{"id":8135,"date":"2024-01-30T10:01:30","date_gmt":"2024-01-30T04:31:30","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=8135"},"modified":"2024-02-03T09:30:02","modified_gmt":"2024-02-03T04:00:02","slug":"ts-10th-class-maths-solutions-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-1-ex-1-1\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1"},"content":{"rendered":"

Students can practice TS 10th Class Maths Solutions<\/a> Chapter 1 Real Numbers Ex 1.1 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.1<\/h2>\n

Question 1.
\nUse Euclid’s algorithm to find the HCF of<\/p>\n

(i) 900 and 270
\nAnswer:
\n900 and 270
\nEuclid Division Lemma
\na = bq + r, q > 0 and 0 \u2264 r < b
\nWhen 900 is divided 270, the remainder is 90 to get 900 = 270 \u00d7 3 + 90
\nNow consider the division of 270 with the remainder 90 in the above and division algorithm to get 270 = 90 \u00d7 3 + 0 The remainder is zero, when we cannot proceed further. We conclude that the HCF of (900, 270) = 90<\/p>\n

(ii) 196 and 38220
\nAnswer:
\n196 and 38220
\nWhen 38220 is divided 196, the remainder is 0 to get 38220 = 196 \u00d7 195 + 0
\nThe remainder is zero, when we cannot proceed further. We conclude that the HCF of (38220, 196) = 196.<\/p>\n

\"TS<\/p>\n

(iii) 1651 and 2032
\nAnswer:
\n1651 and 2032
\nWhen 2032 is divided by 1651, the remainder is 381 to get 2032 = 1651 \u00d7 1 + 381
\nNow consider the division of 1651 with 381 in the above and division algorithm to get 1651 = 381 \u00d7 4 + 127
\nNow consider the division of 381 with the remainder 127 in the above and division algorithm to get 381 = 127 \u00d7 3 + 0
\nThe remainder is zero, when we cannot proceed further we conclude that the HCF (1651, 2032) = 127<\/p>\n

Question 2.
\nUse division algorithm to show that any positive odd integer is ofthe form 6q + 1, or 6 q + 3 or 6 q + 5, where q is some integer.
\nAnswer:
\nLet ‘a’ be any positive odd integer, we apply the division algorithm with a and b = 6. Since 0 \u2264 r \u2264 5, the possible remainders are 0, 1, 2, 3, 4 and 5.
\ni. e., a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where ‘q’ is quotient, However, since ‘a’ is odd, a cannot be 6q, 6q+2, 6q+4 (since they both are divisible by 2)
\n\u2234 any odd integer in the form of 6q + 1, (or) 6q + 3 (or) 6q + 5.<\/p>\n

Question 3.
\nUse division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1.
\nAnswer:
\nConsider ‘a’ be the square of an integer
\nApplying Euclid’s division lemma with ‘a’ and ‘b’ = 3
\n0 \u2264 r < 3 the possible remainders are 0, 1, 2 a = 3p or 3p + 1 (or) 3p + 2
\nAny square number is of the form 3p, 3p + 1, or 3p + 2, Where p is the quotient.<\/p>\n

Question 4.
\nUse division algorithm to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.
\nAnswer:
\nLet ‘a’ be the cube of a positive integer
\nApplying Euclid’s division lemma for ‘a’ and b = 9 a = bm + r, Where’m’ is quotient r is the remainder where 0 \u2264 r < 9 ‘a’ can be of the form
\n9m, 9m + 1, 9m + 2 ………… or 9m + 8
\nThe cube of any positive integer is of the form 9m, 9m + 1, 9m + 2, ….. or 9m + 8.<\/p>\n

\"TS<\/p>\n

Question 5.
\nShow that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
\nAnswer:
\nGiven numbers : n; n + 2 and n + 4<\/p>\n

Case (i) : ‘n’ is divisible by 3, then n is of the form 3k. Now, n + 2 = 3k + 2 – leaves a remainder 2 when divided by 3. n + 4 = 3k + 4 = (3k + 3) + 1 = 3(k + 1) + 1 leaves a remainder 1 when divided by 3.<\/p>\n

Case (ii) : n + 2 is divisible by 3. then n + 2 is of the form 3k.
\nNow n = 3k – 2 leaves a remainder 1 when divided by 3 and n + 4 = n + 2 + 2
\n= 3k + 2 leaves a remainder 2 when divided by 3.<\/p>\n

Case (iii) : n + 4 is divisible by 3.
\nTake n + 4 = 3k
\nNow : n = 3k – 4 = 3(k – 1) + 3 – 4
\n= 3(k – 1) – 1 leaves a remainder 2 when divided by 3.
\nAlso n + 2 = 3k-2 = 3(k- 1) + 3-2
\n= 3(k – 1) + 1 leaves a remainder 1 when divided by 3.
\nIn all the above three cases only one out of n, n + 2 and n + 4 is divisible by 3.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.1 Question 1. Use Euclid’s algorithm to find the HCF of (i) 900 and 270 Answer: 900 and 270 Euclid Division Lemma a … 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\/8135"}],"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=8135"}],"version-history":[{"count":9,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8135\/revisions"}],"predecessor-version":[{"id":8212,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8135\/revisions\/8212"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=8135"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=8135"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=8135"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}