{"id":8202,"date":"2024-02-01T10:09:24","date_gmt":"2024-02-01T04:39:24","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=8202"},"modified":"2024-02-06T17:35:27","modified_gmt":"2024-02-06T12:05:27","slug":"ts-10th-class-maths-solutions-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-1-ex-1-2\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2"},"content":{"rendered":"

Students can practice TS 10th Class Maths Solutions<\/a> Chapter 1 Real Numbers Ex 1.2 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.2<\/h2>\n

Question 1.
\nExpress each of the following numbers as a product of its prime factors.<\/p>\n

(i) 140
\nAnswer:
\n140
\n\"TS
\n\u2234 140 = 2 \u00d7 2 \u00d7 5 \u00d7 7 = 22<\/sup> \u00d7 5 \u00d7 7<\/p>\n

(ii) 156
\nAnswer:
\n156
\n\"TS
\n\u2234 156 = 2 \u00d7 2 \u00d7 3 \u00d7 13 = 22<\/sup> \u00d7 3 \u00d7 13<\/p>\n

\"TS<\/p>\n

(iii) 3825
\nAnswer:
\n3825
\n\"TS
\n\u2234 3825 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 17
\n= 32<\/sup> \u00d7 52<\/sup> \u00d7 17<\/p>\n

(iv) 5005
\nAnswer:
\n5005
\n\"TS
\n\u2234 5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13<\/p>\n

(v) 7429
\nAnswer:
\n7429
\n\"TS
\n\u2234 7429 = 17 \u00d7 19 \u00d7 23<\/p>\n

Question 2.
\nFind the LCM and HCF of the following integers by the prime factorization method.
\n(i) 12, 15 and 21
\nAnswer:
\n12, 15, 21
\n12 = 2 \u00d7 2 \u00d7 3, 15 = 3 \u00d7 5
\n21 = 3 \u00d7 7
\nH.C.F of 12, 15, 21 is 3.
\nL.C.M of 12, 15, 21 = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 7 = 420<\/p>\n

(ii) 17, 23, and 29
\nAnswer:
\n17, 23, 29
\n17 = 1 \u00d7 17, 23 = 1 \u00d7 23
\n29 = 1 \u00d7 29
\nH.C.F of 17, 23 and 29 is 1.
\nL.C.M of 17, 23, 29 is 17 \u00d7 23 \u00d7 29 = 11339<\/p>\n

(iii) 8, 9 and 25
\nAnswer:
\n8, 9, 25
\n8 = 1 \u00d7 2 \u00d7 2 \u00d7 2, 9 = 1 \u00d7 3 \u00d7 3
\n25 = 1 \u00d7 5 \u00d7 5
\nH.C.F. = 1
\nL.C.M = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5
\n= 1800<\/p>\n

\"TS<\/p>\n

(iv) 72 and 108
\nAnswer:
\n72, 108
\n\"TS
\n72 = 2 \u00d7 36
\n= 2 \u00d7 2 \u00d7 18
\n= 2 \u00d7 2 \u00d7 2 \u00d7 9
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3<\/p>\n

108 = 2 \u00d7 54
\n= 2 \u00d7 2 \u00d7 27
\n= 2 \u00d7 2 \u00d7 3 \u00d7 9
\n= 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3
\nH.C.F. = 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36
\nL.C.M. = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 = 216<\/p>\n

(v) 306 and 657
\nAnswer:
\n306, 657
\n\"TS
\nH.C.F. = 9
\nL.C.M. = 9 \u00d7 34 \u00d7 73
\n= 22338<\/p>\n

Question 3.
\nCheck whether 6″ can end with the digit 0 for any natural number n.
\nAnswer:
\nIf the number 6n<\/sup> for any n, were to end with digit ‘0’ then it would be divisible by 5.
\nThe prime factorisation of 6n would contain the prime 5.
\n6n<\/sup> = (2 \u00d7 3)n<\/sup>
\nPrime factorisation of 6n<\/sup> does not contain 5 as a factor, so 6n<\/sup> can never end with the digit 0 for any natural number.<\/p>\n

Question 4.
\nExplain why 7 \u00d7 11 \u00d7 13 + 13 and 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 are composite numbers.
\nAnswer:
\nGiven number is 7 \u00d7 11 \u00d7 13 + 13
\n= 13 (7 \u00d7 11 + 1)
\n= (7 \u00d7 11 + 1) \u00d7 13 distributive law
\n= (77 + 1) \u00d7 13
\n= 78 \u00d7 13
\n= (2 \u00d7 3 \u00d7 13) \u00d7 13
\n= 2 \u00d7 3 \u00d7 132<\/sup>
\n= Product of prime factors
\nHence the given number is a composite number.
\nGiven number is
\n7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5
\n= 5(7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 1)
\n= 5 (1008 + 1)
\n= 5 \u00d7 1009
\n= Product of prime numbers
\nHence the given number is a composite number.<\/p>\n

\"TS<\/p>\n

Question 5.
\nHow will you show that (17 \u00d7 11 \u00d7 2)+ (17 \u00d7 11 \u00d7 5) is a composite number? Explain.
\nAnswer:
\nGiven number is
\n(17 \u00d7 11 \u00d7 2) + (17 \u00d7 11 \u00d7 5)
\n= 17 \u00d7 11 \u00d7 (2 + 5)
\n= 17 \u00d7 11 \u00d7 7 = Product of primes
\nWe know that every composite number can be expressed as a product of primes.<\/p>\n

Question 6.
\nWhich digit would occupy the units place of 6100<\/sup>.
\nAnswer:
\n6100<\/sup> = (2 \u00d7 3)100<\/sup>
\nSo the last digit of 6100<\/sup> is 6.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.2 Question 1. Express each of the following numbers as a product of its prime factors. (i) 140 Answer: 140 \u2234 140 = … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8202"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=8202"}],"version-history":[{"count":6,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8202\/revisions"}],"predecessor-version":[{"id":8226,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8202\/revisions\/8226"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=8202"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=8202"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=8202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}