Download 7 cubes and cube roots

©
CBSEPracticalSkills.com
7
Edulabz International
CUBES AND CUBE ROOTS
Exercise 7.1
Q.1. Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Ans. (i)
216 = 2 × 2 × 2 × 3 × 3 × 3
= 23 × 33
= (2 × 3)3 = (6)3
Hence, 216 is a perfect cube.
(ii)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 2
∴
2 does not appear in a group of three.
Hence, 128 is not a perfect cube.
(iii)
1000 = 2 × 2 × 2 × 5 × 5 × 5
= 23 × 53
= (2 × 5)3
= (10)3
Hence, 1000 is a perfect cube.
(iv)
100 = 2 × 2 × 5 × 5
1
©
CBSEPracticalSkills.com
Edulabz International
= 22 × 52
∴ 2 and 5 do not appear in a groups of three.
Hence, 100 is not a perfect cube.
(v) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3
×3×3×3×3×3
= 23 × 23 × 33 × 33
= (2 × 2 × 3 × 3)3
= 36
Hence, 46656 is a perfect cube.
Q.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
Ans. (i)
243 = 3 × 3 × 3 × 3 × 3
= (3)3 × 32
The second prime factor 3 does not appear in a group of
three. Therefore, 243 is not a perfect cube. To make it a
perfect cube, we need one more 3.
Hence, the smallest number by which 243 should be
multiplied to make a perfect cube is 3.
2
©
CBSEPracticalSkills.com
(ii)
Edulabz International
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 22
The prime factor 2 does not appear
in a group of three. Therefore, 256 is not a
perfect cube. To make it a perfect cube,
we need one more 2. So, we will multiply
256 by 2 to make it a perfect cube.
(iii)
72 = 2 × 2 × 2 × 3 × 3
= 23 × 32
The prime factor 3 does not appear in a
group of three. Therefore, 72 is not a
perfect cube. To make it a perfect cube,
we need one more 3.
So, we will multiply 72 by 3, to make it a
perfect cube.
(iv) 675 = 3 × 3 × 3 × 5 × 5 = 33 × 52
The prime factor 5 does not appear in a
group of three. Therefore, 675 is not a
perfect cube. To make it a perfect cube,
we need one more 5.
So, we will multiply 675 by 5, to make it a perfect cube.
(v)
100 = 2 × 2 × 5 × 5
The prime factors 2 and 5 do not appear
in a group of three. Therefore, 100 is not
a perfect cube. To make it a cube, we need
one more 2 and 5 respectively.
So, we will multiply 100 by 10, to make a perfect cube.
Q.3. Find the smallest number by which each of the following
numbers must be divided to obtain a perfect cube.
3
©
CBSEPracticalSkills.com
(i) 81
Ans. (i)
(ii) 128
(iii) 135
Edulabz International
(iv) 192
(v) 704
81 = 3 × 3 × 3 × 3
= 33 × 3
The prime factor 3 does not appear in a group of three.
So, 81 is not a perfect cube. In the factorisation 3
appears only one time. So if we divide 81 by 3, then the
prime factorisation of the quotient will not contain 3.
81 ÷ 3 = 3 × 3 × 3
Hence the smallest whole number by which
81 should be divided to make it perfect cube is 3.
(ii)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 2
The prime factor 2 does not appear in a group of three.
So, 128 is not a perfect cube. In the factorization 2
appear only one time. So if we divide 128 by 2, then the
prime factorization of the quotient will not contain 2.
128 ÷ 2 = 2 × 2 × 2 × 2 × 2 × 2
Further the perfect cube in that case is
128 ÷ 2 = 64
Hence the smallest whole number by
which 128 should be divide to make
it a perfect cube is 2.
(iii)
135 = 3 × 3 × 3 × 5 = 33 × 5
The prime factor 5 does not appear
in a group of three. So, 135 is not a
perfect cube.
4
©
CBSEPracticalSkills.com
Edulabz International
In the factorisation 5 appears only one time. So if we
divide 135 by 5, then the prime factorisation of the
quotient will not contain 5.
135 ÷ 5 = 3 × 3 × 3
Hence, the smallest whole number by which 135 should
be divided to make it perfect cube is 5.
(iv)
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 23 × 23 × 3
The prime factor 3 does not appear in a group
of three. So, 192 is not a perfect cube. In the
factorization 3 appears only one time. So, if
we divide 192 by 3, then the factorization of the
quotient will not contain 3.
192 ÷ 3 = 2 × 2 × 2 × 2 × 2 × 2
Hence, the smallest whole numbers by which 192
should be divided to make it perfect cube is 3.
(v)
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
= 23 × 23 × 11
The prime factor 11 does not appear in a
group of three. So, 704 is not a perfect cube.
In the factorization 11 appears only one time.
So if we divide 704 by 11, then the
factorization of the quotient will not contain 11.
704 ÷ 11 = 2 × 2 × 2 × 2 × 2 × 2
Hence, the smallest whole number by which 704 should
be divided to make it perfect cube is 11.
5
©
CBSEPracticalSkills.com
Edulabz International
Q.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?
Ans.
Volume of cuboid = l × b × h
= 5 cm × 2 cm × 5 cm = 50 cm3
Now we need to find a smallest number which is divisible
by 50 as well as a perfect cube. Such number is 1000.
So, number of required cuboids =
1000
= 20
50
Exercise 7.2
Q.1. Find the cube root of each of the following numbers by
prime factorisation method.
