Here you can read Class 8th MATHEMATICS "CUBE AND CUBE ROOTS" chapter 7 Solution Based on NCERT and CBSE book. After "SQUARE AND SQUARE ROOTS" chapter 7 EXERCISE 7.1 you can get links to Class 8th and other classes SCIENCE, MATHEMATICS and OTHER NCERT book Notes, Solutions, Practice Papers, etc.
Scroll down for NCERT Book Classes MATHEMATICS Book solution & important study material.
CUBE AND CUBE ROOTS (Chapter-7) SOLUTION
EXERCISE 7.1
1. Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100
(v) 46656
Solution - In the prime factorisation of any number, if each factor appears three times, then, the number is a perfect cube.
(i) 216By prime factorisation, 216 = 2 × 2 × 2 × 3 × 3 × 3
Each factor appears 3 times. 216 = 2³ × 3³ = (2 × 3)³ = 6³
6³ = 216 is a perfect cube.
(ii) 128
By prime factorisation, 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Each factor not appears 3 times.
So, 100 is Not a perfect cube.
(iii) 1000By prime factorisation, 1000 = 2 × 2 × 2 × 5 × 5 × 5
Each factor appears 3 times. 1000 = 2³ × 5³ = (2 × 5)³ = 10³
10³ = 1000 is a perfect cube.
(iv) 100
By prime factorisation, 100 = 2 × 2 × 5 × 5
Each factor not appears 3 times.
So, 100 is Not a perfect cube.
(v) 46656By prime factorisation, 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Each factor appears 3 times. 46656 = 2³ × 2³ × 3³ × 3³ = (2 × 2³ × 3³ × 3)³ = 36³
36³ = 46656 is a perfect cube.
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
Solution-
(i) 243
By prime factorisation, 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 does not appear in a group of three.
Therefore,
243 is not a perfect cube.
To make its a cube, we need one more 3.
In that case 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 which is a perfect cube.
Hence the smallest natural number by which 243 should be multiplied to make a perfect cube is 3.
(ii) 256
By prime factorisation, 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
The prime factor 2 does not appear in a group of three.
Therefore,
256 is not a perfect cube.
To make its a cube, we need one more 2.
In that case 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 which is a perfect cube.
Hence the smallest natural number by which 256 should be multiplied to make a perfect cube is 2.
(iii) 72
By prime factorisation, 72 = 2 × 2 × 2 × 3 × 3
The prime factor 3 does not appear in a group of three.
Therefore,
72 is not a perfect cube.
To make its a cube, we need one more 3.
In that case 72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 which is a perfect cube.
Hence the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.
(iv) 675
By prime factorisation, 675 = 3 × 3 × 3 × 5 × 5
The prime factor 5 does not appear in a group of three.
Therefore,
675 is not a perfect cube.
To make its a cube, we need one more 5.
In that case 675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 2744 which is a perfect cube.
Hence the smallest natural number by which 675 should be multiplied to make a perfect cube is 5.
(v) 100
By prime factorisation, 100 = 2 × 2 × 5 × 5
The prime factor 2 and 5 does not appear in a group of three.
Therefore,
100 is not a perfect cube.
To make its a cube, we need one more 2 and 5 i.e. 10.
In that case 100 × 10 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 which is a perfect cube.
Hence the smallest natural number by which 100 should be multiplied to make a perfect cube is 2 and 5.
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
Solution-
(i) 81
By prime factorisation, 81 = 3 × 3 × 3 × 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.
If we divide the number by 3, then the prime factorisation of the quotient will not contain 3.
So, 81 ÷ 3 = 3 × 3 × 3
Hence the smallest number by which 81 should be divided to make it a perfect cube is 3.
The perfect cube in that case is = 27.
(ii) 128
By prime factorisation, 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
The prime factor 2 does not appear in a group of three.
So, 128 is not a perfect cube.
In the factorisation 2 appears only one time.
If we divide the number by 2, then the prime factorisation of the quotient will not contain 2.
So, 128 ÷ 2 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 128 should be divided to make it a perfect cube is 2.
The perfect cube in that case is = 64.
