{"id":299,"date":"2025-01-08T08:57:09","date_gmt":"2025-01-08T03:27:09","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=299"},"modified":"2025-01-09T10:12:06","modified_gmt":"2025-01-09T04:42:06","slug":"samacheer-kalvi-10th-maths-guide-chapter-2-ex-2-10","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-10th-maths-guide-chapter-2-ex-2-10\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.10"},"content":{"rendered":"

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.10 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.10<\/h2>\n

Multiple choice questions:<\/span><\/p>\n

Question 1.
\nEuclid\u2019s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….
\n(1) 1 < r < b
\n(2) 0 < r < b
\n(3) 0 <<\/span> r < 6
\n(4) 0 < r <<\/span> b
\nAns.
\n(3) 0 <<\/span> r < b<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nUsing Euclid\u2019s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….
\n(1) 0, 1, 8
\n(2) 1, 4, 8
\n(3) 0, 1, 3
\n(4) 1, 3, 5
\nAnswer:
\n(1) 0, 1, 8
\nHint: Let the +ve integer be 1, 2, 3, 4 …………
\n13<\/sup> = 1 when it is divided by 9 the remainder is 1.
\n23<\/sup> = 8 when it is divided by 9 the remainder is 8.
\n33<\/sup> = 27 when it is divided by 9 the remainder is 0.
\n43<\/sup> = 64 when it is divided by 9 the remainder is 1.
\n53<\/sup> = 125 when it is divided by 9 the remainder is 8.
\nThe remainder 0, 1, 8 is repeated.<\/p>\n

Question 3.
\nIf the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is
\n(1) 4
\n(2) 2
\n(3) 1
\n(4) 3
\nAnswer:
\n(2) 2
\nHint:
\nH.C.F. of 65 and 117
\n117 = 65 \u00d7 1 + 52
\n65 = 52 \u00d7 1 + 13
\n52 = 13 \u00d7 4 + 0
\n\u2234 13 is the H.C.F. of 65 and 117.
\n65m – 117 = 65 \u00d7 2 – 117
\n130 – 117 = 13
\n\u2234 m = 2<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nThe sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….
\n(1) 1
\n(2) 2
\n(3) 3
\n(4) 4
\nAnswer:
\n(3) 3
\nHint: 1729 = 7 \u00d7 13 \u00d7 19
\nSum of the exponents = 1 + 1 + 1
\n= 3<\/p>\n

Question 5.
\nThe least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
\n(1) 2025
\n(2) 5220
\n(3) 5025
\n(4) 2520
\nAnswer:
\n(4) 2520
\nHint:
\n\"Samacheer
\nL.C.M. = 23<\/sup> \u00d7 32<\/sup> \u00d7 5 \u00d7 7
\n= 8 \u00d7 9 \u00d7 5 \u00d7 7
\n= 2520<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n74k<\/sup> \u2261 ______ (mod 100)
\n(1) 1
\n(2) 2
\n(3) 3
\n(4) 4
\nAnswer:
\n(1) 1
\nHint:
\n74k<\/sup> \u2261______ (mod 100)
\ny4k<\/sup> \u2261 y4 \u00d7 1<\/sup> = 1 (mod 100)<\/p>\n

Question 7.
\nGiven F1<\/sub> = 1 , F2<\/sub> = 3 and Fn<\/sub> = Fn-1<\/sub> + Fn-2<\/sub> then F5<\/sub> is ………….
\n(1) 3
\n(2) 5
\n(3) 8
\n(4) 11
\nAnswer:
\n(4) 11
\nHint:
\nFn<\/sub> = Fn-1<\/sub> + Fn-2<\/sub>
\nF3<\/sub> = F2<\/sub> + F1<\/sub> = 3 + 1 = 4
\nF4<\/sub> = F3<\/sub> + F2<\/sub> = 4 + 3 = 7
\nF5<\/sub> = F4<\/sub> + F3<\/sub> = 7 + 4 = 11<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nThe first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P
\n(1) 4551
\n(2) 10091
\n(3) 7881
\n(4) 13531
\nAnswer:
\n(3) 7881
\nHint:
\nt1<\/sub> = 1
\nd = 4
\ntn<\/sub> = a + (n – 1)d
\n= 1 + 4n – 4
\n4n – 3 = 4551
\n4n = 4554
\nn = will be a fraction
\nIt is not possible.
\n4n – 3 = 10091
\n4n = 10091 + 3 = 10094
\nn = a fraction
\n4n – 3 = 7881
\n4n = 7881 + 3 = 7884
\nn = \\(\\frac{7884}{4}\\), n is a whole number.
\n4n – 3 = 13531
\n4n = 13531 – 3 = 13534
\nn is a fraction.
\n\u2234 7881 will be 1971st term of A.P.<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nIf 6 times of 6th<\/sup> term of an A.P is equal to 7 times the 7th<\/sup> term, then the 13th<\/sup> term of the A.P. is ………..
\n(1) 0
\n(2) 6
\n(3) 7
\n(4) 13
\nAnswer:
\n(1) 0
\nHint:
\n6 t6<\/sub> = 7 t7<\/sub>
\n6(a + 5d) = 7 (a + 6d) \u21d2 6a + 30d = 7a + 42d
\n30 d – 42 d = 7a – 6a \u21d2 -12d = a
\nt13<\/sub> = a + 12d (12d = -a)
\n= a – a = 0<\/p>\n

