## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1H

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

Other Exercises

Objective questions
Mark against the correct answer in each of the following

Question 1.
Solution:
(c) Place value of 6 in 48632950 is 600000

Question 2.
Solution:
(a) Face value of 4 in the numeral 89247605 is 4
(∴ Face value does not alter)

Question 3.
Solution:
(c) The place of 5 in the numeral 78653421 is 50000 and face value is 5
∴ Difference between 50000 and 5 = 49995

Question 4.
Solution:
(b) The smallest counting number is 1

Question 5.
Solution:
(b) 4-digit numbers are 1000 to 9999
=> 9999 – 999
= 9000

Question 6.
Solution:
(b) 7-digit numbers are from 1000000 to 9999999 or 9999999 – 999999
= 9000000

Question 7.
Solution:
(c) 8-digit numbers are to 99999999 99999999 – 9999999
= 90000000

Question 8.
Solution:
The number before 1000000 will be 1000000 – 1 = 999999 (b)

Question 9.
Solution:
(a) VX is not meaningful as V does not come before X

Question 10.
Solution:
(c) IC is not meaningful as I comes before V and X only

Question 11.
Solution:
(b) XVV is not meaningful as V does not comes more than once.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1G

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G

Other Exercises

Question 1.
Solution:
(i) 2 = II
(ii) 8 = VIII
(iii) 14 = XIV
(iv) 29 = XXIX
(v) 36= XXXVI
(vi) 43 = XLIII
(vii) 54 = LIV
(viii) 61 = LXI
(ix) 73 = LXXIII
(x) 81 = LXXXI
(xi) 91 = XCI
(xii) 95 XCV
(xiii) 99= XCIX
(xiv) 105 CV
(xv) 114 = CXIV

Question 2.
Solution:
(i) 164 = CLXIV
(ii) 195 = CXCV
(iii) 226 = CCXXVI
(iv) 341 = CCCXLI
(v) 475 = CDLXXV
(vi) 596 = DXCVI
(vii) 611 = DCXI
(viii) 759 = DCCLIX

Question 3.
Solution:
(i) XXVII = 27
(ii) XXXIV = 34
(iii) XLV = 45
(iv) LIV = 54
(v) LXXIV = 74
(vi) XCI = 91
(vii) XCVI = 96
(viii) CXI = 111
(ix) CLIV = 154
(x) CCXXIV = 224
(xi) CCCLXV = 365
(xii) CDXIV = 414
(xiii) CDLXIV = 464
(xiv) DVI = 506
(xv) DCCLXVI = 766

Question 4.
Solution:
(i) V is never subtracted
∴ VC is wrong
(ii) I can be subtracted from V and X only
∴ IL is wrong
(iii) V, L, D are never repeated
∴ VVII is wrong
(iv) IX cannot occur to the left of X
∴ IXX is wrong

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1F

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F

Other Exercises

Find the estimated quotient for each of the following :

Question 1.
Solution:
87 ÷ 28
87 is estimated to the nearest ten = 90
28 is estimated to the nearest ten = 30
∴ 90 ÷ 30
= 3 Ans.

Question 2.
Solution:
83 ÷ 17
83 is estimated to the nearest ten = 80
17 is estimated to the nearest ten = 20
∴ 80 ÷ 20
= 4 Ans.

Question 3.
Solution:
75 ÷ 23
75 is estimated to the nearest ten = 80
23 is estimated to the nearest ten = 20
∴ 80 ÷ 20
= 4 Ans.

Question 4.
Solution:
193 ÷ 24
193 is estimated to the nearest ten = 200
24 is estimated to the nearest ten = 20
∴ 200 ÷ 20
= 10 Ans.

Question 5.
Solution:
725 ÷ 23
725 is estimated to the nearest hundred = 700
23 is estimated to the nearest ten = 20
∴700 ÷ 20
= 35 Ans.

Question 6.
Solution:
275 ÷ 25
275 is estimated to the nearest hundred = 300
25 is estimated to the nearest ten = 30
∴ 300 ÷ 30
= 10 Ans.

