Class 6 Maths - Whole Numbers NCERT Solutions

Chapter 2: Whole Numbers (NCERT Solutions)

Exercise 2.1

Q1. Write the next three natural numbers after 10999.

The next three natural numbers are obtained by adding 1 successively:
10,999 + 1 = 11,000
11,000 + 1 = 11,001
11,001 + 1 = 11,002

Q2. Write the three whole numbers occurring just before 10001.

The three whole numbers occurring just before are obtained by subtracting 1 successively:
10,001 - 1 = 10,000
10,000 - 1 = 9,999
9,999 - 1 = 9,998

Q3. Which is the smallest whole number?

The collection of whole numbers starts from 0 (0, 1, 2, 3, ...).
Therefore, the smallest whole number is 0.

Q4. How many whole numbers are there between 32 and 53?

The whole numbers between 32 and 53 do not include 32 and 53 themselves.
Number of whole numbers = (53 - 32) - 1
= 21 - 1 = 20.

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

Q5. Write the successor of:
(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670

The successor is found by adding 1.

(a) Successor of 2440701 = 2440701 + 1 = 24,40,702
(b) Successor of 100199 = 100199 + 1 = 1,00,200
(c) Successor of 1099999 = 1099999 + 1 = 11,00,000
(d) Successor of 2345670 = 2345670 + 1 = 23,45,671

Q6. Write the predecessor of:
(a) 94 (b) 10000 (c) 208090 (d) 7654321

The predecessor is found by subtracting 1.

(a) Predecessor of 94 = 94 - 1 = 93
(b) Predecessor of 10000 = 10000 - 1 = 9,999
(c) Predecessor of 208090 = 208090 - 1 = 2,08,089
(d) Predecessor of 7654321 = 7654321 - 1 = 76,54,320

Q7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001

The smaller number always lies to the left of the larger number on the number line.

(a) 530 and 503
Since 503 < 530, 503 will be on the left of 530.
Statement: 503 < 530

(b) 370 and 307
Since 307 < 370, 307 will be on the left of 370.
Statement: 307 < 370

(d) 9830415 and 10023001
Since 9830415 < 10023001, 98,30,415 will be on the left of 1,00,23,001.
Statement: 98,30,415 < 1,00,23,001

Q8. Which of the following statements are true (T) and which are false (F)?

(a) Zero is the smallest natural number.
False. (1 is the smallest natural number).

(b) 400 is the predecessor of 399.
False. (Predecessor of 399 is 398. 400 is the successor).

(d) 600 is the successor of 599.
True.

(e) All natural numbers are whole numbers.
True.

(f) All whole numbers are natural numbers.
False. (0 is a whole number but not a natural number).

(g) The predecessor of a two digit number is never a single digit number.
False. (Predecessor of 10 is 9).

(h) 1 is the smallest whole number.
False. (0 is the smallest whole number).

(i) The natural number 1 has no predecessor.
True. (There is no natural number before 1).

(j) The whole number 1 has no predecessor.
False. (The predecessor of whole number 1 is 0).

(k) The whole number 13 lies between 11 and 12.
False. (13 lies after 12).

(l) The whole number 0 has no predecessor.
True. (There is no whole number less than 0).

(m) The successor of a two digit number is always a two digit number.
False. (Successor of 99 is 100, which is a 3-digit number).

Exercise 2.2

Q1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647

(a) 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208 = 1408

(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600

Q2. 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) 2 × 1768 × 50 = (2 × 50) × 1768
= 100 × 1768 = 1,76,800

(b) 4 × 166 × 25 = (4 × 25) × 166
= 100 × 166 = 16,600

(d) 625 × 279 × 16 = (625 × 16) × 279
= 10000 × 279 = 27,90,000

(e) 285 × 5 × 60 = 285 × (5 × 60)
= 285 × 300 = 85,500

(f) 125 × 40 × 8 × 25 = (125 × 8) × (40 × 25)
= 1000 × 1000 = 10,00,000

Q3. 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

(Using Distributive Property: a × b + a × c = a × (b + c))

