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Class 10 Class 12

Cubes and Cube Roots

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?

Sides of the cubiod are 5 cm, 2 cm, 5 cm

∴ Volume of the cubiod = 5 cm x 2 cm x 5 cm
To form is as a cube its dimension should be in the group of triples.

∴ Volume of the required cube = [5 cm x 5 cm x 2 cm] x 5 cm x 2 cm x 2 cm
= [ 5 x 5 x 2 cm³] x 20 cm³
Thus, the required number of cubiods = 20.

765 Views

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

(i) We have 243 = 3 x 3 x 3 x 3 x 3

The prime factor 3 is not a group of three.
∴ 243 is not a perfect cube.
Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3
Now, 729 becomes a [perfect cube
Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 256 in triples, we are left over with 2 x 2.
∴ 256 is not a perfect cube.
Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2
or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
i.e. 512 is a perfect cube.
thus, the required smallest number is 2.

(iii) we have 72 = 2 x 2 x 2 x 3 x 3
(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 72 in triples, we are left over with 3 x 3
∴ 72 is not a perfect cube.
Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3
or, 216 = 2 x 2 x 2 x 3 x 3 x 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 x 3 x 3 x 5 x 5
Grouping the prime factors of 675 to triples, we are left over with 5 x 5
(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴ 675 is not a perfect cube.
Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5
Now, 3375 is a perfect cube
Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 x 2 x 5 x 5
The prime factor are not in the groups of triples.

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴ 100 is not a perfect cube.
Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5
or, [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, 1000 is a perfect cube
Thus, the required smallest number is 10

1158 Views

Find the cube root of each of the following numbers by prime factorisation method.

(i) By prime factorisation, we have

$64 space equals space 2 space cross times space 2 space cross times space 2 space space space space space space space cross times space space space 2 space cross times space 2 space cross times space 2 space space space space space space space space space space$
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∴ $cube root of 64 space equals$ 2 x 2

= 4
Thyus, cube root of 64 is 4.

132 Views

State true or false: for any integer m, m² < m³. Why?

It is not always true.
For example, let m = -1
We have m²= (-1)²= 1
and m³= (-1)³= -1
∴ The above statement, i.e. m² < m³ is false.

297 Views

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

(i) We have 81 = 3 x 3 x 3 x 3

Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is not a perfect cube
Now, [81] $divided by$ 3= [3 x 3 x 3 x 3] $divided by$ 3
or 27 = 3 x 3 x 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 128 into triples, we are left with 2
∴ 128 is not a perfect cube
Now, [128] $divided by$ 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2] $divided by$ 2
or 64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
∴ the smallest required number is 2.

(iii) we have 135 = 3 x 3 x 3 x 5

Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [135] $divided by$ 5 = [ 3 x 3 x 3 x 5] $divided by$ 5
or 27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Grouping the prime factors of 192 into triples, 3 is left over.
∴ 192 is not a perfect cube.
Now, [192] $divided by$ 3= [2 x 2 x 2 x 2 x 2 x 2 x 3] $divided by$ 3
or 64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 3.
(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Grouping the prime factors of 704 into triples, 11 is left over
∴ [704] $divided by$ 11 = [2 x 2 x 2 x 2 x 2 x 2 x 11] $divided by$ 11
or 64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
Thus, the required smallest number is 11.