(a) 297 x 17 + 297 x 3 = 297 x (17 + 3) = 297 x 20 = 5940 (b) 54279 x 92 + 8 x 542379 = 54279 x (92 + 8) = 54279 x 100 = 5427900 (c) 81265 x 169 – 81265 x 69 = 81265 x (169 – 69) = 81265 x 100 = 8126500 (d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218) = 3845 x 5 x 782 + 3845Read more
(a) 297 x 17 + 297 x 3
= 297 x (17 + 3)
= 297 x 20
= 5940
(b) 54279 x 92 + 8 x 542379
= 54279 x (92 + 8)
= 54279 x 100
= 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100
= 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218
= 3845 x 5 x 782 + 769 x 5 x 5 x 218)
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225000
Factors of 12 are 1, 2, 3, 4, 6 and 12. Therefore, the number also be divisible by 1, 2, 3 4 and 6. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Therefore, the number also be divisible by 1, 2, 3 4 and 6.
(a) Factors of 18 = 1, 2, 3, 6, 9, 18 Factors of 35 = 1, 5, 7, 35 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. (b) Factors of 15 = 1, 3, 5, 15 Factors of 37 = 1, 37 Common factor = 1 Since, both have only one common factor, i.e., 1, therefRead more
(a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(c) Factors of 30 = 1, 2, 3, 5, 6, 15, 30
Factors of 415 = 1, 5, …….., 83, 415
Common factor = 1, 5
Since, both have more than one common factor, therefore, they are not co-prime numbers.
(d) Factors of 17 = 1, 17
Factors of 68 = 1, 2, 4, 17, 34, 86
Common factor = 1, 17
Since, both have more than one common factor, therefore, they are not co-prime numbers.
(e) Factors of 216 = 1, 2, 3, 4, 6, 8, 36, 72, 108, 216
Factors of 215 = 1, 5, 43, 215
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(f) Factors of 81 = 1, 3, 9, 27, 81
Factors of 16 = 1, 2, 4, 8, 16
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 928389 → Odd places = 9 + 8 + 8 = 25 Even places = 2 + 3 + 9 = 14 Difference = 25 – 14 = 11 (b) We know that a number is divisible by 11Read more
(a) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 928389 → Odd places = 9 + 8 + 8 = 25
Even places = 2 + 3 + 9 = 14
Difference = 25 – 14 = 11
(b) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 869484 → Odd places = 8 + 9 + 8 = 25
Even places = 6 + 4 + 4 = 14
Difference = 25 – 14 = 11
Find the value of the following: (a) 297 x 17 + 297 x 3 (b) 54279 x 92 + 8 x 54279 (c) 81265 x 169 – 81265 x 69 (d) 3845 x 5 x 782 + 769 x 25 x 218
(a) 297 x 17 + 297 x 3 = 297 x (17 + 3) = 297 x 20 = 5940 (b) 54279 x 92 + 8 x 542379 = 54279 x (92 + 8) = 54279 x 100 = 5427900 (c) 81265 x 169 – 81265 x 69 = 81265 x (169 – 69) = 81265 x 100 = 8126500 (d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218) = 3845 x 5 x 782 + 3845Read more
(a) 297 x 17 + 297 x 3
= 297 x (17 + 3)
= 297 x 20
= 5940
(b) 54279 x 92 + 8 x 542379
= 54279 x (92 + 8)
= 54279 x 100
= 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100
= 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218
= 3845 x 5 x 782 + 769 x 5 x 5 x 218)
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225000
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessWhich of the following will not represent zero:
(a) 1 + 0 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
(a) 1 + 0
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessA number is divisible by 12. By what other numbers will that number be divisible?
Factors of 12 are 1, 2, 3, 4, 6 and 12. Therefore, the number also be divisible by 1, 2, 3 4 and 6. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Therefore, the number also be divisible by 1, 2, 3 4 and 6.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessA number is divisible by both 5 and 12. By which other number will that number be always divisible?