Question 10.
\nAn A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
\n(1) 16 m
\n(2) 62 m
\n(3) 31 m
\n(4) \\(\\frac { 31 }{ 2 } \\) m
\nAnswer:
\n(3) 31 m
\nHint:
\nt16<\/sub> = m
\nS31<\/sub> = \\(\\frac { 31 }{ 2 } \\) (2a + 30d)
\n= \\(\\frac { 31 }{ 2 } \\) (2(a + 15d))
\n(\u2235 t16<\/sub> = a + 15d)
\n= 31(t16<\/sub>) = 31m<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nIn an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
\n(1) 6
\n(2) 7
\n(3) 8
\n(4) 9
\nAnswer:
\n(3) 8
\nHere a = 1, d = 4, Sn<\/sub> = 120
\nSn<\/sub> = \\(\\frac { n }{ 2 } \\)[2a + (n – 1)d]
\n120 = \\(\\frac { n }{ 2 } \\) [2 + (n – 1)4] = \\(\\frac { n }{ 2 } \\) [2 + 4n – 4)]
\n= \\(\\frac { n }{ 2 } \\) [4n – 2)] = \\(\\frac { n }{ 2 } \\) \u00d7 2 (2n – 1)
\n120 = 2n2<\/sup> – n
\n\u2234 2n2<\/sup> – n – 120 = 0 \u21d2 2n2<\/sup> – 16n + 15n – 120 = 0
\n2n(n – 8) + 15 (n – 8) = 0 \u21d2 (n – 8) (2n + 15) = 0
\nn = 8 or n = \\(\\frac { -15 }{ 2 } \\) (omitted)
\n\u2234 n = 8<\/p>\n

Question 12.
\nA = 265<\/sup> and B = 264<\/sup> + 263<\/sup> + 262<\/sup> …. + 20<\/sup>\u00a0which of the following is true?
\n(1) B is 264<\/sup>\u00a0more than A
\n(2) A and B are equal
\n(3) B is larger than A by 1
\n(4) A is larger than B by 1
\nAnswer:
\n(4) A is larger than B by
\nA = 265<\/sup>
\nB = 264+63<\/sup> + 262<\/sup> + …….. + 20<\/sup>
\n= 2
\n= 1 + 22<\/sup> + 22<\/sup> + ……. + 264<\/sup>
\na = 1, r = 2, n = 65 it is in G.P.
\nS65<\/sub> = 1 (265<\/sup> – 1) = 265<\/sup>\u00a0– 1
\nA = 265<\/sup>\u00a0is larger than B<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nThe next term of the sequence \\(\\frac { 3 }{ 16 } \\),\\(\\frac { 1 }{ 8 } \\),\\(\\frac { 1 }{ 12 } \\),\\(\\frac { 1 }{ 18 } \\) is ………..
\n(1) \\(\\frac { 1 }{ 24 } \\)
\n(2) \\(\\frac { 1 }{ 27 } \\)
\n(3) \\(\\frac { 2 }{ 3 } \\)
\n(4) \\(\\frac { 1 }{ 81 } \\)
\nAnswer:
\n(2) \\(\\frac { 1 }{ 27 } \\)
\nHint:
\n\\(\\frac { 3 }{ 16 } \\),\\(\\frac { 1 }{ 8 } \\),\\(\\frac { 1 }{ 12 } \\),\\(\\frac { 1 }{ 18 } \\)
\na = \\(\\frac { 3 }{ 16 } \\), r = \\(\\frac { 1 }{ 8 } \\) \u00f7 \\(\\frac { 3 }{ 16 } \\) = \\(\\frac { 1 }{ 8 } \\) \u00d7 \\(\\frac { 16 }{ 3 } \\) = \\(\\frac { 2 }{ 3 } \\)
\nThe next term is = \\(\\frac { 1 }{ 18 } \\) \u00d7 \\(\\frac { 2 }{ 3 } \\) = \\(\\frac { 1 }{ 27 } \\)<\/p>\n

Question 14.
\nIf the sequence t1<\/sub>,t2<\/sub>,t3<\/sub> … are in A.P. then the sequence t6<\/sub>,t12<\/sub>,t18<\/sub> … is
\n(1) a Geometric Progression
\n(2) an Arithmetic Progression
\n(3) neither an Arithmetic Progression nor a Geometric Progression
\n(4) a constant sequence
\nAnswer:
\n(2) an Arithmetic Progression
\nHint:
\nIf t1<\/sub>, t2<\/sub>, t3<\/sub>, … is 1, 2, 3, …
\nIf t6<\/sub> = 6, t12<\/sub> = 12, t18<\/sub> = 18 then 6, 12, 18 … is an arithmetic progression<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nThe value of (13<\/sup> + 23<\/sup> + 33<\/sup> + ……. + 153<\/sup>) – (1 + 2 + 3 + …….. + 15) is …………….
\n(1) 14400
\n(2) 14200
\n(3) 14280
\n(4) 14520
\nAnswer:
\n(3) 14280
\nHint:
\n\"Samacheer
\n1202 – 120 = 120(120 – 1)
\n120 \u00d7 119 = 14280<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.10 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[2],"tags":[],"class_list":["post-299","post","type-post","status-publish","format-standard","hentry","category-class-10"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/299"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=299"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/299\/revisions"}],"predecessor-version":[{"id":39809,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/299\/revisions\/39809"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=299"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=299"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=299"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}