Class 6 maths Exercise 3.7 solutions

NCERT mathematics class 6 chapter 3 Playing with Numbers exercise 3.7 solutions are given. You should study the textbook lesson Playing with Numbers very well. You should also observe and practice all example problems and solutions given in the textbook. You can observe the solutions given below and try them in your own method. NCERT maths class 6 solutions Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 Exercise 3.6 Exercise 3.7 CA foundation maths solutions 

NCERT class 6 maths Chapter 3 exercise 3.7 solutions Playing with Numbers

6th CBSE maths solutions  Chapter 3 Playing with Numbers Exercise 3.7

NCERT class 6 maths chapter 3 Playing with Numbers exercise 3.7 solutions

Std 6 maths textbook solutions  Problem 1 1. Renu purchases two of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times. Solution: The maximum weight of the fertilizer = The HCF of weights of 75 kgs and 69 kgs Prime factorization of 75 = 3 × 5 ×5 Prime factorization of 69 = 3 × 23 Common factors of 75 and 69 = 3 Therefore, HCF of 75 and 69 = 3 So, the maximum weight of required fertilizer is 3 kgs. Problem 2 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? Solution: The required minimum distance = the L C M of 63 cm, 70 cm and 77cm. 2|63, 79, 77 3|63, 35, 77 3|21, 15, 77 5|7, 5, 77 7|7, 1, 77 11|1, 1, 11 |1, 1, 1 The L C M of 63, 70, 77 = 2 × 3 × 3 × 5 × 7 × 11 = 6930 Therefore, the required minimum distance = 6930 cm. Problem 3 3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly. Solution: The measurement of longest tape = the HCF of 825 cm, 675 cm and 450 cm. The prime factorization of 825 = 3 × 5 × 5 × 11 The prime factorization of 675 = 3 × 3 × 3 × 5 × 5 The prime factorization of 450 = 2 × 3 × 3 × 5 × 5 Therefore, the H C F of 825, 675 and 450 = 75 So the required longest tape = 75 cm. Problem 4 4. Determine the smallest 3 – digit number which is exactly divisible by 6, 8 and 12. Solution: The L C M of 6, 8 and 12 = 2| 6, 8, 12 2|3, 4, 6 2|3, 2, 3 3|3, 1, 3 1, 1, 1 Therefore, the L C M of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24 The smallest three-digit number = 100 We divide 100 by 24, then we get the number. 24 × 4 =96 100 ÷ 24 = 96 +4 24 × 5 = 120. It is the three-digit number. So, the required number = 100 – 4 + 24 = 120 Therefore 120 is smallest three-digit number which is exactly divisible by 6, 8, 12. Problem 5 5. Determine the greatest 3 – digit number exactly divisible by 8, 10 and 12. Solution: The L C M of 8, 10 and 12 = 2| 8, 10, 12 2|4, 5, 6 2|2, 5, 3 3|1, 5, 3 5|1, 5, 1 |1, 1, 1 Therefore, the L C M of 8, 10 and 12 = 2 × 2 × 2 × 3 × 5 = 120 The greatest three-digit number = 999 We divide 999 by 120, then we get the number. 120 × 8 = 960 999 ÷ 120 = 960 +39 So, the required number = 999 – 39 = 960 Therefore 960 is greatest three-digit number which is exactly divisible by 8 10, 12. Problem 6 6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively If they change simultaneously at 7 a.m., at what time will they change simultaneously again? Solution: The L C M of 48 seconds, 72 seconds and 108 seconds = 2|48, 72, 108 2|24, 36, 54 2|12, 18, 27 2|6, 9, 27 3|3, 9, 27 3|1, 3, 9 3|1, 1, 3 |1, 1, 1 The L C M = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds After 432 seconds the lights change simultaneously. 432 seconds = 7 minutes and 12 seconds (432 ÷ 60, 1 minute = 60 seconds) Therefore, the time = 7 a.m. + 7 minutes and 12 seconds = 7: 07: 12 a.m. Problem 7 7. Three tankers contain 403 litres, 465 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times. Solution: Maximum capacity of the required tanker = HCF of 403, 434 and 465 403 = 13 × 31 434 = 2 × 7 × 31 465 = 3 × 5 × 31 HCF = 31 Therefore, A container of capacity 31 litres can measure the diesel of 3 containers exact number of times. Problem 8 8. Find the least number which divided by 6, 15 and 18. Solution: The L C M of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90 Therefore, the required number is 5 more than 90. The required least number = 90 + 5 = 95 Problem 9 9. Find the smallest 4-digit number which is divisible by 18, 24 and 32. Solution: The L C M of 18, 24 and 32 Therefore, the L C M of 18, 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288 The smallest four-digit number = 1000 We divide 1000 by 288; we get the number. 288 × 3 = 864, 1000 – 864 = 136 288 × 4 = 1152, it is the 4 – digit number Therefore, the required number is 1000 = 1000 – 136 + 288 = 1152 Problem 10 10. Find the L C M of the following numbers: a. 9 and 4,       b. 12 and 5 c. 6 and 5.       d.15 and 4 Observe a common property in the obtained L C M. Is L C M the product of two numbers in each case? Solutions: a. 9 and 4 The L C M of 9 and 4 = 2|9, 4 2|9, 2 3|9, 1 3|3, 1 |1, 1 Therefore, the L C M of 9 and 4 = 2 × 2 × 3 × 3 = 36 b. 12 and 5 The L C M of 12 and 5 = 2|12, 5 2|6, 5 3|3, 5 5|1, 5 |1, 1 Therefore, the L C M of 12 and 5 = 2 × 2 × 3 × 5 = 60 c. 6 and 5 The L C M of 6 and 5 = 2|6, 5 3|3, 5 5|1, 5 |1, 1 Therefore, the L C M of 6 and 5 = 2 × 3 × 5 = 30 d.15 and 4 The L C M of 15 and 4 = 2 |15, 4 2|15, 2 3|15, 1 5|5, 1 |1, 1 Therefore, the L C M of 15 and 4 = 2 × 2 × 3 × 5 = 60 Here, in each case the L C M is multiples 2, 3 and 6 Yes, in each case the L C M = the product of two numbers. Problem 11 11. Find the LCM of the following numbers in which one number is the factor of the other. a. 5, 20,      b. 6, 18,       c. 12, 48,      d. 9, 45 What do you observe in the results obtained? Solutions: a. 5 and 20 The L C M of 5 and 12 = 2|5, 20 2|5, 10 5|5, 5 |1, 1 Therefore, the L C M of 5 and 12 = 2 × 2 × 5 = 20 b. 6 and 18 The L C M of 6 and 18 = 2|6, 18 3|3, 9 3|1, 3 |1, 1 Therefore, the L C M of 6 and 18 = 2 × 3 × 3 = 18 c. 12 and 48 The L C M of 12 and 48 = 2|12, 48 2|6, 24 2|3, 12 2|3, 6 3|3, 3 |1, 1 Therefore, the L C M of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48 d. 9 and 45 The L C M of 9 and 45 = 3|9, 45 3|3, 15 5|1, 5 |1, 1 Therefore, the L C M of 9 and 45 = 3 × 3 × 5 = 45 The L C M of the given numbers in each case is the larger of the two numbers. Note: Observe the solutions and try them in your own method. Inter maths 1A solutions SSC maths class 10 solutions NCERT maths class 7 solutions NIOS maths 311 book 2 solutions for some chapters