# Frank Learning Maths Class 6 Solutions Chapter 2

## Frank Learning Maths Class 6 Solutions Chapter 2 Whole Numbers

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 2 Whole Numbers. Here students can easily find step by step solutions of all the problems for Whole Numbers. Here students will find solutions for Exercise 2.1, 2.2 and 2.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Whole Numbers Exercise 2.1 Solution

Question no – (1)

Solution :

The whole numbers succeeding 20,009 are

= 201010, 20011 and 20,012

Question no – (2)

Solution :

The whole number occurring just before 3,00,001 is 3,00,000

Question no – (3)

Solution :

The smallest whole number is ‘0’ and there is no greatest whole number.

Question no – (4)

Solution :

The natural number ‘1’ does not have predecessor.

Question no – (5)

Solution :

There are 999 natural numbers upto 1000.

Question no – (6)

Solution :

There are 2000 whole numbers upto 2000.

Question no – (7)

Solution :

The number of whole numbers between 70 and 90 is,

= [(90 – 70) – 1]

= 20 – 1

= 19

Question no – (8)

Solution :

(a) The predecessor of 1 million is 999999 and successor is 1000001.

(b) The predecessor of 10 crore is 99999999 and successor is 100000001.

Question no – (9)

Solution :

(a) Whole numbers between 2568 and 2573 are

= 2569, 2570, 2571, 2572

(b) Whole numbers between 8472 and 8480 are

= 8473, 8474 8475, 8476, 8477, 8478, 8479

(c) Whole numbers between 9999 and 10,002 are

= 10000, 10001

Whole Numbers Exercise 2.2 Solution

Question no – (1)

Solution :

If the product of two whole numbers is zero, I can conclude that at least one of them is zero.

Question no – (2)

Solution :

If product of two whole numbers is 1, then I can conclude that each of them are 1.

Question no – (3)

Solution :

(a) 308 + 563 + 222 + 937

= (308 + 222) + (563 + 937)

= 330 + 1500

= 1830

The sum will be 1830

(b) 753 + 607 + 0 + 94 + 36

= (753 + 607) + (94 + 36)

= 1360 + 130

= 1660

The sum will be 1660

(c) 67 + 42 + 33 + 58

= (67 + 33) + (42 + 58)

= 100 + 100

= 200

The sum will be 200

(d) 564 + 197 + 103

= 564 + (197 + 103)

= 564 + 300

= 868

The sum will be 868

(e) 896 + 208 + 204 + 792

= (896 + 204) + (208 + 792)

= 1100 + 1000

= 2100

The sum will be 2100

Question no – (4)

Solution :

(a) 625 × 275 × 16

= 25 × 25 × 275 × 4 × 4

= (25 × 4) × (25 × 4) × 275

= 100 × 100 × 275

= 2750000

The product will be 2750000

(b) 125 × 400 × 8 × 25

= 536 × (1000 + 2)

= (536 × 1000) + (536 × 2)

= 536000 + 1072

= 5370972

The product will be 5370972

Question no – (5)

Solution :

(c) 165 × 99 + 165

= 165 × (100 – 1) + 165

= (165 × 100) – (165 × 1) + 165

= 16500 – 165 + 186

= 16500 + (165 – 165)

= 16500 + 0

= 16500    …(Simplified)

(d) (639 × 96) + (639 × 4)

= 639 × (96 + 4)

= 639 × 100

= 63900 …(Simplified)

(e) 743 × 999 + 743

= 743 (999 + 1)

= 743 × 1000

= 743000   …(Simplified)

(f) 489 × 17 + 489 × 13 + 489 × 70

= 489 (17 + 13) + 489 × 70

= 489 × 30 + 489 × 70

= 489 × (30 × 70)

= 489 × 100

= 48900    …(Simplified)

(g) 3127 × 124 – 3127 × 24

= 3127 × (124 – 24)

= 3127 × 100

= 312700 …(Simplified)

(h) 541 × 10 × 367 – 267 × 5410

= 5410 × 367 – 5410 × 267

= 5410 × (367 – 267)

= 5410 × 100

= 541000 …(Simplified)

(i) 661 × 1002 – 661 × 2

= 661 × (1002 – 12)

= 661 × 1000

= 661000 …(Simplified)

Question no – (6)

Solution :

(The largest 3 digit number) × (The largest 5 digit number)

= 999 × 99999

= 999 × (100000 – 1)

= (999 × 100000) – (999 × 1)

= 99900000 – 999

= 9989001

Therefore, the product will be 9989001

Question no – (7)

Solution :

(Cost of one scooter) × (number of scooters)

= 24, 350× 999 = 24, 350 × (1000 – 1)

= 24,350,000 – 24,350

= 243215650 rupees

Therefore, the cost of 999 scooters will be 243215650 rupees.

Question no – (8)

Solution :

(Cost of one air conditioner) × (number of air conditioner)

= 16.350 × 101 = 16,350 × (100 + 1)

= (16,3500 × 100) + (16,350 × 1)

= 16,350000 + 16,350

= 16366350 Rs

Therefore, the cost of 101 air-conditioners will be 16366350 Rs.

