Class 6 maths chapter 3 exercise 3.7 solutions

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NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3 Students can learn the concept of “Tests for Divisibility of Numbers” in depth by practising the NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3. Students can thus get well-versed in the various divisibility tests on numbers. These Previous Next Access NCERT Solutions for Class 6 Chapter 3: Playing with Numbers Exercise 3.3 1. Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes or no): Solutions: 2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8: (a) 572 (b) 726352 (c) 5500 (d) 6000 (e) 12159 (f) 14560 (g) 21084 (h) 31795072 (i) 1700 (j) 2150 Solutions: (a) 572 72 are the last two digits. Since 72 is divisible by 4, 572 is also divisible by 4. 572 are the last three digits. Since 572 is not divisible by 8, 572 is not divisible by 8. (b) 726352 52 are the last two digits. Since 52 is divisible by 4, 726352 is divisible by 4. 352 are the last three digits. Since 352 is divisible by 8, 726352 is divisible by 8. (c) 5500 Since the last two digits are 00, 5500 is divisible by 4. 500 are the last three digits. Since 500 is not divisible by 8, 5500 is not divisible by 8. (d) 6000 Since the last two digits are 00, 6000 is divisible by 4. Since the last three digits are 000, 6000 is divisible by 8. (e) 12159 59 are the last two digits....

NCERT Solutions Class 6 Maths Chapter – 3 (Playing with Numbers) The NCERT Solutions in English Language for Class 6 Mathematics Chapter – 3 Playing with Numbers Exercise 3.7 has been provided here to help the students in solving the questions from this exercise. Chapter 3: Playing with Numbers • NCERT Solution Class 6 Maths Exercise – 3.1 • NCERT Solution Class 6 Maths Exercise – 3.2 • NCERT Solution Class 6 Maths Exercise – 3.3 • NCERT Solution Class 6 Maths Exercise – 3.4 • NCERT Solution Class 6 Maths Exercise – 3.5 • NCERT Solution Class 6 Maths Exercise – 3.6 Exercise – 3.7 1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times. Solutions: Given, weight of two bags of fertiliser = 75 kg and 69 kg Maximum weight = HCF of two bags weight i.e (75, 69) 75 = 3 × 5 × 5 69 = 3 × 23 HCF = 3 Hence, 3 kg is the maximum value of weight which can measure the weight of the fertiliser exact number of times. 2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps? Solutions: First boy steps measure = 63 cm Second boy steps measure = 70 cm Third boy steps measure = 77 cm LCM of 63, 70, 77 LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930 Hence, 6930 cm is the distance each should cover so that all can cover the distance in complete ...

NCERT Solutions For Class 6 Maths Playing With Numbers Exercise 3.6 • • • • • • • • NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 Exercise 3.6 Question 1. Find the HCF of the following numbers: (a) 18, 48 (b) 30, 42 (c) 18, 60 (d) 27,63 (e) 36,84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75 Solution: (a) Given numbers are 18 and 48. Prime factorisations of 18 and 48 are: Here, the common factors are 2 and 3. Hence, the HCF = 2 x 3 = 6. (b) The given numbers are 30 and 42. Prime factorisations of 30 and 42, are: Here, the common factors are 2 and 3. Hence, the HCF = 2 x 3 = 6. (c) Given numbers are 18 and 60. Prime factorisations of 18 and 60 are: Here, the common factors are 2 and 3. Hence, the HCF of 18 and 60 = 2 x 3 = 6. (d) Given numbers are 27 and 63. Prime factorisations of 27 and 63 are: Here, the common factor is 3 (occurring twice). Hence, the HCF = 3 x 3 = 9. (e) Given numbers are 36 and 84. Prime factorisations of 36 and 84 are: Here, the common factors are 2, 2 and 3. Hence, the HCF = 2 x 2 x 3 = 12. (f) Given numbers are 34 and 102. Prime factorisations of 34 and 102 are: Here, the common factors are 2 and 17. Thus, HCF is 2 x 17 = 34. (g) The given numbers are 70, 105 and 175. Prime factorisatios of 70, 105 and 175 are: Here, common factors are 5 and 7. Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35. (h) Given numbers are 91, 112 and 49. Prime factorisations of 91, 112 and 49 are: Here, the common facto...