Tag: RS Aggarwal Class 6 Maths Solutions

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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.

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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.

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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 votes received by the first candidate = 687905
Number of votes received by the second candidate = 495086
Number of votes received by the third candidate = 93756
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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1A
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1B
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1C
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1D
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1E
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1F
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1G
• RS Aggarwal Solutions Class 6 Chapter 1 Number System Ex 1H

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2A
• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2B
• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C
• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2D
• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2E
• RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2F

Give the prime factorization of each of the following numbers.

Question 1.
Solution:
We have

∴12 = 2 x 2 x 3 = 2² x 3.

Question 2.
Solution:
We have

∴18 = 2 x 3 x 3 = 2 x 3²

Question 3.
Solution:
We have

∴ 48 = 2 x 2 x 2 x 2 x 3
= 2⁴ x 3.

Question 4.
Solution:
We have

∴ 56 = 2 x 2 x 2 x 7
= 2³ x 7.

Question 5.
Solution:
We have

∴ 90 = 2 x 3 x 3 x 5
= 2 x 3² x 5.

Question 6.
Solution:
We have

136 = 2 x 2 x 2 x 17
= 2³ x 17.

Question 7.
Solution:
We have

∴ 252 = 2 x 2 x 3 x 3 x 7
= 2² x 3² x 7.

Question 8.
Solution:
We have

∴ 420 = 2 x 2 x 3 x 5 x 7
= 2² x 3 x 5 x 7.

Question 9.
Solution:
We have

∴ 637 = 7 x 7 x 13
= 7² x 13.

Question 10.
Solution:
We have

∴ 945 = 3 x 3 x 3 x 5 x 7
= 3³ x 5 x 7.

Question 11.
Solution:
We have

∴ 1224 = 2 x 2 x 2 x 3 x 3 x 17
= 2³ x 3² x 17.

Question 12.
Solution:
We have

∴ 1323 = 3 x 3 x 3 x 7 x 7
= 3³ x 7²

Question 13.
Solution:
We have

∴ 8712 = 2 x 2 x 2 x 3 x 3 x 11 x 11
= 2³ x 3² x 11².

Question 14.
Solution:
We have

∴9317 = 7 x 11 x 11 x 11
= 7 x 11³

Question 15.
Solution:
We have

∴ 1035 = 3 x 3 x 5 x 23
= 3² x 5 x 23

Question 16.
Solution:
We have

∴ 1197 = 3 x 3 x 7 x 19
= 3² x 7 x 19

Question 17.
Solution:
We have

∴ 4641 = 3 x 7 x 13 x 17.

Question 18.
Solution:
We have

∴ 4335 = 3 x 5 x 17 x 17
= 3 x 5 x 17²

Question 19.
Solution:
We have

∴ 2907 = 3 x 3 x 17 x 19
= 3² x 17 x 19.

Question 20.
Solution:
We have

∴ 13915 = 5 x 11 x 11 x 23 = 5 x 11² x 23

Hope given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Online Education for RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

These Solutions are part of Online Education RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A
• RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B

Simplify

Question 1.
Solution:
21 – 12 ÷ 3 x 2
= 21 – 4 x 2
= 21 – 8
= 13. Ans

Question 2.
Solution:
16 + 8 ÷ 4 – 2 x 3
= 16 + 2 – 2 x 3
= 16 + 2 – 6
= 18 – 6
= 12. Ans.

Question 3.
Solution:
13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – (10)
= 13 – 10
= 3 Ans.

Question 4.
Solution:
19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6]
= 19 – 10
= 9. Ans

Question 5.
Solution:
36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
= 36 – [18 – {14 – (15 – 2 x 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
= 21. Ans.

Question 6.
Solution:
\(27-[18-\{ 16-(5-\overline { 4-1 } )\} ] \)
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14]
= 27 – 4
= 23. Ans.

Question 7.
Solution:
\(4\frac { 4 }{ 3 } \div \frac { 3 }{ 5 } of5+\frac { 4 }{ 5 } \times \frac { 3 }{ 10 } -\frac { 1 }{ 5 } \)

Question 8.
Solution:
\(\left( \frac { 2 }{ 3 } +\frac { 4 }{ 9 } \right) of\frac { 3 }{ 5 } \div 1\frac { 2 }{ 3 } \times 1\frac { 1 }{ 4 } -\frac { 1 }{ 3 } \)

Question 9.
Solution:
\(7\frac { 1 }{ 3 } \div \frac { 2 }{ 3 } of2\frac { 1 }{ 5 } +1\frac { 3 }{ 8 } \div 2\frac { 3 }{ 4 } -1\frac { 1 }{ 2 } \)

Question 10.
Solution:
\(5\frac { 1 }{ 7 } -\left\{ 3\frac { 3 }{ 10 } \div \left( 2\frac { 4 }{ 5 } -\frac { 7 }{ 10 } \right) \right\} \)

Question 11.
Solution:
\(9\frac { 3 }{ 4 } \div \left[ 2\frac { 1 }{ 6 } +\left\{ 4\frac { 1 }{ 3 } -\left( 1\frac { 1 }{ 2 } +1\frac { 3 }{ 4 } \right) \right\} \right] \)

Question 12.
Solution:
\(4\frac { 1 }{ 10 } -\left[ 2\frac { 1 }{ 2 } -\left\{ \frac { 5 }{ 6 } -\left( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } -\frac { 4 }{ 15 } \right) \right\} \right] \)

Question 13.
Solution:
\(1\frac { 5 }{ 6 } +\left[ 2\frac { 2 }{ 3 } -\left\{ 3\frac { 3 }{ 4 } \left( 3\frac { 4 }{ 5 } \div 9\frac { 1 }{ 2 } \right) \right\} \right] \)

Question 14.
Solution:
\(4\frac { 4 }{ 5 } \div \left\{ 2\frac { 1 }{ 5 } -\frac { 1 }{ 2 } \left( 1\frac { 1 }{ 4 } -\overline { \frac { 1 }{ 4 } -\frac { 1 }{ 5 } } \right) \right\} \)

Question 15.
Solution:
\(7\frac { 1 }{ 2 } -\left[ 2\frac { 1 }{ 4 } \div \left\{ 1\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \left( \frac { 3 }{ 2 } -\overline { \frac { 1 }{ 3 } -\frac { 1 }{ 6 } } \right) \right\} \right] \)

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.