Question no – (9)

Solution :

The first day he spent = (68 × 45) rupees

The second day he spent = (68 × 35) rupees

So, m two days he spent = (68 × 45) + (68 × 35)

= 68 × (45 + 35)

= 68 × 80

= 5280 rupees

Hence, the taxi driver will spend Rs. 5280 in two days on petrol.

Question no – (10)

Solution :

Total required amount of money,

= {(37 × 67) + (37 × 33)

= {37 × (67 + 33)

= 37 × 100

= 3700 rupees

Thus, Rs. 3700 is due to the vendor per day.

Question no – (11)

Solution :

(a) a × (b + c) = (a × b) + (a × c)

L.H.S, 15 × (7 + 3)

= 15 × 10

= 150

R.H.S, (15 × 7) + (15 × 3)

= 105 + 45

= 150

Since, R.H.S = L.H.S, distributive property is satisfied.

(b) (a × b) × c = a × (b × c)

R.H.S, (15 × 7)

= 105 × 3

= 415

R.H.S, 15 × (7 × 3)

15 × 21

= 415

Since, R.H.S = L.H.S, associative property is satisfied.

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

L.H.S, (15 + 7) + 3

= 22 + 3

= 25

R.H.S, 15 + (7 + 3)

= 15 + 10

= 25

Since, R.H.S = L.H.S, associative property is satisfied.

(d) (a – b) – c = a – (b – c)

L.H.S, (15 – 7) – 3

= 8 – 3

= 5

R.H.S, 15 – (7 – 3)

= 15 – 4

= 11

Since, R.H.S = L.H.S, associative property is satisfied.

Question no – (12)

Solution

(a) 5217 ÷ 1

= 5217

∴ The value will be 5217

(b) 0 ÷ 89

= 0

∴ The value will be 0

(c) 6250 ÷ 10

= 625

∴ The value will be 625

(d) 23 ÷ 0

= Undefined

∴ The value will be Undefined

(e) 240 ÷ 8

= 30

∴ The value will be 30

(f) 2700 ÷ 9

= 300

∴ The value will be 300

Question no – (13)

Solution :

Five pairs of numbers whose sum or difference is 990 are –

(i)  110 + 880 = 900

(ii)  220 + 770 = 990

(iii)  330 + 660 = 990

(iv) 1120 – 230 = 990

(v) 1140 – 250 = 990

Question no – (14)

Solution :

Five pairs of numbers whose difference is less than each number in the pair

(i) 10 – 9 = 1

(ii) 100000 – 55000 = 45000

(iii) 300 – 200 = 100

(iv) 100 – 80 = 20

(v) 20 – 13 = 7

Question no – (15)

Solution :

Three pairs of numbers whose product is 2010 –

(i) 670 × 3 = 2010

(ii)  235 × 6 = 2010

(iii) 47 × 30 = 2010

Question no – (17)

Solution :

(a) 105 + 10 = 10 + 105

= [Commutative Property]

(b) 21 × 75 = 75 × 21

= [Commutative Property]

(c) (91 + 75 ) + 32 = 91 (+ 75 + 32)

= [Associative property]

(d) 85 + (71 + 29) = 100 + 85

= [Associative and commutative]

(e) 967 × 1 = 967

= [Multiplicative identity]

(f) 75 ÷ 75 = 1

= [Division by same number]

(g) 43 × (75 + 25) = 43 × = 75 + 75 × 25

= [Distributive property]

(h) 791 × 98 = 791 × (100 – 2)

= [Distributive property]

Question no – (18)

Solution :

(a) (785 × 27) + (785 × 43) + (785 × 30)

= 785 × (27 + 43) + (785 × 30)

= 785 × 70 + 785 × 30

= 785 × (70 + 300

= 785 × 100

= 78500 …(Simplified)

(b) (94 × 531) + (94 × 469)

= 94 × (531 + 469)

= 94 × 1000

= 94000 …(Simplified)

(c) (63 × 23) + (63 × 17) + (63 × 60)

= 63 × (23 + 17) + (63 × 600

= 63 × 40 + 63 × 60

= 63 × (40 × 60)

= 63 × 100

= 6300 …(Simplified)

(d) (526 × 41) + (526 × 16) + (526 × 43)

= 526 × (41 + 16) + 526 × 43

= (526 × 57) + 526 × 43

= 526 × (57 + 43)

= 526 × 100

= 52600 …(Simplified)

Try These 2 (ii) :

Question no – (1)

Solution :

 8 1 6 3 5 7 4 9 2

Question no – (2)

Solution :

 16 2 12 6 10 14 8 18 4

Question no – (3)

Solution :

 10 5 12 11 9 7 6 13 8

Question no – (4)

Solution :

 8 13 12 15 11 7 10 9 14

Try These 2 (iii) :

Question no – (1)

Solution :

Correct option is – (c)

The first ten numbers 1, 3, 6, 10, 15, 21…whose dot patterns when arranged will form a Triangle.

