Collins Maths Solutions Class 6 Chapter 2 Natural Numbers and Whole Numbers
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 2, Natural Numbers and Whole Numbers. Here students can easily find Exercise wise solution for chapter 2, Natural Numbers and Whole Numbers. Students will find proper solutions for Exercise 2.1, 2.2 and 2.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Natural Numbers and Whole Numbers Exercise 2.1 Solution :
Question no – (1)
Solution :
(a) The smallest natural number is 1 → True.
(b) The smallest whole number is 1 → False.
Because the smallest whole number is 0.
(c) The difference between the smallest whole number and the smallest natural number is 0 → False
Because, the difference between the smallest whole number and the smallest natural number is
= 1 – 0
= 1
(d) The product of the smallest whole number and the smallest natural number is 0 → True.
Because, the product of the smallest whole number and the smallest natural number is
= 1 × 0
= 0
(e) O is a whole number → True.
Question no – (2)
Solution :
Given natural numbers are,
(a) 5 ; (b) 7
(c) 2 ; (d) 8
Now on the number line :
Natural Numbers and Whole Numbers Exercise 2.2 Solution :
Question no – (1)
Solution :
(a) = 6107 + 3049 = 3049 + 6107
(b) = 7021 + 90333 = 90333 + 7021
(c) = 48026 + 0 = 48026
(d) = 6031 + 0 = 6031
Question no – (2)
Solution :
(a) 72091 + 30054 = 30054 + 72091
L.H.S, 72091 + 300054
= 102145
R.H.S, 30054 + 72091
= 102145
∴ L.H.S = R.H.S …(Verified)
(b) 43001 + 362 = 362 + 43001
L.H.S, 43001 + 362
= 43363
R.H.S, 362 + 43001
= 43363
∴ L.H.S = R.H.S …(Verified)
(c) 3003 + (80251 + 6007) = (3003 + 607) + 80251
L.H.S, 300 + (80251 + 6007)
= 3003 + 86258
= 89261
R.H.S, (3003 + 6007) + 80251
= 9010 + 80251
= 89261
∴ L.H.S = R.H.S …(Verified)
(d) (50106 + 4084) + 33707 = 50106 + (4084 + 33707)
L.H.S, (50106 + 4084) + 33707
= 54190 + 33707
= 87897
R.H.S, 50106 + (4084 + 33,707)
= 50106 + 37791
= 87897
∴ L.H.S = R.H.S …(Verified)
Question no – (3)
Solution :
(a) 25, 8, 0
= (25 – 8) – 0 = 17 – 0 = 17
= 25 – (8 – 0)
= 25 – 8
= 17
= Yes
(b) 45, 8, 21
= (45 – 8) – 21
= 37 – 21
= 16
= 45 – (8 – 21)
= 45 + 13
= 58
∴ (45 – 8) – 21 ≠ 45 – (8 – 21)
∴ No
(c) 317, 312, 50
= (317 – 312) – 50
= 5 – 50
= – 45
= 317 – (312 – 50)
= 317 – 262
= 55
∴ No
(d) 87, 082, 2018
= (87 – 082) – 2018
= 5 – 2018
= – 2013
= 87 – (082 – 2018)
= 87 – 1936
= 1849
∴ No
Natural Numbers and Whole Numbers Exercise 2.3 Solution :
Question no – (1)
Solution :
(a) 604 × 3081 = 3081 × 604
(b) 70704 × 0 = 0
(c) 673 × 1 = 673
(d) 4090 × (806 × 31) = 806 × (4090 × 31)
(e) 802 × 205 + 802 × 705 = 802 × (205 + 705)
Question no – (2)
Solution :
(a) 20340 × 10030 × 20340
= Commutative under multiplication
(b) 302 × (604 × 7090) = (302 × 604) × 7090
= Commutative under multiplication
(c) 701080 × 0 = 0
= 701080 × 0 = 0 × 701080
= Commutative under multiplication
(d) 90200 × 1 = 90200
= Multiplicative Identity.
Question no – (3)
Solution :
(a) 400 × 2037 × 2005
= (400 × 2005) × 2037
= 802000 × 2037
= 1633674000
(b) 2000 × 4305 × 5005
= (2000 × 5005) × 4305
= 10010000 × 4305
= 43093050000
(c) 5000 × 10042 × 1004
= (5000 × 1004) × 10042
= 5020000 × 10042
= 50410840000
(d) 40004 × 2009 × 2025
= (4004 × 2025) × 2009
= 8108100 × 2009
= 162901, 72900
(e) 25 × 6025 × 120 × 80
= (25 × 80) × 120 × 6015
= (2000 × 120) × 6025
= 240000 × 6025
= 1446000000
(f) 400 × 500 × 1025 × 800
= (400 × 500) × 800 × 1025
= (200000 × 800) × 1025
= 160000000 × 1025
= 164000000000
Question no – (4)
Solution :
(a) 1846125 ÷ 45
= 41025
(b) 23577392 ÷ 23
= 1025104
(c) 161375832 ÷ 18
= 8965324
(d) 19408519 ÷ 19
= 1021501
Question no – (5)
Solution :
(a) 60742 ÷ 23
Here, divisor = 23
Quotient = 264
remainder = 22
By Division algorithm, (Division × quotient) + remainder
= (23 × 2640) + 22
= 60720 + 22
= 60742 = Dividend …(Proved)
(b) 500505 ÷ 101
Here,
Divisor = 101,
Quotient = 4965
Remainder = 40
∴ By division algorithm = (Division × Quotient) + Remainder
= (101 × 4965) + 40
= 501465 + 40
= 501465 …(Proved)
(c) 701773 ÷ 300
Here,
Divisor = 300
Quotient = 2339
Remainder = 73
By division algorithm (300 × 2339) + 73
= 701700 + 73
= 701773 …(Proved)
(d) 890422 ÷ 312
Here,
Divisor = 312,
Quotient = 2853
Remainder = 286
By Davison algorithm = (Division × Quotient) + Remainder
= (312 × 2853) + 286
= 890136
= 890422 …(Proved)
(e) 5570624 ÷ 571
Here,
Divisor = 571
Quotient = 9773
Remainder = 231
By division algorithm = (Division × Quotient) + Remainder
= (571 × 9773) + 241
= 5580383 + 241
= 5580624 …(Proved)
(f) 8840422 ÷ 230
Here,
Divisor = 230
Quotient = 38436
Remainder = 142
By, Division algorithm = (Division × Quotient) + Remainder
= (230 × 38436) + 142
= 8840280 + 142
= 8840422 …(Proved)
Question no – (6)
Solution :
Given in the question,
The product of two numbers is = 1,41,705.
One of the numbers is = 705
Other number = ?
We know, Another number × 705 = 141,705
∴ Another number,
= 141,705/705
= 201
Therefore, another number will be 201
Question no – (7)
Solution :
(a) 459 = (12 × 38) + 3
Explanation :
= 459 – 3 = (Divisor × 38)
= 456/38 = Divisor
Divisor = 12
∴ 459 = (12 × 38) + 3
(b) 6729 = (22 × 305) + 18
Explanation :
= 6710 + 18
= 6729
(c) 33333 = (22 × 1515) + 3330 + 3
(d) 424284 = (42 × 10102) + 0
Next Chapter Solution :
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