(a) 297 × 17 + 297 × 3
= 297 × (17 + 3) = 297 × 20 = 5,940

(b) 54279 × 92 + 54279 × 8
= 54279 × (92 + 8) = 54279 × 100 = 54,27,900

(d) 3845 × 5 × 782 + 769 × 25 × 218
Notice that 769 × 5 = 3845, so 769 × 25 = 769 × 5 × 5 = 3845 × 5.
Expression becomes: 3845 × 5 × 782 + 3845 × 5 × 218
= (3845 × 5) × (782 + 218)
= 19225 × 1000 = 1,92,25,000

Q4. Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102 (c) 258 × 1008 (d) 1005 × 168

(Using Distributive Property)

(a) 738 × 103
= 738 × (100 + 3)
= (738 × 100) + (738 × 3)
= 73800 + 2214 = 76,014

(b) 854 × 102
= 854 × (100 + 2)
= (854 × 100) + (854 × 2)
= 85400 + 1708 = 87,108

(d) 1005 × 168
= (1000 + 5) × 168
= (1000 × 168) + (5 × 168)
= 168000 + 840 = 1,68,840

Q5. A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs. 44 per litre, how much did he spend in all on petrol?

Petrol filled on Monday = 40 litres
Petrol filled on Tuesday = 50 litres
Total petrol filled = 40 + 50 = 90 litres

Cost of 1 litre of petrol = Rs. 44
Total money spent = 90 × 44
= 44 × (100 - 10) (Using distributive property mentally)
= 4400 - 440 = Rs. 3,960

Q6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs. 15 per litre, how much money is due to the vendor per day?

Milk supplied in morning = 32 litres
Milk supplied in evening = 68 litres
Total milk supplied in a day = (32 + 68) litres = 100 litres

Cost of 1 litre of milk = Rs. 15
Total cost = 100 × 15 = Rs. 1,500

Q7. Match the following:

List A:
(i) 425 × 136 = 425 × (6 + 30 + 100)
(ii) 2 × 49 × 50 = 2 × 50 × 49
(iii) 80 + 2005 + 20 = 80 + 20 + 2005

List B:
(a) Commutativity under multiplication.
(b) Commutativity under addition.
(c) Distributivity of multiplication over addition.

Matching:
(i) → (c) Distributivity of multiplication over addition.
(ii) → (a) Commutativity under multiplication.
(iii) → (b) Commutativity under addition.

Exercise 2.3

Q1. Which of the following will not represent zero:
(a) 1 + 0 (b) 0 × 0 (c) 0 / 2 (d) (10 - 10) / 2

(a) 1 + 0 = 1 (Does not represent 0)
(b) 0 × 0 = 0
(c) 0 / 2 = 0
(d) (10 - 10) / 2 = 0 / 2 = 0

Therefore, (a) 1 + 0 will not represent zero.

Q2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Yes, if the product of two whole numbers is zero, then either one of them is zero, or both of them are zero.

Examples:
If one number is zero: 5 × 0 = 0, or 0 × 8 = 0.
If both are zero: 0 × 0 = 0.

Q3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

To get a product of 1 with whole numbers, both of them must strictly be 1.

Examples:
1 × 1 = 1.
However, if only one of them is 1 (e.g., 5 × 1), the product is 5 (which is not 1).

Q4. Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001 (c) 824 × 25
(d) 4275 × 125 (e) 504 × 35

(a) 728 × 101
= 728 × (100 + 1)
= (728 × 100) + (728 × 1)
= 72800 + 728 = 73,528

(b) 5437 × 1001
= 5437 × (1000 + 1)
= 5437000 + 5437 = 54,42,437

(Alternatively: 824 × 100 / 4 = 82400 / 4 = 20,600)

(d) 4275 × 125
= 4275 × (100 + 20 + 5)
= 427500 + 85500 + 21375 = 5,34,375

(Alternatively: 4275 × 1000 / 8 = 4275000 / 8 = 5,34,375)

(e) 504 × 35
= (500 + 4) × 35
= (500 × 35) + (4 × 35)
= 17500 + 140 = 17,640

Q5. Study the pattern:
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?