Question 7.
Solution:
633 ÷ 33
633 is estimated to the nearest hundred = 600
33 is estimated to the nearest ten = 30
∴ 600 ÷ 30
= 20 Ans.

Question 8.
Solution:
729 ÷ 29
729 is estimated to the nearest hundred = 700
29 is estimated to the nearest ten = 30
∴ 700 ÷ 30
= 70 ÷ 3
= 23 (approximately) Ans.

Question 9.
Solution:
858 ÷ 39
858 is estimated to the nearest hundred = 900
39 is estimated to the nearest ten = 40
∴ 900 ÷ 40
= 90 ÷ 4
= 23 (approximately) Ans

Question 10.
Solution:
868 ÷ 38
868 is estimated to the nearest hundred = 900
38 is estimated to the nearest ten = 40
∴ 900 ÷ 40
= 90 ÷ 4
= 23 (approximately) Ans.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E

Other Exercises

Estimate each of the following products by rounding off each number to the nearest ten.

Question 1.
Solution:
(38 x 63)
38 estimated to the nearest ten = 40
63 estimated to the nearest ten = 60
∴ 40 x 60
= 2400 Ans.

Question 2.
Solution:
(54 x 47)
54 estimated to the nearest ten = 50
47 estimated to the nearest ten = 50
∴ 50 x 50
= 2500 Ans.

Question 3.
Solution:
(28 x 63)
28 estimated to the nearest ten = 30
63 estimated to the nearest ten = 60
∴ 30 x 60
= 1800 Ans.

Question 4.
Solution:
(42 x 75)
42 estimated to the nearest ten = 40
75 estimated to the nearest ten = 80
∴ 40 x 80
= 3200 Ans.

Question 5.
Solution:
(64 x 58)
64 estimated to the nearest ten = 60
58 estimated to the nearest ten = 60
∴ 60 x 60
= 3600 Ans.

Question 6.
Solution:
(15 x 34)
15 estimated to the nearest ten = 20
34 estimated to the nearest ten = 30
∴ 20 x 30
= 600 Ans.

Estimate each of the following products by rounding off each number to the nearest hundred :

Question 7.
Solution:
(376 x 123)
376 estimated to the nearest hundred = 400
123 estimated to the nearest hundred = 100
∴ 400 x 100
= 40000 Ans.

Question 8.
Solution:
(264 x 147)
264 estimated to the nearest hundred = 300
147 estimated to the nearest hundred = 100
∴ 300 x 100
= 30000 Ans.

Question 9.
Solution:
423 x 158)
423 estimated to the nearest hundred = 400
158 estimated to the nearest hundred = 200
∴ 400 x 200
= 80000 Ans.

Question 10.
Solution:
(509 x 179)
509 estimated to the nearest hundred = 500
179 estimated to the nearest hundred = 200
∴ 500 x 200
= 100000 Ans.

Question 11.
Solution:
(392 x 138)
392 estimated to the nearest hundred = 400
138 estimated to the nearest hundred = 100
∴ 400 x 100
= 40000 Ans.

Question 12.
Solution:
(271 x 339)
271 estimated to the nearest hundred = 300
339 estimated to the nearest hundred = 300
∴ 300 x 300
= 90000 Ans.

Estimate each of the following products by rounding off the first number upwards and the second number downwards:

Question 13.
Solution:
(183 x 154)
183 is rounded off upwards = 200
154 is rounded off downwards = 100
∴ 200 x 100
= 20000 Ans.

Question 14.
Solution:
(267 x 146)
267 is rounded off upwards = 300
146 is rounded off downwards = 100
∴ 300 x 100
= 30000 Ans.

Question 15.
Solution:
(359 x 76)
359 is rounded off upwards = 400
76 is rounded off downwards = 70
∴ 400 x 70
= 28000 Ans.

Question 16.
Solution:
(472 x 158)
472 is rounded off upwards = 500
158 is rounded off downwards = 100
∴ 500 x 100
= 50000 Ans.

Question 17.
Solution:
(680 x 164)
680 is rounded off upwards = 700
164 is rounded off downwards = 100
∴ 700 x 100
= 70000 Ans.

