# Ncert Solutions for Class 6 Maths Whole Numbers

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In this page we have Ncert Solutions for Class 6 Maths Whole Numbers. Hope you like them and do not forget to like , social share and comment at the end of the page.

# Exercise – 2.1

Question 1

Write the next three natural numbers after 10999.

Next three natural numbers after 10999 can be obtained by adding 1 to them

 10999+ 1= 11000 11000+1= 11001 11001+1=1 11002

So the numbers are 11000,11001,11002

Question 2:

Write the three whole numbers occurring just before 10001.

3 whole numbers just before 10001 can be obtained by subtracting 1 from them

 10001-1= 10000 10000-1= 9999 9999 -1 = 9998

So the numbers are 10000,9999,9998

Question 3

Which is the smallest whole number?

The smallest whole number is 0.

Question 4

How many whole numbers are there between 32 and 53?

Whole numbers between 32 and 53 can be find using subtracting the smaller number from bigger number, but it will include the highest number also, so subtracting 1 more. So it will be = 53 − 32 − 1 = 20

We can alternatively verify the answer by writing down all the whole number between them and counting them

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

Question 5

Write the successor of:

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

Successor are obtained by adding 1 to the number

 a) 2440701 + 1 = 2440702 b) 100199 + 1 = 100200 c) 1099999+1= 1100000 d) 2345670 + 1= 2345671

Question 6

Write the predecessor of:

(a) 94

(b) 10000

(c) 208090

(d) 7654321

Predecessor is obtained by subtracting 1 from the Number

 a) 94 − 1 = 93 b) 10000 −1= 9999 c) 208090 − 1 = 208089 d) 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

The Number line can be represented as

The smaller is left side of the larger Number

(a) 530, 503

As 530 > 503,

503 is on the left side of 530 on the number line.

(b) 370, 307

As 370 > 307,

307 is on the left side of 370 on the number line.

(c) 98765, 56789

As 98765 > 56789,

56789 is on the left side of 98765 on the number line.

(d) 9830415, 10023001

Since 98, 30, 415 < 1, 00, 23, 001, 98,30,415 is on the left side of 1,00,23,001 on the number line.

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 whole number 1 has no predecessor.

(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.

(a) False, 0 is not a natural number. It is a smallest whole number

(b) False, as predecessor of 399 is 398 (399 − 1 = 398).

(c) True

(d) True, as 599 + 1 = 600

(e) True

(f) False, as 0 is a whole number but it is not a natural number.

(g) False. Example would be two-digit number 10 whose predecessor is 9.

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

(i) True, as 0 is the predecessor of 1 but it is not a natural number.

(j) False, as 0 is the predecessor of 1 and it is a whole number.

(k) False, 13 does not lie in between 11 and 12.

(l) True, predecessor of 0 is −1, which is not a whole number.

(m) False, as successor of 99 is 100.

# Exercise 2.2

Question 1

Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

Rearranging the terms to make the addition simpler

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

(b) 1962 + 453 + 1538 + 647 = (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

We can make the product calculation easier by making it multiples of 10,100 etc.

(a) 2 × 1768 × 50

Here in this product, the product of 2 and 50 makes multiples of 100 which is easier to multiple with other number

= 2 × 50 × 1768 = 100 × 1768 = 176800

(b) 4 × 166 × 25

Here in this product, the product of 4 and 25 makes multiples of 100 which is easier to multiple with other number

= 4 × 25 × 166 = 100 × 166 = 16600

(c) 8 × 291 × 125

Here in this product, the product of 8 and 125 makes multiples of 1000 which is easier to multiple with other number

= 8 × 125 × 291 = 1000 × 291 = 291000

(d) 625 × 279 × 16 = 625 × 16 × 279 = 10000 × 279 = 2790000

(e) 285 × 5 × 60 = 285 × 300 = 85500

(f) 125 × 40 × 8 × 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

Here in these question, we can see a quantity common, so we can take it out of bracket and make the calculation easier

(a) 297 × 17 + 297 × 3

Here 297 is common in both the term

= 297 × (17 + 3) = 297 × 20 = 5940

(b) 54279 × 92 + 8 × 54279 = 54279 × 92 + 54279 × 8

Here 54279 is common in both the term

= 54279 × (92 + 8)

= 54279 × 100 = 5427900

(c) 81265 × 169 − 81265 × 69

Here 81265 is common in both the term

= 81265 × (169 − 69) = 81265 × 100 = 8126500

(d) 3845 × 5 × 782 + 769 × 25 × 218

Here we don’t common term as of now, so rearranging little bit as 769×5=3845

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000 = 19225000

Question 4:

Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

We know that

 Distributive property If a, b and c are any two whole numbers, then a(b+c) = a×b + a×c.

