NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2 cubes

Q(i) Find the one’s digit of the cube of each of the following numbers.

3331

Answer:

The detailed solution for the above-mentioned question is as follows,

Since the given number ends with 1, so the one’s digit of the cube of 3331 will be 1.

Q(ii) Find the one’s digit of the cube of each of the following numbers.

8888

Answer:

The detailed solution for the above-mentioned question is as follows

Since the given number ends with 8, so the one’s digit of the cube of 8888 will be 2.

Q(iii) Find the one's digit of the cube of each of the following numbers.

149

Answer:

The detailed solution for the above-mentioned question is as follows,

Since the given number has 9 at units place, so the one’s digit of the cube of 149 will be 9.

Q(iv) Find the one’s digit of the cube of each of the following numbers.

1005

Answer:

The detailed solution for the above-mentioned questions is as follows

Since the given number ends with 5, so one's digit of its cube will also end with 5.

Q(v) Find the one’s digit of the cube of each of the following numbers.

1024

Answer:

The solution to the above-mentioned question is as follows,

The given digit is ending with 4. So the one’s digit of the cube of 1024 will be 4.

Q(vi) Find the one’s digit of the cube of each of the following numbers.

Answer:

The detailed solution for above-mentioned question is as follows,

The given number is ending with 7, so its cube will end with 3.

Q(vii) Find the one’s digit of the cube of each of the following numbers.

5022

Answer:

The detailed solution for the above-mentioned question is as follows,

Since the given number ends with 2, so its cube will end with 8.

Q(viii) Find the one’s digit of the cube of each of the following numbers.

Answer:

The detailed solution for the above-mentioned question is as follows,

Since the given number has 3 at units place, so, its cube will end with 7.

NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.1 some interesting patterns

Q(a) Express the following numbers as the sum of odd numbers using the above pattern?

$6^{3}$

Answer:

The detailed solution for the above mentioned question is as follows,

216 => 31 + 33 + 35 + 37 + 39 + 41

Q(b) Express the following numbers as the sum of odd numbers using the above pattern?

$8^{3}$

Answer:

The detailed solution for the above-mentioned question is as follows

512 => 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

Q(c) Express the following numbers as the sum of odd numbers using the above pattern?

$7^{3}$

Answer:

The detailed solution for the above-mentioned question is as follows

7 3 = 43 + 45 + 47 + 49 + 51 + 53 + 55

Q(i) The detailed solution of the above-written question is as follows,

Using the above pattern ,

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

find the value of the following.

$7^{3}-6^{3}$

Answer:

The value of the following question is:

$7^{3}-6^{3} =$ $7^{3}-6^{3} = 1 + 7\times6\times3 = 1 + 126 = 127$

Q(ii) The detailed solution for all the above-written question is as follows

Using the above pattern ,

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

find the value of the following.

$12^{3}-11^{3}$

Answer:

$12^{3}-11^{3} = 1 + 12\times11\times3 = 1 + 396 = 397$

Q(iii). Consider the following pattern.

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Using the above pattern , find the value of the following.

$20^{3}-19^{3}$

Answer:

The detailed solution for the above-written question is mentioned below,

$20^{3}-19^{3} = 1 + 20\times19\times3 = 1 + 1140 = 1141$

Q(iv) Consider the following pattern.

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Using the above pattern , find the value of the following.

$51^{3}-50^{3}$

Answer:

The detailed solution for the above written question is mentioned below

$51^{3}-50^{3} = 1 + 51\times50\times3 = 1 + 7650 = 7651$

NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.1 subtopic cubes and their prime factors

Q. Which of the following are perfect cubes?

1. 400

2. 3375

3. 8000

4. 15625

5. 9000

6. 6859

7. 2025

8. 10648

Answer:

We will find it by prime factorization whether they make a pair of three prime numbers or not.

(1) $400 = 2\times2\times2\times2\times5\times5$ . So not a perfect cube.

(2) $3375 = 3\times3\times3\times5\times5\times5$ . So it is a perfect cube.

(3) $8000 = 2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is a perfect cube.

(4) $15625 = 5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(5) $9000 = 2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.

(6) $6859 = 19\times19\times19$ . So it is a perfect cube.

(7) $2025 = 3\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(8) $10648 = 2\times2\times2\times11\times11\times11$ . So it is a perfect cube.

NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.2 smallest multiple that is a perfect cube

Q. Check which of the following are perfect cubes.

(i) 2700

(ii) 16000

(iii) 64000

(iv) 900

(v) 125000

(vi) 36000

(vii) 21600

(viii) 10,000

(ix) 27000000

(x) 1000.

What pattern do you observe in these perfect cubes?

Answer:

The detailed solution for the above-written question is as follows

By prime factorization:

(i) $2700 = 2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(ii) $16000 = 2\times2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is not a perfect cube.

(iii) $64000 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times5\times5\times5= 80\times80\times80$ . So it is a perfect cube.

(iv) $900 = 2\times2\times3\times3\times5\times5$ . So it is not a perfect cube.

(v) $125000 = 2\times2\times2\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(vi) $36000 = 2\times2\times2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.

(vii) $21600 = 2\times2\times2\times2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(viii) $10000 = 2\times2\times2\times2\times5\times5\times5\times5$ . So it is not a perfect cube.

(ix) $27000000 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(x) $1000 = 2\times2\times2\times5\times5\times5$ . So it is a perfect cube.

We observe that the numbers above which are perfect cube have the number of zeros in multiple of 3.

NCERT solutions for class 8 maths chapter 7 cubes and cube roots-Exercise: 7.1

Q.1(i) Which of the following numbers are not perfect cubes?

216

Answer:

The detailed solution for the above-written question is as follows

By prime factorization of 216 gives:

$216 = 2\times2\times2\times3\times3\times3$

Since prime numbers are present in pairs of three, so the given number is a perfect cube.

Q.1(ii) Which of the following numbers are not perfect cubes?

128

Answer:

We have 128. By prime factorization we get,

$128 = 2\times2\times2\times2\times2\times2\times2$

Since the prime numbers are not in pairs of three, so the given number is not a perfect cube.

Q.1(iii) Which of the following numbers are not perfect cubes?

1000

Answer:

The detailed solution for the above written is as follows

By prime factorization of 1000 we get :

$1000 = 2\times2\times2\times5\times5\times5$ .

So the given number is a perfect cube.

Q.1(iv) Which of the following numbers are not perfect cubes?

100

Answer:

The detailed solution for the above-written question is as follows

By prime factorization of 100 :

$100 = 2\times2\times5\times5$ .

Since prime numbers are not in pair of three so given number is not a perfect cube.

Q.1(v) Which of the following numbers are not perfect cubes?

46656

Answer:

We have 46656, by prime factorisation:

$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$ .

Since prime numbers are in group of three. So the given number 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

Answer:

This can be found by knowing about the prime factors of the number.

(i) 243 : $3\times3\times3\times3\times3$ .

So it must be multiplied by 3.

(ii) 256 : $2\times2\times2\times2\times2\times2\times2\times2$

So the given number must be multiplied by 2 to make it a perfect cube.

(iii) 72 : $2\times2\times2\times3\times3$

So 72 must be multiplied by 3 to make it a perfect cube.

(iv) 675 : $3\times3\times3\times5\times5$

So it should be multiplied by 5.

(v) 100 : $2\times2\times5\times5$

So it should be multiplied by 10.

Q.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:

By prime factorization of given numbers :

(i) 81 : $3\times3\times3\times3$

So given number needs to be divided by 3 to get a perfect cube.

(ii) 128 : $2\times2\times2\times2\times2\times2\times2$ .

So the given number needs to be divided by 2 to get a perfect cube.

(iii) 135 : $3\times3\times3\times5$

So the given number needs to be divided by 5 to get a perfect cube.

(iv) 192 : $2\times2\times2\times2\times2\times2\times3$

So the given number needs to be divided by 3 to get a perfect cube.

(v) 704 : $2\times2\times2\times2\times2\times2\times11$

So the given number needs to be divided by 11 to get a perfect cube.

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?

Answer:

Volume of cuboid is $5\times2\times5 = 2\times5\times5$

To make it a cube need to make this a pefect cube number.

So we need $2\times2\times5$ cuboids

or 20 cuboids.

Q. State true or false:

for any integer $m,m^{2}< m^{3}.$ Why?

Answer:

The detailed solution for the above-written question is as follows.

False.

or $m^2\left ( m-1 \right )>0$

Now put any number less than 1, we see that this relation doesn't hold.

So for m<1 this condition is not true.

NCERT solutions for class 8 maths chapter 7 cubes and cube roots-Exercise: 7.2

Q.1(i) Find the cube root of each of the following numbers by prime factorisation method.

