –

(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670

The successors the following are –

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

### Question 5. Write the predecessor of the following –

(a) 94 (b) 10000 (c) 208090 (d) 7654321

The predecessors of the following are –

(a) 94 – 1 = 93

(b) 10000 – 1 = 9999

(c) 208090 – 1 = 208089

(d) 7654321 – 1 = 7654320

### Question 6. Arrange the following numbers is descending order:

925,786,1100,141,325,886,0,270

Numbers in descending order:

1100,925,886,786,325,270,141,0

### Question 7. In each one of the following pairs of numbers given below, a state in which the whole number is on the left of another number on the number line. Also, write them with the suitable sign (>, <) between them.

(a) 530 and 503

(b) 370 and 307

(c) 98765 and 56789

(d) 9830415 and 10023001

(a) Since, we know 530 is greater than 503 (530 > 503)

Hence, 503 is on the left-hand side of 530 on the number line.

(b) Since, we know 370 is greater than 307 (370 > 307)

Hence, 307 is on the left-hand side of 370 on the number line.

(c) Since we know 98765 is greater than 56789 (98765 > 56789)

Hence, 56789 is on the left-hand side of 98765 on the number line.

(d) Since we know 9830415 is less than 10023001 (9830415 < 10023001)

Hence, 9830415 is on the left-hand side of 10023001 on the number line.

### Question 8. Which of the following statements given are true (T) and which are false (F)?

(1) Zero is the smallest natural number.

False.

0 is not a natural number.

(2) 400 is the predecessor of 399.

False.

The predecessor of 399 is 398 as (399 – 1 = 398).

(3) 0 is the smallest whole number.

True.

0 is the smallest whole number.

(4) 600 is the successor of 599.

True.

Since, (599 + 1 = 600)

(5) All the natural numbers on the number line are whole numbers.

True.

All natural numbers on the number line are whole numbers.

(6) All whole numbers on the number line are natural numbers.

False.

Zero is a whole number but is not a natural number.

(7) The predecessor of a 2-digit number is never a single-digit number.

False.

For example, the predecessor of 10 is 9.

(8) One is the smallest whole number.

False.

Zero is the smallest whole number.

(9) The natural number one has no predecessor.

True.

The predecessor of 1 is 0, but it is not a natural number.

(10) The whole number one has no predecessor.

False.

0 is the predecessor of one and is a whole number

(11) The whole number 13 comes between 11 and 12.

False

13 does not come between 11 and 12

(12) The whole number 0 has no predecessor.

True

The predecessor of 0 is -1 and is not a whole number.

(13) The successor of a two-digit number is always a two-digit number.

False

As the successor of 99 is 100

Question 8. Find the product by suitable rearrangement:

(a) 2 × 1768 × 50

(b) 4 × 166 × 25

(c) 8 × 291 × 125

(d) 625 × 279 × 16

(e) 285 × 5 × 60

(f) 125 × 40 × 8 × 25

(a) Given 2 × 1768 × 50

= 2 × 50 × 1768

= 100 × 1768

= 176800

(b) Given 4 × 166 × 25

= 4 × 25 × 166

= 100 × 166

= 16600

(c) Given 8 × 291 × 125

= 8 × 125 × 291

= 1000 × 291

= 291000

(d) Given 625 × 279 × 16

= 625 × 16 × 279

= 10000 × 279

= 2790000

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 1000000

### Question 9. Find the value of the following:

(a) 297 × 17 + 297 × 3

(b) 54279 × 92 + 8 × 54279

(c) 81265 × 169 – 81265 × 69

(d) 3845 × 5 × 782 + 769 × 25 × 218

(a) Given 297 × 17 + 297 × 3

= 297 × (17 + 3)

= 297 × 20

= 5940

(b) Given 54279 × 92 + 8 × 54279

= 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

(c) Given 81265 × 169 – 81265 × 69

= 81265 × (169 – 69)

= 81265 × 100

= 8126500

(d) Given 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000

= 19225000

### Question 10. Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

(a) Given 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (using distributive property)

= 168000 + 840

= 168840

### Question 11. A vendor supplies thirty-two litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 45 per litre, how much money is due to the vendor per day?