(a) 738 × 103 = 738 × (100 + 3)

= 738 × 100 + 738 × 3 (Distributive property)

= 73800 + 2214

= 76014

(b) 854 × 102 = 854 × (100 + 2)

= 854 × 100 + 854 × 2 (Distributive property)

= 85400 + 1708 = 87108

(c) 258 × 1008 = 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (Distributive property)

= 258000 + 2064 = 260064

(d) 1005 × 168 = (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (Distributive property)

= 168000 + 840 = 168840

Question 5

A taxi driver 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?

Quantity of petrol filled on Monday = 40 litres

Quantity of petrol filled on Tuesday = 50 litres

Total quantity filled = (40 + 50) litres =90 Litres

Cost of petrol (per l) = Rs 44

Total money spent = 44 × 90 = Rs 3960

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 Rs 15 per litre, how much money is due to the vendor per day?

Quantity of milk supplied in the morning = 32 l

Quantity of milk supplied in the evening = 68 l

Total of milk per litre = (32 + 68) l

Cost of milk per litre = Rs 15

Total cost per day = 15 × (32 + 68)

= 15 × 100 = Rs 1500

Question 7

Match the following:

 (i) 425 × 136 = 425 × (6 + 30 +100) (a) Commutativity under multiplication. (ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition. (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.

We Know that

 Closure Property If a and b are any two whole numbers, then a+b, axb are also whole numbers. Commutative property If a and b are any two whole numbers, then a+b = b+a and a×b = b×a. Associative property If a, b and c are any two whole numbers, then (a+b)+c = a+(b+c) and (a×b)×c = a×(b×c). Distributive property If a, b and c are any two whole numbers, then a(b+c) = a×b + a×c.

(i) 425 × 136 = 425 × (6 + 30 + 100)   which is Distributivity of multiplication over addition Hence, (c)

(ii) 2 × 49 × 50 = 2 × 50 × 49 which is Commutativity under multiplication Hence, (a)

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 which is Commutativity under addition Hence, (b)

 (i) 425 × 136 = 425 × (6 + 30 +100) (c) Distributivity of multiplication over addition. (ii) 2 × 49 × 50 = 2 × 50 × 49 (a) Commutativity under multiplication. (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (b) 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

 a) 1+0=1 So it does not represent zero b) 0 × 0=0 So it represents zero c) 0/2=0 So it represents zero d) (10-10)/2=0/2=0 So it represents 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.

If the product of 2 whole numbers is zero, then one of them is definitely zero.

We can take examples to justify it

For example,

0 × 11 = 0

968 × 0 = 0

0 × 0 = 0

If the product of 2 whole numbers is zero, then both of them may be zero.

Now if we take non zero numbers,

6 × 8 = 48

(Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.)

Question 3

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

If the product of 2 numbers is 1, then both the numbers have to be equal to 1.

For example, 1 × 1 = 1

However, 1 × 9 = 9 or 12 × 1=12

Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.

Question 4

Find using distributive property:

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

We know that

 Distributive property If a, b and c are any two whole numbers, then a(b+c) = a×b + a×c.

(a) 728 × 101= 728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728 = 73528

(b) 5437 × 1001 = 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437 = 5442437

(c) 824 × 25 = (800 + 24) × 25

= (800 + 25 − 1) × 25

= 800 × 25 + 25 × 25 − 1 × 25

= 20000 + 625 − 25

= 20000 + 600 = 20600

(d) 4275 × 125 = (4000 + 200 + 100 − 25) × 125

= 4000 × 125 + 200 × 125 + 100 × 125 − 25 × 125

= 500000 + 25000 + 12500 − 3125

= 534375

(e) 504 × 35 = (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140 = 17640

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?

(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

1 × 8 + 1 = 9

12 × 8 + 2 = 98
i.e. (11 + 1) × 8 + 2 = 11 × 8 + 1 × 8 + 2 = 88 + 8 + 2 = 98

123 × 8 + 3 = 987;

i.e. (111 + 11 + 1) × 8 + 3 = 111 × 8 + 11 × 8 + 1 × 8 + 2 = 888 + 88 + 8 + 3 = 987

1234 × 8 + 4 = 9876;
i.e. (1111 + 111 + 11 + 1) × 8 + 4 = 1111 × 8 + 111 × 8 + 11× 8 + 1 × 8 + 4
= 8888 + 888 + 88 + 8 + 4 = 9876

12345 × 8 + 5 = 98765
i.e. (11111 + 1111 + 111 + 11 + 1) × 8 + 5 = 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 5
= 88888 + 8888 + 888 + 88 + 8 + 5 = 98765

Extending the pattern, we have

123456 × 8 + 6 =
(111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 5
= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 5
= 888888 + 88888 + 8888 + 888 + 88 + 8 + 6 = 987654

1234567 × 8 + 7 =
(1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 7
= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 7
= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8 + 7 = 9876543