{"id":25376,"date":"2020-03-07T17:20:49","date_gmt":"2020-03-07T11:50:49","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25376"},"modified":"2021-07-02T15:21:17","modified_gmt":"2021-07-02T09:51:17","slug":"kseeb-solutions-for-class-10-maths-chapter-8-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-8-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 8 Real Numbers Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 8 Real Numbers Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Additional Questions<\/h2>\n

I. Multiple Choice Questions:<\/span><\/p>\n

Question 1.
\nThe H.C.F of 2 and 3 by Euclid’s division lemma
\na. 2
\nb. 3
\nc. 5
\nd. 1
\nAnswer:
\nd. 1<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe H.C.F of 10 and 5 is
\na. 5
\nb. 10
\nc. 15
\nd. 150
\nAnswer:
\na. 5<\/p>\n

Question 3.
\nThe relation of Euclid’s division lemma is
\na. a = bq + r
\nb. a = b + q \u00d7 r
\nc. a = bqr
\nd. a = (b – q) r
\nAnswer:
\na. a = bq + r<\/p>\n

Question 4.
\nWhich of the fololwing is a rational number.
\na. \\(\\sqrt{2}\\)
\nb. \u03c0
\nc. \\(\\frac{1}{3}\\)
\nd. \\(\\sqrt{3}\\)
\nAnswer:
\nc. \\(\\frac{1}{3}\\)<\/p>\n

Question 5.
\nWhat are the quotient and remainder when 10 is divided by 3
\na. 3, 1
\nb. 1, 3
\nc. 1, 1
\nd. 3, 3
\nAnswer:
\na. 3, 1<\/p>\n

Question 6.
\nWhat is the H.C.F of two consecutive natural numbers?
\na. 1
\nb. 2
\nc. 3
\nd. 4
\nAnswer:
\na. 1<\/p>\n

Question 7.
\nThe prime factors of 256 is
\na. 26<\/sup>
\nb. 27<\/sup>
\nc. 28<\/sup>
\nd. 29<\/sup>
\nAnswer:
\nc. 28<\/sup><\/p>\n

Question 8.
\nIf a and b are two positive integers then H.C.F (a.b) \u00d7 L.C.M (a, b) =
\na. a + b
\nb. a \u00d7 b
\nc. a2<\/sup> + b2<\/sup>
\nd. 2ab
\nAnswer:
\nb. a \u00d7 b<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\n\\(\\sqrt{2}\\) is
\na. a rational number
\nb. an irrational number
\nc. an integer
\nd. a natural number
\nAnswer:
\nb. an irrational number<\/p>\n

Question 10.
\n5 – \\(\\sqrt{3}\\) is
\na. an irrational number
\nb. a rational number
\nc. a composite number
\nd. a prime number
\nAnswer:
\na. an irrational number<\/p>\n

Question 11.
\nWhich of the following is not an irrational number?
\na. \\(\\sqrt{2}\\)
\nb. \\(\\sqrt{3}\\)
\nc. \\(\\sqrt{9}\\)
\nd. \\(\\sqrt{5}\\)
\nAnswer:
\nc. \\(\\sqrt{9}\\)<\/p>\n

Question 12.
\nThe number of prime number between 1 and 10 is
\na. 2
\nb. 3
\nc. 4
\nd. 12
\nAnswer:
\nc. 4<\/p>\n

Question 13.
\n\u03c0 is
\na. an irrational number
\nb. a rational number
\nc. a prime number
\nd. a composite number
\nAnswer:
\na. an irrational number<\/p>\n

Question 14.
\nH.C.F (8, 9, 25) \u00d7 L.C.M (8, 9, 25)
\na. 1800
\nb. 2.25
\nc. 2.375
\nd. 3600
\nAnswer:
\na. 1800<\/p>\n

Question 15.
\nThe decimal expansion of \\(\\frac{17}{8}\\) is
\na. 2.1
\nb. 2.25
\nc. 2.375
\nd. 2.0125
\nAnswer:
\na. 2.1<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nWhat is the use of Euclid’s division lemma?
\nAnswer:
\nEuclid’s division lemma can be used only for calculating H.C.F.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nWrite the even prime number.
\nAnswer:
\n2.<\/p>\n

Question 3.
\n(7 \u00d7 11 \u00d7 13 + 13) is a composite number. Justify the statement.
\nAnswer:
\n(7 \u00d7 11 \u00d7 13) + 13
\n= 13(77 + 1) =13 \u00d7 78.
\n\u2234 78 is composite number.
\n\u2234 It is a composite number.<\/p>\n

Question 4.
\nWhat is the HCF of 33<\/sup>\u00d7 5 and 32<\/sup> \u00d7 52<\/sup>
\nAnswer:
\nHCF of 33<\/sup> \u00d7 5 and 32<\/sup> \u00d7 52<\/sup>
\n= 32<\/sup> \u00d7 5 = 45<\/p>\n

