Class 8 maths ch 7 ex 7.1

(ii) 128 128 = 2 3 × 2 3 × 2 2 remains after grouping in triplets, so 128 is not a perfect cube. (iii) 1000 1000 = 2 3 × 5 3 Each prime factor appears in triplets so 1000 is a perfect cube. (iv) 100 100 = 2 2 × 5 2 As we do not get any triplets, So 100 is not a perfect cube. (v) 46656 46656 = 2 3 × 2 3 × 3 3 × 3 3 Each prime factor appears in triplets, so 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 Solution: (i) 243 243 = 3 3 × 3 2 On grouping the factors in triplets, the prime factor 3 does not appear in a group of three. To make it a perfect cube, we need one more 3. The number 243 should be multiplied by 3 to get a perfect cube. (ii) 256 256 = 2 3 × 2 3 × 2 2 On grouping the factors in triplets, the prime factor 2 does not appears in a group of three. To make it a perfect cube, we need one more 2. The number 256 should be multiplied by 2 to get a perfect cube. (iii) 72 72 = 2 3 × 3 2 On grouping, the factors in triplets the prime factor 3 does not appear in a group of 3. To make it a perfect cube, we multiply it by 3. The number 72 should be multiplied by 3 to get a perfect cube. (iv) 675 675 = 3 3 × 5 2 On grouping the factors in triplets, the prime factor 5 does not appear in a group of three. To make it a perfect cube, we multiply it by 5. The number 675 should be multiplied by 5 to get a perfect cube. (v) 100 100 = 2 2 × 5 ...

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 are part of • Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 7 Chapter Name Cubes and Cube Roots Exercise Ex 7.1 Number of Questions Solved 4 Category NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 Ex 7.1 Class 8 Maths Question 1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Solution: (i) Resolving 216 into prime factors, we find that 2 16 = 2x2x2 x 3x3x3 Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. ∴ 216 is a perfect cube. (ii) Resolving 128 into prime factors, we find that 128 = 2x2x2 x 2x2x2 x 2 Now, if we try to group together triples of equal factors, we are left with a single factor, 2. ∴ 128 is not a perfect cube. (iii) Resolving 1000 into prime factors, we find that 1000 = 10 x 10 x 10 = 2x5x2x5x2x5 = 2x2x2x 5x5x5 Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. ∴ 1000 is a perfect cube. (iv) Resolving 100 into prime factors, we find that 100 = 10×10=2x5x2x5=2x2x5x5 Clearly, the prime factors of 100 cannot be grouped into triples of equal factors. ∴ 100 is not a perfect cube. (v) Resolving 46656 into prime factors, we find that 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x3 x 3x3x3 Clearly, the prime factors of 46656 can be grouped into prime factors and no factor is left over. ∴...

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.1 Question 1. Express the following percentages as fractions: (i) 356% (ii) \(2 \frac\) of the students failed both in English and Maths, 35% of students failed in Maths and 30% failed in English. (i) Find the percentage of students who failed in any of the subjects. (ii) Find the percentage of students who passed in both subjects. (iii) If the number of students who failed only in English was 25, find the total number of students. Solution: Question 19. On increasing the price of an article by 16%, it becomes ₹1479. What was its original price? Solution: Question 20. Pratibha reduced her weight by 15%. If now she weighs 59.5 kg, what was her earlier weight? Solution: Question 21. In a sale, a shop reduces all its prices by 15%. Calculate: (i) the cost of an article which was originally priced at ₹40. (ii) the original price of an article which was sold for ₹20.40. Solution: Question 22. Increase the price of ₹200 by 10% and then decrease the new price by 10%. Is the final price same as the original one ? Solution: Question 23. Chandani purchased some parrots. 20% flew away and 5% died. Of the remaining, 45% were sold. Now 33 parrots remain. How many parrots had Chandani purchased? Solution: Question 24. A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks, gets 18 more marks than the minimum pass marks. Find the maximum marks and the percentage of...

