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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 :