Milk quantity supplied in the morning = 32 litres

Milk quantity supplied in the evening = 68 litres

Cost of milk per litre = ₹ 45

Total cost of milk per day = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

Hence, the money which is due to the vendor per day is ₹ 4500.

### Question 12. Match the following:

(i) 425 × 136 = 425 × (6 + 30 + 100) (1) Commutativity under multiplication.

(ii) 2 × 49 × 50 = 2 × 50 × 49 (2) Commutativity under addition.

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (3) Distributivity of multiplication over addition.

(i) 425 × 136 = 425 × (6 + 30 + 100) (3) Distributivity of multiplication over addition.

(ii) 2 × 49 × 50 = 2 × 50 × 49 (1) Commutativity under multiplication

Hence, commutative under multiplication is the correct answer.

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (2) Commutativity under addition

### Question 13. If multiplication of two whole numbers is 0, can we say that one or both of them will be zero? Justify through examples.

If the multiplication of 2 whole numbers is 0, one of them is definitely zero.

Example: 0 × 3 = 0 and 15 × 0 = 0

If multiplication of two whole numbers is 0, both of them may be zero.

Example: 0 × 0 = 0

Yes, if the multiplication of two whole numbers is zero, both will be zero.

### Question 14. If multiplication of 2 whole numbers is one, can we say that one or both of them will be 1? Justify through examples.

If multiplication of two whole numbers is 1, both the numbers should be equal to 1

Example: 1 × 1 = 1

But 1 × 5 = 5

Hence, it is clear that the product of two whole numbers will be 1, only in a situation when both numbers to be multiplied are 1.

### Question 15. Find using distributive property:

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

(a) Given 728 × 101

= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728

= 73528

(b) Given 5437 × 1001

= 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c )Given 824 × 25

= (800 + 24) × 25

= (800 + 25 – 1) × 25

= 800 × 25 + 25 × 25 – 1 × 25

= 20000 + 625 – 25

= 20000 + 600

= 20600

(d) Given 4275 × 125

= ( 4000 + 200 + 100 – 25) × 125)

= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)

= 500000 + 25000 + 12500 – 3125

= 534375

(e) Given 504 × 35

= (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140

= 17640

### Question 16. Which property does the following statements hold?

(1) 6 + 4 = 4 + 6

(2) 3 + 2 = whole number

(1) 6 + 4 = 4 + 6 holds commutative property of addition

(2) 3 + 2 = whole number holds closure property.

### Question 17.

Add the following in three different ways. Indicate the property used.

(1) 25 + 36 + 15

(2) 30 + 18 + 22

(1) 25 + 36 + 15

Approach I: 25 + (36 + 15) = 25 + 51 = 76

Approach II: (25 + 36) + 15 = 61 + 15 = 76

Approach III: (25 + 15) + 36 = 40 + 36 = 76

Here, we have used the associative property.

(2) 30 + 18 + 22

Approach I: 30 + (18 + 22) = 30 + 40 = 70

Approach II: (30 + 18) + 22 = 48 + 22 = 70

Approach III: (30 + 22) + 18 = 52 + 18 = 70

Here, we have used the associative property.

### Question 18. Fill in the blanks.

(1) The smallest whole number is ……….

(2) The smallest natural number is ……….

(3) Difference between the 5-digit smallest number and the 4-digit largest number is ……….

(4) Any number divided by 0 is not ……….

(5) The property used in 84 x 25 = 25 x 84 is ……… .

(6) The property used in 80 x (60 + 3) = 80 x 60 + 80 x 3 is ……… .

(7) The smallest number, shown by two dotted rectangles, is ……….

(8) Every whole number except ……… is a natural number.

(9) When any counting number is multiplied by zero, the product is ……….

(10) When zero is divided by any non-zero whole number, the quotient is ……….

(1) The smallest whole number is 0.

