class 6th math chapter 2 ncert

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Chapter 2: Whole Numbers - NCERT Solutions

Exercise 2.1

Question 1: Write the next three natural numbers after 10999.

Solution 1: The next three natural numbers after 10999 are:

  • 11000
  • 11001
  • 11002

Question 2: Write the three whole numbers occurring just before 10001.

Solution 2: The three whole numbers occurring just before 10001 are:

  • 10000
  • 9999
  • 9998

Question 3: Which is the smallest whole number?

Solution 3: The smallest whole number is 0 (Zero).

Question 4: How many whole numbers are there between 0 and 100?

Solution 4: There are 100 whole numbers between 0 and 100. (These are the whole numbers from 1 to 99. We exclude 0 and 100 because we are looking for numbers between them).

Question 5: Write the successor of: (a) 2440701 (b) 100199 (c) 1099999 (d) 2345670

Solution 5: The successor of a number is the number immediately following it, which is obtained by adding 1 to the number. (a) Successor of 2440701 is 2440701 + 1 = 2440702 (b) Successor of 100199 is 100199 + 1 = 100200 (c) Successor of 1099999 is 1099999 + 1 = 1100000 (d) Successor of 2345670 is 2345670 + 1 = 2345671

Question 6: Write the predecessor of: (a) 94 (b) 10000 (c) 208090 (d) 7654321

Solution 6: The predecessor of a number is the number immediately before it, which is obtained by subtracting 1 from the number. (a) Predecessor of 94 is 94 - 1 = 93 (b) Predecessor of 10000 is 10000 - 1 = 9999 (c) Predecessor of 208090 is 208090 - 1 = 208089 (d) Predecessor of 7654321 is 7654321 - 1 = 7654320

Question 7: 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  

Solution 7: On a number line, the smaller number is always to the left of the larger number. (a) 530, 503: 503 is smaller than 530. So, 503 is on the left of 530. We write this as 503 < 530. (b) 370, 307: 307 is smaller than 370. So, 307 is on the left of 370. We write this as 307 < 370. (c) 98765, 56789: 56789 is smaller than 98765. So, 56789 is on the left of 98765. We write this as 56789 < 98765. (d) 9830415, 10023001: 9830415 is smaller than 10023001. So, 9830415 is on the left of 10023001. We write this as 9830415 < 10023001.

Question 8: Which of the following statements are true (T) and which are false (F)? (a) Zero is the smallest natural number. (b) 400 is the predecessor of 399. (c) Zero is the smallest whole number. (d) 600 is the successor of 599. (e) All natural numbers are whole numbers. (f) All whole numbers are natural numbers. (g) The predecessor of a two digit number is never a single digit number. (h) 1 is the smallest whole number. (i) The natural number 1 has no predecessor. (j) The(k) The whole number 13 lies between 11 and 12. (l) The whole number 0 has no predecessor. (m) The successor of a two digit number is always a two digit number.  

Solution 8: Here are the truth values for each statement: (a) False (F): Natural numbers start from 1. The smallest natural number is 1. (b) False (F): The predecessor of 399 is 399 - 1 = 398. (c) True (T): Whole numbers start from 0. Zero is the smallest whole number. (d) True (T): The successor of 599 is 599 + 1 = 600. (e) True (T): Natural numbers (1, 2, 3, ...) are included within whole numbers (0, 1, 2, 3, ...). (f) False (F): Whole numbers include 0, but natural numbers do not. So, not all whole numbers are natural numbers. (g) False (F): For example, the predecessor of the two-digit number 10 is the single-digit number 9. (h) False (F): The smallest whole number is 0. (i) True (T): In natural numbers, 1 is the starting number and has no predecessor within natural numbers. (j) False (F): The predecessor of the whole number 1 is 1 - 1 = 0, which is a whole number. (k) False (F): 12 lies between 11 and 13, not 13 between 11 and 12. (l) True (T): There is no whole number smaller than 0 that can be its predecessor in the set of whole numbers. (m) False (F): For example, the successor of the two-digit number 99 is the three-digit number 100.  

