{"id":25004,"date":"2021-06-21T18:34:10","date_gmt":"2021-06-21T13:04:10","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25004"},"modified":"2021-06-21T18:40:33","modified_gmt":"2021-06-21T13:10:33","slug":"ncert-solutions-for-class-6-maths-chapter-3-ex-3-7","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-3-ex-3-7\/","title":{"rendered":"NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7"},"content":{"rendered":"

These\u00a0NCERT Solutions for Class 6 Maths<\/a> Chapter 3 Playing With Numbers Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.7<\/h2>\n

Question 1.
\nRenu purchases two bags 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.
\nAnswer:
\nFor finding maximum weight, we have to find H.C.F. of 75 and 69.
\nFactors of 75 = 3 \u00d7 5 \u00d7 5
\nFactors of 69 = 3 \u00d7 23
\nH.C.F. = 3
\nTherefore, 3 kg is the maximum value of weight which can measure the weight of the fertiliser exact number of times.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThree 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?
\nAnswer:
\nFor finding minimum distance, we have to find L.C.M of 63, 70 and 77.
\n\"NCERT<\/p>\n

L.C.M. of 63, 70 and 77
\n= 7 \u00d7 9 \u00d7 10 \u00d7 11 = 6930 cm.
\nTherefore, the minimum distance is 6930 cm.<\/p>\n

Question 3.
\nThe 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.
\nAnswer:
\nThe measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm. Factors of 825 = 3 \u00d7 5 \u00d7 5 \u00d7 11 Factors of 675 = 3 \u00d7 5 \u00d7 5 \u00d7 3 \u00d7 3
\nFactors of 450 = 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5
\nH.C.F. = 3 \u00d7 5 \u00d7 5 = 75 cm
\nTherefore, the longest tape is 75 cm.<\/p>\n

Question 4.
\nDetermine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
\nAnswer:
\n\"NCERT
\nL.C.M. of 6, 8 and 12
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 = 24
\nThe smallest 3-digit number =100
\nTo find the number, we have to divide 100 by 24
\n100 = 24 \u00d7 4 + 4
\nTherefore, the required number = 100 + (24 – 4) = 120.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nDetermine the greatest 3-digit number which is exactly divisible by 8,10 and 12.
\nAnswer:
\n\"NCERT
\nL.C.M. of 8,10, 12 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 5 = 120
\nThe largest three digit number = 999
\nTo find the number, we have to divide 999 by 120
\n999 = 120 \u00d7 8 + 39
\nTherefore, the required number = 999 – 39 = 960<\/p>\n

Question 6.
\nThe 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?
\nAnswer:
\n\"NCERT
\nL.C.M. of 48, 72, 108
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 = 432 seconds
\nAfter 432 seconds, the lights change simultaneously.
\n432 seconds = 7 minutes 12 seconds
\nTherefore, the time = 7 a.m. + 7 minutes 12 seconds = 7:07:12 a.m.<\/p>\n

Question 7.
\nThree tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.
\nAnswer:
\nThe maximum capacity of container = H.C.F. (403, 434, 465)
\nFactors of 403 = 13 \u00d7 31
\nFactors of 434 = 2 \u00d7 7 \u00d7 31
\nFactors of 465 = 3 \u00d7 5 \u00d7 31
\nH.C.F. = 31
\nTherefore, 31 litres of container is required to measure the quantity<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nFind the least number which when divided by 6, 15 and 18, leave remainder 5 in each case.
\nAnswer:
\n\"NCERT
\nL.C.M. of 6, 15 and 18
\n= 2 \u00d7 3 \u00d7 3 \u00d7 5 = 90
\nTherefore, the required number.
\n= 90 + 5 = 95.<\/p>\n

Question 9.
\nFind the smallest 4-digit number which is divisible by 18, 24 and 32.
\nAnswer:
\n\"NCERT
\nL.C.M. of 18, 24 and 32
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 = 288
\nThe smallest four-digit number = 1000
\nTo find the number, we have to divide 1000 by 288
\n1000 = 288 \u00d7 3 + 136
\nTherefore, the required number is 1000+ (288 – 136) = 1152.<\/p>\n

Question 10.
\nFind the L.C.M. of the following numbers:
\n(a) 9 and 4
\n(b) 12 and 5
\n(c) 6 and 5
\n(d) 15 and 4
\nObserve a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?
\nAnswer:
\n(a) L.C.M. of 9 and 4
\n= 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36
\n\"NCERT<\/p>\n

(b) L.C.M. of 12 and 5 = 2 \u00d7 2 \u00d7 3 \u00d7 5 = 60
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(c) L.C.M. of 6 and 5 = 2 \u00d7 3 \u00d7 5 = 30 6,5
\n\"NCERT<\/p>\n

(d) L.C.M. of 15 and 4 = 2 \u00d7 2 \u00d7 3 \u00d7 5 = 60
\n\"NCERT<\/p>\n

Yes, L.C.M. is equal to the product of rbers in each case. And L.C.M. is multiple of 3.<\/p>\n

Question 11.
\nFind the L.C.M. of the following numbers in which one number is the factor of other:
\n(a) 5,20
\n(b) 6,18
\n(c) 12,48
\n(d) 9,45
\nWhat do you observe in the result obtained?
\nAnswer:
\n(a) L.C.M. of 5 and 20 = 2 \u00d7 2 \u00d7 5 = 20
\n\"NCERT<\/p>\n

(b) L.C.M. of 6 and 18 = 2 \u00d7 3 \u00d7 3 = 18
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(c) L.C.M. of 12 and 48 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 = 48
\n\"NCERT<\/p>\n

(d) L.C.M. of 9 and 45 = 3 \u00d7 3 \u00d7 5 = 45
\n\"NCERT<\/p>\n

From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of these two numbers is equal to that of larger number.<\/p>\n","protected":false},"excerpt":{"rendered":"

These\u00a0NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.7 Question 1. Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum …<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"yoast_head":"\nNCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-3-ex-3-7\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These\u00a0NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts. 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