Question 5.
\nFind the value of ‘a’ in the factor tree.
\n\"KSEEB
\nAnswer:
\n2 \u00d7 5 = 10
\n10 \u00d7 2 = 20
\n20 \u00d7 5 = 100
\n\u2234 a= 100<\/p>\n

Question 6.
\nGiven that L.C.M (91, 26) = 182, then Find H.C.F (91,26)
\nAnswer:
\n\"KSEEB<\/p>\n

Question 7.
\nWhat the two numbers called if their HCF is 1.
\nAnswer:
\nCo-Prime numbers<\/p>\n

Question 8.
\nIf P is prime number then, what is the LCM of P, P2<\/sup> and P5<\/sup>.
\nAnswer:
\nP5<\/sup>.<\/p>\n

Question 9.
\nDefine twin prime numbers.
\nAnswer:
\nA pair of prime numbers is said to be twin primes it they differ by 2. Eg: (3,5) & (11,13)<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nWrite the prime factors of 210.
\nAnswer:
\n\"KSEEB
\n210 = 2 \u00d7 3 \u00d7 5 \u00d7 7<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nIf n is odd positive integer. Show that (n5<\/sup> – 1) divisible by 8?
\nAnswer:
\nWe know that an odd positive integer n is of the form (4q + 1) or (4q + 3) for some integer q.
\nCase – I
\nwhen n = (4q + 1)
\nIn case n2<\/sup> – 1 = (4q + 1)2<\/sup> – 1
\n= 162<\/sup> q2<\/sup> + 8q = 8q(2q + 1)
\nwhich is clearly divisible by 8.<\/p>\n

Case – II
\nWhen n = (4q + 3)
\nIn this case, we have.
\nn2<\/sup> – 1 = (4q + 3)2<\/sup> – 1
\n= 16 q2<\/sup> + 24q + 8(2q2<\/sup> + 3q + 1)
\nwhich is clearly divisible by 8.
\nHence (n2<\/sup> – 1) is divisible by 8.<\/p>\n

Question 2.
\nUsing prime factorisation method, find the HCF and LCM of 30, 72 and 432. Also show that HCF \u00d7 LCM \u2260 product of three no. ?
\nAnswer:
\nGiven Numbers = 30, 72, 432.
\n30 = 2 \u00d7 3 \u00d7 5; 72 = 23<\/sup> \u00d7 32<\/sup> and 432 = 24<\/sup> \u00d7 32<\/sup>
\nHere 21<\/sup> and 31<\/sup> are the smallest powers of the common factors 2 and 3 respectively.
\nSo, HCF (30, 72, 432) = 21<\/sup> \u00d7 31<\/sup>
\n= 2 \u00d7 3 = 6.
\nAgain 24<\/sup>, 33<\/sup> and 51<\/sup> are the greatest powers of the prime factors 2, 3 and 5 respectively.
\nSo, LCM (30, 72, 432)
\n= 24<\/sup> \u00d7 33<\/sup> \u00d7 51<\/sup> = 2160
\nHCF \u00d7 LCM = 6 \u00d7 2160 = 12960
\nProduct of numbers = 30 \u00d7 72 \u00d7 432 = 933120
\n\u2234 HCF \u00d7 LCM \u2260 product of the number.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.
\nAnswer:
\nIt is given on dividing 398 by the required number, that is a remainder of 7. This means that 398 – 7 = 391 is exactly divisible by the required number. In other words, required number is a factor of 391.<\/p>\n

Similarly, required positive integer is a factor of (436 – 11= 425) and (542 – 15 = 527). Clearly, required number is the HCF of 391, 425 and 527.<\/p>\n

Using the factors tree, we get the prime factorisations of 391, 425 and 527 as follows. 391 = 17 \u00d7 23, 425 = 52<\/sup> \u00d7 17 and 527 = 17 \u00d7 31
\n\u2234 HCF of 391, 425 and 527 is 17
\nHence, required number = 17<\/p>\n

Question 4.
\nThere are 75 roses and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets that can be formed and the number of flowers in them?
\nAnswer:
\nThe number of bouquets = HCF of (75, 45). Now by Euclid’s algorithm from 75 and 45
\n\"KSEEB
\nHCF of 75 and 45 is 15
\nEach bouquets contains = 15 flowers
\n\u2234 Number of roses in bouquets = \\(\\frac{75}{15}\\) = 5
\n\u2234 Number of lilly’s in bouquets = \\(\\frac{45}{15}\\) = 3
\n\u2234 Number of bouquets = 8<\/p>\n

