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**Mathsight Class 6 Solutions Chapter 4 Natural Numbers and Whole Numbers**

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students to solve MathSight Class 6 Mathematics Book, Chapter 4, Natural Numbers and Whole Numbers. Here students can easily get all the exercise questions solution for Chapter 4, Natural Numbers and Whole Numbers Exercise 4.1 and 4.2

**Natural Numbers and Whole Numbers Exercise 4.1 Solution : **

**Question no – (1)**

**Solution : **

**(a)** a + b = b

Or, a = b – b = 0

Or, a = 0

**(b)** 1

**(c)** 1

**(d)** 0

**(e)** 0

**Question no – (2) **

**Solution : **

**(a)** 3157 + 8713 = 11870

= 8713 + 3157

= 11870

**(b)** 3605 + 7727 = 17332

= 7727 + 9605

= 17322

**Question no – (3)**

**Solution :**

Associate property,

(72 + 49) + 28

= 121 + 28

= 149

72 + (49 + 28)

= 72 + 77

= 149

**Question no – (4)**

**Solution : **

Sum of 1st 100 natural numbers = 5050

Sum of 1st 100 Whole numbers = 5050

∴ Difference of them

= 5050 – 5050

= 0

**Question no – (5)**

**Solution : **

Let, natural numbers = 5

Whole numbers = 0

∴ Sum of them,

= 5 + 0

= 5

**Question no – (6)**

**Solution :**

Let, a = 1 , b = 2 , c = 3

L.H.S, (a – b) – c

= (1 – 2 ) – 3

= – 1 – 3

= -4

R.H.S, a – (b – c)

= 1 – (2 – 3)

= 1 – (-1)

= 1 + 1

= 2

**∴** L.H.S ≠ R.H.S

**Question no – (7)**

**Solution : **

Kusum 1st scored = 53 runs

Kusum 2nd scored = 45 rune

Kusum total scored,

= 53 + 45

= 98 rune

Mitalis total scored,

= 45 + 53

= 98 rune

Both Kusum and Mitalis scored same runes.

**Question no – (8)**

**Solution : **

The system of system will be,

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18,

19, 21, 22, 23, and so on .

**Question no – (9)**

**Solution :**

Using commutative property of whole numbers –

(a + b) + (c + d) = (a + d) + (c + b)

Let, a = 36, b = 34, c = 67, d = 89

**L.H.S,** (a + b) + (c + d)

= (36 + 34) + (67 + 89)

= 70 + 156

= 226

**R.H.S,** (a + c) + (b + d)

= 36 + 67) + (34 + 89)

= 103 + 123

= 226

∴ It is showed that whole numbers can be added using more them 2 arrangements.

**Question no – (10)**

**Solution : **

Let, one numbers = x

∴ x = 490/2

= 245

Another numbers = y

∴ y = 376 – 245 = 131

∴ Two numbers = 245, 131

**Natural Numbers and Whole Numbers Exercise 4.2 Solution : **

**Question no – (1)**

**Solution : **

(a) → (iv) Multiplicative identity

(b) → (v) Distributive property of multiplication over addition.

(c) → (i) Communicative property of multiplication.

(d) → (ii) Multiplicative property of zero.

(e) → (iii) Distributive property of multiplication over subtraction.

**Question no – (2)**

**Solution : **

**(a)** Using multiplication identity –

= 95 × 102

= 102 × 95

= 9690

**(c) **Using distributive property of multiplication over identity,

(8 × 6) + (8 × 4)

= 8 × (6 × 4)

= 80

**Question no – (3)**

**Solution :**

(a) 5 × 1 = 5

(b) 5 + 5 = 10

(c) 5 – 5 = 0

**Required photo : **

**Question no – (4)**

**Solution :**

Whole number “1” doesn’t change after division by any number.

**Question no – (5)**

**Solution :**

**Additive identity :**

if zero is added to any whole numbers then, the sum remain zero, is known as additive identity.

**Multiplicative identity :**

if ‘1’ is multiplicative to any whole numbers then the product remain same itself is know as multiplicative identity.

**Question no – (6)**

**Solution :**

0 and 1 are the two whole numbers that when added or multiplied with themselves gives same result.

**Question no – (7)**

**Solution :**

Number of switches of one floor,

= 18 × 20

= 360

Number of switches of 5 floor,

= 360 × 5

= 1800

**Question no – (8)**

**Solution : **

Number of pages laving puzzle,

= 176/8

= 22 page.

∴ Numbers of puzzle of page = 1

∴ Numbers of puzzle of 22 page = 22

**Question no – (9)**

**Solution : **

Cost of 1 pair of jeans = 1500 Rs

∴ Cost of 2 pair of jeans,

= 1500 × 5

= 7500 Rs

∴ Cost of 1 T-shirt = 750

∴ Cost of 5 T-shirt

= 750 × 5

= 3750 Rs.

∴ Total cost need,

= 7500 + 3750

= 11250 Rs.

**Question no – (10)**

**Solution :**

= (6 + 5 + 5)

= 16 m is covered by athlete.

**Question no – (11)**

**Solution :**

It is correct.

But, when multiplied by ‘0’, then the product would be ‘0’. When, multiplied by ‘1’ then the product would be remain same as after numbers.

**Question no – (12)**

**Solution :**

**(a)** Total amount of watermelons,

= ( 17 × 12) + (17 × 8)

**(b)** Payment,

= (17 × 12) + (17 × 8)

= 204 + 136

= 340 Rs

**Previous Chapter Solutions : **