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots • • • NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.1 Ex 7.1 Class 8 Maths Question 1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Solution: (i) Prime factorisation of 216 is: 216 = 2 × 2 × 2 × 3 × 3 × 3 In the above factorisation, 2 and 3 have formed a group of three. Thus, 216 is a perfect cube. (ii) Prime factorisation of 128 is: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Here, 2 is left without making a group of three. Thus 128 is not a perfect cube. (iii) Prime factorisation of 1000, is: 1000 = 2 × 2 × 2 × 5 × 5 × 5 Here, no number is left for making a group of three. Thus, 1000 is a perfect cube. (iv) Prime factorisation of 100, is: 100 = 2 × 2 × 5 × 5 Here 2 and 5 have not formed a group of three. Thus, 100 is not a perfect cube. (v) Prime factorisation of 46656 is: 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 Here 2 and 3 have formed the groups of three. Thus, 46656 is a perfect cube. Ex 7.1 Class 8 Maths 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 Solution: (i) Prime factorisation of 243, is: 243 = 3 × 3 × 3 × 3 × 3 = 3 3 × 3 × 3 Here, number 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3 Thus, the required smallest number to be multiplied is 3. (ii) Prime factorisation of 256, is: 25...

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2 • • • NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2 Ex 7.2 Class 8 Maths Question 1. Find the cube root of each of the following numbers by prime factorisation method. (i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125 Solution: Ex 7.2 Class 8 Maths Question 2. State True or False. (i) Cube of an odd number is even. (ii) A perfect cube does not end with two zeros. (iii) If the 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. Solution: (i) False – Cube of any odd number is always odd, e.g., (7) 3 = 343 (ii) True – A perfect cube does not end with two zeros. (iii) True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5) 2 = 25 and (5) 3 = 625 (iv) False – (12) 3 = 1728 (ends with 8) (v) False – (10) 3 = 1000 (4-digit number) (vi) False – (99) 3 = 970299 (6-digit number) (vii) True – (2) 3 = 8 (1-digit number) Ex 7.2 Class 8 Maths Question 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. Solution: The given perfect cube = 1331 F...

Rajasthan Board RBSE Class 8 Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 Question 1. Which of the following numbers are not perfect cubes? (i) 216 Answer: 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 3 × 3 3 = (6) 3 Which is a perfect cube. (ii) 128 Answer: We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor. ∴ 128 is not a perfect cube. (iii) 1000 Answer: 1000 = 2 × 2 × 2 × 5 × 5 × 5 = (2) 3 × (5) 3 = (10) 3 Which is a perfect cube. (iv) 100 Answer: 100 = 2 × 2 × 5 × 5 Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and and 5 × 5 are not in triples. ∴ 100 is not a perfect cube. (v) 46656 Answer: 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 Clearly the prime factors of 46656 can be grouped into prime factors and no factor is left over. Question 2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 Answer: Writing 243 as a product of prime factors, we have 243 = 3 × 3 × 3 × 3 × 3 Clearly, to make it a perfect cube, it must be multiplied by 3. (ii) 256 Answer: Resolving 256 into prime factors, we find that 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Clearly, to make it a perfect cube, it must be multiplied by 2. (iii) 72 Answer: Resolving 72 as a product of prime factors, we have 72 = 2 × 2 × 2 × 3 × 3 Clearly, to make it a perfect cube, it must be multiplied by 3. (iv) 675 Answer: Resolving 675 as a ...

• Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 7 Chapter Name Cubes and Cube Roots Exercise Ex 7.1 Number of Questions Solved 4 Category NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 Question 1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Solution. (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 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 Solution. (i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100 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 Solution. (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704 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? Solution. Volume of a cuboid = 5 x 2 x 5 \(\). Since there is only one 2 and only two 5’s in the prime factorization, so, we need 2 x 2 x 5, i.e., 20 to make a perfect cube. Therefore, we need 20 such cuboids to make a cube. We hope the NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1, drop a comment below and we will get back to you at the earliest. Filed Under: Tagged With: Primary Sidebar