Question no – (2)

Solution :

Some numbers can be shown by two rectangles. five other such examples are –

(1) Triangle

Whole Numbers Exercise 2.3 Solution

Question no – (1)

Solution :

(a) 137 + 9

= 137 + 10 – 1

= 147 – 1

= 146

The value will be 146

(b) 637 + 99

= 637 + 100 – 1

= 737 – 1

= 736

The value will be 736

(c) 217 – 99

= 217 + 100 – 1

= 317 – 1

= 316

The value will be 316

(d) 516 + 999

= 516 + 1000 – 1

= 516 – 1

= 1515

The value will be 1515

(e) 524 – 999

= 524 + 1000 – 1

= 1524 – 1

= 1523

The value will be 1523

Question no – (2)

Solution :

(a) 48 × 5; 48 × 25; 48 × 125

= 48 × 5 = 48 × 10/2 = 480/2 = 240

= 48 × 25 = 48 × 100/4 = 4800/4 = 1200

= 48 × 1 = 48

(b) 56 × 5; 56 × 25; 56 × 125

= 56 × 5 = 56 × 10/2 = 560/2 = 280

= 56 × 25 = 56 × 100/4 = 5600/4 = 1400

= 56 × 125 = 56 × 100/8 = 56000/8 = 7000

25 = 48 × 1000/8 = 48000/8 = 6000

(c) 824 × 5; 824 × 25; 824 × 12

= 824 × 5 = 824 × 10/2 = 8240/2 = 4120

= 824 × 25 = 824 × 100/4 = 82400/4 = 20600

= 824 × 125 = 824 × 1000/8 = 824000/8 = 103000

(d) 72 × 5; 72 × 25; 72 × 125

= 72 × 5 = 72 × 10/2 = 720/2 = 360

= 72 × 25 = 720 × 100/4 = 7200/4 = 18000

= 72 × 125 = 720 × 1000/8 = 7200008 = 90000

Question no – (3)

Solution :

(a) 1 + 3 = 4 = 2 × 2

(b) 1 + 3 + 5 = 9 = 3 × 3

(c) 1 + 2 + 5 + 7 = 16 = 4 × 4

(d) 1 + 3 + 5 + 7 + 9 = 25 = 5

(e) 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6

(f)  1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 7

(g)  1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 82

Question no – (5)

Solution :

1 × 8 + 1 = 9

12 × 8 + 2 = 98

123 × 8 + 3 = 9876

1234 × 8 + 4 = 9876

12345 × 8 + 5 = 98765

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

12345678 × 8 = 8 = 98765432

123456789 × 9 + 9 = 987654321

Question no – (6)

Solution :

1 × 1 = 1

11 × 11 = 121

111 × 111 = 12321

1111 × 1111 = 1234321

11111 × 11111 = 123454321

111111 × 111111 = 12345654321

1111111 × 1111111 = 1234567654321

11111111 × 11111111 = 123456787654321

111111111 × 1111111111 = 12345678987654321

Question no – (7)

Solution :

(a) 1 ÷ 0

= does not represent ‘0’

Question no – (8)

Solution :

The complete adjacent magic square, as follows

 6 11 7 9 8 7 9 5 10

Question no – (9)

Solution :

(a) Given number, 9999

Successor,

= (9999 + 1)

= 1000

So, the Successor of 9999 is 1000.

Predecessor,

= (9999 – 1)

= 9998

So, the predecessor of 9999 is 9998.

(b) Given number, 10862

Successor,

= (10862 + 1)

= 10863

So, the Successor of 10862 is 10863 .

Predecessor,

= (10861 – 1)

= 10861

So, the predecessor of 10862 is 10861.

(c) We know, The smallest 4 digit number is = 1000

Successor,

= (1000 + 1)

= 1001

So, the Successor of 1000 is 1001 .

Predecessor,

= (1000 – 1)

= 999

So, the predecessor of 1000 is 999.

(d) We know, The largest 3 digit number is = 999

Successor,

= (999 + 1)

= 1000

So, the Successor of 999 is 1000.

Predecessor,

= (999 – 1)

= 998

So, the predecessor of 9999 is 9998.

Question no – (10)

Solution :

The number of whole numbers between 38 and 65 is

= [(65 – 38) – 1]

= 26

Question no – (11)

Solution :

(a) 4895 – 2367 < 2538

(b) 17896 + 4095 < 22000

(c) 2345 × 10 < 40000

(d) 985600 ÷ 100 = 9856

Question no – (12)

Solution :

The total amount given by Tina is

= (50 × 1250) + (10 × 6250)

= (1250 × 50) + (6250 × 10) [Commutative property]

= (1250 × 50) + (1250 × 5 × 50)

= (1250 × 50) + (1250 × 50) [Associative property]

= 1250 (50 + 50) [Distributive property]

= 1250 × 100

= 125000

Question no – (13)

Solution :

[The greatest number of four digits] + [The smallest number of four digits]

= 9999 + 1000 = 10999

Now, 10000 + 999 = 10999

I observe that, (9999 + 1000)

= (10000 + 999)

Previous Chapter Solution :

Updated: June 5, 2023 — 2:04 pm
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