Exercise 7.2
Q 1 (i, ii, iii, iv, v, vi) - Ex 7.2 - Cube and Cube Roots - NCERT Maths Class 8th - Chapter 7
Q 1 (vii, viii, ix, x) - Ex 7.2 - Cube and Cube Roots - NCERT Maths Class 8th - Chapter 7
Question 1
Find the cube root of each of the following numbers by prime factorization method.
(i) 64
Sol :
(i) Prime factorization of 64=2×2×2×2×2×2
∴
(ii) 512
Sol :
(ii) Prime factorization of 512=2×2×2×2×2×2×2×2×2
∴
(iii) 10648
Sol :
(iii) Prime factorization of
∴
(iv) 27000
Sol :
(iv) Prime factorization of
∴
(v) 15625
Sol :
(v) Prime factorization of
∴
(vi) 13824
Sol :
(vi) Prime factorization of
∴
(vii) 110592
Sol :
(vii) Prime factorization of
∴
(viii) 46656
Sol :
(viii) Prime factorization of
∴
(ix) 175616
Sol :
(ix) Prime factorization of
∴
(x) 91125
Sol :
(x)Prime factorization of
∴
Q 2 - Ex 7.2 - Cube and Cube Roots - NCERT Maths Class 8th - Chapter 7
Question 2
State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zero's.
(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.
Sol :
For finding the cube of any number, the number is first multiplied with itself and this product is again multiplied with this number.
(i) False. When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will be again an odd number.
For example, the cube of 3 (i.e., an odd number) is 27, which is again an odd number.
(ii) True. Perfect cube will end with a certain number of zero's that are always a perfect multiple of 3.
Foe example, the cube of 10 is 1000 and there are 3 zero's at the end of it.
The cube of 100 is 1000000 and there are 6 zero's at the end of it.
(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.
For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.
(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8.
The cube of 12 is 1728 and the cube of 22 is 10648.
(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.
(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.
(vii)True, as the cube of 1 and 2 are 1 and 8 respectively.
Q 3 - Ex 7.2 - Cube and Cube Roots - NCERT Maths Class 8th - Chapter 7
Question 3
You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768
Sol :
Firstly, we will make groups of three digits starting from the rightmost digit of the number as
.There are 2 groups, 1 and 331, in it.
Considering 331,
The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.
Taking the other group i.e., 1,
The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.
Hence,
The cube root of 4913 has to be calculated.
We will make groups of three digits starting from the rightmost digit of 4913, as
. The groups are 4 and 913.Considering the group 913,
The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.
Taking the other group i.e., 4,
We know that, 13 = 1 and 23 = 8
Also, 1 < 4 < 8
Therefore, 1 will be taken at the tens place of the required cube root.
Thus,
The cube root of 12167 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 12167, as
. The groups are 12 and 167.Considering the group 167,
167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.
Taking the other group i.e., 12,
We know that, 23 = 8 and 33 = 27
Also, 8 < 12 < 27
2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.
Thus,
The cube root of 32768 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 32768, as
.Considering the group 768,
768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.
Taking the other group i.e., 32,
We know that, 33 = 27 and 43 = 64
Also, 27 < 32 < 64
3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.
Thus,
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