Question 18.
Solution:
(255 x 350)
255 is rounded off upwards = 300
350 is rounded off downwards = 300
∴ 300 x 300
= 90000 Ans.

Estimate each of the following products by rounding off the first number downwards and the second number upwards:

Question 19.
Solution:
(356 x 278)
356 is rounded off downwards = 300
278 is rounded off upwards = 300
∴ 300 x 300
= 90000 Ans.

Question 20.
Solution:
(472 x 76)
472 is rounded off downwards = 400
76 is rounded off upwards = 80
∴ 400 x 80
= 32000 Ans.

Question 21.
Solution:
(578 x 369)
578 is rounded off downwards = 500
369 is rounded off upwards = 400
∴ 500 x 400
= 200000 Ans.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D

Other Exercises

Question 1.
Solution:
(a) 40
(b) 170
(c) 3870
(d) 16380

Question 2.
Solution:
(a) 800
(b) 1300
(c) 43100
(d) 98200

Question 3.
Solution:
(a) 1000
(b) 5000
(c) 17000
(d) 28000

Question 4.
Solution:
(a) 20000
(b) 30000
(c) 30000
(d) 270000

Estimate each sum to the nearest ten :

Question 5.
Solution:
(57 + 34)
57 estimated to the nearest ten = 60
34 estimated to the nearest ten = 30
Required sum = 60 + 30
= 90 Ans.

Question 6.
Solution:
(43 + 78)
43 estimated to the nearest ten = 40
78 estimated to the nearest ten = 80
Required sum = 40 + 80
= 120 Ans.

Question 7.
Solution:
(14 + 69)
14 estimated to the nearest ten = 10
69 estimated to the nearest ten = 70
Required sum = 10 + 70
= 80 Ans.

Question 8.
Solution:
(86 +19)
86 estimated to the nearest ten = 90
19 estimated to the nearest ten = 20
Required sum = 90 + 20
= 110 Ans.

Question 9.
Solution:
(95 + 58)
95 estimated to the nearest ten =100
58 estimated to the nearest ten = 60
Required sum = 100 + 60
= 160 Ans

Question 10.
Solution:
77 estimated to the nearest ten = 80
63 estimated to the nearest ten = 60
Required sum = 80 + 60
= 140 Ans.

Question 11.
Solution:
(356 + 275)
356 estimated to the nearest ten = 360
275 estimated to the nearest ten = 280
Required sum = 360 + 280
= 640 Ans.

Question 12.
Solution:
463 + 182
463 estimated to the nearest ten = 460
182 estimated to the nearest ten = 180
Required sum = 460 + 180
= 640 Ans.

Question 13.
Solution:
(538 + 276)
538 estimated to the nearest ten = 540
276 estimated to the nearest ten = 280
Required sum = 540 + 280
= 820 Ans.

Estimate each sum to the nearest hundred:

Question 14.
Solution:
(236 + 689)
236 estimated to the nearest hundred = 200
689 estimated to the nearest hundred = 700
Required sum = 200 + 700
= 900 Ans.

Question 15.
Solution:
(458 + 324)
458 estimated.to the nearest hundred = 500
324 estimated to the nearest hundred = 300
Required sum = 500 + 300
= 800 Ans.

Question 16.
Solution:
(170 + 395)
170 estimated to the nearest hundred = 200
395 estimated to the nearest hundred = 400
Required sum = 200 + 400
= 600 Ans.

Question 17.
Solution:
(3280 + 4395)
3280 estimated to the nearest hundred = 3300
4395 estimated to the nearest hundred = 4400
Required sum = 3300 + 4400
= 7700 Ans.

Question 18.
Solution:
(5130 + 1410)
5130 estimated to the nearest hundred = 5100
1410 estimated to the nearest hundred = 1400
Required sum = 5100 + 1400
= 6500 Ans.

Question 19.
Solution:
(10083 + 29380)
10083 estimated to the nearest hundred =10100
29380 estimated to the nearest hundred = 29400
Required sum = 10100 + 29400
= 39500 Ans.

