Class 8 - Mathematics
Cubes and Cube Roots - Exercise 7.1
Top Block 1
Question :1. Which of the following numbers are not perfect cubes:
(i)216
(ii)128
(iii)1000
(iv)100
(v)46656
(i)216
(ii)128
(iii)1000
(iv)100
(v)46656
Answer :
(i)216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
(i)216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
Therefore, 216 is a perfect cube number.
(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.
(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 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2 x2 x 2 x 5 x 5 x 5
(iii) 1000
Prime factors of 1000 = 2 x2 x 2 x 5 x 5 x 5
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Mddle block 1
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656
= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
(v) 46656
Prime factors of 46656
= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Therefore, 46656 is a perfect cube.
Question :2. 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
(ii) 256
(iii) 72
(iv) 675
(v) 100
Answer :
(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.
(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.
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.
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.
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 = 10 to make it a perfect cube.
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 = 10 to make it a perfect cube.
Question :3. 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
(ii) 128
(iii) 135
(iv) 192
(v) 704
Answer :
(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.
(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
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.
Prime factors of 128 = 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.
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 x2 x2 x2 x2 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.
Prime factors of 192 = 2 x2 x2 x2 x2 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 x2 x2 x2 x2 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.
Prime factors of 704= 2 x 2 x2 x2 x2 x2 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.
Question :4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Answer :
Given numbers = 5 x 2 x 5
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 x 5 x 2 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.
Given numbers = 5 x 2 x 5
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 x 5 x 2 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.