Exercise 2.2

Question 1: Find the sum by suitable rearrangement: (a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647

Solution 1: We use the property of commutativity and associativity of addition to rearrange the numbers for easier calculation. (a) 837 + 208 + 363 Rearrange to group numbers that add up to round numbers: = (837 + 363) + 208 = 1200 + 208 = 1408  

(b) 1962 + 453 + 1538 + 647 Rearrange to group numbers that add up to round numbers: = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600

Question 2: 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

Solution 2: We use the property of commutativity and associativity of multiplication to rearrange the numbers to make multiplication easier. Remember that 2 × 50 = 100, 4 × 25 = 100, and 8 × 125 = 1000. (a) 2 × 1768 × 50 Rearrange to multiply 2 and 50 first: = (2 × 50) × 1768 = 100 × 1768 = 176800

(b) 4 × 166 × 25 Rearrange to multiply 4 and 25 first: = (4 × 25) × 166 = 100 × 166 = 16600

(c) 8 × 291 × 125 Rearrange to multiply 8 and 125 first: = (8 × 125) × 291 = 1000 × 291 = 291000

(d) 625 × 279 × 16 Rearrange to multiply 625 and 16 first (625 × 16 = 10000): = (625 × 16) × 279 = 10000 × 279 = 2790000

(e) 285 × 5 × 60 Rearrange to multiply 5 and 60 first: = 285 × (5 × 60) = 285 × 300 = 85500

(f) 125 × 40 × 8 × 25 Rearrange to group (125 × 8) and (40 × 25): = (125 × 8) × (40 × 25) = 1000 × 1000 = 1000000

Question 3: 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

Solution 3: We use the distributive property: a × b + a × c = a × (b + c) and a × b - a × c = a × (b - c). (a) 297 × 17 + 297 × 3 Take 297 common: = 297 × (17 + 3) = 297 × 20 = 5940

(b) 54279 × 92 + 8 × 54279 Take 54279 common: = 54279 × (92 + 8) = 54279 × 100 = 5427900

(c) 81265 × 169 – 81265 × 69 Take 81265 common: = 81265 × (169 – 69) = 81265 × 100 = 8126500

(d) 3845 × 5 × 782 + 769 × 25 × 218 Notice that 25 = 5 × 5 and 769 × 5 = 3845. Rewrite 769 × 25 as (769 × 5) × 5 = 3845 × 5. = (3845 × 5) × 782 + (3845 × 5) × 218 Now take (3845 × 5) common: = (3845 × 5) × (782 + 218) = (3845 × 5) × 1000 = 19225 × 1000 = 19225000  

Question 4: Find the product using suitable properties: (a) 738 × 103 (b) 854 × 102 (c) 258 × 1008 (d) 1005 × 168 (e) 504 × 35

Solution 4: We use the distributive property to make multiplication easier by breaking down one number into a sum or difference of easier numbers (like 100, 1000, etc.). (a) 738 × 103 Write 103 as (100 + 3): = 738 × (100 + 3) = 738 × 100 + 738 × 3 (Distributive property) = 73800 + 2214 = 76014

(b) 854 × 102 Write 102 as (100 + 2): = 854 × (100 + 2) = 854 × 100 + 854 × 2 (Distributive property) = 85400 + 1708 = 87108

(c) 258 × 1008 Write 1008 as (1000 + 8): = 258 × (1000 + 8) = 258 × 1000 + 258 × 8 (Distributive property) = 258000 + 2064 = 260064

(d) 1005 × 168 Write 1005 as (1000 + 5): = (1000 + 5) × 168 = 1000 × 168 + 5 × 168 (Distributive property) = 168000 + 840 = 168840

(e) 504 × 35 Write 504 as (500 + 4): = (500 + 4) × 35 = 500 × 35 + 4 × 35 (Distributive property) = 17500 + 140 = 17640  

Question 5: A taxi-driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled 50 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?

Solution 5: First, find the total amount of petrol filled in both days: Total petrol filled = Petrol on Monday + Petrol on Tuesday Total petrol filled = 40 litres + 50 litres = 90 litres

The cost of petrol is ₹ 44 per litre. To find the total money spent, multiply the total petrol by the cost per litre: Total money spent = Total petrol filled × Cost per litre Total money spent = 90 litres × ₹ 44/litre Total money spent = ₹ 3960

Answer: The taxi-driver spent ₹ 3960 in all on petrol.

Question 6: 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 ₹ 15 per litre, how much money is due to the vendor per day?

Solution 6: First, find the total amount of milk supplied per day: Total milk supplied per day = Milk in morning + Milk in evening Total milk supplied per day = 32 litres + 68 litres = 100 litres

The cost of milk is ₹ 15 per litre. To find the money due to the vendor per day, multiply the total milk supplied by the cost per litre: Money due per day = Total milk supplied per day × Cost per litre Money due per day = 100 litres × ₹ 15/litre Money due per day = ₹ 1500

Answer: ₹ 1500 is due to the vendor per day.