Estimate each sum to the nearest thousand :

Question 20.
Solution:
(32836 + 16466)
32836 estimated to the nearest thousand = 33000
16466 estimated to the nearest thousand = 16000
Required sum = 33000 + 16000
= 49000 Ans.

Question 21.
Solution:
(46703 + 11375)
46703 estimated to the nearest thousand = 47000
11375 estimated to the nearest thousand = 11000
Required sum = 47000 + 11000
= 58000 Ans.

Question 22.
Solution:
54 balls + 79 balls
54 balls estimated to the nearest 10 = 50
79 balls estimated to the nearest 10 = 80
Required total number of balls = 50 + 80 + 130 Ans.

Estimate each difference to the nearest ten :

Question 23.
Solution:
(53 – 18)
53 estimated to the nearest ten = 50
18 estimated to the nearest ten = 20
Difference of 50 and 20
= 50 – 20
= 30 Ans.

Question 24.
Solution:
(97 – 38)
97 estimated to the nearest ten =100
38 estimated to the nearest ten = 40
Difference of 100 and 40
= 100 – 40
= 60 Ans.

Question 25.
Solution:
(409 – 148)
409 estimated to the nearest ten = 410
148 estimated to the nearest ten = 150
Difference of 410 and 150
= 410 – 150
= 260 Ans.

Estimate each difference to the nearest hundred :

Question 26.
Solution:
(678 – 215)
678 estimated to the nearest hundred = 700
215 estimated to the nearest hundred = 200
Difference between 700 and 200
= 700 – 200
= 500 Ans.

Question 27.
Solution:
(957 – 578)
957 estimated to the nearest hundred = 1000
578 estimated to the nearest hundred = 600
Difference between 1000 and 600
= 1000 – 600
= 400 Ans.

Question 28.
Solution:
(7258 – 2429)
7258 estimated to the nearest hundred = 7300
2429 estimated to the nearest hundred = 2400
Difference between 7300 and 2400
= 7300 – 2400
= 4900 Ans.

Question 29.
Solution:
5612 estimated to the nearest hundred = 5600
3095 estimated to the nearest hundred = 3100
Difference between 5600 and 3100
= 5600 – 3100
= 2500 Ans.

Estimate each difference to the nearest thousand :

Question 30.
Solution:
35863 estimated to the nearest thousand = 36000
27677 estimated to the nearest thousand = 28000
Difference between 36000 and 28000
= 36000 – 28000
= 8000 Ans.

Question 31.
Solution:
(47005 – 39488)
47005 estimated to the nearest thousand = 47000
39488 estimated to the nearest thousand = 39000
Difference between 47000 and 39000
= 47000 – 39000
= 8000 Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D are helpful to complete your math homework.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C

Other Exercises

Question 1.
Solution:
Number of persons in the first year = 13789509
Number of persons in the second year = 12976498
Total number of persons in the two years =13789509 +12976498
= 26766007 Ans.

Question 2.
Solution:
Number of sugar bags in the first factory = 24809565
Number of sugar bags in the second factory = 18738576
Number of sugar bags in the third factory = 9564568
Total number of sugar bags in the three factories
= 24809565 + 18738576 + 9564568
= 53112709 bags Ans.

Question 3.
Solution:
The given number = 37684955
The number which exceeds the given number by 3615045 will be
= 37684955 + 3615045
= 41300000 Ans.

Question 4.
Solution:
Number of invalid votes = 13849
Number of persons who did not vote = 25467
Total number of registered votes = 687905 + 495086 + 93756 + 13849 + 25467
= 1316063 Ans.

Question 5.
Solution:
Number of people who got primary education = 1623546
Number of people who got secondary education = 9768678
Number of people who got higher education = 6837954
Number of illiterate people = 2684536
Number of children below the age of admission = 698781
Total population of the state = 1623546 + 9768678 + 6837954 + 2684536 + 698781
= 21613495 Ans.

Question 6.
Solution:
In the first year, number of cycles produced = 8765435
In the second year, number of cycles produced = 8765435 + 1378689
= 10144124
The number of bicycles produced in the two years = 8765435 + 10144124
= 18909559 Ans.

