# Collins Maths Solutions Class 6 Chapter 2

## 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 :

Updated: June 14, 2023 — 8:59 am