(iii) 135
By prime factorisation, 135 = 3 × 3 × 3 × 5
The prime factor 5 does not appear in a group of three.
So, 135 is not a perfect cube.
In the factorisation 5 appears only one time.
If we divide the number by 5, then the prime factorisation of the quotient will not contain 5.
So, 135 ÷ 5 = 3 × 3 × 3
Hence the smallest number by which 135 should be divided to make it a perfect cube is 5.
The perfect cube in that case is = 27.
(iv) 192
By prime factorisation, 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
The prime factor 3 does not appear in a group of three.
So, 192 is not a perfect cube.
In the factorisation 3 appears only one time.
If we divide the number by 3, then the prime factorisation of the quotient will not contain 3.
So, 192 ÷ 3 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 192 should be divided to make it a perfect cube is 3.
The perfect cube in that case is = 64.
(v) 704
By prime factorisation, 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
The prime factor 11 does not appear in a group of three.
So, 704 is not a perfect cube.
In the factorisation 11 appears only one time.
If we divide the number by 11, then the prime factorisation of the quotient will not contain 11.
So, 704 ÷ 11 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 704 should be divided to make it a perfect cube is 11.
The perfect cube in that case is = 64.
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?
Solution- According to the question,
The sides of the Cuboid = 5 cm, 2 cm and 5 cm. (given)
Volume of the Cuboid = 5 cm × 2 cm × 5 cm = 50 cm³
For the prime factorisation of 50,
we have
50 = 2 × 5 × 5
To make it a Perfect Cube,
we must have, 2 × 2 × 2 × 5 × 5 × 5
= 20 × (2 × 5 × 5)
= 20 × volume of the given Cuboid
Thus, the required number of Cuboids = 20.
THANKS FOR READING
please follow and comment about the content which you Read above.
To read more chapter of 8th Class and other classes NCERT book based Notes and solution, Please click below link as per Your Choice.
From this students study the NCERT SOLUTION to get knowledge of the type of questions asked from the chapter, "CUBE AND CUBE ROOTS". This notes and solution of class 8th is very helpful to understand the Science subject and chapter 7 in a Better manner. We provided the chapter wise link below of class 8th MATHEMATICS book. These NCERT SOLUTIOS are Based on the latest CBSE and RBSE syllabus. All the study material was prepared to help you understand the topic easy and better way. If you like our resources, please share the post. NCERT SOLUTION for Class 8th MATHEMATICS Chapter 7 "CUBE AND CUBE ROOTS" is an outstanding study Materials which is needed for the students of CBSE and RBSE of Class 8th. The NCERT Solutions for Class 8th Science Chapter 7 has good weightage. Thorough knowledge and good practice will help you score full marks on the questions asked from this chapter.
Learn more about "CUBE AND CUBE ROOTS" by exploring CBSE/RBSE based NCERT Notes for Class 8th MATHEMATICS Chapter 7. These CBSE/RBSE based NCERT notes are comprehensive and detailed yet concise enough to glance through for exam preparations.
NCERT Solutions For Class 8th MATHEMATICS Chapter 6 SQUARE AND SQUARE ROOTS
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution
You can Read the class 7th NCERT SCIENCE notes based on CBSE and RBSE pattern.
NUTRITION IN PLANTS Class 7th SCIENCE chapter 1 notes
NUTRITION IN ANIMALS Class 7th SCIENCE chapter 2 notes
FIBRE TO FABRIC Class 7th SCIENCE chapter 3 notes
HEAT Class 7th SCIENCE chapter 4 notes
ACIDS, BASES AND SALTS Class 7th SCIENCE chapter 5 notes
PHYSICAL AND CHEMICAL CHANGES Class 7th SCIENCE chapter 6 notes
WEATHER, CLIMATE AND ADAPTATIONS OF ANIMALS TO CLIMATE Class 7th SCIENCE chapter 7 notes
Class 8th Study Material Solution based on CBSE and RBSE.
coming soon
For Class 8th Science subject you can get Revision Notes, Important Solutions.
coming soon
Here students can get classes & chapters wise Class 8th and other classes NCERT books notes based on CBSE and RBSE pattern,
coming soon