(2) The smallest natural number is 1.

(3) Difference between the 5-digit smallest number and the 4-digit largest number is 1.

(4) Any number divided by 0 is not defined.

(5) The property used in 84 x 25 = 25 x 84 is commutative property.

(6) The property used in 80 x (60 + 3) = 80 x 60 + 80 x 3 is distributive property.

(7) The smallest number, shown by two dotted rectangles, is 6.

(8) Every whole number except 0 is a natural number.

(9) When any counting number is multiplied by zero, the product is 0.

(10) When zero is divided by any non-zero whole number, the product is 0.

### Write down three consecutive whole numbers just preceding 8510001.

We have:

First number = 8,510,001 – 1 = 8,510,000

Second number = 8,510,000 – 1 = 8,509,999

Third number = 8,509,999 – 1 = 8,509,998

Hence, the three consecutive whole numbers just preceding 8,510,001 are 8,510,000, 8,509,999 and 85,09,998.

### Solve the following and establish a pattern:

(1) 84 x 9

(2) 84 x 99

(3) 84 x 999

(4) 84 x 9999

(1) 84 x 9 = 84 x (10 – 1) = 84 x 10 – 84 x 1 = 840 – 84 = 756

(2) 84 x 99 = 84 x (100 – 1) = 84 x 100 – 84 x 1 = 8400 – 84 = 8316

(3) 84 x 999 = 84 x (1000 – 1) = 84 x 1000 – 84 x 1 = 84000 – 84 = 83916

(4) 84 x 9999 = 84 x (10000 – 1) = 84 x 10000 – 84 x 1 = 840000 – 84 = 839916

### Write the largest number of six digits and the smallest number of seven digits. Which one of these two is greater and by how much?

Largest 6 digit number = 999,999

Smallest 7 digit number = 1,000,000

Thus, the smallest 7 digit number is larger than the largest 6 digit number.

Again,

Difference between these two numbers = 1,000,000 – 999,999 = 1

Hence, the smallest seven-digit number is larger than the largest six-digit number by 1.

### Find the product of the greatest three-digit number and the greatest two-digit number.

Greatest three-digit number = 999

Greatest two-digit number = 99

∴ Product of both numbers = 999 x 99 = 999 x (100 – 1)

= 999 x 100 – 999 x 1

= 99900 – 999 = 98901

### Question 23. I A housing complex built by DLF consists of 25 large buildings and 40 small buildings. Each of the large buildings has 15 floors with four apartments on each floor, and each small building has nine floors with three apartments on each floor. How many apartments are there in total?

Number of the large buildings = 25

Number of the floors = 15

Number of the apartments on each floor = 4

∴ Total number of the apartments in large buildings = 25 x 15 x 4

Number of small buildings = 40

Number of the floors = 9

Number of the apartments on each floor = 3

∴ Total number of the apartments in small buildings = 40 x 9 x 3

Hence, the number of apartments in total = 25 x 15 x 4 + 40 x 9 x 3 = 1500 + 1080 = 2580.

### A school principal placed orders for 85 chairs and 25 tables with a dealer. Each chair costs ₹180, and each table costs ₹140. If the principal has given ₹2500 to the dealer as an advance money, what amount is to be given to the dealer now?

Number of total chairs = 85

Cost of one chair = ₹ 180

Cost of 85 chairs = ₹ (85 x 180)

Number of total tables = 25

Cost of one table = ₹ 140

Cost of 25 tables = ₹ (25 x 140)

Total amount of all chairs and tables = ₹(85 x 180 + 25 x 140)

= ₹ (15300 + 3500) = ₹ 18800

Money given in advance by principal = ₹ 2500

∴ Remaining money to be paid to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300

### A dealer purchased 124 LED sets. If the cost of one set is ₹38,540, determine their total cost.

Total cost of 124 LED sets = ₹(38,540 x 124)

= ₹ [38,540 x (100 + 20 + 4)]

= ₹ [38,540 x 100 + 38,540 x 20 + 38,540 x 4]

