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  1. Asked: In:

    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?

    Best Answer
    Varnu Singh
    This answer was edited.

    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 2 x 5 = 20 to make it a perfect cube. Hence he needs 20 cuboids. Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video for more answers vist to: https://www.tiwariacadRead more

    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 2 x 5 = 20 to make it a perfect cube.
    Hence he needs 20 cuboids.

    Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/

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  2. Asked: In:

    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

    Best Answer
    Varnu Singh
    This answer was edited.

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/

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  3. Asked: In:

    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

    Best Answer
    Varnu Singh
    This answer was edited.

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/

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  4. Asked: In:

    Which of the following numbers are not perfect cubes:(i) 216 (ii) 128(iii) 1000 (iv) 100 (v) 46656

    Varnu Singh
    This answer was edited.

    (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. Therefore, 128 is not a perfect cube. (iii) 10Read more

    (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.

    Therefore, 128 is not a perfect cube.

    (iii) 1000

    Prime factors of 1000 = 2 x 2 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, 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

    Here all factors appear in 3’s group.

    Therefore, 46656 is a perfect cube.

    Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video

    for more answers vist to:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/

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