We know that 10³ = 1000 and Possible cube of 11³ = 1331 Since, cube of unit’s digit 1³ = 1 Therefore, cube root of 1331 is 11. 4913 We know that 7³ = 343 Next number comes with 7 as unit place 17³ = 4913 Hence, cube root of 4913 is 17. 12167 We know that 3³ = 27 Here in cube, ones digit is 7 Now nexRead more
We know that 10³ = 1000 and Possible cube of 11³ = 1331
Since, cube of unit’s digit 1³ = 1
Therefore, cube root of 1331 is 11.
4913
We know that 7³ = 343
Next number comes with 7 as unit place 17³ = 4913
Hence, cube root of 4913 is 17.
12167
We know that 3³ = 27
Here in cube, ones digit is 7
Now next number with 3 as ones digit 13³ = 2197
And next number with 3 as ones digit 23³ = 12167
Hence cube root of 12167 is 23.
32768
We know that 2³ = 8
Here in cube, ones digit is 8
Now next number with 2 as ones digit 12³ = 1728
And next number with 2 as ones digit 22³ = 10648
And next number with 2 as ones digit 32³ = 32768
Hence cube root of 32768 is 32.
Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video
(i) False Since, 1³ = 1,3³ =27,5³ = 125,............ are all odd. (ii) True Since, a perfect cube ends with three zeroes. e.g. 10³ =1000,20³ = 8000,30³ = 27000,....... so on (iii) False Since, 5² = 25,5³ = 125,15² 225,15³ 3375 (Did not end with 25) (iv) False Since 12³ = 1728 [Ends with 8] And 22³ =Read more
(i) False
Since, 1³ = 1,3³ =27,5³ = 125,………… are all odd.
(ii) True
Since, a perfect cube ends with three zeroes.
e.g. 10³ =1000,20³ = 8000,30³ = 27000,……. so on
(iii) False
Since, 5² = 25,5³ = 125,15² 225,15³ 3375 (Did not end with 25)
(iv) False
Since 12³ = 1728 [Ends with 8]
And 22³ = 10648 [Ends with 8]
(v) False
Since 10³ = 1000 [Four digit number]
And 11³ = 1331 [Four digit number]
(vi) False
Since 3 99³ = 970299 [Six digit number]
(vii) True
1³ = 1 [Single digit number]
2³ = 8 [Single digit number]
Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video
(i) 64 ∛64 = ∛2X2X2X2X2X2 ∛64 = 2x2 = 4 (ii) 512 ∛512 = ∛2X2X2X2X2X2X2X2x2 = 2x2x2 = 8 Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
(i) 216 Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3 Here all factors are in groups of 3’s (in triplets) Therefore, 216 is a perfect cube number. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 is not a perfect cube. (iii) 10Read more
(i) 216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
Therefore, 216 is a perfect cube number.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
(i) 216 Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3 Here all factors are in groups of 3’s (in triplets) Therefore, 216 is a perfect cube number. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 is not a perfect cube. (iii) 10Read more
(i) 216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
Therefore, 216 is a perfect cube number.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
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.
We know that 10³ = 1000 and Possible cube of 11³ = 1331 Since, cube of unit’s digit 1³ = 1 Therefore, cube root of 1331 is 11. 4913 We know that 7³ = 343 Next number comes with 7 as unit place 17³ = 4913 Hence, cube root of 4913 is 17. 12167 We know that 3³ = 27 Here in cube, ones digit is 7 Now nexRead more
We know that 10³ = 1000 and Possible cube of 11³ = 1331
Since, cube of unit’s digit 1³ = 1
Therefore, cube root of 1331 is 11.
4913
We know that 7³ = 343
Next number comes with 7 as unit place 17³ = 4913
Hence, cube root of 4913 is 17.
12167
We know that 3³ = 27
Here in cube, ones digit is 7
Now next number with 3 as ones digit 13³ = 2197
And next number with 3 as ones digit 23³ = 12167
Hence cube root of 12167 is 23.
32768
We know that 2³ = 8
Here in cube, ones digit is 8
Now next number with 2 as ones digit 12³ = 1728
And next number with 2 as ones digit 22³ = 10648
And next number with 2 as ones digit 32³ = 32768
Hence cube root of 32768 is 32.
Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
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.
(i) False Since, 1³ = 1,3³ =27,5³ = 125,............ are all odd. (ii) True Since, a perfect cube ends with three zeroes. e.g. 10³ =1000,20³ = 8000,30³ = 27000,....... so on (iii) False Since, 5² = 25,5³ = 125,15² 225,15³ 3375 (Did not end with 25) (iv) False Since 12³ = 1728 [Ends with 8] And 22³ =Read more
(i) False
Since, 1³ = 1,3³ =27,5³ = 125,………… are all odd.
(ii) True
Since, a perfect cube ends with three zeroes.
e.g. 10³ =1000,20³ = 8000,30³ = 27000,……. so on
(iii) False
Since, 5² = 25,5³ = 125,15² 225,15³ 3375 (Did not end with 25)
(iv) False
Since 12³ = 1728 [Ends with 8]
And 22³ = 10648 [Ends with 8]
(v) False
Since 10³ = 1000 [Four digit number]
And 11³ = 1331 [Four digit number]
(vi) False
Since 3 99³ = 970299 [Six digit number]
(vii) True
1³ = 1 [Single digit number]
2³ = 8 [Single digit number]
Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
Find the cube root of each of the following numbers by prime factorization method: (i) 64 (ii) 512
(i) 64 ∛64 = ∛2X2X2X2X2X2 ∛64 = 2x2 = 4 (ii) 512 ∛512 = ∛2X2X2X2X2X2X2X2x2 = 2x2x2 = 8 Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
(i) 64
∛64 = ∛2X2X2X2X2X2
∛64 = 2×2
= 4
(ii) 512
∛512 = ∛2X2X2X2X2X2X2X2x2
= 2x2x2
= 8
Class 8 Maths Chapter 7 Exercise 7.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
Which of the following numbers are not perfect cubes:(i) 216 (ii) 128(iii) 1000 (iv) 100 (v) 46656
(i) 216 Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3 Here all factors are in groups of 3’s (in triplets) Therefore, 216 is a perfect cube number. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 is not a perfect cube. (iii) 10Read more
(i) 216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
Therefore, 216 is a perfect cube number.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/
Which of the following numbers are not perfect cubes:(i) 216 (ii) 128(iii) 1000 (iv) 100 (v) 46656
(i) 216 Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3 Here all factors are in groups of 3’s (in triplets) Therefore, 216 is a perfect cube number. (ii) 128 Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 is not a perfect cube. (iii) 10Read more
(i) 216
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets)
Therefore, 216 is a perfect cube number.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
(iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Class 8 Maths Chapter 7 Exercise 7.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-7/