Question 7.
Solution:
Sale receipt during first year = Rs. 20956480
Sale receipt during the second year = Rs. 20956480 + Rs. 6709570
= Rs. 27666050
Total sale receipt during the two years = Rs. 20956480 + Rs.27666050
= Rs. 48622530 Ans

Question 8.
Solution:
Total population of a city = 28756304
Number of males = 16987059
Number of females = 28756304 – 16987059
= 11769245 Ans.

Question 9.
Solution:
The number 13246510 is larger than 4658642
= 13246510 – 4658642
= 8587868 Ans.

Question 10.
Solution:
5643879 is smaller than one crore
= 10000000 – 5643879
= 4356121 Ans.

Question 11.
Solution:
To, get the required number, we should subtract 2635967 from 11010101
= 11010101 – 2635967
= 8374134 Ans.

Question 12.
Solution:
Sum of two numbers = 10750308
First number = 8967519
Second number = 10750308 – 8967519
= 1782789 Ans.

Question 13.
Solution:
Total money, a man had = Rs 20000000
Amount spent on buying a school building = Rs. 13607085
Amount left with him
= Rs. 20000000 – Rs. 13607085
= Rs. 6392915 Ans.

Question 14.
Solution:
The total requirement of a society = Rs. 18536000
Amount of fee collection = Rs. 7253840
Amount of loan taken = Rs. 5675450
Amount of donation = Rs. 2937680
Total amount collected = Rs. 7253840 + Rs. 5675450 + Rs. 2937680
= Rs. 15866970
Short amount
= Rs. 18536000 – Rs. 15866970
= Rs. 2669030 Ans.

Question 15.
Solution:
Total amount a man had = Rs. 10672540
Amount given to his wife = Rs. 4836980
Amount given to his son = Rs 3964790
Total amount given to wife and son = Rs. 4836980 + Rs 3964790
= Rs. 8801770

Balance amount given to his daughter
= Rs. 10672540 – Rs. 8801770
= Rs. 1870770 Ans.

Question 16.
Solution:
Cost of one chair = Rs. 1485
Cost of 469 chairs = Rs. 1485 x 469
= Rs. 696465 Ans.

Question 17.
Solution:
Collection from one student = Rs. 625
Collection from 1786 students = Rs. 1786 x 625
= Rs. 1116250 Ans.

Question 18.
Solution:
Number of screws produced in one day = 6985
Number of screws produced in 358 days = 6985 x 358
= 2500630 Ans.

Question 19.
Solution:
Number of months in one year = 12
Number of months in 13, years = 12 x 13
= 156
months Saving in one month = Rs. 8756
Saving in 156 months = Rs. 8756 x 156
= Rs. 1365936 Ans.

Question 20.
Solution:
Cost of 1 scooter = Rs. 36725
Cost of 487 scooters = Rs. 36725 x 487
= Rs. 17885075 Ans.

Question 21.
Solution:
Distance covered in 1 hour = 1485 km
Distance covered in 72 hours = 1485 x 72 km
= 106920 km Ans.

Question 22.
Solution:
Product of two numbers = 13421408
First number = 364
Second number = 13421408 ÷ 364
= 36872 Ans.

Question 23.
Solution:
Cost of 36 flats = Rs. 68251500
Cost of one flat
= Rs. 68251500 ÷ 36
= Rs. 1895875 Ans.

Question 24.
Solution:
Mass of cylinder with gas = 30 kg 250 g and mass of empty cylinder = 14 kg 480 g
Mass of gas = 30 kg, 250 g – 14 kg, 480 g
= 15 kg, 770 g Ans.

Question 25.
Solution:
Total length of cloth = 5 m
Length of piece cut off = 2 m 85 cm
Length of remaining piece of cloth = 5 m – 2 m 85 cm
= 2 m 15 cm Ans.

Question 26.
Solution:
Cloth required for 1 shirt = 2 m 75 cm
Cloth required for 16 shirts = 2 m 75 cm x 16
= 44 m Ans.