Answer:

The detailed solution for the above-written question is as follows

Prime factorization of 64 gives :

$64 = 2\times2\times2\times2\times2\times2$

So its cube root is $2\times2$ = 4

Q.1(ii) Find the cube root of each of the following numbers by prime factorisation method.

512

Answer:

By prime factorisation of 512 :

$512 = 2\times2\times2\times2\times2\times2\times2\times2\times2$

So its cube root is $2\times2\times2 = 8$

Q.1(iii) Find the cube root of each of the following numbers by prime factorisation method.

10648

Answer:

The detailed solution for the above-written question is as follows

Prime factorization of 10648 gives :

$10648 = 2\times2\times2\times11\times11\times11$

So its cube root is 22.

Q.1(iv) Find the cube root of each of the following numbers by prime factorisation method.

27000

Answer:

The detailed solution for the above-written question is as follows

By prime factorization method, we get :

$27000 = 2\times2\times2\times3\times3\times3\times5\times5\times5$

So its cube root is 30.

Q.1(v) Find the cube root of each of the following numbers by prime factorisation method.

15625

Answer:

The detailed solution for the above-written question is as follows

By prime factorization:

$15625 = 5\times5\times5\times5\times5\times5$

So its cube root is 25.

Q.1(vi) Find the cube root of each of the following numbers by prime factorisation method.

13824

Answer:

The detailed solution for the above-written question is as follows

By prime factorization:

$13824 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$

So its cube root is 24.

Q.1(vii) Find the cube root of each of the following numbers by prime factorisation method.

110592

Answer:

The detailed solution for the above-written question is as follows

By prime factorization:

$110592 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$

So its cube root is $2\times2\times2\times2\times3$ = 48.

Q.1(viii) Find the cube root of each of the following numbers by prime factorisation method.

46656

Answer:

The detailed solution for the above-written question is as follows

By prime factorization, we get :

$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$

So its cube root is $2\times2\times3\times3$ = 36.

Q.1(ix) Find the cube root of each of the following numbers by prime factorisation method.

175616

Answer:

The detailed solution for the above-written question is as follows

By prime factorization we get :

$175616 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7$

So its cube root is $2\times2\times2\times7$ = 56.

Q.1(x) Find the cube root of each of the following numbers by prime factorisation method.

91125

Answer:

The detailed solution for the above-written question is as follows

By prime factorization, we get :

$91125 = 3\times3\times3\times3\times3\times3\times5\times5\times5$

So its cube root is $3\times3\times5$ = 45.

Q2. 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.

Answer:

(i) False. Cube of an odd number can never be even.

(ii) True. Perfect cube number ends with zeros multiple of three.

(iii) False. We can say only about units place.

(iv) False. Cube of numbers which ends with 2 end with 8.

(v) False. Can never be.

(vi) False. Can never be. It can be proved by taking examples.

(vii) True. e.g. 1,2

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.

Answer:

We have 1331.

Divide number in two parts: The first part is 1 and second is 331.

Since the given number is ending with 1 so the last digit of cube root will be 1.

In the first part, we have 1.

By estimation, the cube root of 1331 is 11.

Similarly for all other parts.

4913:- First part is 4 and the second part is 913.

The number is ending with 3 so its cube root will have 7 at units place.

In the first part, the nearest cube root is 1.

So the cube root of 4913 is 17.

12167:- First part is 12 and the second part is 167.

The number is ending with 7 so its cube root will have 3 at units place.

In the first part, the nearest cube root is 2.

So the cube root of 12167 is 23.

32768:- First part is 32 and the second part is 768.

The number is ending with 8, so its cube root will have 2 at units place.

In the first part, the nearest cube root is 3.

So the cube root of 32768 is 32.

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

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NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2 cubes

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Q(iii) Find the one's digit of the cube of each of the following numbers.

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NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.1 some interesting patterns

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Q(ii) The detailed solution for all the above-written question is as follows

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NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.1 subtopic cubes and their prime factors

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NCERT solutions for class 8 maths chapter 7 cubes and cube roots topic 7.2.2 smallest multiple that is a perfect cube

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NCERT solutions for class 8 maths chapter 7 cubes and cube roots-Exercise: 7.1

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NCERT solutions for class 8 maths chapter 7 cubes and cube roots-Exercise: 7.2

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Q.1(v) Find the cube root of each of the following numbers by prime factorisation method.

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NCERT Class 8 Mathematics Solutions

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