{"id":5060,"date":"2021-03-03T15:10:52","date_gmt":"2021-03-03T09:40:52","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=5060"},"modified":"2021-09-15T11:40:34","modified_gmt":"2021-09-15T06:10:34","slug":"ap-ssc-10th-class-maths-solutions-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-1-ex-1-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2<\/h2>\n

10th Class Maths 1st Lesson Real Numbers Ex 1.2 Textbook Questions and Answers<\/h3>\n

Question 1.
\nExpress each of the following numbers as a product of its prime factors.
\ni) 140
\nii) 156
\niii) 3825
\niv) 5005
\nv) 7429
\nAnswer:
\ni) 140
\n\"AP
\n\u2234 140 = 2 \u00d7 2 \u00d7 5 \u00d7 7 = 22<\/sup> \u00d7 5 \u00d7 7<\/p>\n

\"AP<\/p>\n

ii) 156
\n\"AP
\n\u2234 156 = 2 \u00d7 2 \u00d7 3 \u00d7 13 = 22<\/sup> \u00d7 3 \u00d7 13<\/p>\n

iii) 3825
\n\"AP
\n\u2234 3825 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 17 = 32<\/sup> \u00d7 52<\/sup> \u00d7 17<\/p>\n

iv) 5005
\n\"AP
\n\u2234 5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13<\/p>\n

v) 7429
\n\"AP
\n\u2234 7429 = 17 \u00d7 19 \u00d7 23<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the L.C.M and H.C.F of the following integers by the prime factorization method.
\ni) 12, 15 and 21
\nii) 17, 23 and 29
\niii) 8, 9 and 25
\niv) 72 and 108
\nv) 306 and 657
\nAnswer:
\ni) 12, 15 and 21
\n12 = 2 \u00d7 2 \u00d7 3 = 22 \u00d7 3
\n15 = 3 \u00d7 5
\n21 = 3 \u00d7 7
\nL.C.M = 22<\/sup> \u00d7 3 \u00d7 5 \u00d7 7 = 420
\nH.C.F = 3<\/p>\n

ii) 17, 23 and 29
\nThe given numbers 17, 23 and 29 are all primes.
\nL.C.M = their product
\n= 17 \u00d7 23 \u00d7 29 = 11339
\n\u2234 H.C.F = 1<\/p>\n

iii) 8, 9 and 25
\n8 = 2 \u00d7 2 \u00d7 2 = 23<\/sup>
\n9 = 3 \u00d7 3 = 32<\/sup>
\n25 = 5 \u00d7 5 = 52<\/sup>
\nL.C.M = 23<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup> = 1800
\n(or)
\n8, 9 and 25 are relatively prime, therefore L.C.M is equal to their product,
\n(i.e.,) L.C.M = 8 \u00d7 9 \u00d7 25 = 1800
\nH.C.F = 1<\/p>\n

iv) 72 and 108
\n\"AP
\n72 = 23<\/sup> \u00d7 32<\/sup>
\n108 = 22<\/sup> \u00d7 33<\/sup>
\nL.C.M = 23<\/sup> \u00d7 33<\/sup> = 8 \u00d7 27 = 216
\nH.C.F = 22<\/sup> \u00d7 32<\/sup> = 4 \u00d7 9 = 36<\/p>\n

\"AP<\/p>\n

v) 306 and 657
\n\"AP
\n306 = 2 \u00d7 32<\/sup> \u00d7 17
\n657 = 32 \u00d7 73
\nL.C.M = 2 \u00d7 32<\/sup> \u00d7 17 \u00d7 73 = 22338
\nH.C.F = 32<\/sup> = 9<\/p>\n

Question 3.
\nCheck whether 6n<\/sup> can end with the digit \u20180\u2019 for any natural number n.
\nAnswer:
\nGiven number = 6n<\/sup> = (2 \u00d7 3)n<\/sup>
\nThe prime factors here are 2 and 3 only.
\nTo be end with 0; 6n<\/sup> should have a prime factor 5 and also 2.
\nSo, 6n<\/sup> can\u2019t end with zero.<\/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 numbers are 7 \u00d7 11 \u00d7 13
\n7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5
\n\u21d2 13(7 \u00d7 11 + 1) and
\n5(7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 1)
\n\u21d2 13 K and 5 L, where K = 78 and L = 7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 1 = 1009
\nAs the given numbers can be written as product of two numbers, they are composite.<\/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:
\n(17 \u00d7 11 \u00d7 2) + (17 \u00d7 11 \u00d7 5)
\n= (17 \u00d7 11) (2 + 5)
\n= (17 \u00d7 11) (7)
\n= 187 \u00d7 7
\nNow the given expression is written as a product of two integers and hence it is a composite number.<\/p>\n

\"AP<\/p>\n

Question 6.
\nWhat is the last digit of 6100?
\nAnswer: We know that
\n61<\/sup> = 6
\n62<\/sup> = 36
\n63<\/sup> = 216
\n64<\/sup> = 1296
\n65<\/sup> = 7776
\nWe see that 6n<\/sup> for any positive integer n ends is 6.
\ni.e., unit digit is always 6.
\n\u2234 Unit digit of 6100<\/sup> is 6.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2 10th Class Maths 1st Lesson Real Numbers Ex 1.2 Textbook Questions and Answers Question 1. Express each of the following numbers …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5060"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=5060"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5060\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=5060"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=5060"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=5060"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}