Question 27.
Solution:
Total length of cloth for 8 trousers = 14 m 80 cm
Length of cloth for 1 trouser = 14 m 80 cm ÷ 8
= 1 m 85 cm Ans.

Question 28.
Solution:
Mass of 1 brick = 2 kg 750 g
Total mass of 14 bricks = 2 kg 750 g x 14
= 38 kg 500 g Ans.

Question 29.
Solution:
Total mass of 8 packets = 10 kg 600 g
Mass of one packet = 10 kg 600 ÷ 8
= 1 kg 325 g Ans.

Question 30.
Solution:
Total length of rope = 10 m
No of pieces = 8
Length of each piece = 10 m ÷ 8
= 1 m 25 cm Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C are helpful to complete your math homework.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B.

Other Exercises

Fill in each of the following boxes with the correct symbol > or < :

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Arrange the following numbers in the descending order :

Question 7.
Solution:
102345680 > 63521047 > 63514759 > 7355014 > 7354206

Question 8.
Solution:
23794206 > 23756819 > 5032790 > 5032786 > 987876

Question 9.
Solution:
16060666 > 16007777 > 1808090 > 1808088 > 190909 > 181888

Question 10.
Solution:
1712040 > 1704382 > 1702497 > 201200 > 200175 > 199988.

Arrange the following numbers in the ascending order

Question 11.
Solution:
990357 < 9873426 < 9874012 < 24615019 < 24620010

Question 12.
Solution:
5694437 < 5695440 < 56943201 < 56943300 < 56944000

Question 13.
Solution:
700087 < 8014257 < 8014306 < 8015032 < 10012458

Question 14.
Solution:
893245 < 893425 < 980134 < 1020216 < 1020304 < 1021403

Hope given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B are helpful to complete your math homework.

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## RS Aggarwal Class 6 Solutions Chapter 1 Number System Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A

Other Exercises

Question 1.
Solution:
(i) Nine thousand eighteen = 9,018.
(ii) Fifty four thousand seventy three = 54,073
(iii) Three lakh two thousand five hundred six = 3,02,506.
(iv) Twenty lakh ten thousand eight = 20,10,008.
(v) Six crore five lakh fifty-seven = 6,05,00,057.
(vi) Two crore two lakh two thousand two hundred two = 2,02,02,202.
(vii) Twelve crore twelve lakh twelve thousand twelve = 12,12,12,012.
(viii) Fifteen crore fifty lakh twenty thousand sixty-eight = 15,50,20,068.

Question 2.
Solution:
(i) 63,005 = Sixty three thousand five.
(ii)7,07,075 = Seven lakh, seven thousand seventy five.
(iii) 34,20,019 = Thirty-four lakh twenty thousand nineteen.
(iv) 3,05,09,0.12 = Three crore, five lakh, nine thousand twelve.
(v) 5,10,03,604 = Five crore ten lakh three thousand six hundred four.
(vi) 6,18,05,008 = Six crore eighteen lakh five thousand eight.
(vii) 19,09,09,900 = Nineteen crore nine lakh, nine thousand nine hundred.
(viii) 6,15.30,807 = Six crore fifteen lakh, thirty thousand eight hundred seven.
(ix) 6,60,60,060 = Six crore sixty lakh sixty thousand sixty. Ans.

Question 3.
Solution:
(i) 15,768 = (1 x 10000) + (5 x 1000) + (7 x 100) + (6 x 10) + (8 x 1)
(ii) 3,08,927 = (3 x 100000) + (8 x 1000) + (9 x 100) +(2 x 10) +(7 x 1)
(iii) 24,05,609 = (2 x 1000000) + (4 x 100000) + (5 x 1000) + (6 x 100) + (9 x 1)
(iv) 5,36,18,493 = (5 x 10000000) + (3 x 1000000) + (6 x 100000) + (1 x 10000) + (8 x 1000) + (4 x 100) + (9 x 10) + (3 x 1)
(v) 6,06,06,006 = (6 x 10000000) + (6 x 100000) + (6 x 1000) + (6 x 1)
(vi) 9,10,10,510 = (9 x 10000000) + (1 x 1000000) + (1 x 10000) + (5 x 100) + (1 x 10) Ans,

