Exercise 6.1 (Revised) - Chapter 7 - Cube & Cube Roots - Ncert Solutions class 8 - Maths

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Chapter 6 - Cube & Cube Roots | NCERT Solutions for Class 8 Maths

Ex 6.1 Question 1.

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

Answer.

(i) 216

Prime factors of $216=2 \times 2 \times 2 \times 3 \times 3 \times 3$
Here all factors are in groups of 3's (in triplets) Therefore, 216 is a perfect cube number.
(ii) 128

Prime factors of $128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Here one factor 2 does not appear in a 3's group.

Therefore, 128 is not a perfect cube.
(iii) 1000

Prime factors of $1000=2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 5 \mathrm{X} 5 \mathrm{X} 5$
Here all factors appear in 3's group.
Therefore, 1000 is a perfect cube.
(iv) 100

Prime factors of $100=2 \times 2 \times 5 \times 5$
Here all factors do not appear in 3's group. Therefore, 100 is not a perfect cube.
(v) 46656

Prime factors of $46656=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$
Here all factors appear in 3 's group.
Therefore, 46656 is a perfect cube.
Ex 6.1 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

Answer.

(i) 243

Prime factors of $243=3 \times 3 \times 3 \times 3 \times 3$
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 $256=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
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 $72=2 \times 2 \times 2 \times 3 \times 3$
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 $675=3 \times 3 \times 3 \times 5 \times 5$
Here factor 5 does not appear in 3 's group.
Therefore 675 must be multiplied by 5 to make it a perfect cube.
(v) 100

Prime factors of $100=2 \times 2 \times 5 \times 5$
Here factor 2 and 5 both do not appear in 3's group.
Therefore 100 must be multiplied by $2 \times 5=10$ to make it a perfect cube.
Ex 6.1 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

Answer.
(i) 81

Prime factors of $81=3 \times 3 \times 3 \times 3$
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 $128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
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 $135=3 \times 3 \times 3 \times 5$
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 $192=2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 3$
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 $704=2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 2 \mathrm{X} 11$
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.
Ex 6.1 Question 4.

Parikshit makes a cuboid of plasticine of sides $5 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$. How many such cuboids will he need to form a cube?

Answer.

Given numbers $=5 \times 2 \times 5$
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by $2 \times 5 \times 2=20$ to make it a perfect cube. Hence he needs 20 cuboids.