Next two steps:
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543

How it works:
Observe 12 × 8 + 2.
12 can be written as (11 + 1).
So, (11 + 1) × 8 + 2 = 88 + 8 + 2 = 98.

For 123 × 8 + 3:
123 = 111 + 11 + 1.
(111 + 11 + 1) × 8 + 3 = 888 + 88 + 8 + 3 = 987.

Class 6 Maths - Whole Numbers Practice Questions

Chapter 2: Whole Numbers (Practice Questions)

RD Sharma / Extra Practice

Q1. How many whole numbers are there between 1064 and 1201?

Number of whole numbers = (Greater number - Smaller number) - 1
= (1201 - 1064) - 1
= 137 - 1 = 136.

Q2. Write the predecessor of the smallest 5-digit number.

The smallest 5-digit number is 10,000.
Predecessor = 10,000 - 1 = 9,999

(Which is the greatest 4-digit number).

Q3. Find the product using suitable rearrangement: 4 × 258 × 25.

Rearranging the numbers to make calculation easier:
= (4 × 25) × 258
= 100 × 258 = 25,800

Q4. Determine the sum using suitable rearrangement: 953 + 407 + 647.

Rearranging the numbers ending in 3 and 7:
= (953 + 647) + 407
= 1600 + 407 = 2,007

Q5. Name the property used here: 432 × 105 = 432 × 100 + 432 × 5.

This shows the multiplication distributing over addition.
Property used: Distributive property of multiplication over addition.

Q6. Find the value to confirm the property: 54279 × 92 + 8 × 54279.

Using distributive property, a × b + a × c = a × (b + c).
= 54279 × (92 + 8)
= 54279 × 100 = 54,27,900

Q7. True or False: If a and b are two distinct whole numbers, then a ÷ b = b ÷ a.

False.

Division of whole numbers is not commutative. For example, 10 ÷ 5 = 2, but 5 ÷ 10 is not a whole number (it is 0.5).

Q8. A dealer purchased 124 LED television sets. If the cost of one set is Rs. 38,540, determine their total cost.

Cost of one set = Rs. 38,540
Number of sets = 124

Total cost = 38540 × 124
Using distributive property: = 38540 × (100 + 20 + 4)
= (38540 × 100) + (38540 × 20) + (38540 × 4)
= 3854000 + 770800 + 154160
= Rs. 47,78,960

Q9. The product of two numbers is 504347. If one of the numbers is 317, find the other.

Product = 5,04,347
One number = 317
Other number = 504347 ÷ 317

By long division, 504347 ÷ 317 = 1591.

Q10. On dividing 55,390 by 299, the remainder is 75. Find the quotient.

Using the division algorithm: Dividend = (Divisor × Quotient) + Remainder

55,390 = (299 × Quotient) + 75
299 × Quotient = 55390 - 75 = 55315
Quotient = 55315 ÷ 299 = 185.

Q11. Find the largest 4-digit number that is exactly divisible by 35.

The largest 4-digit number is 9999.
Divide 9999 by 35:
9999 ÷ 35 gives quotient 285 and a remainder of 24.

To find the largest exactly divisible number, subtract the remainder from 9999.
Largest 4-digit number = 9999 - 24 = 9975.

Q12. What least number must be subtracted from 13,601 to get a number exactly divisible by 87?

Divide 13,601 by 87.
13601 ÷ 87 gives quotient 156 and remainder 29.

The remainder is the extra amount. So, we must subtract the remainder.
Least number to be subtracted = 29.

Q13. In a school, there are 1,250 chairs. Each classroom requires 45 chairs. How many completely filled classrooms can there be, and how many chairs will be left over?

Divide the total chairs by chairs per classroom.
1250 ÷ 45 gives quotient 27 and remainder 35.

Classrooms completely filled = 27.
Chairs left over = 35.

Q14. Find the value of 81,265 × 169 - 81,265 × 69.

Using distributive property: a × b - a × c = a × (b - c).
= 81265 × (169 - 69)
= 81265 × 100 = 81,26,500

Q15. Write the smallest 3-digit number which does not change if the digits are written in reverse order.

