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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 = 52
(e) 1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
(f) 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 72
(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 :
