Cube root of 729÷216 cube root × cube root of 6÷cube root of 9
Answers
Answer:
cube root of 729/216 cube root *cube
Step-by-step explanation:
Q1. Which of the following numbers are not perfect cubes?
(i) 216(ii) 128(iii) 1000 (iv) 100(v) 46656
Sol. (i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3
Grouping the prime factors of 216 into triples, no factor is left over.
∴ 216 is a perfect cube.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.
∴ 128 is not a perfect cube.
(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5
Grouping the prime factors of 1000 into triples, we are not left over with any factor.
∴ 1000 is a perfect cube.
(iv) We have 100 = 2 × 2 × 5 × 5
Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and 5 × 5 are not in triples.
∴ 100 is not a perfect cube.
(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Grouping the prime factors of 46656 in triples we are not left over with any prime factor.
∴ 46656 is a perfect cube.
Q2. 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
Sol. (i) We have 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 is not a group of three.
∴ 243 is not a perfect cube.
Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3
or 729 =3 × 3 × 3 × 3 × 3 × 3
Now, 729 becomes a perfect cube.
Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.
(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 256 in triples, we are left over with 2 × 2.
∴ 256 is not a perfect cube.
Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2
or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
i.e. 512 is a perfect cube.
Thus, the required smallest number is 2.
(iii) We have 72 = 2 × 2 × 2 × 3 × 3
Grouping the prime factors of 72 in triples, we are left over with 3 × 3.
∴ 72 is not a perfect cube.
Now, [72] × 3 =[2 × 2 × 2 × 3 × 3] × 3
or 216 = 2 × 2 × 2 × 3 × 3 × 3
i.e. 216 is a perfect cube.
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.
(iv) We have 675 = 3 × 3 × 3 × 5 × 5
Grouping the prime factors of 675 to triples, we are left over with 5 × 5.
∴ 675 is not a perfect cube.
Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5
or 3375 = 3 × 3 × 3 × 5 × 5 × 5
Now, 3375 is a perfect cube.
Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.
(v) We have 100 = 2 × 2 × 5 × 5
The prime factor are not in the groups of triples.
∴100 is not a perfect cube.
Now [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5
or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now, 1000 is a perfect cube.
Thus, the required smallest number is 10.
Q3. 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
Sol. (i) We have 81 = 3 × 3 × 3 × 3
Grouping the prime factors of 81 into triples, we are left with 3.
∴ 81 is not a perfect cube.
Now, [81] 3 = [3 × 3 × 3 × 3] + 3
or 27 = 3 × 3 × 3
i.e. 27 is a prefect cube
Thus, the required smallest number is 3.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 128 into triples, we are left with 2.
∴ 128 is not a perfect cube
Now, [128] 2 = [2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
∴ The smallest required number is 2.
(iii) We have 135 = 3 × 3 × 3 × 5
Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [l35] 5 = [3 × 3 × 3 × 5] 5
or 27 = 3 × 3 × 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5.
(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Grouping the prime factors of 192 into triples, 3 is left over.
∴ 192 is not a perfect cube.
Now, [192] 3 =[2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 3.
(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Grouping the prime factors of 704 into triples, 11 is left over.
∴ [704] 11 =[2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 11.
Q4. Parikshit makes a cuboid of plasticine of sidec 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Sol: Sides of the cuboid arc: 5 cm, 2 cm, 5 cm
∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm
To form it as a cube its dimension should be in the group of triples.
∴ Volume of the required cube = [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm
=[5 × 5 × 2 cm3] = 20 cm3
Thus, the required number of cuboids = 20.
Step-by-step explanation:
(∛ 729 ÷ ∛216) x (∛6 ÷ ∛9)
= (9 ÷ 6) x ( 1.81 ÷ 2.08)
= 1.5 x 0.87
= 1.305
Hope it helps you!