(i) 64
(ii) 512
(iii) 10648 (iv) 27000
(v) 15625
(vi) 13824 (vii) 110592 (viii) 46656
(ix) 175616 (x) 91125
Ans. (i)
64 = 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23
∴
3
(ii)
64 = 2 × 2 = 4
512 = 2 × 2 × 2 × 2 × 2 × 2
×2×2×2
= 23 × 23 × 23
= (2 × 2 × 2)3
∴
3
512 = 2 × 2 × 2 = 8
6
©
CBSEPracticalSkills.com
(iii)
Edulabz International
10648 = 2 × 2 × 2 × 11 × 11 × 11
= 23 × 113
= (2 × 11)3
∴
3
(iv)
10648 = 2 × 11 = 22
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
= 23 × 33 × 53
= (2 × 3 × 5)3
∴
3
(v)
27000 = 2 × 3 × 5 = 30
15625 = 5 × 5 × 5 × 5 × 5 × 5
= 53 × 53
= (5 × 5)3
∴
3
15625 = 5 × 5 = 25
7
©
CBSEPracticalSkills.com
(vi)
Edulabz International
13824 = 2 × 2 × 2 × 2 × 2 × 2
×2×2×2×3×3×3
= 23 × 23 × 23 × 33
= (2 × 2 × 2 × 3)3
∴
(vii)
3
13824 = 2 × 2 × 2 × 3 = 24
110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
×2×3×3×3
= 23 × 23 × 23 × 23 × 33
= (2 × 2 × 2 × 2 × 3)3
= 2×2×2×2×3
∴ 3 110592 = 48
8
©
CBSEPracticalSkills.com
(viii)
Edulabz International
46656 = 2 × 2 × 2 × 2 × 2 × 2
×3×3×3×3×3×3
= 23 × 23 × 33 × 33
= (2 × 2 × 3 × 3)3
∴
3
(ix)
46656 = 2 × 2 × 3 × 3 = 36
175616 = 2 × 2 × 2 × 2 × 2 × 2
×2×2×2×7×7×7
= 23 × 23 × 23 × 73
= (2 × 2 × 2 × 7)3
∴
3
(x)
46656 = 2 × 2 × 2 × 7 = 56
91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
= 33 × 33 × 53
= (3 × 3 × 5)3
∴
3
91125 = 3 × 3 × 5 = 45
9
©
CBSEPracticalSkills.com
Edulabz International
Q.2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube
ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit
number.
(vi) The cube of a two digit number may have seven or
more digits.
(vii) The cube of a single digit number may be a single
digit number.
Ans. (i)
False
(ii)
True
(iii) False
(v)
False
(vi) False
(vii) True
(iv) False
Q.3. You are told that 1,331 is a perfect cube. Can you guess
without factorisation what is its cube root? Similarly,
guess the cube roots of 4913, 12167, 32768.
Ans. The given number is 1331
Step I
Form groups of three starting from the rightmost
digit of 1331.
1331. In this case one group i.e., 331 has three
digits whereas 1 has only one digit.
Step II
Take 331
The digit 1 is at its one’s place. We know that I
comes at the unit’s place of a number only when
its cube root ends in 1. So, we take the one’s place
of the required cube root as 1.
Step III Take the other group, i.e., 1.
13 = 1, and 23 = 8, so, 1 lies between 0 and 8.
10
©
CBSEPracticalSkills.com
Edulabz International
The smaller number among 1 and 2 is 1. The
one’s place of 1 is 1 itself. Take 1 as ten’s place of
the cube root of 1331.
Hence,
3
1331 = 11
Similarly, 4913
Step I
Form groups of three starting from the right most
digit of 4913
4913. In this case one group i.e., 913 has three
digits whereas 4 has only one digit.
Step II
Take 913
The digit 3 is at its one’s place we know that 3
comes at the unit place of a number when it’s
cube root ends in 7.
Step III 13 = 1 and 23 = 8, 4 lies between 1 and 8.
The smaller number among 1 and 2 is 1. The
one’s place of 1 is 1 itself. Take 1 as ten’s place of
the cube root of 4913.
Hence,
3
4913 = 17.
Similarly, 12167
Step I
Form groups of three starting from the right most
digit of 12167
12 167. In this case one group i.e., 167 has three
digits whereas 12 has only two digits.
Step II
Take 167
The digit 7 is at one’s place. We know that, 7
comes at the unit’s place only when it’s cube root
ends in 3 as, 33 = 27
So, 3 will be at the unit’s place of the cube root.
11
©
CBSEPracticalSkills.com
Edulabz International
Step III Take 12
The digits 2 is at units place, 23 = 8 and 33 = 27
So, 12 lies between 8 and 27
The smaller number among 2 and 3 is 2.
The one’s place of 2 is 2 itself. Take 2 as ten’s
place of the cube root of 12167.
Hence,
3
12167 = 23
Similarly, 32768
Step I
Form groups of three starting from the right most
digit of 32768
In 32768, one group is 768 and other is 32
Step II
768, the digit 8 is at units place.
Cube of 8 is 512.
So, one’s place is 2.
Step II
Take 32, 33 = 27 and 43 = 64
So, 33 < 32 < 43.
The smaller number among 3 and 4 is 4.
The one’s place of 3 is 3 itself. Then, we can take
2 at the tens place of the cube of root of 32768
Hence,
3
32768 = 32
12