Question 7: Match the following:

(Column A) (Column B) (i) 425 × 136 = 425 × (6 + 30 + 100) (a) Commutativity under multiplication (ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Distributivity of multiplication over addition (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Commutativity under addition (d) Associativity under multiplication (e) Identity for multiplication (f) Associativity under addition

Solution 7: Let's analyze each matching:

(i) 425 × 136 = 425 × (6 + 30 + 100) This shows that 136 is broken down into (6 + 30 + 100) and then multiplied with 425. This illustrates the distributive property of multiplication over addition. Match: (i) - (b) Distributivity of multiplication over addition

(ii) 2 × 49 × 50 = 2 × 50 × 49 In this, the order of factors 49 and 50 is changed while multiplying. This shows the commutativity of multiplication. Match: (ii) - (a) Commutativity under multiplication  

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 Here, the order of addends 2005 and 20 is changed while adding. This shows the commutativity of addition. Match: (iii) - (c) Commutativity under addition

Exercise 2.3

Question 1: Which of the following will not represent zero? (a) 1 + 0 (b) 0 × 0 (c) 0 / 2 (d) (10 – 10) / 2

Solution 1: Let's evaluate each option: (a) 1 + 0 = 1 (This is equal to 1, not zero) (b) 0 × 0 = 0 (This is equal to zero) (c) 0 / 2 = 0 (Zero divided by any non-zero number is zero) (d) (10 – 10) / 2 = 0 / 2 = 0 (This is equal to zero)

Answer: Option (a) 1 + 0 will not represent zero.

Question 2: If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Solution 2: Yes, if the product of two whole numbers is zero, then at least one of them must be zero, or both of them can be zero.

Justification with examples:

  • Example 1: One number is zero. Let's take whole numbers 0 and 5. Their product is 0 × 5 = 0. Here, one of the numbers (0) is zero, and the product is zero.
  • Example 2: Both numbers are zero. Let's take whole numbers 0 and 0. Their product is 0 × 0 = 0. Here, both numbers are zero, and the product is zero.
  • Example 3: Neither number is zero. Let's take whole numbers 2 and 3. Their product is 2 × 3 = 6. Here, neither number is zero, and the product is not zero.

From these examples, we can see that if the product of two whole numbers is zero, it is necessary that at least one of the numbers must be zero.

Question 3: If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

Solution 3: Yes, if the product of two whole numbers is 1, then both of them must be 1.

Justification with examples:

  • Example 1: Both numbers are 1. Let's take whole numbers 1 and 1. Their product is 1 × 1 = 1. Here, both numbers are 1, and the product is 1.
  • Example 2: One number is 1, and another is not. Let's try to take one number as 1 and another whole number that is not 1, say 2. Their product is 1 × 2 = 2. The product is not 1.
  • Example 3: Neither number is 1. Let's take two whole numbers that are not 1, say 2 and 3. Their product is 2 × 3 = 6. The product is not 1.
  • Example 4: Consider 0. If we take 0 as one of the numbers, say 0 and 1, their product is 0 × 1 = 0. If we take 0 and 0, product is 0 × 0 = 0. If we take 0 and any other whole number, the product will be 0.  

From these examples, we can see that the only way the product of two whole numbers can be 1 is if both whole numbers are 1.

Question 4: Find using distributive property: (a) 728 × 101 (b) 5437 × 1001 (c) 258 × 1008 (d) 1005 × 168

Solution 4: We apply the distributive property of multiplication over addition: a × (b + c) = a × b + a × c. (a) 728 × 101 Write 101 as (100 + 1): = 728 × (100 + 1) = 728 × 100 + 728 × 1 (Distributive property) = 72800 + 728 = 73528

(b) 5437 × 1001 Write 1001 as (1000 + 1): = 5437 × (1000 + 1) = 5437 × 1000 + 5437 × 1 (Distributive property) = 5437000 + 5437 = 5442437

(c) 258 × 1008 Write 1008 as (1000 + 8): = 258 × (1000 + 8) = 258 × 1000 + 258 × 8 (Distributive property) = 258000 + 2064 = 260064  

(d) 1005 × 168 Write 1005 as (1000 + 5): = (1000 + 5) × 168 = 1000 × 168 + 5 × 168 (Distributive property) = 168000 + 840 = 168840

Question 5: 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?

Solution 5:

Next two steps in the pattern: Following the pattern, we can write the next two steps:

  • 123456 × 8 + 6 = 987654
  • 1234567 × 8 + 7 = 9876543

How the pattern works: Observe the pattern:

  • In each step, the first number on the left side is formed by sequentially adding digits starting from 1 (1, 12, 123, 1234, ...).
  • This number is multiplied by 8.
  • The number added to the product is the last digit of the first number (in 1 × 8 + 1, we add 1; in 12 × 8 + 2, we add 2; and so on).
  • The result on the right side is a number formed by sequentially decreasing digits starting from 9 (9, 98, 987, 9876, ...).

General Rule: If we take a number formed by the first 'n' natural numbers in sequence (1, 12, 123, ..., up to n digits), multiply it by 8, and add 'n', the result will be a number formed by the first 'n' digits in reverse sequence starting from 9 (9, 98, 987, ..., up to n digits).

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