NCERT Solutions for Class 8 Maths<\/a> Chapter 7 Cube and Cube Roots Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.1<\/h2>\n Question 1. \nWhich of the following numbers are not perfect cubes? \n(i) 216 \n(ii) 128 \n(iii) 1000 \n(iv) 100 \n(v) 46656 \nSolution: \n(i) 216 \n216 = 23<\/sup> \u00d7 33<\/sup> \nEach prime factors appear in triplets, so 216 is a perfect cube. \n <\/p>\n(ii) 128 \n128 = 23<\/sup> \u00d7 23<\/sup> \u00d7 2 \n2 remains after grouping in triplets, so 128 is not a perfect cube. \n <\/p>\n <\/p>\n
(iii) 1000 \n1000 = 23<\/sup> \u00d7 53<\/sup> \nEach prime factor appears in triplets so 1000 is a perfect cube. \n <\/p>\n(iv) 100 \n100 = 22<\/sup> \u00d7 52<\/sup> \nAs we do not get any triplets, So 100 is not a perfect cube. \n <\/p>\n(v) 46656 \n46656 = 23<\/sup> \u00d7 23<\/sup> \u00d7 33<\/sup> \u00d7 33<\/sup> \nEach prime factor appears in triplets, so 46656 is a perfect cube. \n <\/p>\nQuestion 2. \nFind the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. \n(i) 243 \n(ii) 256 \n(iii) 72 \n(iv) 675 \n(v) 100 \nSolution: \n(i) 243 \n243 = 33<\/sup> \u00d7 32<\/sup> \nOn grouping the factors in triplets, the prime factor 3 does not appear in a group of three. \nTo make it a perfect cube, we need one more 3. \nThe number 243 should be multiplied by 3 to get a perfect cube. \n <\/p>\n(ii) 256 \n256 = 23<\/sup> \u00d7 23<\/sup> \u00d7 22<\/sup> \nOn grouping the factors in triplets, the prime factor 2 does not appears in a group of three. \nTo make it a perfect cube, we need one more 2. \nThe number 256 should be multiplied by 2 to get a perfect cube. \n <\/p>\n <\/p>\n
(iii) 72 \n72 = 23<\/sup> \u00d7 32<\/sup> \nOn grouping, the factors in triplets the prime factor 3 does not appear in a group of 3. \nTo make it a perfect cube, we multiply it by 3. \nThe number 72 should be multiplied by 3 to get a perfect cube. \n <\/p>\n(iv) 675 \n675 = 33<\/sup> \u00d7 52<\/sup> \nOn grouping the factors in triplets, the prime factor 5 does not appear in a group of three. \nTo make it a perfect cube, we multiply it by 5. \nThe number 675 should be multiplied by 5 to get a perfect cube. \n <\/p>\n(v) 100 \n100 = 22<\/sup> \u00d7 52<\/sup> \nOn grouping the factors in triplets, the prime factors 2 and 5 do not appear in a group of three. \nTo make it a perfect cube we need one 2 and one 5 (2 \u00d7 5 = 10) \nThe number 100 should be multiplied by 10 to get a perfect cube. \n <\/p>\nQuestion 3. \nFind the smallest number by which each of the following numbers must be divided to obtain a perfect cube. \n(i) 81 \n(ii) 128 \n(iii) 135 \n(iv) 192 \n(v) 704 \nSolution: \n(i) 81 \n81 = 33<\/sup> \u00d7 3 \nOn grouping the factors in triplets, we find that three is no triplet of 3. \nTo make it a perfect cube, we divide it by 3. \n <\/p>\n(ii) 128 \n128 = 23<\/sup> \u00d7 23<\/sup> \u00d7 2 \nOn grouping the factors in triplets, we find that there is no triplet of 2. \nTo make it a perfect cube, we divide it by 2. \n <\/p>\n <\/p>\n
(iii) 135 \n135 = 33<\/sup> \u00d7 5 \nOn grouping the factors in triplets, the prime factor 5 does not appear in a group of three. \nTo make it a perfect cube, 135 should be divided by 5. \n <\/p>\n(iv) 192 \n192 = 23<\/sup> \u00d7 23<\/sup> \u00d7 3 \nOn grouping the factors in triplets, the prime factor 3 does not appear in a group of three. \nTo make it a perfect cube, 192 should be divided by 3. \n <\/p>\n(v) 704 \n704 = 23<\/sup> \u00d7 23<\/sup> \u00d7 11 \nOn grouping the factors in triplets, the prime factor 11 does not appear in a group of three. \nTo make it a perfect cube, 704 should be divided by 11. \n <\/p>\n <\/p>\n
Question 4. \nParikshit makes a cuboid of plasticine on sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube? \nSolution: \nSides of the cuboid are 5 cm, 2 cm, and 5 cm \nVolume of the cuboid = 5 cm \u00d7 2 cm \u00d7 5 cm \nTo form it as a cube, its dimension should be in the group of triplets. \nSince there is only one 2 and only two 5 in the prime factorization. \nWe need 2 \u00d7 2 \u00d7 5 = 20 such cuboids to make a cube.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Exercise 7.1 Question 1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n