Maths Class 6 RS Aggarwal Solutions

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

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

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 2A.

Other Exercises

Question 1.
Solution:
(i) A factor of a number is an exact divisor of that number.
Examples : 1. 2 is a factor of 8
2. 5 is a factor of 15
3. 9 is a factor of 27
4. 4 is a factor of 20
5. 3 is a factor of 12.
(ii) Multiple. A number is said to be a multiple of any of its factors.
Examples : 1. 15 is a multiple of 3
2. 8 is a multiple of 4
3. 10 is a multiple of 2
4. 25 is a multiple of 5
5.18 is a multiple of 9.

Question 2.
Solution:
(i) We know that
20 = 1 x 20, 20 = 2 x 10, 20 = 4 x 5
which shows that the numbers 1, 2, 4, 5, 10, 20 exactly divide 20.
1, 2, 4, 5, 10 and 20 are all factors of 20
(ii) We know that
36 = 1 x 36, 36 = 2 x 18, 36 = 3 x 12, 36 = 4 x 9, 36 = 6 x 6
This shows that each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36 exactly divides 36.
1, 2, 3, 4, 6, 9, 12, 18, 36 are the factors of 36.
(iii) We know that
60 = 1 x 60, 60 = 2 x 30, 60 = 3 x 20, 60 = 4 x 15, 60 = 5 x 12, 60 = 6 x 10
This shows that each of the numbers 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 exactly divides 60.
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 are all the factors of 60.
(iv) We know that
75 = 1 x 73, 75 = 3 x 25, 75 = 5 x 15
This shows that each of the numbers 1, 3, 5, 15, 25, 75 exactly divides 75.
1, 3, 5, 15, 25, 75 are all the factors of 75.

Question 3.
Solution:
(i) First five multiples of 17 are :
17 x 1 = 17
17 x 2 = 34
17 x 3 = 51
17 x 4 = 68
17 x 5 = 85
(ii) First five multiples of 23 are :
23 x 1 = 23
23 x 2 = 46
23 x 3 = 69
23 x 4 = 92
23 x 5 = 115
(iii) First five multiples of 65 are :
65 x 1 = 65
65 x 2 = 130
65 x 3 = 195
65 x 4 = 260
65 x 5 = 325
(iv) First five multiples of 70 are :
70 x 1 = 70
70 x 2 = 140
70 x 3 = 210
70 x 4 = 280
70 x 5 = 350

Question 4.
Solution:
(i) 32 is a multiple of 2, so it is an even number.
(ii) 37 is not a multiple of 2, so it is an odd number.
(iii) 50 is a multiple of 2, so it is an even number.
(iv) 58 is a multiple of 2, so it is an even number.
(v) 69 is not a multiple of 2, so it is an odd number.
(vi) 144 is a multiple of 2, so it is an even number.
(vii) 321 is not a multiple of 2, so it is an odd number.
(viii) 253 is not a multiple of 2, so it is an odd number.

Question 5.
Solution:
Prime Numbers. Each of the numbers which has exactly two factors, namely 1 and itself, is called a prime number.
Examples. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 are all prime numbers.

Question 6.
Solution:
(i) Prime numbers between 10 and 40 are: 11, 13, 17, 19, 23, 29, 31, 37.
(ii) Prime numbers between 80 and 100 are : 83, 89, 97.
(iii) Prime numbers between 40 and 80 are : 41, 43, 47, 53, 59, 61, 67, 71, 73, 79
(iv) Prime numbers between 30 and 40 are : 31, 37.

Question 7.
Solution:
(i) 2 is the smallest prime number.
(ii) 2 is the only even prime number.
(iii) 3 is the smallest odd prime number.

Question 8.
Solution:
(i) We know that
87 = 1 x 87, 87 = 3 x 29
This shows that 1, 3, 29, 87 are the factors of 87.
The number 87 is not a prime number as it has more than 2 factors.
(ii) We have 89 = 1 x 89
The number 89 is a prime number as it has only 2 factors.
(iii) We have 63 = 1 x 63, 63 = 3 x 21,
63 = 7 x 9
This shows that the number 63 has more than 2 factors namely 1, 3, 7, 9, 21,63. So, it is not a prime number.
(iv) We have 91 = 1 x 91, 91 = 7 x 13 This shows that the number 91 has more than 2 factors namely 1, 7, 13, 91.
So, it is not a prime number.

Question 9.
Solution:
From the Sieve of Eratosthenes, we see that the seven consecutive numbers are 90, 91, 92, 93, 94, 95 and 96

Question 10.
Solution:
(i) There is no counting number having no factor at all.
(ii) The number 1 has exactly one factor.
(iii) The numbers between 1 and 100 having exactly three factors are : 4, 9, 25, 49.

Question 11.
Solution:
Composite Numbers. Numbers having more than two factors are called composite numbers. A composite number can be an odd number. The smallest odd composite number is 9.

Question 12.
Solution:
Twin-primes. Two consecutive odd prime numbers are known as twin- primes.
The prime numbers between 50 and 100 are:
53, 59, 61, 67, 71, 73, 79, 83, 89, 97
From above pairs of twin-primes are (59, 61), (71, 73)

Question 13.
Solution:
Co-primes. Two numbers are said to
be co-prime if they do not have a common factor.
Examples. Five pairs of co-primes are:
(i) 2, 3
(ii) 3, 4
(iii) 4, 5
(iv) 8, 15
(v) 9, 16
Co-primes are not always prime.
Illustration. In the pair (3, 4) of co-primes, 3 is a prime number whereas 4 is a composite number.

Question 14.
Solution:
(i) 36 = 7 + 29
(ii) 42 = 5 + 37
(iii) 84= 17 + 67
(iv) 98 = 19 + 79

Question 15.
Solution:
(i) 31 = 5 + 7 + 19
(ii) 35 = 5 + 7 + 23
(iii) 49 = 3 + 5 + 41
(iv) 63 = 7+ 13 +43

Question 16.
Solution:
(i) 36 = 17 + 19
(ii) 84 = 41 + 43
(iii) 120 = 59 + 61
(iv) 144 = 71+73

Question 17.
Solution:
(i) to (iv). None of the given statements is true.

 

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

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

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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

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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

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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

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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.

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

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