Exercise 7.1
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 = 2x2x2x3x3x3
Here all factors are in groups of 3’s [in triplets) Therefore, 216 is a perfect cube number.
(ii)
Prime factors of 128 =2x2x2x2x2x2x2
Here one factor 2 does not appear in a 3's group.
Therefore, 128 is not a perfect cube.
(iii)
Prime factors of 1000 =2x2x2x3x3x3
Here all factors appear in 3's group.
Therefore, 1000 is a perfect cube.
(iv)
Prime factors of 100 =2x2x5x5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v)
Prime factors of 46656 = 2x2x2 x2x2x2x3x3x3x3x3x3
Here all factors appear in 3's group.
Therefore, 46656 is a perfect cube.
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 2:
(i)
Prime factors of 243 =3x3x3x3x3
Here 3 does not appear in 3's group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii)
Prime factors of 256 = 2 x 2 x 2 x 2x2x 2x2x2
Here one factor 2 is required to make a 3's group.
Therefore, 256 must be multipl ied by 2 to make it a perfect cube.
(iii)
Prime factors of 72 = 2x2x2x3x3
Here 3 does not appear in 3's group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
(iv)
Prime factors of 675 =3x3x3x5x5
Here factor 5 does not appear in 3's group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
(v)
Prime factors of l00 = 2x2x5x5
Here factor 2 and 5 both do not appear in 3's group.
Therefore 100 must be multiplied by 2 x 5 = 10 to make it a perfect cube
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 3:
(i)
Prime factors of 81 = 3x3x3x3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
(ii)
Prime factors of 128 =2x2x2x2x2x2x2
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)
Prime factors of 135 = 3x3x3x5
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
(iv)
Prime factors of 192 =2x2x2x2x2x2x3
Here one factor 3 does not appear in a triplet
Therefore, 192 must be divided by 3 to make it a perfect cube.
(v)
Prime factors of 704 = 2x2x2x2x2x2x11
Here one factor 11 does not appear in a triplet
Therefore, 704 must be divided by 11 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 4:
Given numbers = 5 x 2 x 5
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 x 2 x 5 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.
Exercise 7.2
Question 1: Find the cube root of each of the following numbers by prime factorization method:
(i) 64
(ii) 512
(iii) 10648
(iv] 27000
(v) 15625
[vi] 13824
(vii) 110592
[viii] 46656
(ix) 175616
(x) 91125
Answer 1:
Question 2:
State true or false:
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeroes.
(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.
Answer :
(i) False
Since, 1^{3 }'=1,3^{2 }= 27,5^{3} = 125,. ......... are all odd.
(ii) True
Since, a perfect cube ends with three zeroes.
e.g. 10^{3} = 1000,20^{3} =8000,30^{3} = 27000,...... soon
(iii) False
Since, 5^{2} = 25,5^{3} = 125, 15^{2} = 225,15^{3} = 3375 [Did not end with 2 5)
(iv) False
Since 12^{3 }=1728 [Ends with 8]
And 22^{3 }=10648 [Ends with 8]
(v) False
Since 10^{3 }=1000 [Four digit number]
And 11^{3} =1331 [Four digit number]
[vi] False
Since 99^{3} = 970299 [Six digit number]
(vii) True
1^{3} =1 [Single digit number]
2^{3} =8 [Single digit number]
Question 3:
You are told that 1,331 is a perfect cube. Can you guess with factorization what is its cube root? Similarly guess the cube roots of 4913, 12167, 32768.
Answer :
We know that 10^{3} = 1000 and Possible cube of 11^{3} = 1331
Since, cube of unit's digit I^{3} = 1
Therefore, cube root of 1331 is 11,
4913
We know that 7^{3} = 343
Next number comes with 7 as unit place 17^{3} =4913
Hence, cube root of 4913 is 17.
12167
We know that 3^{3} = 27
Here in cube, ones digit is 7
Now next number with 3 as ones digit 13^{3} = 2197
And next number with 3 as ones digit 23^{3} =12167 Hence cube root of 12167 is 23.
32768
We know that 2^{3} = 8
Here in cube, ones digit is 8
Now next number with 2 as ones digit I2^{3} = 1728
And next number with 2 as ones digit 22^{3} = 10648
And next number with 2 as ones digit 32^{3} = 32768
Hence cube root of 32768 is 32.
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