(i) 81 Prime factors of 81 = 3 x 3 x 3 x 3 Here one factor 3 is not grouped in triplets. Therefore 81 must be divided by 3 to make it a perfect cube. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 must be divided by 2 to makRead more
(i) 81
Prime factors of 81 = 3 x 3 x 3 x 3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 must be divided by 2 to make it a perfect cube.
(iii) 135
Prime factors of 135 = 3 x 3 x 3 x 5
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
(iv) 192
Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
(v) 704
Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
(i) 243 Prime factors of 243 = 3 x 3 x 3 x 3 x 3 Here 3 does not appear in 3’s group. Therefore, 243 must be multiplied by 3 to make it a perfect cube. (ii) 256 Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 is required to make a 3’s group. Therefore, 256 must be multiplied bRead more
(i) 243
Prime factors of 243 = 3 x 3 x 3 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii) 256
Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
(iii) 72
Prime factors of 72 = 2 x 2 x 2 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
(iv) 675
Prime factors of 675 = 3 x 3 x 3 x 5 x 5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
(v) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by 2 x 5 = 10 to make it a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube: (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704
(i) 81 Prime factors of 81 = 3 x 3 x 3 x 3 Here one factor 3 is not grouped in triplets. Therefore 81 must be divided by 3 to make it a perfect cube. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 must be divided by 2 to makRead more
(i) 81
Prime factors of 81 = 3 x 3 x 3 x 3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 must be divided by 2 to make it a perfect cube.
(iii) 135
Prime factors of 135 = 3 x 3 x 3 x 5
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
(iv) 192
Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
(v) 704
Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube: (i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100
(i) 243 Prime factors of 243 = 3 x 3 x 3 x 3 x 3 Here 3 does not appear in 3’s group. Therefore, 243 must be multiplied by 3 to make it a perfect cube. (ii) 256 Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 is required to make a 3’s group. Therefore, 256 must be multiplied bRead more
(i) 243
Prime factors of 243 = 3 x 3 x 3 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii) 256
Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
(iii) 72
Prime factors of 72 = 2 x 2 x 2 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
(iv) 675
Prime factors of 675 = 3 x 3 x 3 x 5 x 5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
(v) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by 2 x 5 = 10 to make it a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/