5 x 12 = 60. The number must be divisible by 60. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
5 x 12 = 60. The number must be divisible by 60.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessWhich of the following numbers are co-prime: (a) 18 and 35 (b) 15 and 37 (c) 30 and 415 (d) 17 and 68 (e) 216 and 215 (f) 81 and 16
(a) Factors of 18 = 1, 2, 3, 6, 9, 18 Factors of 35 = 1, 5, 7, 35 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. (b) Factors of 15 = 1, 3, 5, 15 Factors of 37 = 1, 37 Common factor = 1 Since, both have only one common factor, i.e., 1, therefRead more
(a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(c) Factors of 30 = 1, 2, 3, 5, 6, 15, 30
Factors of 415 = 1, 5, …….., 83, 415
Common factor = 1, 5
Since, both have more than one common factor, therefore, they are not co-prime numbers.
(d) Factors of 17 = 1, 17
Factors of 68 = 1, 2, 4, 17, 34, 86
Common factor = 1, 17
Since, both have more than one common factor, therefore, they are not co-prime numbers.
(e) Factors of 216 = 1, 2, 3, 4, 6, 8, 36, 72, 108, 216
Factors of 215 = 1, 5, 43, 215
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
(f) Factors of 81 = 1, 3, 9, 27, 81
Factors of 16 = 1, 2, 4, 8, 16
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessWrite all the numbers less than 100 which are common multiples of 3 and 4.
Multiple of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 Multiple of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100 Common multiples of 3 and 4 = 12, 24, 36,Read more
Multiple of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99
Multiple of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100
Common multiples of 3 and 4 = 12, 24, 36, 48, 60, 72, 84, 96
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessFind the first three common multiples of: (a) 6 and 8 (b) 12 and 18
(a) Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72, ………… Multiple of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, ……………………. Common multiples of 6 and 8 = 24, 48, 72 (b) Multiple of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ……… Multiple of 18 = 18, 36, 54, 72, 90, 108, ……………………………… Common mRead more
(a) Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72, …………
Multiple of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, …………………….
Common multiples of 6 and 8 = 24, 48, 72
(b) Multiple of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ………
Multiple of 18 = 18, 36, 54, 72, 90, 108, ………………………………
Common multiples of 12 and 18 = 36, 72, 108
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessFind the common factors of: (a) 4, 8 and 12 (b) 5, 15 and 25
(a) Factors of 4 = 1, 2, 4 Factors of 8 = 1, 2, 4, 8 Factors of 12 = 1, 2, 3, 4, 6, 12 Common factors of 4, 8 and 12 = 1, 2, 4 (b) Factors of 5 = 1, 5 Factors of 15 = 1, 3, 5, 15 Factors of 25 = 1, 5, 25 Common factors of 5, 15 and 25 = 1, 5 https://www.tiwariacademy.com/ncert-solutions/class-6/mathRead more
(a) Factors of 4 = 1, 2, 4
Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 3, 4, 6, 12
Common factors of 4, 8 and 12 = 1, 2, 4
(b) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors of 5, 15 and 25 = 1, 5
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessFind the common factors of: (a) 20 and 28 (b) 15 and 25 (c) 35 and 50 (d) 56 and 120
(a) Factors of 20 = 1, 2, 4, 5, 10, 20 Factors of 28 = 1, 2, 4, 7, 14, 28 Common factors = 1, 2, 4 (b) Factors of 15 = 1, 3, 5, 15 Factors of 25 = 1, 5, 25 Common factors = 1, 5 (c) Factors of 35 = 1, 5, 7, 35 Factors of 50 = 1, 2, 5, 10, 25, 50 Common factors = 1, 5 (d) Factors of 56 = 1, 2, 4, 7,Read more
(a) Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factors = 1, 2, 4
(b) Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors = 1, 5
(c) Factors of 35 = 1, 5, 7, 35
Factors of 50 = 1, 2, 5, 10, 25, 50
Common factors = 1, 5
(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 60, 120
Common factors = 1, 2, 4, 8
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessWrite the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 11: (a) 92 __________ 389 (b) 8 __________ 9484
(a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 928389 → Odd places = 9 + 8 + 8 = 25 Even places = 2 + 3 + 9 = 14 Difference = 25 – 14 = 11 (b) We know that a number is divisible by 11Read more
(a) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 928389 → Odd places = 9 + 8 + 8 = 25
Even places = 2 + 3 + 9 = 14
Difference = 25 – 14 = 11
(b) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 869484 → Odd places = 8 + 9 + 8 = 25
Even places = 6 + 4 + 4 = 14
Difference = 25 – 14 = 11
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See less