Such numbers are called palindromes.
The smallest 3-digit number is 100. Let's check: reversed is 001 (not equal).
To be a palindrome, the first and last digits must be the same.
The smallest non-zero first digit is 1. The smallest middle digit is 0.
So, the number is 101.

Q16. Find the product of the successor and predecessor of 99.

Successor of 99 = 100.
Predecessor of 99 = 98.

Product = 100 × 98 = 9,800.

Q17. A car travels 43 km on 1 litre of petrol. How far would it travel on 250 litres? Use property to calculate.

Distance = 43 × 250

Using property:
= 43 × (1000 ÷ 4)
= 43000 ÷ 4
= 10,750 km.

Q18. Fill in the blank: 888 + 777 + 555 = 111 × _____.

Factor out 111 from each term.
= (111 × 8) + (111 × 7) + (111 × 5)
= 111 × (8 + 7 + 5)
= 111 × 20.

So, the blank should be filled with 20.

Q19. Divide 535 by 31 and check your answer using the division algorithm.

535 ÷ 31 gives quotient 17 and remainder 8.

Check:
Dividend = (Divisor × Quotient) + Remainder
= (31 × 17) + 8
= 527 + 8 = 535 (which matches the dividend).

Q20. What is the identity element for multiplication of whole numbers?

The identity element for multiplication is the number which, when multiplied by any whole number, leaves the number unchanged.

a × 1 = a.
Therefore, the multiplicative identity is 1.

Class 6 Maths - Whole Numbers Summary

Chapter 2: Whole Numbers (Concepts & Summary)

1. Natural Numbers & Whole Numbers

Natural Numbers: Counting numbers (1, 2, 3, 4, ...). The smallest natural number is 1. There is no largest natural number.
Whole Numbers: The collection of natural numbers together with 0. The set is (0, 1, 2, 3, ...). The smallest whole number is 0.
Note: All natural numbers are whole numbers, but not all whole numbers are natural numbers (0 is the exception).

2. Successor & Predecessor

Successor: The number that comes just after a given number. It is obtained by adding 1 to the number.
Example: Successor of 15 is 15 + 1 = 16.
Predecessor: The number that comes just before a given number. It is obtained by subtracting 1 from the number.
Example: Predecessor of 20 is 20 - 1 = 19.
Every whole number has a successor. Every whole number EXCEPT 0 has a predecessor in the set of whole numbers.

3. The Number Line

A straight line marked with whole numbers at equal intervals from a starting point 0.

Numbers on the right are greater than numbers on the left (e.g., 5 is to the right of 2, so 5 > 2).
Addition: Move to the right on the number line.
Subtraction: Move to the left on the number line.
Multiplication: Repeated jumps to the right, starting from 0.

4. Properties of Whole Numbers

a) Closure Property

Addition & Multiplication: Whole numbers are closed under addition and multiplication. Provide two whole numbers; their sum or product will ALWAYS be a whole number.
Note: Not closed under subtraction or division (e.g., 5 - 7 or 5 ÷ 2 do not yield whole numbers).

b) Commutativity

Addition & Multiplication: You can add or multiply whole numbers in any order.
a + b = b + a and a × b = b × a
Note: Subtraction and division are NOT commutative.

c) Associativity

Addition & Multiplication: When adding or multiplying three or more numbers, the grouping of numbers does not change the result.
(a + b) + c = a + (b + c) and (a × b) × c = a × (b × c)
Note: Subtraction and division are NOT associative.

d) Distributivity of Multiplication over Addition/Subtraction

Multiplying a number by a sum (or difference) is the same as multiplying the number by each part and then adding (or subtracting) the products.
a × (b + c) = (a × b) + (a × c)
a × (b - c) = (a × b) - (a × c)
This property makes calculations significantly easier.

5. Identity Elements

Additive Identity (0): Adding 0 to any whole number leaves it unchanged. a + 0 = a.
Multiplicative Identity (1): Multiplying any whole number by 1 leaves it unchanged. a × 1 = a.
Multiplication by Zero: The product of any whole number and 0 is 0. a × 0 = 0.
Division by Zero: Division of a whole number by zero is not defined.

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