Question 5.
\nIn a school, the strength in 8th, 9th and 10th standards are respectively 48, 42 and 60. Find the least number of books required to be distributed equally among the students of 8th, 9th and 10th standard?
\nAnswer:
\nThe least number of books required.
\n= LCM(48, 42 and 60)
\nNow 48 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 =24<\/sup> \u00d7 3
\n42 = 2 \u00d7 3 \u00d7 7
\n60 = 2 \u00d7 2 \u00d7 3 \u00d7 5
\nHence LCM (48, 42 and 60)
\n= 24<\/sup> \u00d7 3 \u00d7 7 \u00d7 5 = 1680.
\n\u2234 Number of books required is 1680.<\/p>\n

Question 6.
\nShow that any positive odd integers is of the form 4q + 1 or 4q + 3, where q is some integer.
\nAnswer:
\nBy Euclid’s division algorithm,
\na + bq + r
\nTake b = 4 \u2234 r = 0, 1, 2, 3
\nSo, a = 4q, 4q + 1, 4q + 2, 4q + 3
\nClearly a = 4q ; 4q + 2 are even, as they are divisible by 2.
\nBut 4q + 1, 4q + 3 are odd, as they are not divisible by 2.
\n\u2234 Any positive odd integer is of the form (4q + 1) or (4q + 3)<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nx, y and z start at the same time in the same direction to run around a circular stadium x completes a round in 126 sec y in 154 sec and z in 231 sec all starting at the same point. After what time will they meet again at the starting point. How many rounds would have x, y and z completed by this time?
\nAnswer:
\nTo find LCM(126, 154, 231)
\n126 = 2 \u00d7 3 \u00d7 3 \u00d7 7
\n154 = 2 \u00d7 7 \u00d7 11
\n231 = 3 \u00d7 7 \u00d7 11
\nLCM of 126, 154, 231.
\n= 2 \u00d7 3 \u00d7 3 \u00d7 7 \u00d7 11
\n= 1386,
\nHence, after 1386 sec they meet again the starting point.
\nNow x completed the rounds = \\(\\frac{1386}{126}\\)
\n= 11 rounds
\nY completed the rounds = \\(\\frac{1386}{154}\\)
\n= 9 rounds
\nz completed the rounds = \\(\\frac{1386}{231}\\)
\n= 6 rounds<\/p>\n

Question 8.
\nShow that any positive even integers of the form 4q or 4q + 2, where is q is a whole number.
\nAns:
\n(a) Let ‘a’ be an even positive integer,
\na = 4q + r
\na = 4q + r, where 0 \u2264 r \u2264 4
\na = 2 (2q) + r where r = 0, 1, 2, 3
\nSince a is an even positive integer, 2
\ndivides ‘a’ \u2234 r = 0
\na = 2(2q) + 0
\na = 4q.<\/p>\n

(b) Let ‘a’ be an in even positive integer
\nApply division algorithm with ‘a’ and ‘b’ where b = 4.
\na = 4q + r where 0 \u2264 r < 4
\na = 4q + 2 = 2(2q) + 2
\n= 2(2q + 1).
\nSince ‘a’ is an even positive integer, 2 divides ‘a’
\nr = 2.
\na = 2(2q + 1)
\na = 4q + 2
\nHence a = 4q + 2, when ‘a’ is an even positive integer.<\/p>\n

Question 9.
\nThe HCF of 65 and 117 is expressible in the form 65m – 117, find the value of m. Also find the LCM of 65 and 117 using prime factorisation method?
\nAnswer:
\nWe have 117 = 65 \u00d7 1 + 52
\n65 = 52 \u00d7 1 + 13
\n52 = 13 \u00d7 4 + 0
\nHence HCF (65, 117) = 13
\n\u2234 65m – 117 = HCF = 13
\n\u21d2 65m = 117 + 13 = 130.
\nm = \\(\\frac{130}{65}\\) = 2
\nNow = 65 = 13 \u00d7 5
\n117 = 32<\/sup> \u00d7 13
\nLCM (65, 117) = 113 \u00d7 5 \u00d7 32<\/sup>
\n= 585.<\/p>\n

Question 10.
\nThe length and breadth of a rectangular field is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth of field exactly.
\nAnswer:
\nLength = 110 m; Breadth = 30 m.
\nWe have to find HCF of 110 and 30 by Euclid’s Lemma.
\n110 = (30 \u00d7 3) + 20
\n30 = (20 \u00d7 1) + 10; 20 = (10 \u00d7 2) + 0
\nHCF of 110 and 30 is 10.
\n\u2234 Length of longest rod is 10 m.<\/p>\n

Question 11.
\nProve that \\(\\frac{\\sqrt{7}}{5}\\) is an irrational number.
\nAnswer:
\n\"KSEEB
\nBut \\(\\sqrt{7}\\) is not a rational number. This leads us to a contradiction.
\nOur as supposition that \\(\\frac{\\sqrt{7}}{4}\\) number is wrong.
\n\u21d2 \\(\\frac{\\sqrt{7}}{4}\\) is a rational number.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 8 Real Numbers Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Additional Questions I. Multiple Choice Questions: Question 1. The …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25376"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25376"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25376\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25376"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25376"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25376"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}