NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part-3 (For CBSE, ICSE, IAS, NET, NRA 2022)

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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

Table Shows the Prime Factorization of 81
381
327
39
33
1

Prime Factors of

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Thus, is a perfect cube.

Hence, 81 divided by 3 to make it a perfect cube.

(ii) 128

Table Shows the Prime Factorization of 128
2128
264
232
216
28
24
22
1

Prime Factors of

Here, one 2is left which is not in a triplet.

If we divide by 2, then it will become a perfect cube.

Thus, is a perfect cube.

Hence, 128 divided by 2 to make it a perfect cube.

(iii) 135

Table Shows the Prime Factorization of 135
3135
345
315
55
1

Prime Factors of

Here, one 2 is left which is not in a triplet.

If we divide 135 by 5, then it will become a perfect cube

Thus, is a perfect cube.

Hence, 135 divided by 5 to make it a perfect cube.

(iv) 192

Table Shows the Prime Factorization of 192
2192
296
248
224
212
26
33
1

Prime Factors of

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube

Thus, is a perfect cube.

Hence, 192 divided by 3 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

Since, factors of 5 and 2 both are not in group of three.

Therefore, the number must be multiplied by to make it a perfect cube. Hence he needs 20 cuboids.

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