Question 4.
Solution:
(i) 6 x 10000 + 2 x 1000 + 5 x 100 + 8 x 10 + 4 x 1
= 60000 + 2000 + 500 + 80 + 4
= 62,584.
(ii) 5 x 100000 + 8 x 10000 + 1 x 1000 + 6 x 100 + 2 x 10 + 3 x 1
= 500000 + 80000 + 1000 + 600 + 20 + 3
= 5,81,623
(iii) 2 x 10000000 + 5 x 100000 + 7 x 1000 + 9 x 100 + 5 x 1
= 20000000 + 500000 + 7000 + 900 + 5
= 2,05,07,905
(iv) 3 x 1000000 + 4 x 100000 + 6 x 1000 + 5 x 100 + 7 x 1
= 3000000 + 400000 + 6000 + 500 + 7
= 34,06,507 Ans.

Question 5.
Solution:
In 79520986
Value of first 9 = 9000000
and value of second 9 = 900
Difference = 9000000 – 900
= 89,99,100 Ans.

Question 6.
Solution:
In 27650934
place value of 7 = 7000000
and face value of 7 = 7
Difference = 7000000 – 7
= 6999993 Ans.

Question 7.
Solution:
There are 900000 6-digits numbers in all
i.e. 999999 – 99999
= 900000 Ans

Question 8.
Solution:
There are 9999999 – 999999
= 9000000 7-digits numbers in all.

Question 9.
Solution:
In 1,00,000, there are 100 thousands. Ans.

Question 10.
Solution:
In 10000,000, there are 10000 thousands.

Question 11.
Solution:
Given Number = 738
Reversing its digits = 837
Difference between 738 and 837
= 837 – 738
= 99 Ans.

Question 12.
Solution:
Numbers after 9547999
= 9547999 + 1
= 9548000 Ans.

Question 13.
Solution:
Number first before 9900000
= 9900000 – 1
= 9899999 Ans.

Question 14.
Solution:
Number first before 10000000
= 10000000 – 1
= 9999999 Ans.

Question 15.
Solution:
3-digits numbers using 2, 3, 4 taking each digit only once will be
234, 243, 324, 342, 423, 432 Ans.

Question 16.
Solution:
The smallest number by using different digits 3, 1, 0, 5 and 7 will be
= 10357 Ans.

Question 17.
Solution:
The largest number formed by using different digits 2, 4, 0, 3, 6 and 9 will be
= 964320. Ans.

Question 18.
Solution:

(i) Seven hundred thirty five thousand eight hundred twenty-one.
(ii) Six million fifty seven thousand eight hundred ninety-four.
(iii) Fifty-six million nine hundred forty-three thousand eight hundred twenty-one.
(iv) Thirty-seven million five hundred two thousand ninety-three.
(v) Eighty-nine million three hundred fifty thousand sixty-four.
(vi) Ninety million seven hundred three thousand and six. Ans.

Question 19.
Solution:

OBJECTIVE QUESTIONS
Tick the correct answer in each of the following :

Question 20.
Solution:
(c) Because 6 is at lakh place.

Question 21.
Solution:
(a) Because face value is always same.

Question 22.
Solution:
(b) Because place value and face value of 5 in 87653421 are 50000 and 5.
difference
= 50000 – 5
= 49995.

Question 23.
Solution:
(b) Smallest counting number or natural number is 1.

Question 24.
Solution:
(b) Number of 4-digits numbers
= 9999 – 999 (i.e. these are 1000 to 9999)
= 9000

Question 25.
Solution:
(b) Number of 7-digit numbers (from 1000000 to 9999999)
= 9999999 – 999999
= 9000000

Question 26.
Solution:
(c) Numbers of 8-digit numbers (from 10000000 to 99999999)
= 99999999 – 9999999
= 90000000

Question 27.
Solution:
(b) Because 1000000 – 1
= 999999 Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A are helpful to complete your math homework.

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