= ₹ [38,54,000 + 7,70,800 + 1,54,160]

= ₹ 47,789,60

## Benefits of Solving Class 6 Maths Chapter 2 Extra Questions

Maths is a very important subject taught in school, but many students fear Maths. It is because they may not have developed clear concepts and face problems in understanding the subject. Students must practise Maths as much as possible to strengthen their knowledge of the subject. It will also help them to boost their confidence. The Maths experts at   Extramarks have prepared the Important Questions Class 6 Maths Chapter 2 to help the students to develop their Mathematical abilities through guided practice which provides enough conceptual clarity to students. There are multiple benefits of studying the question series, such as.

• Students must practise Maths from lower classes to build the habit of solving questions. Solving problems will help to strengthen concepts and give them the confidence to solve problems on their own.  Extramarks subject experts know the importance of solving problems. For this purpose, they have prepared the Chapter 2 Class 6 Maths Important Questions to help students. They collected several questions from different sources. They have taken help from the CBSE past years’ question papers so that students can get a fair idea of  questions expected  in exams.
• Solving questions may not be enough for practice. Students must check the answers too. If they cannot solve any question, they require the following solutions. Keeping this fact in mind, the experts have also provided the solutions to the questions in the Important Questions Class 6 Maths Chapter 2. They have solved the questions following the pattern students should use in exams. Experienced subject experts  have prepared  the answers to ensure the best preparation for students.
• Practising questions will help the students in multiple ways. The experts have tried to include different  types of questions possible from the chapter. Thus, the Maths Class 6 Chapter 2 Important Questions covers all the important concepts. It will also help them to solve the questions within the time limit. Therefore, it will help them to develop interest in Maths and motivate them to get a 100% score  in exams.

Extramarks subject experts understand the importance of solving important questions and we take our role seriously to provide the best resource to the students and help them excel in academics.The experts have prepared chapter-wise important question answers for all subjects. Like the Important Questions Class 6 Maths Chapter 2, students can also find important questions and answers for other chapters. Extramarks is one of the best  educational platforms  providing students with plenty of  study materials. You may register on our official website and download the CBSE syllabus, NCERT books, CBSE revision notes, CBSE extra questions, chapter-wise important questions, CBSE past years’ question papers, vital formulas, NCERT solutions etc. Click on the  links to download the following study materials-

Q.1 98 × 5 = 490,

98 × 15 = 490 × 3,

98 × 25 = 490 × 5,

Basis on this pattern, 98 × 35 will be

490 × 2

490 × 4

490 × 7

p>490 × 9

Marks:1

Ans

Since, 98×5 = 490

98×15 = 98×5×3 = 490×3
98×25 = 98×5×5 = 490×5

Therefore, 98×35 = 98×5×7 = 490×7

Q.2 What is the value of 384 × 12 + 384 × 8?

(a) 7680

(b) 7660

(c) 9690

(d) 7690

Marks:1

Ans

7680

384 × 12 + 384 × 8

= 384 × (12 + 8) (Using distributive property)

= 384 × 20

= 7680

Q.3 Which of the following is incorrect regarding whole numbers?

(a) Whole numbers are closed under subtraction.

(b) Whole numbers are closed under addition.

(c) Whole numbers are closed under multiplication.

(d)Whole numbers are associative under addition.

Marks:1

Ans

Whole numbers are closed under subtraction.

The collection of whole numbers is not closed under subtraction.
Example, 2 ? 4 = ?2, which is not a whole number.

Q.4 The digits 5 and 7 are interchanged in the number 27658. Find the difference between the original number and the new number.

Original number  =27658
New number =25678
Difference

Marks:2

Ans

Original number =27658
New number =25678
Difference

Q.5 Multiply using suitable rearrangements.

$25×7896×4×50×2$ $\left(25×4\right)×7896×\left(50×2\right)$ $=100×7896×100$ $=78960000$

Marks:3

Ans

$\left(25—4\right)—7896—\left(50—2\right)$ $=100—7896—100$ $=78960000$